Math 395 Category TheoryNorthwestern University Lecture Notes

Written by Santiago Canez

These are lecture notes for an undergraduate seminar covering Category Theory taught by theauthor at Northwestern University The book we roughly follow is ldquoCategory Theory in Contextrdquoby Emily Riehl These notes outline the specific approach wersquore taking in terms the order in whichtopics are presented and what from the book we actually emphasize We also include things welook at in class which arenrsquot in the book but otherwise various standard definitions and examplesare left to the book Watch out for typos Comments and suggestions are welcome

Contents

Introduction to Categories 1

Special Morphisms Products 3

Coproducts Opposite Categories 7

Functors Fullness and Faithfulness 9

Coproduct Examples Concreteness 12

Natural Isomorphisms Representability 14

More Representable Examples 17

Equivalences between Categories 19

Yoneda Lemma Functors as Objects 21

Equalizers and Coequalizers 25

Some Functor Properties An Equivalence Example 28

Segalrsquos Category Coequalizer Examples 29

Limits and Colimits 29

More on LimitsColimits 29

More LimitColimit Examples 30

Continuous Functors Adjoints 30

Limits as Equalizers Sheaves 30

Fun with Squares Pullback Examples 30

More Adjoint Examples 30

Stone-Cech 30

Group and Monoid Objects 30

Monads 30

Algebras 30

Ultrafilters 30

Introduction to Categories

Category theory provides a framework through which we can relate a constructionfact in one areaof mathematics to a constructionfact in another The goal is an ultimate form of abstractionwhere we can truly single out what about a given problem is specific to that problem and whatis a reflection of a more general phenomenom which appears elsewhere Practically this is done byphrasing as many constructionsfacts as possible in terms of ldquoarrowsrdquo where the point is that bydoing so relations between different objects of mathematics are simpler to see

Definition of a Category The precise definition of a category is spelled-out in the book inSection 11 The key points is that a category consists of a collection of ldquoobjectsrdquo a collection ofldquomorphismsrdquo between objects and a way to ldquocomposerdquo morphisms to get other morphisms in away which is associative and admits identities

Standard examples The main examples of categories wersquoll care about can also be found in thebook such as

bull Set the category of sets where morphisms are given by ordinary functions

bull Grp the category of groups where morphisms are given by group homomorphisms

bull Top the category of topological spaces where morphisms are given by continuous maps

bull Vect the category of vector spaces over some fixed field (which should be included as partof the notation) where morphisms are given by linear transformations

Another example All of the above are examples of what are called concrete categories This is aterm wersquoll define later but it essentially means that the objects of the categories above are all setspossibly equipped with additional structure and that the morphisms are just ordinary functionsbetween sets possibly satisfying some additional condition (For instance an object in Grp is a setequipped with a group operation and a morphism in Grp is simply a function f G rarr H betweengroups which satisfies the additional condition that it be a group homomorphism)

But even though these types of examples are the main ones wersquoll care about a category canbe a much more general type of object For instance let us describe the category called BZ Thiscategory has only one object which we will denote by lowast and think of as a single ldquopointrdquo Accordingto the definition of a category we should then have for this one object a collection of morphismsMor(lowast lowast) which in this case we declare to be Z

Mor(lowast lowast) = Z

That is we interpret each element of Z as being a ldquomorphismrdquo from lowast to lowast Now of course thereis only one possible function from lowast to lowastmdashnamely the one that sends the single element lowast of thedomain to the single element lowast of the codomainmdashso here we are considering ldquomorphismsrdquo whichdo not represent ordinary functions Visually we interpret each morphism as an arrow beginningand ending at the single object lowast

lowast

0

1

2

3

2

Now in order for this to be category we also need a notion of ldquocompositionrdquo which in this casesince there is only one object amounts to a map of the form

Mor(lowast lowast)timesMor(lowast lowast) rarr Mor(lowast lowast)

which in our case becomesZtimes Z rarr Z

We take this map to be ordinary addition + on Z (So ldquocomposingrdquo the arrow labeled 1 in thepicture above with the arrow labeled 2 results in the arrow labeled 3) The point is that this isNOT ldquocomposition of functionsrdquo in the usual sense but will give a valid ldquocompositionrdquo operationin the category BZ as long as it satisfies the properties required in the definition of a category Onesuch requirement is that composition be associative which is satisfied since + is indeed associativeThe second requirement is that exist a morphism idlowast isin Mor(lowast lowast) which acts as an identity for theldquocompositionrdquo wersquove defined In this case this identity morphism is idlowast = 0 that is the integer0 in Mor(lowast lowast) = Z Saying that this is the ldquoidentity morphismrdquo with respect to ldquocompositionrdquo inthis case simply says that is the identity element for addition

Thus we get a category BZ as claimed albeit one where the notion of ldquomorphismrdquo and ldquocompo-sitionrdquo are more general than simply ldquofunctionrdquo and ordinary composition In fact we can alreadynow give a ldquocategoricalrdquo characterization of a well-known algebraic object a (locally small) cat-egory one with a single object is precisely what is known as a monoid which is a set equippedwith an associative binary operation which admits an identity element The monoid in questionis precisely the set Mor(lowast lowast) and the associative operation which turns this into a monoid is theldquocompositionrdquo of the category in question (A locally small category is one where each collectionof morphisms is actually a set which is needed here in order to guarantee that Mor(lowast lowast) is a set onwhich we can define a binary operation) So for instance any group can be viewed as a categorywith one object where the group elements give the ldquomorphismsrdquo from that one object to itself(Of course a group gives a special type of category where every morphism is in fact ldquoinvertiblerdquoA small category where every morphism is invertible is called a groupoid so a group is nothing buta groupoid with one object Wersquoll talk more about groupoids later on)

A final example Here is one final example wersquoll consider from time to time which is again acategory where the ldquomorphismsrdquo are not simply functions The category Rel of relations has asobjects ordinary sets but as morphisms relations between sets

f isin Mor(AB) is a relation from A to B which is a subset f of AtimesB

The usual graph of a function from A to B gives an example of such a relation so we can say thatRel contains Set as a subcategory but not all relations come from graphs of functions In generalwe can think of a relation as a ldquopartially-defined possibly multivalued functionrdquo if we interpret(a b) being in the relation f as saying that ldquof sends a to brdquo

Composition in Rel comes from ordinary composition of relations given a relation f A rarr Band a relation g B rarr C we define their composition g f A rarr C to be

g f = (a c) isin Atimes C | there exists b isin B such that (a b) isin f and (b c) isin g

So g f consists of all pairs of elements of A and C which are ldquorelatedrdquo to a common elementof B In the case where f and g are graphs of ordinary functions this composition of relationsreproduces the ordinary notion of composing functions if (a b) isin f and (b c) isin g then b = f(a)

3

and c = g(b) (here we abuse notation by denoting the function of which f is graph by the symbolf itself) so

c = g(b) = g(f(a)) = (g f)(a)

meaning that (a c) isin g f precisely when c = (g f)(a) Composition of relations is associativeand you can check that graphs of ordinary identity functions

graph(idA) = (a a) isin AtimesA | a isin A

give the identity morphisms (I spent much time as graduate student trying to do ldquogeometryrdquo witha certain category of relations so this is an example near and dear to my heart)

Special Morphisms Products

Isomorphisms A morphism f isin Mor(AB) between two objects A and B in a category is anisomorphism or is invertible if it has an inverse there exists a morphism g isin Mor(BA) such thatgf = idA and fg = idB where idA isin Mor(AA) and idB isin Mor(BB) are the identity morphismswhich are assumed to exist as part of the definition of a category We say that A and B areisomorphic and we write A sim= B if there exists an isomorphism between them

This general categorical notion of isomorphism reduces to the one you would expect in thestandard examples isomorphisms in Set are bijective functions in Grp they are bijective grouphomomorphisms in Vect they are invertible linear transformations and in Top they are homeo-moprhisms ie continuous bijections with continuous inverses

A category of metric spaces Now consider the category MetC of metric spaces where mor-phisms are given by continuous functions In this category the notion of ldquoisomorphismrdquo againmeans homeomorphism so two metric spaces are isomorphic in this category when they are home-omorphic However often when considering whether or not two metric spaces are the ldquosamerdquo wehave another definition in mind that of two metric spaces being isometric X is isometric to Y ifthere exists a bijective isometry between them which is a bijective map f X rarr Y which preservesdistance in the sense that

dY (f(p) f(q)) = dX(p q) for all p q isin X

After all in a metric space you have more data than simply ldquoopen setsrdquo (which is all that ahomeomorphism detects) so it would be good to have a notion of ldquosamenessrdquo which incorporatedthe extra data of distance as well

The goal is then to construct a category of metric spaces where ldquoisomorphicrdquo indeed meansisometric and we can do so simply by restricting the types of continuous maps we consider Thecategory Met has metric spaces as objects but now a morphism f X rarr Y is taken to be afunction which does not ldquoenlargerdquo distance in the sense that

dY (f(p) f(q)) le dX(p q) for all p q isin X

Maps with this property are often called metric maps Any metric map is actually continuous soMet is a subcategory of MetC (The subscript C used in MetC is meant to denote ldquocontinuousrdquoindicating that we take arbitrary continuous maps as morphisms) In this category now for f X rarr Y to be an isomorphism requires that f and its inverse fminus1 both be metric maps so

dY (f(p) f(q)) le dX(p q) and dX(fminus1(s) fminus1(t)) le dY (s t)

4

for all p q isin X and s t isin Y But writing s and t as s = f(p) and t = f(q) the second conditionbecomes dX(p q) le dY (f(s) f(t)) and the two inequalities together thus imply

dY (f(p) f(q)) = dX(p q) for all p q isin X

so that f X rarr Y is actually an isometry The moral is that certain properties one might wantcan be achieved by changing the morphisms you consider

Monomorphisms and epimorphisms The definitions of monomorphism and epimorphismcan be found in Section 12 of the book The standard examples are those in Set where themonomorphisms are the injective functions and the epimorphisms are the surjective functionsProofs of these facts are left to the homework as well as the interesting problem of determining theepimorphisms in Met or even MetC or Haus (the category of Hausdorff spaces where morphismsare arbitrary continuous maps) where the answer is that epimorphism means more than simplyldquocontinuous and surjectiverdquo (As opposed to all of Top where epimorphism does indeed meancontinuous and surjective)

So monomorphisms and epimorphisms should be treated as categorical analogues of ldquoinjectionsrdquoand ldquosurjectionsrdquo although this analogy only goes so far The thing to note is that we are thusable to give a complete definition of ldquoinjectiverdquo for ordinary functions which makes no referenceto elements and is stated only in terms of the relation between an injective function and otherfunctions certainly proving that ldquomonomorphismrdquo means ldquoinjectionrdquo in Set requires working withelements but stating the definition of an injection as a monomorphism can be done solely usingldquoarrowsrdquo

Products The definition of a product in a category shows up in Section 31 of the book inthe context of the more general notion known of a limit Wersquoll discuss this more general notioneventually but for now we will only focus on products of two objects at a time Here is what themore general notion boils down to in this case which wersquoll take as our definition

A product of two objects A and B in a category is an object P together with morphismspA P rarr A and pB P rarr B (which we call projections) which satisfy the followingproperty given any morphisms f C rarr A and g C rarr B into A and B from acommon object C there exists a unique morphism h C rarr P such that f = pA h andg = pB h We express this by saying that the following diagram commutes

C

A B

Pf g

pA pB

existh

which means that composing any two composable arrows in the diagram results in theother arrow with the same domain and codomain Concretely there are two ways in thisdiagram to get from C to A via f or via the composition pA h and the requirementis that both give the same morphisms (ie f = pA h) and similarly for the two waysto get from C to B The dashed arrow is the one which is required to exist (uniquelyas denoted by the exclamation mark ) in the definition of product

5

So the upshot is that for any C and any f and g in the diagram above there exists a uniqueh which makes the diagram commute The point is that a product gives a way to turn the twopieces of data f C rarr A and g C rarr B into the single piece of data h C rarr P in a uniqueway Being able to turn a possibly large amount of data into a single piece of data in this way isone of the benefits the ldquocategoricalrdquo perspective provides Products in categories are often denotedusing the usual A times B notation even though it is not necessarily true that a product is literallya set-theoretic Cartesian product although it is in the standard examples The map h C rarr Prequired in the definition is then often denoted by h = f times g

Examples In Set the product of two objects A and B always exists and is indeed the ordinaryCartesian product A times B But of course the definition of product also requires that data of theprojection morphisms which in this case are simply the usual ones

pA (a b) 983041rarr a and pB (a b) 983041rarr b

Given f C rarr A and g C rarr B the map h C rarr AtimesB is h(c) = (f(c) g(c)) and you can checkthat these constructions satisfy the properties required in the definition (Note that you would alsohave to argue that h defined in this way is the only map which can make the required diagramcommute)

Products in Grp are given by the usual direct product group structure GtimesH with the groupoperation defined as

(g1 h1) middot (g2 h2) = (g1g2 h1h2)

The projection morphisms are the ordinary ones you would expect as well But let us actuallynow verify that this group structure indeed is among other possible group structures on G times Hthe one required in the definition of a product Suppose f K rarr G and g K rarr H are grouphomomorphisms The key point is that the map h(k) = (f(k) g(k)) which should satisfy therequired product property should itself be a morphism in Grp meaning a group homomorphismand we claim that this is only true for the direct product group structure

Indeed we want it to be true that

h(k1k2) = (f(k1k2) g(k1k2)) equal h(k1)h(k2) = (f(k1) g(k1)) middot (f(k2)g(k2))

for all k1 k2 isin K and some to-be-determined group structure middot on G times H Since f and g arehomomorphisms this requirement becomes

(f(k1)f(k2) g(k1)g(k2)) = (f(k1) g(k1)) middot (f(k2)g(k2))

which says that middot should indeed be componentwise multiplication at least on the image of f andg But recall that the properties required of a product should hold for any K and any f K rarr Gand g K rarr H in particular then it should hold for instances when f and g are surjective whichshows that the componentwise-multiplication property above should in fact hold on all of G timesHso that middot is indeed the ordinary direct product structure

Uniqueness Notice that in the definition of a product in a category we used the word ldquoardquo asopposed to ldquotherdquo we spoke about a product of A and B instead of the product of A and B Sowe now ask are products unique The answer is ldquonordquo if we interpret ldquouniquerdquo in the strictestpossible sense but ldquoyesrdquo if we interpret it correctly For instance in the example of Grp takenow N to be any group which is isomorphic to the direct product G timesH say with isomorphismℓ N rarr G timesH We claim that we can also turn N into a categorical product of G of H as long

6

as we define appropriate ldquoprojectionsrdquo N rarr G and N rarr H take N rarr G to be the compositionpG ℓ and N rarr H to be pH ℓ Given f K rarr G and g K rarr H as in the definition of a productthe map hprime K rarr N defined by

hprime = ℓminus1 h

where h K rarr GtimesH is the usual h = f timesg will satisfy the requirement needed to able to concludethat N is also a product of G and H in Grp A similar trick works in any category so that anyobject isomorphic a product can also itself be turned into a product

The correct answer is that products are unique up to unique isomorphism which means thatgiven a product P of A and B any other product P prime of A and B will in fact be isomorphic to P and in a unique way in the sense that there can exist only one isomorphism P rarr P prime The phraseldquounique up to unique isomorphismrdquo is a common one wersquoll see again and again objects definedby some categorical property are almost never genuinely unique in a strict sense but they willessentially be as close to unique as possible

So suppose that P and P prime are both products of A and B with associated projection morphismspA pB for P and pprimeA p

primeB for P prime Consider the diagram

P

A B

P primepA pB

pprimeA pprimeB

existh

The morphism h P rarr P prime is the unique one guaranteed to exist by the fact that P prime is a productof A and B Commutativity of this diagram says that pA = pprimeA h and pB = pprimeB h A similardiagram with the roles of P and P prime reversed which uses the fact that P is a product results in aunique morphism hprime P prime rarr P satisfying pprimeA = pA hprime and pprimeB = pB hprime

Now consider the following diagram

P

A B

PpA pB

pA pB

hprime h

This commutes which we verify using the properties h and hprime satisfy as follows

pA (hprime h) = (pA hprime) h = pprimeA h = pA

and similar for the morphisms involving B This shows that hprimeh P rarr P satisfies the requirementin the definition of P being a product But of course idP P rarr P satisfies this requirement as wellso since the map satisfying this requirement is assumed to be unique we must have hprime h = idP Similar reasoning considering h hprime P prime rarr P prime shows that h hprime = idprimeP so that h and hprime are indeedinverses and hence that P and P prime are isomorphic The uniqueness of the isomorphism betweenthem comes from the uniqueness of h in the construction above

7

Abstract nonsense It is common to say in the setting above that the fact that products areunique up to unique isomorphism follows from ldquoabstract nonsenserdquo which is a succinct way ofsaying that it follows solely from the definition of a categorical product itself and not any deepermathematical content In particular many of the ldquouniquenessrdquo properties which ldquoproductsrdquo havein the settings we might care aboutmdashCartesian products in Set direct products in Grp producttopologies in Topmdashreally have nothing to do with how those specific products are defined but arejust consequences of the ldquoabstract nonsenserdquo of category theory

But of course we do not use the phrase ldquoabstract nonsenserdquo in a negative or dismissive waysince recognizing what types of properties are consequences of ldquoabstract nonsenserdquo goes a long waytowards understanding well the given problem at hand Soon you too will come to appreciate thisnotion of abstract nonsense

Coproducts Opposite Categories

Initial and terminal objects The notions of initial and terminal objects in a category aredefined in Section 16 of the book where you can also find standard examples Here I want topoint out that initial and terminal objects if they exist are unique up to unique isomorphism asanother example of arguing via ldquoabstract nonsenserdquo

Suppose I and I prime are both initial objects in some category By the fact that I is initial thereexists a unique morphism f I rarr I prime and by the fact that I prime is initial there exists a unique morphismg I prime rarr I But then the composition g f I rarr I must equal idI I rarr I since the latter is theonly morphism from I to itself because I is initial and similarly f g = idIprime Thus I and I prime areunique and there is a unique isomorphism between them

Coproducts The dual concept to that of a product is the notion of a coproduct (The termldquodualrdquo refers to the phenomena you get when you reverse all arrows in a given construction Forinstance the notion of a terminal object is dual to that of an initial object and the notion of anepimorphism is dual to that of a monomorphism) Here is the precise definition a coproduct of anobject A and an object B in a category is an object C together with morphisms iA A rarr C andiB B rarr C such that for any object D and morphisms f A rarr D and g B rarr D there exists aunique morphism h C rarr D which makes the following diagram commute

D

A B

Cf g

iA iB

existh

By abstract nonsense if a coproduct of A and B exists it is unique up to unique isomorphism

Examples In Set the coproduct of two sets A and B is the disjoint union A⊔B (If you havenrsquotseen the notion of a disjoint union before it is essentially the same as the union only that we donrsquotcare about whether A and B have any overlap if they do we treat the common elements as beingldquodifferentrdquo For instance Z ⊔ Z is not Z but consists of two copies of each integer one 0 comesfrom the first copy of Z and another comes from the second but these are different ldquozeroesrdquo)Given f A rarr D and g rarr D the unique map A ⊔B rarr D is defined by applying f to elements of

8

A and g to elements of B (Note that if we simply took the ordinary union AcupB as the candidatefor the coproduct when A and B were not disjoint the required map h A cup B rarr D would notexist since h would have to send any element x isin A cap B to both f(x) and g(x) in order to makethe required diagram commutative)

The coproduct of two objects X and Y in Top is again the disjoint union X ⊔ Y equippedwith the disjoint union topology an open subset of X ⊔ Y is defined to be the (disjoint) union ofan open subset of X with an open subset of Y The coproduct of two groups G and H in Grp isthe free product G lowastH which is further elaborated on in the homework

The coproduct of two vector spaces V and W in Vect is the direct sum V oplusW which is definedto be the set of formal sums

v + w where v isin V and w isin W

where addition and scalar multiplication are defined ldquocomponentwiserdquo

(v1 + w1) + (v2 + w2) = (v1 + v2) + (w1 + w2) and r(v + w) = rv + rw

This is isomorphic to the product V times W with operations also defined componentwise but it iscustomary to speak about the ldquodirect sumrdquo V oplusW when referring to the coproduct as opposed tothe product wersquoll say why in a bit The maps iV V rarr V oplusW and iW W rarr V oplusW required aspart of the data of a coproduct are

iV v 983041rarr v + 0W and iW w 983041rarr 0V + w

To verify that V oplusW is the correct coproduct take any linear transformations S V rarr U andT W rarr U where U is some other vector space We then need a (unique) linear transformationh V oplusW rarr U such that

S = h iV and T = h iW

Let us determine what this map must be thereby not only showing that it exists but also that itis unique Let v + w isin V oplusW which we can write as

v + w = (v + 0W ) + (0V + w)

Since h is required to be linear we have

h(v + w) = h([v + 0W ] + [0V + w]) = h(v + 0W ) + h(0V + w)

The first term at the end is (h iV )v which must thus be Sv and the second term is (h iW )wwhich must be Tw Thus h = S + T is the map

h = S + T v + w 983041rarr Sv + Tw

which you can verify is indeed linear Thus V oplusW does satisfy the property required of a coproductSo the product of two objects in Vect is the same as their coproduct only that we write the

latter using ldquosumrdquo notation Later we will discuss products and coproducts of an arbitrary numberof elements We will see these as special cases of the general notions of limits and colimits butthey are defined via similar requirements as products and coproducts of two objects at a timeIn the category of vector spaces arbitrary products are given by the usual product

983124α Vα with

ldquocomponentwiserdquo addition and scalar multiplication but it turns out that an arbitrary coproductis given by the direct sum

983119α Vα which is defined to be the subspace of

983124α Vα where only finitely

many components are nonzero (This is in a way analogous to how the product topology on an

9

arbitrary number of topological spaces is defined) Wersquoll go over this later but is the underlyingreason why we use oplus instead of times when discussing coproducts

Encoding data via productscoproducts We can nicely summarize the definition of productsand coproducts in the following way the product of A and B is the object with the property thatfor any object D we have

Mor(DA)timesMor(DB) = Mor(D product)

and the coproduct of A and B is the object such that for any object D we have

Mor(AD)timesMor(BD) = Mor(coproductx D)

The first reflects the fact that products give a unique way to turn two morphisms (f and g in thedefinition) mapping into A and B into a single morphism (h in the definition) and the second saysthat coproducts give a unique way to turn two morphisms mapping out of A and B into a singlemorphism

Soon enough we will discuss the idea of universality and the point is that products are ldquouni-versalrdquo for maps into A and B and coproducts are universal for maps from A and B

Opposite categories Finally we give a definition which might seem silly but is actually quiteuseful Given a category C the opposite category is the category Cop is the category with the sameobjects as C but where we defined Mor(AB) in Cop to be Mor(BA) in C To be more concrete weconsider each morphism f A rarr B in C to instead be a morphism fop B rarr A in Cop (Perhaps itwould be better to write this as A larr B fop) The point is that we donrsquot actually care about whatf actually is as say a functioncontinuous maphomomorphismwhatever-we only care about itsbehavior as an ldquoarrowrdquo The opposite category Cop only retains information about how arrowsrelate to one another nothing more

Since arrows are reversed products in C become coproducts in Cop and coproduts in C becomecoproducts in Cop Similarly initialterminal objects and monomorphismsepimorphisms in C be-come terminalinitial objects and epimorphismsmonomorphisms in Cop Again the notion of anopposite category might seem like a strange thing to look at but wersquoll see that it does have uses

Functors Fullness and Faithfulness

One more productcoproduct example Let X be a topological space and define its categoryof open sets Op(X) as follows The objects of Op(X) are simply the open subsets of X and themorphisms are given by inclusions so to be concrete

Mor(U V ) =

983083the unique inclusion U rarr V if U sube V

empty otherwise

The fact that U sube V and V sube W implies U sube W says that the composition of two morphismsin this category is still a morphism in this category The product of U and V in this categoryturns out to be their intersection (with ldquoprojection morphismsrdquo U cap V rarr U and U cap V rarr V beinginclusions) and the coproduct of U and V turns out to be their (ordinary) union Note that weneed X to be a topological space in order to guarantee that U cap V and U cup V are still objects inthis category More generally the product of finitely many objects in Op(X) exists and is theirintersection and the coproduct of an arbitrary number of objects exists and is their union (Wesay that Op(X) has finite products and arbitrary coproducts)

10

Given a set X and a collection T of subsets of X we can define a similar category with elementsof T as objects (I guess we should assume that empty X isin T) and inclusions as morphisms We thenhave that this category has finite products and arbitrary coproducts if and only if T is a topologyon X so that we can rephrase the definition of a topological space itself solely in categorical termsThis perspective on what a ldquotopologyrdquo is makes it simpler to state the definition of whatrsquos calleda sheaf on a topological space which is an important construction in various areas of analysis andgeometry We might discuss this a bit later on

Functors As with most things in mathematics we should care not only about categories asmathematical structures on their own but also about ldquomapsrdquo between these structures whichldquopreserverdquo said structure The notion of a (covariant) functor between categories provides suchldquomapsrdquo in the case at hand where a functor is a way of sending objects to objects and morphismto morphisms in a way which preserves compositions and identities The precise definition is inSection 13 of the book as is the definition of a contravariant functor between categories whichis a functor which reverses the direction of arrows as opposed to ldquocovariantrdquo ones which preservedirections (It is common to use the term ldquofunctorrdquo without qualification to mean one which iscovariant Note that any functor can be made covariant by working with the opposite category acontravariant functor C rarr D is the same as a covariant functor Cop rarr D)

All of the standard examples we looked at are in the book forgetful functors power set functorsthe dual space functor in Vect etc Of particular interest is Example 139 in the book whichshows that various types of group actions in different contexts-the usual notion of a group actionon a set the notion of a continuous action of a group on a topological space and the notion ofa linear action of a group on a vector space (also known as a representation of the group)-are allencoded by a functor with domain category BG

Adjoint functors Let F Top rarr Set be the forgetful functor and let D Set rarr Top be thefunctor which sends a set S to the space S equipped with the discrete topology and a functionS rarr Sprime between sets to the same function only now viewed as a continuous map S rarr Sprime betweendiscrete spaces These two functors satisfy the following property in relation to one another givinga morphism D(S) rarr Y in Top is the same as giving a morphism S rarr F (Y ) in Set or moresymbolically

MorTop(D(S) Y ) = MorSet(S F (Y ))

We say that D is left adjoint to F and that F is right adjoint to D The point is that adjointfunctors can be used to move data back and forth between two categories

Wersquoll talk more about adjoint functors later on but for now here is one more example Nowtake F Grp rarr Set to again be the forgetful functor and take G Grp rarr Set to be the freegroup functor which sends a set S to the free group generated by S G(S) has as elements ldquowordsrdquos1 sn made up of elements of S with the group operation being ldquoconcatenationrdquo A basic factis that giving a group homomorphism from G(S) to some other group H is the same as giving anordinary function from S to F (H) which is the required adjoint property

MorGrp(G(S) H) = MorSet(S F (H))

Faithful and full functors The definitions of what it means for a functor to be faithful and fullare in Section 15 of the book where faithful functors are ones which are injective on morphisms andfull functors are ones which are surjective on morphisms but where in both cases we refer only tomorphisms which occur between the objects F (A) and F (B) in the target category corresponding tofixed objects A and B in the source category (In general a functor could send two objects AB in

11

the source category to the same object F (A) = F (B) in the target category and thus there could betwo morphisms A rarr C and B rarr C which are sent to the same morphism F (A) = F (B) rarr F (C)Such a thing could still possibly be ldquofaithfulrdquo since the non-injectivity here is arising from differentobjects in the source)

Faithfullness and fullness will be nice properties going forward and here is one hint at how theymight be useful Say we have a functor F C rarr D The problem is to determine properties onF which will imply it send products in C to products in D or coproducts in C to coproducts inD So given the former product diagram below we want the second diagram to also be a productdiagram

C

A B

P Fminusminusminusrarr

F (C)

F (A) F (B)

F (P )f g

pA pB

existh

Ff Fg

F (pA) F (pB)

Fh

Of course in order F (P ) to be an actual product of F (A) and F (B) on the right a similarcommutative diagram property would have to hold for all objects in D in place of F (C) and allmorphisms D rarr F (A) and D rarr F (B) not just those of the form Ff and Fg respectively Oneway to guarantee this is to require that all objects in D are of the form F (C) and then for allmorphisms F (C) rarr F (A) and F (C) rarr F (B) to be of the form Ff Fg the former is true whenthe functor F is surjective on objects and the latter is true when F is full We are not saying thatthese are the only conditions under which F might preserve products but it definitely seems togive a sufficient condition at least

Almost there is one more thing to require the uniqueness of the morphism Fh F (C) rarr F (P )making the diagram on the right commute Suppose there were two such morphisms Fh and Fhprimewhere h and hprime are both morphisms C rarr P in C The commutativity of the diagram on the rightgives for instance

Ff = F (pA)Fh and Ff = F (pA)Fhprime

which the functorial properites of F turns into

Ff = F (pA h) and Ff = (pA hprime)

We want h and hprime to make the first diagram commute in order to be able to apply uniqueness ofthe morphism C rarr P But this requires know that f = pA h and f = pA hprime which would followfrom F being faithful In this case it follows that h and hprime both make the first diagram commute(after we go through a similar argument with g) so the fact that P is a product guarantees thath = hprime and hence that Fh = Fhprime so that F (P ) is an actual product on the right

So we conclude that a functor which is full faithful and surjective on objects will send productsto products (Actually we can relax this a bit by requiring not that every object in D be in theliteral image of F but only that it be isomorphic to something in the image A functor with thisproperty is said to be essentially surjective which is a concept will see in a related context soonenough) Again this is not an ldquoif and only ifrdquo statement since functors can preserve products (orcoproducts) without being full faithful and surjective on objects but it at least gives a sufficientcondition Later we will see other ways to guarantee that functors preserve products which arerelated to the existence of adjoints for that given functor

12

Coproduct Examples Concreteness

Products and coproducts for metric spaces The product of two objects in Met the categoryof metric spaces with morphisms given by metric maps is the Cartesian product X times Y equippedwith the box metric and the coproduct of X and Y does not exist if both X and Y are nonemptyThis was the topic of a student presentation and here are some proof sketches

Given metric spaces (X dX) and (Y dY ) the box metric d on X times Y is defined by

d((x1 y1) (x2 y2)) = maxdX(x1 x2) dY (y1 y2))

(The name ldquoboxrdquo metric comes from the fact that in RtimesR = R2 or R3 open balls indeed looks likeldquoboxesrdquo) The key requirement needed in the definition of a product is given morphisms S rarr Xand S rarr Y that the map

h S rarr X times Y defined by h(s) = (f(s) g(s))

actually be a morphism in Met This turns into the requirement that

d((f(s) g(s)) ((f(t) g(t))) = maxdX(f(s) f(t)) dY (g(s) g(t)) le dS(s t) for all s t isin S

given the fact that

dX(f(s) f(t)) le dS(s t) and dY (g(s) g(t)) le dS(s t) for all s t isin S

For the latter inequalities to imply the former d must (at least up to isometry) be given by thebox metric Note that the two other ldquostandardrdquo metrics one might put on a productmdashthe taxicab

dX + dY and Euclidean983156

d2X + d2Y metricsmdashdo not satisfy this required ldquometric maprdquo property

and so do not give the product in MetTo see that coproducts do not exist if XY are nonempty suppose instead (P dP ) was a co-

product for X and Y with inclusion morphisms iX X rarr P and iY Y rarr P For any n isin Ntake 0 n to be a metric space with the absolute value metric d and let f X rarr 0 n andg Y rarr 0 n be the constant function 0 and the constant function n respectively Since P is acoproduct there exists h P rarr 0 n such that

h iX = f and h iY = g

But in order for h to actually be a morphism in Met we must then have

d(h(iX(x) iY (y)) le dP (iX(x) iY (y)) for any x isin X y isin Y

which translates inton = |0minus n| le dP (iX(x) iY (y))

But this should then be true for any n isin N which would imply that the distance in P betweenan element coming from X and one coming from Y should be infinite which is nonsense (Notethat if you altered your definition of ldquometricrdquo to allow for infinite distance then a coproduct wouldexist the disjoint union X ⊔Y where you defined the distance between elements of X and Y to beinfinite)

Coproducts for groups The coproduct of two groups G and H in Grp is the free product GlowastHwhich is defined to be the set of all words consisting of elements of G andH in an alternating fashion

13

with concatenation as the group operation (When we have to elements of G or two elements of Hnext to each other with use their respective group operations to turn these into single elements ofG or H) This was also the topic of a student presentation and here is the basic idea Given grouphomomorphisms g G rarr N and k H rarr N that G lowastH is the coproduct comes down to showingthat the map

h G lowastH rarr N defined by h(g1h1 gnhn) = f(g1)k(h1) middot middot middot f(gn)k(hn)

and similarly on other possible words in GlowastH is a group homomorphism which follows essentiallyfrom the definition of the group operation on G lowastH indeed this group operation arises preciselyfrom this requirement

Now if we instead tried to use GtimesH with its standard inclusions G rarr GtimesH and H rarr GtimesHas the coproduct the issue is that the analogous map

h GtimesH rarr N defined by h((g h)) = f(g)k(h)

would NOT be a homomorphism because h((g1 h1)(g2 h2)) is not equal to h(g1 h1)h(g2 h2) bythe way in which the group operation is defined on the direct product G times H However if westrict our category to only consist of abelian groups so that the N rsquos we consider in the defini-tion of coproduct are also abelian then G times H does serve as a coproduct we can always rewritef(g1)k(g1) middot middot middot f(gn)k(gn) as f(g1) middot middot middot f(gn)k(h1) middot middot middot k(hn) in N Irsquoll leave it to you to flesh out allof these details

Concrete categories The categories Set Top Vect and Grp all have the property thattheir objects are simply sets possibly equipped with extra structure and that their morphisms areordinary set-theoretic functions possibly satisfying some additional requirement Such categoriesare nice since we can often use intuition about sets and functions in their study A concrete categoryis intuitively one where we can think of objects as being ldquosetsrdquo and morphisms as being ldquofunctionsrdquoonly that we have to interpret this in the right way

Here is the definition a concrete category is a category C equipped with a faithful functorF C rarr Set which is part of the data The idea is that such a functor gives us way to turnan object A isin Ob(C) into a set F (A) and a morphism f isin MorC(AB) into a function Ff isinMorSet(F (A) F (B)) where the ldquofaithfulrdquo requirement says that we do not lose information aboutf when doing so so that we can (essentially) learn everything we need to know about f by studyingFf instead In the four examples above the required faithful functor is simply the forgetful functorand indeed in general we think of F as being a ldquoforgetful functorrdquo for the concrete category (C F )

Examples Here are two more examples of concrete categories Let G be a group and considerthe category BG with a single object and elements of G morphisms As written this does notappear to be concrete since morphisms are not honest functions lowast rarr lowast but the point is that thiscategory can indeed by ldquoconcretizedrdquo by specifying an appropriate ldquoforgetful functorrdquo A functorF BG rarr Set is the data of a (left) action of G on a set S = F (lowast) and saying this functor isfaithful is what it means to say that this action is faithful different g h isin G act on S in differentways or equivalently that there exists s isin S such that gs ∕= hs Any group has at least onefaithful left actionmdashnamely the action on itself given by left multiplicationmdashto BG can always beequipped with a faithful functor to Set This is all just a way of phrasing the fact that any groupis isomorphic to a subgroup of a permutation group

The category Rel of relations can also be ldquoconcretizedrdquo even though as written morphismsA rarr B are not honest functions Take F Rel rarr Set to be the power set functor which sends an

14

object in Rel to its power set and a relation f A rarr B to the induced function Ff P (A) rarr P (B)defined by

(Ff)(S) = b isin B | there exists a isin S such that (a b) isin f

In other words if you think of the relation f sube A times B as a ldquopossibly not everywhere-definedpossibly multivalued functionrdquo (Ff)(S) is analogous to taking the image of S That F is faithfulwill be left to the homework

In fact most categories yoursquoll see can be made concrete by specifying an appropriate forgetfulfunctor but not all categories can be made concrete in this way Here is the standard example of anon-concrete category which we mention only to say there is such a thing but which we wonrsquot doanything more with since understanding it better requires knowledge of algebraic topology Thehomotopy category of topological spaces is the category hTop whose objects are topological spacesand whose morphisms are given by equivalence classes of homotopic maps

MorhTop(XY ) = homotopy classes of continuous maps X rarr Y

(Saying that two maps are homotopic means that you can ldquodeformrdquo one into the other but wersquollleave the precise definition to a course in algebraic topology) It turns out that no functor fromhTop to Set can be faithful so that hTop cannot be made concrete but this is actually a deepand difficult thing to show so we make no attempt to do so here

Natural Isomorphisms Representability

Natural transformations The definition of a natural transformation η F rArr G between two(both covariant or both contravariant) functors FG C rarr D is given in Section 14 of the bookThe key idea is that a natural transformation gives a way to turn data about F into data aboutG in a way which is compatible with all possible morphisms as expressed by the commutativity ofthe diagrams

F (A) F (B)

G(A) G(B)

Ff

ηA ηB

Gf

arising from morphisms f A rarr B (or f B rarr A in the contravariant case) in C In the case of anatural isomoprhism η F rArr G so that each ηA F (A) rarr G(A) is actually an isomorphism in Dthe point is not only that F and G produce isomorphic objects but they do so in a way which iscompatible with all possible category-theoretic data

The first standard example of a natural isomorphism is the one between the identity functor onthe category of finite-dimensional and the functor which sends a finite-dimensional vector space Vto V lowastlowast the dual of its dual induced by the isomorphism V rarr V lowastlowast defined by

v 983041rarr (the linear map on V lowast which sends f to f(v))

This was the topic of a student presentation and can be found in Section 14 of the book Theessential reason why this isomorphism is ldquonaturalrdquo is that V rarr V lowastlowast can be defined without referenceto a basis Contrast this with the the functor which sends V to its (single) dual V lowast it is still truethat finite-dimensional vector space is isomorphic to its dual but specifying such an isomorphism

15

requires additional data which prevents the identity functor from being ldquonaturally isomorphicrdquo tothe single dual functor

Some history At this point it makes sense to say a bit about the interesting origins of categorytheory in the 1950rsquos Eilenberg and Mac Lane where studying cohomology (or possible homologyone of those two) in algebraic topology which associates algebraic data (a group ring or similar)to a topological space in a way which encodes some of the topology They had two cohomologicalconstructions which ended up proving isomorphic results but they noticed that not only werethe resulting objects the same but the actual constructions themselves were in some sense theldquosamerdquo They thus then needed a way to formalize this notion and invented the concept of anatural transformation to do so where they interpreted two constructions as being the same asbeing compatible with all morphisms

But this then leads to the question what types of objects should a natural transformationoccur between Eilenberg and Mac Lane then had to invent the notion of a ldquofunctorrdquo to describewhat a natural transformation went from and what it went to In their motivating setting theyinvented the notion of a functor to describe more formally properties which their cohomologicalconstructions were supposed to satisfy But this then leads to the question what types of objectsshould a functor map between In other words what type of data should a functor (cohomologicalconstruction in their example) take as input and should it output They finally had to thus inventthe notion of a ldquocategoryrdquo to describe the source and target of a functor so that their cohomologicalconstructions could be interpreted as functors from a category of topological spaces to a categoryof algebraic objects

Thus historically natural transformations came first then functors and then categories Aswith most things in mathematics categories thus arose from attempts to encode what was beingseen in some concrete examples in a more general and formal setting

Representable functors Representable functors are defined in Section 21 of the book but wersquollgive the required definitions here in a hopefully simpler to digest manner The point is that anyobject A in a (locally small) category C gives two ldquonaturalrdquo functors C rarr Set one covariant andthe other contravariant which as wersquoll see pretty much encode all data about A itself We definehA C rarr Set to be the covariant functor defined by

hA(B) = Mor(AB) on objects and

hA(Bfminusrarr C) = the map Mor(AB)

hAfminusminusrarr Mor(AC) defined by g 983041rarr f g

We define hA C rarr Set to the contravariant functor defined by

hA(B) = Mor(BA) on objects and

hA(Bfminusrarr C) = the map Mor(CA)

hAfminusminusrarr Mor(BA) defined by g 983041rarr g f

(The functor hA is called the functor of points of A and wersquoll usually think of it as being a covariantfunctor hA Cop rarr Set The functor hA does not have a standard name) So hA is defined bymorphisms from A and hA is defined by morphisms into A Wersquoll see later the sense in which suchfunctors encode all information about the object A

A functor F C rarr Set is representable if is naturally isomorphic to such a functor so F sim= hA

for some A in the covariant case and F sim= hA for some A in the contravariant case The idea wersquollbuild towards is that we can essentially treat such a functor as being A itself so that representable

16

functors can be thought as being actual objects in C For now here is a key example wersquove seenbefore which we can now formulate using the language of representability

Given AB isin Ob(C) let F C rarr Set be the contravariant functor defined by

F (C) = Mor(CA)timesMor(CB) and F (Cfminusrarr D) = [(g k) 983041rarr (g f k f)]

where (g k) 983041rarr (g f k f) gives a map Mor(DA) timesMor(DB) rarr Mor(CA) timesMor(CB) ieF (D) rarr F (C) In order for this functor to be representable requires the existence of an object inC satisfying

Mor(CA)timesMor(CB) sim= Mor(C representing object)

in a way which is compatible with all morphisms but we have actually already seen this notionelsewhere it is essentially the definition of a product AtimesB for A and B Indeed we claim that Fis naturally isomorphic to hAtimesB as we now show

First to define a natural isomorphism hAtimesBsim= F requires for each C isin Ob(C) a choice of a

bijection (ie isomorphism in Set)

ηC hAtimesB(C)sim=minusrarr F (C)

which in our case looks like

ηC Mor(CAtimesB))sim=minusrarr Mor(CA)timesMor(CB)

Take this to be the map

(Cfminusrarr AtimesB) 983041rarr (pA f pB f)

where pA A times B rarr A and pB A times B rarr B are the two projection morphisms in the definitionof a product The definition of product implies that this map is invertible since given k C rarr Aand ℓ C rarr B there exists a unique h C rarr A times B such that k = pA h and ℓ = pB h Forthese bijections to define a natural isomorphism hAtimesB

sim= F requires that they be compatible withall morphisms in C which means that given any g C rarr D in C the following diagram shouldcommute

hAtimesB(C) hAtimesB(D)

F (C) F (D)

hAtimesB(g)

ηC ηD

Fg

which in the case at hand looks like

Mor(CAtimesB) Mor(DAtimesB)

Mor(CA)timesMor(CB) Mor(DA)timesMor(DB)

minus g

ηC ηD

(minus gminus g)

Take an element k D rarr AtimesB in the upper right Applying the map on top gives the elementk g in the upper left and then applying ηC on the left side gives

(pA (k g) pB (k g))

17

in the lower left Instead starting with k D rarr AtimesB in the upper right applying ηD on the rightgives (pA k pB k) and applying the map on the bottom gives

((pA k) g (pB k) g)

in the lower left which is the same as the previous element in the lower left by the associativity ofcomposition Hence the isomorphisms ηC do define a natural isomorphism between hAtimesB and F Again the point is that you can then interpret the functor F as being the ldquosamerdquo as the productAtimesB since both encode the same data

More Representable Examples

Coproducts As in the case of products the definition of coproducts can also be phrased in termsof a representable functor Given objects A and B in C let F C rarr Set be the covariant functordefined by

F (C) = Mor(AC)timesMor(BC) and F (Cfminusrarr D) = [(g k) 983041rarr (f g f k)]

where the latter is a function Mor(AC)timesMor(BC) rarr Mor(AD)timesMor(BD) This functor isrepresentable if and only if a coproduct A ⊔ B for A and B exists and the representing object isthat coproduct the key is the requirement that we have bijections

Mor(AC)timesMor(BC) sim= Mor(A ⊔BC)

for all C which is essentially the defining property a coproduct for A and B should have

Power set functor contravariant version Let P Set rarr Set be the contravariant powerset functor defined on objects by sending a set to its power set and on morphisms by sendingf A rarr B to the preimage function Pf P (B) rarr P (A) given by S 983041rarr fminus1(S) In order for this tobe representable requires the existence of a set X with natural bijections

P (A) sim= Mor(AX)

for all sets A that is a set X so that mapping into X is the same data as specifying a subset ofthe domain We take X = 0 1 to be a two-element set and the required bijections

ηA P (A) rarr Mor(A 0 1) to be given by S 983041rarr χS

where XS A rarr 0 1 is the indicator or characteristic function S defined by sending elementsof S to 1 and elements of A minus S to 0 The inverse of ηA sends a function g A rarr 0 1 to thepreimage gminus1(1) sube A

To verify that the bijections ηA define the data of a natural isomorphism η P rArr h01 wecheck the commutativity of some diagrams Given a function f A rarr B consider

P (A) P (B)

h01(A) = Mor(A 0 1) h01(B) = Mor(B 0 1)

Pf = fminus1

ηA ηB

minus f

18

Take S isin P (B) in the upper right Applying the map on top gives fminus1(S) isin P (A) and thenapplying the map on the left gives χfminus1(S) isin Mor(A 0 1) On the other hand applying the mapon the right to S isin P (B) gives χS isin Mor(B 0 1) and then applying the map on the bottomgives χS f isin Mor(A 0 1) The two resulting functions χfminus1(S)χS f in the lower left are

χfminus1(S)(x) =

9830831 x isin fminus1(S)

0 otherwiseand χS(f(x)) =

9830831 f(x) isin S

0 otherwise

so these are the same since x isin fminus1(S) if and only if f(x) isin S Thus the diagram above commutesso P is naturally isomorphic to h01 and hence P is representable

Power set functor covariant version Now consider the covariant power set functor stilldenoted by P Set rarr Set which again sends a set to its power set but now sends a functionf A rarr B to the image function Pf P (A) rarr P (B) defined by S 983041rarr f(S) In order for this tobe representable there would have to exist a set X such that P sim= hX which translates into havingnatural bijections

P (S) sim= Mor(XS)

for all sets S However in this case there is no natural choice for X as there is no natural way todetect subsets of S by mapping into S as opposed to the case of P (S) sim= Mor(S 0 1)

That no such X exists can be made precise using a simple argument when S = empty P (S) hascardinality 1 soX would have to be empty as well since otherwise Mor(XS) would have cardinality0 but if X is empty then Mor(empty S) always cardinality 1 regardless of S which is clearly not trueof P (S) Thus the covariant power set functor is not representable

Forgetful on Top The forgetful functor F Top rarr Set is representable Indeed a representingobject would have to satisfy

X sim= Map(objectX)

as sets for any topological space X (where Map denotes the set of continuous maps) and it iseasy to see that a single point satisfies this property a continuous map f pt rarr X is completelydetermined by the element f(pt) of X and any such element gives a continuous map in this wayOf course checking that F sim= hpt requires checking that various diagrams commute but this issimple to work out in this case so we omit it here

Extracting a topology With an eye towards the idea that the functors hA hA should captureall information about an object A in a category note that the result above shows that in TophX first of all captures all information about X as a set since the set X can be extracted fromhX(pt) = Map(ptX)

We claim that hX actually captures all information about X as a topological space since thetopology on X can be extracted from hX in the following way As we saw in the power set examplewe have a bijection

P (X) sim= Mor(X 0 1)

Now equip 0 1 with the topology in which 1 is open but 0 is not Then a continuous mapf X rarr 0 1 is fully determined by the preimage fminus1(1) since to be continuous only requiresthat this single preimage be open in X Moreoever any open subset of X arises in this way bytaking f to be the indicator function of that open set Thus we get a bijection

Op(X) sim= Map(X 0 1)

19

where Op(X) denotes the set of open subsets of X ie the topology on X Hence the topology onX can be extracted from hX as claimed so hX and hX do encode all information about X

This fact can also be interpreted as saying that the contravariant functor which sends a topo-logical space X to its collection of open subsets (and which sends a continuous map to the mapinduced by taking preimages) is representable with representing object 0 1 equipped with thetopology given above This will be worked out in a homework problem

Forgetful on Grp The forgetful functor Grp rarr Set is also representable In this case arepresenting object should be a group giving set-theoretic bijections

G sim= Hom(objectG)

for all groups G where Hom denotes the set of group homomorphisms The group Z works

G sim= Hom(Z G)

since a homomorphism φ Z rarr G is completely determined by φ(1) since Z is generated by 1 andthere are no restrictions on what φ(1) isin G can actually be The inverse of the desired bijection isthus φ 983041rarr φ(1) and it is simple to check that morphisms behave appropriately so that the forgetfulfunctor in this setting is indeed isomoprhic to hZ

Forgetful on Ring The forgetful functor Ring rarr Set is also representable where by Ring wemean the category of rings with identity elements and where morphisms (ie ring homomorphisms)are required to preserve identities Here we need an object satisfying

R sim= Hom(object R)

for all R isin Ob(Ring) and we can see that Z[x] the ring of polynomials in the variable x withcoefficients in Z works Indeed given a ring homomorphism φ Z[x] rarr R the fact that φ isrequired to preserve identities fully determines φ(n) for any n isin Z (since φ(n) = nφ(1)) and hencesince φ preserves multiplication its behavior is fully determined by φ(x) which can be any elementof R The inverse of the bijection we want is thus φ 983041rarr φ(x) and it will follow that this forgetfulfunctor is isomorphic to hZ[x]

Equivalences between Categories

Definitions iso and equiv Now that we have a notion of a ldquomorphismrdquo between categorieswe can talk about the sense in which two categories are the ldquosamerdquo The first possible definitionis the ldquoobviousrdquo one you might expect given the definition of ldquoisomorphismrdquo wersquove seen in everyother context until now we say that C is isomorphic to D if there exist functors F C rarr D andG D rarr C such that G F = idC and F G = idD where idC and idD are identity functors

However this definition turns out to be too restrictive Indeed the requirement that the givencompositions equal identity functors says that given an object A in C the object G(F (A)) beliterally the same as A itself and similarly for objects in D But we know that two objects in acategory can be thought of as being the ldquosamerdquo without being literally the same as long as theyare isomorphic For instance productscoproducts are not unique in the literal sense but only inthe sense that they are isomorphic in a unique way To take another example the definition ofa functor being representable does not say that F (B) literally be the same as hB(A) only thatthey be isomorphic for sure the power set P (S) of a set S is not literally the same as the set offunctions S rarr 0 1mdashit is only the ldquosamerdquo in the sense that they encode the same data

20

Thus it makes sense to relax the requirement that A andG(F (A)) in the definition of isomorphiccategories be equal to the requirement that they be isomorphic This leads to the following betterdefinition of two categories being the ldquosamerdquo C is equivalent to D if there exist functors F C rarr D

and G D rarr C such that G F sim= idC and F G sim= idD So we require only that for instance thefunctor G F be naturally isomorphic to the identity functor on C which says that for any objectA G(F (A)) be isomorphic to A in a natural way Wersquoll see that this truly is the better notionof what it means for two categories to be the ldquosamerdquo and it reflects the idea that all categoricalproperties of one category can be transported to the other

Alternate characterization of equivalences A functor F C rarr D implementing an equiv-alence between categories is called unsurprisingly and equivalence and the ldquoinverserdquo functorG D rarr C is still referred to as being its inverse even though it is not technically an ldquoinverserdquoin the sense that compositions give literal identities But as in the case of Set where ldquoinvertiblerdquocan be characterized without reference to an inverse as ldquoinjectiverdquo and ldquosurjectiverdquo so too canequivalences between categories be characterized without explicit reference to an inverse

A functor F C rarr D is an equivalence if and only it is full faithful and essentiallysurjective

We have seen the notions of full and faithful before and essentially surjective means that any objectin D is isomorphic to something in the image of F for all D isin Ob(D) there exits C isin Ob(C) suchthat F (C) sim= D Justifying this result was the topic of a student presentation and can be foundin the book in Section 15 (Technically to construct an ldquoinverserdquo of F requires that we work withsmall categories but whatever) The bulk of the real work goes into showing that the ldquoinverserdquofunctor constructed is actually a functor which comes down to some abstract nonsense argumentvia commutative diagrams

Equivalences preserve everything We previously saw the notions of full faithful and es-sentially surjective back when thinking about what properties a functor could have which wouldguarantee it sent products to products so the result we proved back then was that equivalencespreserve products Similarly equivalences preserve coproducts and the proof is basically the sameafter reversing arrows A clean way of deriving this is as follows if F C rarr D is an equivalencethen so is F op Cop rarr Dop so since F op preserves products F preserves coproducts becausecoproducts in a category become products in the opposite category

In general an equivalence between categories will preserve whatever categorical notions youwant monomorphisms epimorphisms initial objects terminal objects etc so that equivalentcategories really do have essentially the ldquosamerdquo properties

Equivalence example Here is a first example (also found in the book) of equivalent categorieswhich are not isomorphic Let FVectR be the category whose objects are finite-dimensional realvector spaces and whose morphisms are linear transformations Let MatR be the category ofmatrices defined as follows

bull objects in MatR are nonnegative integers

bull a morphism n rarr m is an mtimes n matrix with real entries

bull composition in MatR is given by matrix multiplication given B n rarr m and A m rarr kmdashsoB is an mtimes n matrix and A is a k timesm matrixmdashthe composition A B n rarr k is the k times nmatrix AB

bull identity morphisms in MatR are given by identity matrices

21

We claim that FVectR and MatR are equivalent First note that they certainly cannot be isomo-prhic isomorphisms in MatR exist only between two objects which are literally the same (ie anmtimesn matrix can be invertible only when m = n) but isomorphisms in FVectn can exist betweenobjects which are not literally the same So there are in a sense not enough objects in MatR toallow it to be isomorphic to FVectR

To construct an equivalence choose one and for all an isomorphism TV V rarr RdimV forevery object in FVectR or equivalently choose once and for all a basis for every V DefineF FVectR rarr MatR by

V 983041rarr dimV on objects and

(T V rarr W ) 983041rarr (the standard matrix of SW T Sminus1V RdimV rarr RdimW ) on morphisms

Equivalently F sends T to the matrix of T relative to the chosen bases for V and W It isstraightforward to check that this is a functor The inverse functor G MatR rarr VectR is givenby

n 983041rarr Rn on objects and

(mtimes n matrix A) 983041rarr (the linear transformation A Rn rarr Rm induced by A) on morphisms

One can check that G is a functor and that F G and G F are naturally isomorphic to identitiesor instead one can argue that F alone is full faithful and essentially surjective The point is thatG(F (V )) gives back RdimV so not literally V itself but only something isomorphic to V which iswhy this gives an equivalence of categories and not an isomorphism

Yoneda Lemma Functors as Objects

Compact Hausdorff spaces and Clowast-algebras Last time we gave an example of equivalentcategories which were not isomorphic but in many ways that example is unsatisfactory when doinglinear algebra people for sure often work with matrix representations of linear transformations asthe example from last would suggest but no one in their right mind literally thinks about suchthings in terms of the equivalence FVectR rarr MatR itself In other words this equivalence isessentially ldquocooked uprdquo for the sake of giving an example but it does not really give any insightinto the actual mathematics of linear algebra

So with that in mind we will describe another equivalence which actually does have some realmeaning in the sense of producing new ways of thinking about some mathematics Given a compactHausdorff space X we let C(X) denote the space of continuous complex-valued functions on X

C(X) = f X rarr C | f is continuous

We can equip the set C(X) with various additional structures First it is a complex vector spaceunder ordinary addition and scalar multiplication of functions Second we can multiply two func-tions together

(fg)(p) = f(p)g(p)

which turns C(X) into whatrsquos called an algebra over C (We wonrsquot give the definition of anldquoalgebrardquo but the point is that this multiplication is in some sense ldquocompatiblerdquo with the vectorspace structure) Next any element f of C(X) has a conjugate element f obtained by takingcomplex conjugates of the values of f

f(p) = f(p)

22

And finally we can equip C(X) with a norm via

983042f983042 = suppisinX

|f(p)|

where |middot| denotes complex absolute value (This supremum exists sinceX is compact by the ExtremeValue Theorem) All of these structures (vector space algebra multiplication conjugation norm)turn C(X) into whatrsquos called a Clowast-algebra We wonrsquot define this precisely but the definition encodesvarious ways in which these structures are compatible Actually C(X) is a unital commutative Clowast-algebra where ldquounitalrdquo means that the multiplication has an identity element (namely the constantfunction 1) and commutative means that fg = gf There is a natural notion of homomorphismbetween Clowast-algebras which are maps between them which behave nicely with all the structuresinvolved In particular given a continuous map f X rarr Y between compact Hausdorff spaces weget a Clowast-algebra homomorphism flowast C(Y ) rarr C(X) given by pre-composition with f flowast sendsg isin C(Y ) to g f isin C(X)

We thus have a contravariant functor CHaus rarr ucClowast-Alg from the category of compactHausdorff spaces and the category of unital commutative Clowast-algebras It turns out that anyunital commutative Clowast-algebra arises in this way in the sense that given any unital commutativeClowast-algebra B there exists a compact Hausdorff space X such that C(X) sim= B as Clowast-algebrasand that morphisms between these algebra arises from continuous maps in the manner describeabove This says that the functor above is essentially surjective full and it turns out that it isalso faithful so that it is an equivalence Thus all information about compact Hausdorff spaces iscompletely encoded in information about unital commtuative Clowast-algebras and indeed one couldlearn everything there is to know about a compact Hausdorff space X from its Clowast-algebra C(X)of functions The fact that the categories CHaus and ucClowast-Alg are equivalent is the startingpoint in the subject known as noncommutative geometry which wersquoll say something about a bitlater on If we drop the requirement that our algebras be unital it turns out that the categoryof commutative Clowast-algebras in general is equivalent to the category of locally compact Hausdorffspaces via essentially the same functor where we take as morphisms in the category of locallycompact Hausdorff spaces to be continuous functions which ldquovanish at infinityrdquo

We are only introducing Clowast-algebras to give an example of an interesting and actually usefulequivalence between categories but just for the fun of it wersquoll note the following Examples ofnoncommutative Clowast-algebras come from what are known as ldquoalgebras of bounded operators onHilbert spacesrdquo which show up in quantum mechanics as the things which describe physicallyobservable quantities like position momentum and energy (In fact all Clowast-algebras arise fromHilbert spaces in this way) Thus Clowast-algebras show up in mathematical formulations of quantummechanics and the notion of noncommutative geometry mentioned above is at the core of someattempts to understand the elusive concept known as ldquoquantum geometryrdquo

Yoneda Lemma The Yoneda Lemma states that given a (covariant) functor F C rarr Set whereC is locally small for any A isin Ob(C) there exists a bijection

F (A) sim= Nat(hA F )

where the right side denotes the set of natural transformations from hA to F A similar result holdsin the contravariant case where we get bijections F (A) sim= Nat(hA F ) (There is also a sense inwhich these bijections are themselves natural but wersquoll ignore this bit) The proof of the YonedaLemma was the topic of a student presentation and can be found in Section 22 of the book

Here we will get a bit philosophical and talk about what the point of the Yoneda Lemmaactually is and later the sense in which it says that we can view arbitrary functors as ldquonon-existent

23

objectsrdquo in a category First we mention what we will call the ldquofunctor of pointsrdquo approach tomathematics a mathematical object can be fully characterized by all morphisms into it That isknowledge of Mor(SA) for all S is equivalent to knowledge of A itself Indeed this is an idea wementioned previously in relation to the functors hAmdashwhich we previously said is referred to as beingthe functor of points of Amdashand now we seek to push this idea further In some examples literalpoints of A can be extracted from hA as in the case of the bijection X sim= hX(pt) = Map(ptX) fora topological space X or G sim= hG(Z) = Mor(Z G) for a group G Moreover in these examples therest of the (topological) structure onX or (group) structure on G can also be extracted from hX andhG respectively In general we can view Mor(SA) as in a sense being ldquopoints of A parameterizedby Srdquo which comes from the idea that in say Set Mor(SA) = AS is ldquothe product of S-manycopies of Ardquo so literally points of A ldquoparameterizedrdquo by S In various areas of mathematics anelement of Mor(SA) is thought of as being a ldquogeneralizedrdquo point of A so that the functor of pointshA detects generalized points of A

So an object A is fully encoded by the functor hA Now we claim that morphisms betweenobjects are also encoded by such functors Indeed in the case where F = hB is itself a functor ofpoints the Yoneda result Nat(hA F ) sim= F (A) becomes

Nat(hA hB) sim= hB(A) = Mor(AB)

which says precisely that morphisms between the objects A and B can be obtained solely fromthe functors hA and hB (This fact is one reason why one should prefer the contravariant functorhA to the covariant functor hA in the covariant case Nat(hA F ) sim= F (A) for F = hB becomesNat(hA hB) sim= hB(A) = Mor(BA) so that morphisms between two objects get reversed whencharacterizing them in terms of representable functors)

Pushing our philosophy further we now know that not only objects but also morphisms betweenobjects are fully encoded by the functors hA The final step is to note that we should not only careabout objects and morphisms between them but that an important part of the data of an objectA is also how arbitrary contravariant functors F C rarr Set behave on A and that this too can beobtained from hA precisely via the Yoneda result that F (A) sim= Nat(hA F ) So objects morphismsbetween objects and how arbitrary functors behave on objects are all encoded in knowledge ofthe functors hA so that these functors truly do capture in the most general sense possible allinformation about A

Functors as nonexistent objects The discussion above is all well and good but it still doesnrsquotquite say why we should care about the fact that we can study on object A by studying hA insteadin that we still havenrsquot spoken about what type of insight about A can be gleamed from hA thatwasnrsquot readily apparent in A itself already Now we argue that the true point of the YonedaLemma is that it gives a way to think about arbitrary functors F Cop rarr Set as ldquoobjectsrdquo in C

itself or really as objects in an ldquoenlargementrdquo of C Certainly for those functors F sim= hA which arerepresentable we can indeed think about F as simply being the object A but now we claim thateven non-representable F can be thought of as being ldquononexistentrdquo objects in C The upshot is thatoften in mathematics we come across things which do not actually exist but the Yoneda Lemmagives a way to (indirectly) study such things anyway by interpreting them as functors instead

The starting point of course is that an actual object A is the same as hA as we have explainedabove Thus by analogy we can interpret a non-representable functor F as being an ldquoobjectrdquo in C

as long as we know what it means to ldquomaprdquo into F from other things which are honestly objects in CThat is if F were indeed representable then F (A) sim= Nat(hA F ) sim= Mor(AF ) would characterizethe functor of points of F (here we abuse notation by using the same letter F to denote both arepresentable functor and the object which represents it) so now we take Mor(AF ) = F (A) as

24

the definition of the ldquofunctor of pointsrdquo of the non-existent object F We are thus taking elementsof F (A) as being what it means to ldquomaprdquo from A into F even when F is not representable Thefact that the Yoneda Lemma gives a way to define maps from A into F thus says that F itself isequipped with the same data that a literal object of A would be equipped with so why shouldnrsquotwe interpret F as a being a type of ldquoobjectrdquo in C as well

To phrase this another way note that the Yoneda Lemma gives rise to a functor

C rarr Fun(CopSet)

where Fun(CopSet) denotes the category whose objects are functors from Cop to Set and morphismsare natural transformation defined on objects by A 983041rarr hA and on morphisms by the bijectionMor(AB) sim= Nat(hA hB) arising from the Yoneda Lemma That this is actually bijection saysthat this functor is full and faithful so that we can view C as being a subcategory of the functorcategory Fun(CopSet) (We cay say that C is isomorphic to the subcategory of Fun(CopSet)consisting of the functors hA or that it is equivalent to the subcategory of Fun(CopSet) consistingof representable functors This functor C rarr Fun(CopSet) is usually called the Yoneda embeddingof C into Fun(CopSet)) Thus we can view Fun(CopSet) as an ldquoenlargmentrdquo of C and the ideawe are trying to make sense of is that other objects in this ldquolargerrdquo category should be interpretedas ldquonon-existent objectsrdquo of C

Just how far can we actually push this idea that arbitrary functors Cop rarr Set should be thoughtof as being ldquoobjectsrdquo of C Pretty far in fact For instance often times in geometry one comesacross a type of ldquosmoothrdquo space on which some group acts and one is interested in doing ldquocalculusrdquoon the resulting quotient space which is the space of orbits under the given group action Theproblem is that such quotient spaces are very often (usually in fact) not ldquosmoothrdquo themselves sothat the tools from calculus are not readily available The solution is to nonetheless treat thisquotient as if it were ldquosmoothrdquo and attempt to do calculus anyway To make this precise thisldquonon-smoothrdquo space actually turns out to have a natural interpretation as a ldquosmooth functorrdquo andit is through this functor which we can view this non-smooth space as actually being smooth afterall so we are forced to expand our notion of what we mean by ldquosmooth spacerdquo by allowing thingswhich are not spaces in the usual sense which certainly leads to more abstraction but the benefitis that it actually makes sense to do calculus on such ldquospacesrdquo interpreted as functors

The discussion above is pretty vague so here is one final more concrete example Recall thatequivalence

CHausop rarr ucClowast-Alg

between the category of compact Hausdorff spaces and the category of unital commutative Clowast-algebras mentioned at the start We now claim that a non-commutative (unital) Clowast-algebra shouldbe interpreted as the algebra of functions on some non-existent ldquononcommutative compact Haus-dorff spacerdquo and it is this idea which the subject of noncommutative geometry makes precise Solet B be a non-commutative Clowast-algebra Then the functor of points hB gives a functor

ucClowast-Alg rarr Set

which sends C(X) (where X is some compact Hausdorff space) to Hom(C(X) B) where Hom de-notes homomorphisms of Clowast-algebras whatever that means Composing with the previous functorwe get a functor wersquoll denote by B itself

B CHausop rarr Set

which on objects at least is given by X 983041rarr Hom(C(X) B) This functor is not representable sinceB is not commutative but the philosophy espoused above suggests we should think of it as giving a

25

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

Introduction to Categories

Category theory provides a framework through which we can relate a constructionfact in one areaof mathematics to a constructionfact in another The goal is an ultimate form of abstractionwhere we can truly single out what about a given problem is specific to that problem and whatis a reflection of a more general phenomenom which appears elsewhere Practically this is done byphrasing as many constructionsfacts as possible in terms of ldquoarrowsrdquo where the point is that bydoing so relations between different objects of mathematics are simpler to see

Definition of a Category The precise definition of a category is spelled-out in the book inSection 11 The key points is that a category consists of a collection of ldquoobjectsrdquo a collection ofldquomorphismsrdquo between objects and a way to ldquocomposerdquo morphisms to get other morphisms in away which is associative and admits identities

Standard examples The main examples of categories wersquoll care about can also be found in thebook such as

bull Set the category of sets where morphisms are given by ordinary functions

bull Grp the category of groups where morphisms are given by group homomorphisms

bull Top the category of topological spaces where morphisms are given by continuous maps

bull Vect the category of vector spaces over some fixed field (which should be included as partof the notation) where morphisms are given by linear transformations

Another example All of the above are examples of what are called concrete categories This is aterm wersquoll define later but it essentially means that the objects of the categories above are all setspossibly equipped with additional structure and that the morphisms are just ordinary functionsbetween sets possibly satisfying some additional condition (For instance an object in Grp is a setequipped with a group operation and a morphism in Grp is simply a function f G rarr H betweengroups which satisfies the additional condition that it be a group homomorphism)

But even though these types of examples are the main ones wersquoll care about a category canbe a much more general type of object For instance let us describe the category called BZ Thiscategory has only one object which we will denote by lowast and think of as a single ldquopointrdquo Accordingto the definition of a category we should then have for this one object a collection of morphismsMor(lowast lowast) which in this case we declare to be Z

Mor(lowast lowast) = Z

That is we interpret each element of Z as being a ldquomorphismrdquo from lowast to lowast Now of course thereis only one possible function from lowast to lowastmdashnamely the one that sends the single element lowast of thedomain to the single element lowast of the codomainmdashso here we are considering ldquomorphismsrdquo whichdo not represent ordinary functions Visually we interpret each morphism as an arrow beginningand ending at the single object lowast

lowast

0

1

2

3

2

Now in order for this to be category we also need a notion of ldquocompositionrdquo which in this casesince there is only one object amounts to a map of the form

Mor(lowast lowast)timesMor(lowast lowast) rarr Mor(lowast lowast)

which in our case becomesZtimes Z rarr Z

We take this map to be ordinary addition + on Z (So ldquocomposingrdquo the arrow labeled 1 in thepicture above with the arrow labeled 2 results in the arrow labeled 3) The point is that this isNOT ldquocomposition of functionsrdquo in the usual sense but will give a valid ldquocompositionrdquo operationin the category BZ as long as it satisfies the properties required in the definition of a category Onesuch requirement is that composition be associative which is satisfied since + is indeed associativeThe second requirement is that exist a morphism idlowast isin Mor(lowast lowast) which acts as an identity for theldquocompositionrdquo wersquove defined In this case this identity morphism is idlowast = 0 that is the integer0 in Mor(lowast lowast) = Z Saying that this is the ldquoidentity morphismrdquo with respect to ldquocompositionrdquo inthis case simply says that is the identity element for addition

Thus we get a category BZ as claimed albeit one where the notion of ldquomorphismrdquo and ldquocompo-sitionrdquo are more general than simply ldquofunctionrdquo and ordinary composition In fact we can alreadynow give a ldquocategoricalrdquo characterization of a well-known algebraic object a (locally small) cat-egory one with a single object is precisely what is known as a monoid which is a set equippedwith an associative binary operation which admits an identity element The monoid in questionis precisely the set Mor(lowast lowast) and the associative operation which turns this into a monoid is theldquocompositionrdquo of the category in question (A locally small category is one where each collectionof morphisms is actually a set which is needed here in order to guarantee that Mor(lowast lowast) is a set onwhich we can define a binary operation) So for instance any group can be viewed as a categorywith one object where the group elements give the ldquomorphismsrdquo from that one object to itself(Of course a group gives a special type of category where every morphism is in fact ldquoinvertiblerdquoA small category where every morphism is invertible is called a groupoid so a group is nothing buta groupoid with one object Wersquoll talk more about groupoids later on)

A final example Here is one final example wersquoll consider from time to time which is again acategory where the ldquomorphismsrdquo are not simply functions The category Rel of relations has asobjects ordinary sets but as morphisms relations between sets

f isin Mor(AB) is a relation from A to B which is a subset f of AtimesB

The usual graph of a function from A to B gives an example of such a relation so we can say thatRel contains Set as a subcategory but not all relations come from graphs of functions In generalwe can think of a relation as a ldquopartially-defined possibly multivalued functionrdquo if we interpret(a b) being in the relation f as saying that ldquof sends a to brdquo

Composition in Rel comes from ordinary composition of relations given a relation f A rarr Band a relation g B rarr C we define their composition g f A rarr C to be

g f = (a c) isin Atimes C | there exists b isin B such that (a b) isin f and (b c) isin g

So g f consists of all pairs of elements of A and C which are ldquorelatedrdquo to a common elementof B In the case where f and g are graphs of ordinary functions this composition of relationsreproduces the ordinary notion of composing functions if (a b) isin f and (b c) isin g then b = f(a)

3

and c = g(b) (here we abuse notation by denoting the function of which f is graph by the symbolf itself) so

c = g(b) = g(f(a)) = (g f)(a)

meaning that (a c) isin g f precisely when c = (g f)(a) Composition of relations is associativeand you can check that graphs of ordinary identity functions

graph(idA) = (a a) isin AtimesA | a isin A

give the identity morphisms (I spent much time as graduate student trying to do ldquogeometryrdquo witha certain category of relations so this is an example near and dear to my heart)

Special Morphisms Products

Isomorphisms A morphism f isin Mor(AB) between two objects A and B in a category is anisomorphism or is invertible if it has an inverse there exists a morphism g isin Mor(BA) such thatgf = idA and fg = idB where idA isin Mor(AA) and idB isin Mor(BB) are the identity morphismswhich are assumed to exist as part of the definition of a category We say that A and B areisomorphic and we write A sim= B if there exists an isomorphism between them

This general categorical notion of isomorphism reduces to the one you would expect in thestandard examples isomorphisms in Set are bijective functions in Grp they are bijective grouphomomorphisms in Vect they are invertible linear transformations and in Top they are homeo-moprhisms ie continuous bijections with continuous inverses

A category of metric spaces Now consider the category MetC of metric spaces where mor-phisms are given by continuous functions In this category the notion of ldquoisomorphismrdquo againmeans homeomorphism so two metric spaces are isomorphic in this category when they are home-omorphic However often when considering whether or not two metric spaces are the ldquosamerdquo wehave another definition in mind that of two metric spaces being isometric X is isometric to Y ifthere exists a bijective isometry between them which is a bijective map f X rarr Y which preservesdistance in the sense that

dY (f(p) f(q)) = dX(p q) for all p q isin X

After all in a metric space you have more data than simply ldquoopen setsrdquo (which is all that ahomeomorphism detects) so it would be good to have a notion of ldquosamenessrdquo which incorporatedthe extra data of distance as well

The goal is then to construct a category of metric spaces where ldquoisomorphicrdquo indeed meansisometric and we can do so simply by restricting the types of continuous maps we consider Thecategory Met has metric spaces as objects but now a morphism f X rarr Y is taken to be afunction which does not ldquoenlargerdquo distance in the sense that

dY (f(p) f(q)) le dX(p q) for all p q isin X

Maps with this property are often called metric maps Any metric map is actually continuous soMet is a subcategory of MetC (The subscript C used in MetC is meant to denote ldquocontinuousrdquoindicating that we take arbitrary continuous maps as morphisms) In this category now for f X rarr Y to be an isomorphism requires that f and its inverse fminus1 both be metric maps so

dY (f(p) f(q)) le dX(p q) and dX(fminus1(s) fminus1(t)) le dY (s t)

4

for all p q isin X and s t isin Y But writing s and t as s = f(p) and t = f(q) the second conditionbecomes dX(p q) le dY (f(s) f(t)) and the two inequalities together thus imply

dY (f(p) f(q)) = dX(p q) for all p q isin X

so that f X rarr Y is actually an isometry The moral is that certain properties one might wantcan be achieved by changing the morphisms you consider

Monomorphisms and epimorphisms The definitions of monomorphism and epimorphismcan be found in Section 12 of the book The standard examples are those in Set where themonomorphisms are the injective functions and the epimorphisms are the surjective functionsProofs of these facts are left to the homework as well as the interesting problem of determining theepimorphisms in Met or even MetC or Haus (the category of Hausdorff spaces where morphismsare arbitrary continuous maps) where the answer is that epimorphism means more than simplyldquocontinuous and surjectiverdquo (As opposed to all of Top where epimorphism does indeed meancontinuous and surjective)

So monomorphisms and epimorphisms should be treated as categorical analogues of ldquoinjectionsrdquoand ldquosurjectionsrdquo although this analogy only goes so far The thing to note is that we are thusable to give a complete definition of ldquoinjectiverdquo for ordinary functions which makes no referenceto elements and is stated only in terms of the relation between an injective function and otherfunctions certainly proving that ldquomonomorphismrdquo means ldquoinjectionrdquo in Set requires working withelements but stating the definition of an injection as a monomorphism can be done solely usingldquoarrowsrdquo

Products The definition of a product in a category shows up in Section 31 of the book inthe context of the more general notion known of a limit Wersquoll discuss this more general notioneventually but for now we will only focus on products of two objects at a time Here is what themore general notion boils down to in this case which wersquoll take as our definition

A product of two objects A and B in a category is an object P together with morphismspA P rarr A and pB P rarr B (which we call projections) which satisfy the followingproperty given any morphisms f C rarr A and g C rarr B into A and B from acommon object C there exists a unique morphism h C rarr P such that f = pA h andg = pB h We express this by saying that the following diagram commutes

C

A B

Pf g

pA pB

existh

which means that composing any two composable arrows in the diagram results in theother arrow with the same domain and codomain Concretely there are two ways in thisdiagram to get from C to A via f or via the composition pA h and the requirementis that both give the same morphisms (ie f = pA h) and similarly for the two waysto get from C to B The dashed arrow is the one which is required to exist (uniquelyas denoted by the exclamation mark ) in the definition of product

5

So the upshot is that for any C and any f and g in the diagram above there exists a uniqueh which makes the diagram commute The point is that a product gives a way to turn the twopieces of data f C rarr A and g C rarr B into the single piece of data h C rarr P in a uniqueway Being able to turn a possibly large amount of data into a single piece of data in this way isone of the benefits the ldquocategoricalrdquo perspective provides Products in categories are often denotedusing the usual A times B notation even though it is not necessarily true that a product is literallya set-theoretic Cartesian product although it is in the standard examples The map h C rarr Prequired in the definition is then often denoted by h = f times g

Examples In Set the product of two objects A and B always exists and is indeed the ordinaryCartesian product A times B But of course the definition of product also requires that data of theprojection morphisms which in this case are simply the usual ones

pA (a b) 983041rarr a and pB (a b) 983041rarr b

Given f C rarr A and g C rarr B the map h C rarr AtimesB is h(c) = (f(c) g(c)) and you can checkthat these constructions satisfy the properties required in the definition (Note that you would alsohave to argue that h defined in this way is the only map which can make the required diagramcommute)

Products in Grp are given by the usual direct product group structure GtimesH with the groupoperation defined as

(g1 h1) middot (g2 h2) = (g1g2 h1h2)

The projection morphisms are the ordinary ones you would expect as well But let us actuallynow verify that this group structure indeed is among other possible group structures on G times Hthe one required in the definition of a product Suppose f K rarr G and g K rarr H are grouphomomorphisms The key point is that the map h(k) = (f(k) g(k)) which should satisfy therequired product property should itself be a morphism in Grp meaning a group homomorphismand we claim that this is only true for the direct product group structure

Indeed we want it to be true that

h(k1k2) = (f(k1k2) g(k1k2)) equal h(k1)h(k2) = (f(k1) g(k1)) middot (f(k2)g(k2))

for all k1 k2 isin K and some to-be-determined group structure middot on G times H Since f and g arehomomorphisms this requirement becomes

(f(k1)f(k2) g(k1)g(k2)) = (f(k1) g(k1)) middot (f(k2)g(k2))

which says that middot should indeed be componentwise multiplication at least on the image of f andg But recall that the properties required of a product should hold for any K and any f K rarr Gand g K rarr H in particular then it should hold for instances when f and g are surjective whichshows that the componentwise-multiplication property above should in fact hold on all of G timesHso that middot is indeed the ordinary direct product structure

Uniqueness Notice that in the definition of a product in a category we used the word ldquoardquo asopposed to ldquotherdquo we spoke about a product of A and B instead of the product of A and B Sowe now ask are products unique The answer is ldquonordquo if we interpret ldquouniquerdquo in the strictestpossible sense but ldquoyesrdquo if we interpret it correctly For instance in the example of Grp takenow N to be any group which is isomorphic to the direct product G timesH say with isomorphismℓ N rarr G timesH We claim that we can also turn N into a categorical product of G of H as long

6

as we define appropriate ldquoprojectionsrdquo N rarr G and N rarr H take N rarr G to be the compositionpG ℓ and N rarr H to be pH ℓ Given f K rarr G and g K rarr H as in the definition of a productthe map hprime K rarr N defined by

hprime = ℓminus1 h

where h K rarr GtimesH is the usual h = f timesg will satisfy the requirement needed to able to concludethat N is also a product of G and H in Grp A similar trick works in any category so that anyobject isomorphic a product can also itself be turned into a product

The correct answer is that products are unique up to unique isomorphism which means thatgiven a product P of A and B any other product P prime of A and B will in fact be isomorphic to P and in a unique way in the sense that there can exist only one isomorphism P rarr P prime The phraseldquounique up to unique isomorphismrdquo is a common one wersquoll see again and again objects definedby some categorical property are almost never genuinely unique in a strict sense but they willessentially be as close to unique as possible

So suppose that P and P prime are both products of A and B with associated projection morphismspA pB for P and pprimeA p

primeB for P prime Consider the diagram

P

A B

P primepA pB

pprimeA pprimeB

existh

The morphism h P rarr P prime is the unique one guaranteed to exist by the fact that P prime is a productof A and B Commutativity of this diagram says that pA = pprimeA h and pB = pprimeB h A similardiagram with the roles of P and P prime reversed which uses the fact that P is a product results in aunique morphism hprime P prime rarr P satisfying pprimeA = pA hprime and pprimeB = pB hprime

Now consider the following diagram

P

A B

PpA pB

pA pB

hprime h

This commutes which we verify using the properties h and hprime satisfy as follows

pA (hprime h) = (pA hprime) h = pprimeA h = pA

and similar for the morphisms involving B This shows that hprimeh P rarr P satisfies the requirementin the definition of P being a product But of course idP P rarr P satisfies this requirement as wellso since the map satisfying this requirement is assumed to be unique we must have hprime h = idP Similar reasoning considering h hprime P prime rarr P prime shows that h hprime = idprimeP so that h and hprime are indeedinverses and hence that P and P prime are isomorphic The uniqueness of the isomorphism betweenthem comes from the uniqueness of h in the construction above

7

Abstract nonsense It is common to say in the setting above that the fact that products areunique up to unique isomorphism follows from ldquoabstract nonsenserdquo which is a succinct way ofsaying that it follows solely from the definition of a categorical product itself and not any deepermathematical content In particular many of the ldquouniquenessrdquo properties which ldquoproductsrdquo havein the settings we might care aboutmdashCartesian products in Set direct products in Grp producttopologies in Topmdashreally have nothing to do with how those specific products are defined but arejust consequences of the ldquoabstract nonsenserdquo of category theory

But of course we do not use the phrase ldquoabstract nonsenserdquo in a negative or dismissive waysince recognizing what types of properties are consequences of ldquoabstract nonsenserdquo goes a long waytowards understanding well the given problem at hand Soon you too will come to appreciate thisnotion of abstract nonsense

Coproducts Opposite Categories

Initial and terminal objects The notions of initial and terminal objects in a category aredefined in Section 16 of the book where you can also find standard examples Here I want topoint out that initial and terminal objects if they exist are unique up to unique isomorphism asanother example of arguing via ldquoabstract nonsenserdquo

Suppose I and I prime are both initial objects in some category By the fact that I is initial thereexists a unique morphism f I rarr I prime and by the fact that I prime is initial there exists a unique morphismg I prime rarr I But then the composition g f I rarr I must equal idI I rarr I since the latter is theonly morphism from I to itself because I is initial and similarly f g = idIprime Thus I and I prime areunique and there is a unique isomorphism between them

Coproducts The dual concept to that of a product is the notion of a coproduct (The termldquodualrdquo refers to the phenomena you get when you reverse all arrows in a given construction Forinstance the notion of a terminal object is dual to that of an initial object and the notion of anepimorphism is dual to that of a monomorphism) Here is the precise definition a coproduct of anobject A and an object B in a category is an object C together with morphisms iA A rarr C andiB B rarr C such that for any object D and morphisms f A rarr D and g B rarr D there exists aunique morphism h C rarr D which makes the following diagram commute

D

A B

Cf g

iA iB

existh

By abstract nonsense if a coproduct of A and B exists it is unique up to unique isomorphism

Examples In Set the coproduct of two sets A and B is the disjoint union A⊔B (If you havenrsquotseen the notion of a disjoint union before it is essentially the same as the union only that we donrsquotcare about whether A and B have any overlap if they do we treat the common elements as beingldquodifferentrdquo For instance Z ⊔ Z is not Z but consists of two copies of each integer one 0 comesfrom the first copy of Z and another comes from the second but these are different ldquozeroesrdquo)Given f A rarr D and g rarr D the unique map A ⊔B rarr D is defined by applying f to elements of

8

A and g to elements of B (Note that if we simply took the ordinary union AcupB as the candidatefor the coproduct when A and B were not disjoint the required map h A cup B rarr D would notexist since h would have to send any element x isin A cap B to both f(x) and g(x) in order to makethe required diagram commutative)

The coproduct of two objects X and Y in Top is again the disjoint union X ⊔ Y equippedwith the disjoint union topology an open subset of X ⊔ Y is defined to be the (disjoint) union ofan open subset of X with an open subset of Y The coproduct of two groups G and H in Grp isthe free product G lowastH which is further elaborated on in the homework

The coproduct of two vector spaces V and W in Vect is the direct sum V oplusW which is definedto be the set of formal sums

v + w where v isin V and w isin W

where addition and scalar multiplication are defined ldquocomponentwiserdquo

(v1 + w1) + (v2 + w2) = (v1 + v2) + (w1 + w2) and r(v + w) = rv + rw

This is isomorphic to the product V times W with operations also defined componentwise but it iscustomary to speak about the ldquodirect sumrdquo V oplusW when referring to the coproduct as opposed tothe product wersquoll say why in a bit The maps iV V rarr V oplusW and iW W rarr V oplusW required aspart of the data of a coproduct are

iV v 983041rarr v + 0W and iW w 983041rarr 0V + w

To verify that V oplusW is the correct coproduct take any linear transformations S V rarr U andT W rarr U where U is some other vector space We then need a (unique) linear transformationh V oplusW rarr U such that

S = h iV and T = h iW

Let us determine what this map must be thereby not only showing that it exists but also that itis unique Let v + w isin V oplusW which we can write as

v + w = (v + 0W ) + (0V + w)

Since h is required to be linear we have

h(v + w) = h([v + 0W ] + [0V + w]) = h(v + 0W ) + h(0V + w)

The first term at the end is (h iV )v which must thus be Sv and the second term is (h iW )wwhich must be Tw Thus h = S + T is the map

h = S + T v + w 983041rarr Sv + Tw

which you can verify is indeed linear Thus V oplusW does satisfy the property required of a coproductSo the product of two objects in Vect is the same as their coproduct only that we write the

latter using ldquosumrdquo notation Later we will discuss products and coproducts of an arbitrary numberof elements We will see these as special cases of the general notions of limits and colimits butthey are defined via similar requirements as products and coproducts of two objects at a timeIn the category of vector spaces arbitrary products are given by the usual product

983124α Vα with

ldquocomponentwiserdquo addition and scalar multiplication but it turns out that an arbitrary coproductis given by the direct sum

983119α Vα which is defined to be the subspace of

983124α Vα where only finitely

many components are nonzero (This is in a way analogous to how the product topology on an

9

arbitrary number of topological spaces is defined) Wersquoll go over this later but is the underlyingreason why we use oplus instead of times when discussing coproducts

Encoding data via productscoproducts We can nicely summarize the definition of productsand coproducts in the following way the product of A and B is the object with the property thatfor any object D we have

Mor(DA)timesMor(DB) = Mor(D product)

and the coproduct of A and B is the object such that for any object D we have

Mor(AD)timesMor(BD) = Mor(coproductx D)

The first reflects the fact that products give a unique way to turn two morphisms (f and g in thedefinition) mapping into A and B into a single morphism (h in the definition) and the second saysthat coproducts give a unique way to turn two morphisms mapping out of A and B into a singlemorphism

Soon enough we will discuss the idea of universality and the point is that products are ldquouni-versalrdquo for maps into A and B and coproducts are universal for maps from A and B

Opposite categories Finally we give a definition which might seem silly but is actually quiteuseful Given a category C the opposite category is the category Cop is the category with the sameobjects as C but where we defined Mor(AB) in Cop to be Mor(BA) in C To be more concrete weconsider each morphism f A rarr B in C to instead be a morphism fop B rarr A in Cop (Perhaps itwould be better to write this as A larr B fop) The point is that we donrsquot actually care about whatf actually is as say a functioncontinuous maphomomorphismwhatever-we only care about itsbehavior as an ldquoarrowrdquo The opposite category Cop only retains information about how arrowsrelate to one another nothing more

Since arrows are reversed products in C become coproducts in Cop and coproduts in C becomecoproducts in Cop Similarly initialterminal objects and monomorphismsepimorphisms in C be-come terminalinitial objects and epimorphismsmonomorphisms in Cop Again the notion of anopposite category might seem like a strange thing to look at but wersquoll see that it does have uses

Functors Fullness and Faithfulness

One more productcoproduct example Let X be a topological space and define its categoryof open sets Op(X) as follows The objects of Op(X) are simply the open subsets of X and themorphisms are given by inclusions so to be concrete

Mor(U V ) =

983083the unique inclusion U rarr V if U sube V

empty otherwise

The fact that U sube V and V sube W implies U sube W says that the composition of two morphismsin this category is still a morphism in this category The product of U and V in this categoryturns out to be their intersection (with ldquoprojection morphismsrdquo U cap V rarr U and U cap V rarr V beinginclusions) and the coproduct of U and V turns out to be their (ordinary) union Note that weneed X to be a topological space in order to guarantee that U cap V and U cup V are still objects inthis category More generally the product of finitely many objects in Op(X) exists and is theirintersection and the coproduct of an arbitrary number of objects exists and is their union (Wesay that Op(X) has finite products and arbitrary coproducts)

10

Given a set X and a collection T of subsets of X we can define a similar category with elementsof T as objects (I guess we should assume that empty X isin T) and inclusions as morphisms We thenhave that this category has finite products and arbitrary coproducts if and only if T is a topologyon X so that we can rephrase the definition of a topological space itself solely in categorical termsThis perspective on what a ldquotopologyrdquo is makes it simpler to state the definition of whatrsquos calleda sheaf on a topological space which is an important construction in various areas of analysis andgeometry We might discuss this a bit later on

Functors As with most things in mathematics we should care not only about categories asmathematical structures on their own but also about ldquomapsrdquo between these structures whichldquopreserverdquo said structure The notion of a (covariant) functor between categories provides suchldquomapsrdquo in the case at hand where a functor is a way of sending objects to objects and morphismto morphisms in a way which preserves compositions and identities The precise definition is inSection 13 of the book as is the definition of a contravariant functor between categories whichis a functor which reverses the direction of arrows as opposed to ldquocovariantrdquo ones which preservedirections (It is common to use the term ldquofunctorrdquo without qualification to mean one which iscovariant Note that any functor can be made covariant by working with the opposite category acontravariant functor C rarr D is the same as a covariant functor Cop rarr D)

All of the standard examples we looked at are in the book forgetful functors power set functorsthe dual space functor in Vect etc Of particular interest is Example 139 in the book whichshows that various types of group actions in different contexts-the usual notion of a group actionon a set the notion of a continuous action of a group on a topological space and the notion ofa linear action of a group on a vector space (also known as a representation of the group)-are allencoded by a functor with domain category BG

Adjoint functors Let F Top rarr Set be the forgetful functor and let D Set rarr Top be thefunctor which sends a set S to the space S equipped with the discrete topology and a functionS rarr Sprime between sets to the same function only now viewed as a continuous map S rarr Sprime betweendiscrete spaces These two functors satisfy the following property in relation to one another givinga morphism D(S) rarr Y in Top is the same as giving a morphism S rarr F (Y ) in Set or moresymbolically

MorTop(D(S) Y ) = MorSet(S F (Y ))

We say that D is left adjoint to F and that F is right adjoint to D The point is that adjointfunctors can be used to move data back and forth between two categories

Wersquoll talk more about adjoint functors later on but for now here is one more example Nowtake F Grp rarr Set to again be the forgetful functor and take G Grp rarr Set to be the freegroup functor which sends a set S to the free group generated by S G(S) has as elements ldquowordsrdquos1 sn made up of elements of S with the group operation being ldquoconcatenationrdquo A basic factis that giving a group homomorphism from G(S) to some other group H is the same as giving anordinary function from S to F (H) which is the required adjoint property

MorGrp(G(S) H) = MorSet(S F (H))

Faithful and full functors The definitions of what it means for a functor to be faithful and fullare in Section 15 of the book where faithful functors are ones which are injective on morphisms andfull functors are ones which are surjective on morphisms but where in both cases we refer only tomorphisms which occur between the objects F (A) and F (B) in the target category corresponding tofixed objects A and B in the source category (In general a functor could send two objects AB in

11

the source category to the same object F (A) = F (B) in the target category and thus there could betwo morphisms A rarr C and B rarr C which are sent to the same morphism F (A) = F (B) rarr F (C)Such a thing could still possibly be ldquofaithfulrdquo since the non-injectivity here is arising from differentobjects in the source)

Faithfullness and fullness will be nice properties going forward and here is one hint at how theymight be useful Say we have a functor F C rarr D The problem is to determine properties onF which will imply it send products in C to products in D or coproducts in C to coproducts inD So given the former product diagram below we want the second diagram to also be a productdiagram

C

A B

P Fminusminusminusrarr

F (C)

F (A) F (B)

F (P )f g

pA pB

existh

Ff Fg

F (pA) F (pB)

Fh

Of course in order F (P ) to be an actual product of F (A) and F (B) on the right a similarcommutative diagram property would have to hold for all objects in D in place of F (C) and allmorphisms D rarr F (A) and D rarr F (B) not just those of the form Ff and Fg respectively Oneway to guarantee this is to require that all objects in D are of the form F (C) and then for allmorphisms F (C) rarr F (A) and F (C) rarr F (B) to be of the form Ff Fg the former is true whenthe functor F is surjective on objects and the latter is true when F is full We are not saying thatthese are the only conditions under which F might preserve products but it definitely seems togive a sufficient condition at least

Almost there is one more thing to require the uniqueness of the morphism Fh F (C) rarr F (P )making the diagram on the right commute Suppose there were two such morphisms Fh and Fhprimewhere h and hprime are both morphisms C rarr P in C The commutativity of the diagram on the rightgives for instance

Ff = F (pA)Fh and Ff = F (pA)Fhprime

which the functorial properites of F turns into

Ff = F (pA h) and Ff = (pA hprime)

We want h and hprime to make the first diagram commute in order to be able to apply uniqueness ofthe morphism C rarr P But this requires know that f = pA h and f = pA hprime which would followfrom F being faithful In this case it follows that h and hprime both make the first diagram commute(after we go through a similar argument with g) so the fact that P is a product guarantees thath = hprime and hence that Fh = Fhprime so that F (P ) is an actual product on the right

So we conclude that a functor which is full faithful and surjective on objects will send productsto products (Actually we can relax this a bit by requiring not that every object in D be in theliteral image of F but only that it be isomorphic to something in the image A functor with thisproperty is said to be essentially surjective which is a concept will see in a related context soonenough) Again this is not an ldquoif and only ifrdquo statement since functors can preserve products (orcoproducts) without being full faithful and surjective on objects but it at least gives a sufficientcondition Later we will see other ways to guarantee that functors preserve products which arerelated to the existence of adjoints for that given functor

12

Coproduct Examples Concreteness

Products and coproducts for metric spaces The product of two objects in Met the categoryof metric spaces with morphisms given by metric maps is the Cartesian product X times Y equippedwith the box metric and the coproduct of X and Y does not exist if both X and Y are nonemptyThis was the topic of a student presentation and here are some proof sketches

Given metric spaces (X dX) and (Y dY ) the box metric d on X times Y is defined by

d((x1 y1) (x2 y2)) = maxdX(x1 x2) dY (y1 y2))

(The name ldquoboxrdquo metric comes from the fact that in RtimesR = R2 or R3 open balls indeed looks likeldquoboxesrdquo) The key requirement needed in the definition of a product is given morphisms S rarr Xand S rarr Y that the map

h S rarr X times Y defined by h(s) = (f(s) g(s))

actually be a morphism in Met This turns into the requirement that

d((f(s) g(s)) ((f(t) g(t))) = maxdX(f(s) f(t)) dY (g(s) g(t)) le dS(s t) for all s t isin S

given the fact that

dX(f(s) f(t)) le dS(s t) and dY (g(s) g(t)) le dS(s t) for all s t isin S

For the latter inequalities to imply the former d must (at least up to isometry) be given by thebox metric Note that the two other ldquostandardrdquo metrics one might put on a productmdashthe taxicab

dX + dY and Euclidean983156

d2X + d2Y metricsmdashdo not satisfy this required ldquometric maprdquo property

and so do not give the product in MetTo see that coproducts do not exist if XY are nonempty suppose instead (P dP ) was a co-

product for X and Y with inclusion morphisms iX X rarr P and iY Y rarr P For any n isin Ntake 0 n to be a metric space with the absolute value metric d and let f X rarr 0 n andg Y rarr 0 n be the constant function 0 and the constant function n respectively Since P is acoproduct there exists h P rarr 0 n such that

h iX = f and h iY = g

But in order for h to actually be a morphism in Met we must then have

d(h(iX(x) iY (y)) le dP (iX(x) iY (y)) for any x isin X y isin Y

which translates inton = |0minus n| le dP (iX(x) iY (y))

But this should then be true for any n isin N which would imply that the distance in P betweenan element coming from X and one coming from Y should be infinite which is nonsense (Notethat if you altered your definition of ldquometricrdquo to allow for infinite distance then a coproduct wouldexist the disjoint union X ⊔Y where you defined the distance between elements of X and Y to beinfinite)

Coproducts for groups The coproduct of two groups G and H in Grp is the free product GlowastHwhich is defined to be the set of all words consisting of elements of G andH in an alternating fashion

13

with concatenation as the group operation (When we have to elements of G or two elements of Hnext to each other with use their respective group operations to turn these into single elements ofG or H) This was also the topic of a student presentation and here is the basic idea Given grouphomomorphisms g G rarr N and k H rarr N that G lowastH is the coproduct comes down to showingthat the map

h G lowastH rarr N defined by h(g1h1 gnhn) = f(g1)k(h1) middot middot middot f(gn)k(hn)

and similarly on other possible words in GlowastH is a group homomorphism which follows essentiallyfrom the definition of the group operation on G lowastH indeed this group operation arises preciselyfrom this requirement

Now if we instead tried to use GtimesH with its standard inclusions G rarr GtimesH and H rarr GtimesHas the coproduct the issue is that the analogous map

h GtimesH rarr N defined by h((g h)) = f(g)k(h)

would NOT be a homomorphism because h((g1 h1)(g2 h2)) is not equal to h(g1 h1)h(g2 h2) bythe way in which the group operation is defined on the direct product G times H However if westrict our category to only consist of abelian groups so that the N rsquos we consider in the defini-tion of coproduct are also abelian then G times H does serve as a coproduct we can always rewritef(g1)k(g1) middot middot middot f(gn)k(gn) as f(g1) middot middot middot f(gn)k(h1) middot middot middot k(hn) in N Irsquoll leave it to you to flesh out allof these details

Concrete categories The categories Set Top Vect and Grp all have the property thattheir objects are simply sets possibly equipped with extra structure and that their morphisms areordinary set-theoretic functions possibly satisfying some additional requirement Such categoriesare nice since we can often use intuition about sets and functions in their study A concrete categoryis intuitively one where we can think of objects as being ldquosetsrdquo and morphisms as being ldquofunctionsrdquoonly that we have to interpret this in the right way

Here is the definition a concrete category is a category C equipped with a faithful functorF C rarr Set which is part of the data The idea is that such a functor gives us way to turnan object A isin Ob(C) into a set F (A) and a morphism f isin MorC(AB) into a function Ff isinMorSet(F (A) F (B)) where the ldquofaithfulrdquo requirement says that we do not lose information aboutf when doing so so that we can (essentially) learn everything we need to know about f by studyingFf instead In the four examples above the required faithful functor is simply the forgetful functorand indeed in general we think of F as being a ldquoforgetful functorrdquo for the concrete category (C F )

Examples Here are two more examples of concrete categories Let G be a group and considerthe category BG with a single object and elements of G morphisms As written this does notappear to be concrete since morphisms are not honest functions lowast rarr lowast but the point is that thiscategory can indeed by ldquoconcretizedrdquo by specifying an appropriate ldquoforgetful functorrdquo A functorF BG rarr Set is the data of a (left) action of G on a set S = F (lowast) and saying this functor isfaithful is what it means to say that this action is faithful different g h isin G act on S in differentways or equivalently that there exists s isin S such that gs ∕= hs Any group has at least onefaithful left actionmdashnamely the action on itself given by left multiplicationmdashto BG can always beequipped with a faithful functor to Set This is all just a way of phrasing the fact that any groupis isomorphic to a subgroup of a permutation group

The category Rel of relations can also be ldquoconcretizedrdquo even though as written morphismsA rarr B are not honest functions Take F Rel rarr Set to be the power set functor which sends an

14

object in Rel to its power set and a relation f A rarr B to the induced function Ff P (A) rarr P (B)defined by

(Ff)(S) = b isin B | there exists a isin S such that (a b) isin f

In other words if you think of the relation f sube A times B as a ldquopossibly not everywhere-definedpossibly multivalued functionrdquo (Ff)(S) is analogous to taking the image of S That F is faithfulwill be left to the homework

In fact most categories yoursquoll see can be made concrete by specifying an appropriate forgetfulfunctor but not all categories can be made concrete in this way Here is the standard example of anon-concrete category which we mention only to say there is such a thing but which we wonrsquot doanything more with since understanding it better requires knowledge of algebraic topology Thehomotopy category of topological spaces is the category hTop whose objects are topological spacesand whose morphisms are given by equivalence classes of homotopic maps

MorhTop(XY ) = homotopy classes of continuous maps X rarr Y

(Saying that two maps are homotopic means that you can ldquodeformrdquo one into the other but wersquollleave the precise definition to a course in algebraic topology) It turns out that no functor fromhTop to Set can be faithful so that hTop cannot be made concrete but this is actually a deepand difficult thing to show so we make no attempt to do so here

Natural Isomorphisms Representability

Natural transformations The definition of a natural transformation η F rArr G between two(both covariant or both contravariant) functors FG C rarr D is given in Section 14 of the bookThe key idea is that a natural transformation gives a way to turn data about F into data aboutG in a way which is compatible with all possible morphisms as expressed by the commutativity ofthe diagrams

F (A) F (B)

G(A) G(B)

Ff

ηA ηB

Gf

arising from morphisms f A rarr B (or f B rarr A in the contravariant case) in C In the case of anatural isomoprhism η F rArr G so that each ηA F (A) rarr G(A) is actually an isomorphism in Dthe point is not only that F and G produce isomorphic objects but they do so in a way which iscompatible with all possible category-theoretic data

The first standard example of a natural isomorphism is the one between the identity functor onthe category of finite-dimensional and the functor which sends a finite-dimensional vector space Vto V lowastlowast the dual of its dual induced by the isomorphism V rarr V lowastlowast defined by

v 983041rarr (the linear map on V lowast which sends f to f(v))

This was the topic of a student presentation and can be found in Section 14 of the book Theessential reason why this isomorphism is ldquonaturalrdquo is that V rarr V lowastlowast can be defined without referenceto a basis Contrast this with the the functor which sends V to its (single) dual V lowast it is still truethat finite-dimensional vector space is isomorphic to its dual but specifying such an isomorphism

15

requires additional data which prevents the identity functor from being ldquonaturally isomorphicrdquo tothe single dual functor

Some history At this point it makes sense to say a bit about the interesting origins of categorytheory in the 1950rsquos Eilenberg and Mac Lane where studying cohomology (or possible homologyone of those two) in algebraic topology which associates algebraic data (a group ring or similar)to a topological space in a way which encodes some of the topology They had two cohomologicalconstructions which ended up proving isomorphic results but they noticed that not only werethe resulting objects the same but the actual constructions themselves were in some sense theldquosamerdquo They thus then needed a way to formalize this notion and invented the concept of anatural transformation to do so where they interpreted two constructions as being the same asbeing compatible with all morphisms

But this then leads to the question what types of objects should a natural transformationoccur between Eilenberg and Mac Lane then had to invent the notion of a ldquofunctorrdquo to describewhat a natural transformation went from and what it went to In their motivating setting theyinvented the notion of a functor to describe more formally properties which their cohomologicalconstructions were supposed to satisfy But this then leads to the question what types of objectsshould a functor map between In other words what type of data should a functor (cohomologicalconstruction in their example) take as input and should it output They finally had to thus inventthe notion of a ldquocategoryrdquo to describe the source and target of a functor so that their cohomologicalconstructions could be interpreted as functors from a category of topological spaces to a categoryof algebraic objects

Thus historically natural transformations came first then functors and then categories Aswith most things in mathematics categories thus arose from attempts to encode what was beingseen in some concrete examples in a more general and formal setting

Representable functors Representable functors are defined in Section 21 of the book but wersquollgive the required definitions here in a hopefully simpler to digest manner The point is that anyobject A in a (locally small) category C gives two ldquonaturalrdquo functors C rarr Set one covariant andthe other contravariant which as wersquoll see pretty much encode all data about A itself We definehA C rarr Set to be the covariant functor defined by

hA(B) = Mor(AB) on objects and

hA(Bfminusrarr C) = the map Mor(AB)

hAfminusminusrarr Mor(AC) defined by g 983041rarr f g

We define hA C rarr Set to the contravariant functor defined by

hA(B) = Mor(BA) on objects and

hA(Bfminusrarr C) = the map Mor(CA)

hAfminusminusrarr Mor(BA) defined by g 983041rarr g f

(The functor hA is called the functor of points of A and wersquoll usually think of it as being a covariantfunctor hA Cop rarr Set The functor hA does not have a standard name) So hA is defined bymorphisms from A and hA is defined by morphisms into A Wersquoll see later the sense in which suchfunctors encode all information about the object A

A functor F C rarr Set is representable if is naturally isomorphic to such a functor so F sim= hA

for some A in the covariant case and F sim= hA for some A in the contravariant case The idea wersquollbuild towards is that we can essentially treat such a functor as being A itself so that representable

16

functors can be thought as being actual objects in C For now here is a key example wersquove seenbefore which we can now formulate using the language of representability

Given AB isin Ob(C) let F C rarr Set be the contravariant functor defined by

F (C) = Mor(CA)timesMor(CB) and F (Cfminusrarr D) = [(g k) 983041rarr (g f k f)]

where (g k) 983041rarr (g f k f) gives a map Mor(DA) timesMor(DB) rarr Mor(CA) timesMor(CB) ieF (D) rarr F (C) In order for this functor to be representable requires the existence of an object inC satisfying

Mor(CA)timesMor(CB) sim= Mor(C representing object)

in a way which is compatible with all morphisms but we have actually already seen this notionelsewhere it is essentially the definition of a product AtimesB for A and B Indeed we claim that Fis naturally isomorphic to hAtimesB as we now show

First to define a natural isomorphism hAtimesBsim= F requires for each C isin Ob(C) a choice of a

bijection (ie isomorphism in Set)

ηC hAtimesB(C)sim=minusrarr F (C)

which in our case looks like

ηC Mor(CAtimesB))sim=minusrarr Mor(CA)timesMor(CB)

Take this to be the map

(Cfminusrarr AtimesB) 983041rarr (pA f pB f)

where pA A times B rarr A and pB A times B rarr B are the two projection morphisms in the definitionof a product The definition of product implies that this map is invertible since given k C rarr Aand ℓ C rarr B there exists a unique h C rarr A times B such that k = pA h and ℓ = pB h Forthese bijections to define a natural isomorphism hAtimesB

sim= F requires that they be compatible withall morphisms in C which means that given any g C rarr D in C the following diagram shouldcommute

hAtimesB(C) hAtimesB(D)

F (C) F (D)

hAtimesB(g)

ηC ηD

Fg

which in the case at hand looks like

Mor(CAtimesB) Mor(DAtimesB)

Mor(CA)timesMor(CB) Mor(DA)timesMor(DB)

minus g

ηC ηD

(minus gminus g)

Take an element k D rarr AtimesB in the upper right Applying the map on top gives the elementk g in the upper left and then applying ηC on the left side gives

(pA (k g) pB (k g))

17

in the lower left Instead starting with k D rarr AtimesB in the upper right applying ηD on the rightgives (pA k pB k) and applying the map on the bottom gives

((pA k) g (pB k) g)

in the lower left which is the same as the previous element in the lower left by the associativity ofcomposition Hence the isomorphisms ηC do define a natural isomorphism between hAtimesB and F Again the point is that you can then interpret the functor F as being the ldquosamerdquo as the productAtimesB since both encode the same data

More Representable Examples

Coproducts As in the case of products the definition of coproducts can also be phrased in termsof a representable functor Given objects A and B in C let F C rarr Set be the covariant functordefined by

F (C) = Mor(AC)timesMor(BC) and F (Cfminusrarr D) = [(g k) 983041rarr (f g f k)]

where the latter is a function Mor(AC)timesMor(BC) rarr Mor(AD)timesMor(BD) This functor isrepresentable if and only if a coproduct A ⊔ B for A and B exists and the representing object isthat coproduct the key is the requirement that we have bijections

Mor(AC)timesMor(BC) sim= Mor(A ⊔BC)

for all C which is essentially the defining property a coproduct for A and B should have

Power set functor contravariant version Let P Set rarr Set be the contravariant powerset functor defined on objects by sending a set to its power set and on morphisms by sendingf A rarr B to the preimage function Pf P (B) rarr P (A) given by S 983041rarr fminus1(S) In order for this tobe representable requires the existence of a set X with natural bijections

P (A) sim= Mor(AX)

for all sets A that is a set X so that mapping into X is the same data as specifying a subset ofthe domain We take X = 0 1 to be a two-element set and the required bijections

ηA P (A) rarr Mor(A 0 1) to be given by S 983041rarr χS

where XS A rarr 0 1 is the indicator or characteristic function S defined by sending elementsof S to 1 and elements of A minus S to 0 The inverse of ηA sends a function g A rarr 0 1 to thepreimage gminus1(1) sube A

To verify that the bijections ηA define the data of a natural isomorphism η P rArr h01 wecheck the commutativity of some diagrams Given a function f A rarr B consider

P (A) P (B)

h01(A) = Mor(A 0 1) h01(B) = Mor(B 0 1)

Pf = fminus1

ηA ηB

minus f

18

Take S isin P (B) in the upper right Applying the map on top gives fminus1(S) isin P (A) and thenapplying the map on the left gives χfminus1(S) isin Mor(A 0 1) On the other hand applying the mapon the right to S isin P (B) gives χS isin Mor(B 0 1) and then applying the map on the bottomgives χS f isin Mor(A 0 1) The two resulting functions χfminus1(S)χS f in the lower left are

χfminus1(S)(x) =

9830831 x isin fminus1(S)

0 otherwiseand χS(f(x)) =

9830831 f(x) isin S

0 otherwise

so these are the same since x isin fminus1(S) if and only if f(x) isin S Thus the diagram above commutesso P is naturally isomorphic to h01 and hence P is representable

Power set functor covariant version Now consider the covariant power set functor stilldenoted by P Set rarr Set which again sends a set to its power set but now sends a functionf A rarr B to the image function Pf P (A) rarr P (B) defined by S 983041rarr f(S) In order for this tobe representable there would have to exist a set X such that P sim= hX which translates into havingnatural bijections

P (S) sim= Mor(XS)

for all sets S However in this case there is no natural choice for X as there is no natural way todetect subsets of S by mapping into S as opposed to the case of P (S) sim= Mor(S 0 1)

That no such X exists can be made precise using a simple argument when S = empty P (S) hascardinality 1 soX would have to be empty as well since otherwise Mor(XS) would have cardinality0 but if X is empty then Mor(empty S) always cardinality 1 regardless of S which is clearly not trueof P (S) Thus the covariant power set functor is not representable

Forgetful on Top The forgetful functor F Top rarr Set is representable Indeed a representingobject would have to satisfy

X sim= Map(objectX)

as sets for any topological space X (where Map denotes the set of continuous maps) and it iseasy to see that a single point satisfies this property a continuous map f pt rarr X is completelydetermined by the element f(pt) of X and any such element gives a continuous map in this wayOf course checking that F sim= hpt requires checking that various diagrams commute but this issimple to work out in this case so we omit it here

Extracting a topology With an eye towards the idea that the functors hA hA should captureall information about an object A in a category note that the result above shows that in TophX first of all captures all information about X as a set since the set X can be extracted fromhX(pt) = Map(ptX)

We claim that hX actually captures all information about X as a topological space since thetopology on X can be extracted from hX in the following way As we saw in the power set examplewe have a bijection

P (X) sim= Mor(X 0 1)

Now equip 0 1 with the topology in which 1 is open but 0 is not Then a continuous mapf X rarr 0 1 is fully determined by the preimage fminus1(1) since to be continuous only requiresthat this single preimage be open in X Moreoever any open subset of X arises in this way bytaking f to be the indicator function of that open set Thus we get a bijection

Op(X) sim= Map(X 0 1)

19

where Op(X) denotes the set of open subsets of X ie the topology on X Hence the topology onX can be extracted from hX as claimed so hX and hX do encode all information about X

This fact can also be interpreted as saying that the contravariant functor which sends a topo-logical space X to its collection of open subsets (and which sends a continuous map to the mapinduced by taking preimages) is representable with representing object 0 1 equipped with thetopology given above This will be worked out in a homework problem

Forgetful on Grp The forgetful functor Grp rarr Set is also representable In this case arepresenting object should be a group giving set-theoretic bijections

G sim= Hom(objectG)

for all groups G where Hom denotes the set of group homomorphisms The group Z works

G sim= Hom(Z G)

since a homomorphism φ Z rarr G is completely determined by φ(1) since Z is generated by 1 andthere are no restrictions on what φ(1) isin G can actually be The inverse of the desired bijection isthus φ 983041rarr φ(1) and it is simple to check that morphisms behave appropriately so that the forgetfulfunctor in this setting is indeed isomoprhic to hZ

Forgetful on Ring The forgetful functor Ring rarr Set is also representable where by Ring wemean the category of rings with identity elements and where morphisms (ie ring homomorphisms)are required to preserve identities Here we need an object satisfying

R sim= Hom(object R)

for all R isin Ob(Ring) and we can see that Z[x] the ring of polynomials in the variable x withcoefficients in Z works Indeed given a ring homomorphism φ Z[x] rarr R the fact that φ isrequired to preserve identities fully determines φ(n) for any n isin Z (since φ(n) = nφ(1)) and hencesince φ preserves multiplication its behavior is fully determined by φ(x) which can be any elementof R The inverse of the bijection we want is thus φ 983041rarr φ(x) and it will follow that this forgetfulfunctor is isomorphic to hZ[x]

Equivalences between Categories

Definitions iso and equiv Now that we have a notion of a ldquomorphismrdquo between categorieswe can talk about the sense in which two categories are the ldquosamerdquo The first possible definitionis the ldquoobviousrdquo one you might expect given the definition of ldquoisomorphismrdquo wersquove seen in everyother context until now we say that C is isomorphic to D if there exist functors F C rarr D andG D rarr C such that G F = idC and F G = idD where idC and idD are identity functors

However this definition turns out to be too restrictive Indeed the requirement that the givencompositions equal identity functors says that given an object A in C the object G(F (A)) beliterally the same as A itself and similarly for objects in D But we know that two objects in acategory can be thought of as being the ldquosamerdquo without being literally the same as long as theyare isomorphic For instance productscoproducts are not unique in the literal sense but only inthe sense that they are isomorphic in a unique way To take another example the definition ofa functor being representable does not say that F (B) literally be the same as hB(A) only thatthey be isomorphic for sure the power set P (S) of a set S is not literally the same as the set offunctions S rarr 0 1mdashit is only the ldquosamerdquo in the sense that they encode the same data

20

Thus it makes sense to relax the requirement that A andG(F (A)) in the definition of isomorphiccategories be equal to the requirement that they be isomorphic This leads to the following betterdefinition of two categories being the ldquosamerdquo C is equivalent to D if there exist functors F C rarr D

and G D rarr C such that G F sim= idC and F G sim= idD So we require only that for instance thefunctor G F be naturally isomorphic to the identity functor on C which says that for any objectA G(F (A)) be isomorphic to A in a natural way Wersquoll see that this truly is the better notionof what it means for two categories to be the ldquosamerdquo and it reflects the idea that all categoricalproperties of one category can be transported to the other

Alternate characterization of equivalences A functor F C rarr D implementing an equiv-alence between categories is called unsurprisingly and equivalence and the ldquoinverserdquo functorG D rarr C is still referred to as being its inverse even though it is not technically an ldquoinverserdquoin the sense that compositions give literal identities But as in the case of Set where ldquoinvertiblerdquocan be characterized without reference to an inverse as ldquoinjectiverdquo and ldquosurjectiverdquo so too canequivalences between categories be characterized without explicit reference to an inverse

A functor F C rarr D is an equivalence if and only it is full faithful and essentiallysurjective

We have seen the notions of full and faithful before and essentially surjective means that any objectin D is isomorphic to something in the image of F for all D isin Ob(D) there exits C isin Ob(C) suchthat F (C) sim= D Justifying this result was the topic of a student presentation and can be foundin the book in Section 15 (Technically to construct an ldquoinverserdquo of F requires that we work withsmall categories but whatever) The bulk of the real work goes into showing that the ldquoinverserdquofunctor constructed is actually a functor which comes down to some abstract nonsense argumentvia commutative diagrams

Equivalences preserve everything We previously saw the notions of full faithful and es-sentially surjective back when thinking about what properties a functor could have which wouldguarantee it sent products to products so the result we proved back then was that equivalencespreserve products Similarly equivalences preserve coproducts and the proof is basically the sameafter reversing arrows A clean way of deriving this is as follows if F C rarr D is an equivalencethen so is F op Cop rarr Dop so since F op preserves products F preserves coproducts becausecoproducts in a category become products in the opposite category

In general an equivalence between categories will preserve whatever categorical notions youwant monomorphisms epimorphisms initial objects terminal objects etc so that equivalentcategories really do have essentially the ldquosamerdquo properties

Equivalence example Here is a first example (also found in the book) of equivalent categorieswhich are not isomorphic Let FVectR be the category whose objects are finite-dimensional realvector spaces and whose morphisms are linear transformations Let MatR be the category ofmatrices defined as follows

bull objects in MatR are nonnegative integers

bull a morphism n rarr m is an mtimes n matrix with real entries

bull composition in MatR is given by matrix multiplication given B n rarr m and A m rarr kmdashsoB is an mtimes n matrix and A is a k timesm matrixmdashthe composition A B n rarr k is the k times nmatrix AB

bull identity morphisms in MatR are given by identity matrices

21

We claim that FVectR and MatR are equivalent First note that they certainly cannot be isomo-prhic isomorphisms in MatR exist only between two objects which are literally the same (ie anmtimesn matrix can be invertible only when m = n) but isomorphisms in FVectn can exist betweenobjects which are not literally the same So there are in a sense not enough objects in MatR toallow it to be isomorphic to FVectR

To construct an equivalence choose one and for all an isomorphism TV V rarr RdimV forevery object in FVectR or equivalently choose once and for all a basis for every V DefineF FVectR rarr MatR by

V 983041rarr dimV on objects and

(T V rarr W ) 983041rarr (the standard matrix of SW T Sminus1V RdimV rarr RdimW ) on morphisms

Equivalently F sends T to the matrix of T relative to the chosen bases for V and W It isstraightforward to check that this is a functor The inverse functor G MatR rarr VectR is givenby

n 983041rarr Rn on objects and

(mtimes n matrix A) 983041rarr (the linear transformation A Rn rarr Rm induced by A) on morphisms

One can check that G is a functor and that F G and G F are naturally isomorphic to identitiesor instead one can argue that F alone is full faithful and essentially surjective The point is thatG(F (V )) gives back RdimV so not literally V itself but only something isomorphic to V which iswhy this gives an equivalence of categories and not an isomorphism

Yoneda Lemma Functors as Objects

Compact Hausdorff spaces and Clowast-algebras Last time we gave an example of equivalentcategories which were not isomorphic but in many ways that example is unsatisfactory when doinglinear algebra people for sure often work with matrix representations of linear transformations asthe example from last would suggest but no one in their right mind literally thinks about suchthings in terms of the equivalence FVectR rarr MatR itself In other words this equivalence isessentially ldquocooked uprdquo for the sake of giving an example but it does not really give any insightinto the actual mathematics of linear algebra

So with that in mind we will describe another equivalence which actually does have some realmeaning in the sense of producing new ways of thinking about some mathematics Given a compactHausdorff space X we let C(X) denote the space of continuous complex-valued functions on X

C(X) = f X rarr C | f is continuous

We can equip the set C(X) with various additional structures First it is a complex vector spaceunder ordinary addition and scalar multiplication of functions Second we can multiply two func-tions together

(fg)(p) = f(p)g(p)

which turns C(X) into whatrsquos called an algebra over C (We wonrsquot give the definition of anldquoalgebrardquo but the point is that this multiplication is in some sense ldquocompatiblerdquo with the vectorspace structure) Next any element f of C(X) has a conjugate element f obtained by takingcomplex conjugates of the values of f

f(p) = f(p)

22

And finally we can equip C(X) with a norm via

983042f983042 = suppisinX

|f(p)|

where |middot| denotes complex absolute value (This supremum exists sinceX is compact by the ExtremeValue Theorem) All of these structures (vector space algebra multiplication conjugation norm)turn C(X) into whatrsquos called a Clowast-algebra We wonrsquot define this precisely but the definition encodesvarious ways in which these structures are compatible Actually C(X) is a unital commutative Clowast-algebra where ldquounitalrdquo means that the multiplication has an identity element (namely the constantfunction 1) and commutative means that fg = gf There is a natural notion of homomorphismbetween Clowast-algebras which are maps between them which behave nicely with all the structuresinvolved In particular given a continuous map f X rarr Y between compact Hausdorff spaces weget a Clowast-algebra homomorphism flowast C(Y ) rarr C(X) given by pre-composition with f flowast sendsg isin C(Y ) to g f isin C(X)

We thus have a contravariant functor CHaus rarr ucClowast-Alg from the category of compactHausdorff spaces and the category of unital commutative Clowast-algebras It turns out that anyunital commutative Clowast-algebra arises in this way in the sense that given any unital commutativeClowast-algebra B there exists a compact Hausdorff space X such that C(X) sim= B as Clowast-algebrasand that morphisms between these algebra arises from continuous maps in the manner describeabove This says that the functor above is essentially surjective full and it turns out that it isalso faithful so that it is an equivalence Thus all information about compact Hausdorff spaces iscompletely encoded in information about unital commtuative Clowast-algebras and indeed one couldlearn everything there is to know about a compact Hausdorff space X from its Clowast-algebra C(X)of functions The fact that the categories CHaus and ucClowast-Alg are equivalent is the startingpoint in the subject known as noncommutative geometry which wersquoll say something about a bitlater on If we drop the requirement that our algebras be unital it turns out that the categoryof commutative Clowast-algebras in general is equivalent to the category of locally compact Hausdorffspaces via essentially the same functor where we take as morphisms in the category of locallycompact Hausdorff spaces to be continuous functions which ldquovanish at infinityrdquo

We are only introducing Clowast-algebras to give an example of an interesting and actually usefulequivalence between categories but just for the fun of it wersquoll note the following Examples ofnoncommutative Clowast-algebras come from what are known as ldquoalgebras of bounded operators onHilbert spacesrdquo which show up in quantum mechanics as the things which describe physicallyobservable quantities like position momentum and energy (In fact all Clowast-algebras arise fromHilbert spaces in this way) Thus Clowast-algebras show up in mathematical formulations of quantummechanics and the notion of noncommutative geometry mentioned above is at the core of someattempts to understand the elusive concept known as ldquoquantum geometryrdquo

Yoneda Lemma The Yoneda Lemma states that given a (covariant) functor F C rarr Set whereC is locally small for any A isin Ob(C) there exists a bijection

F (A) sim= Nat(hA F )

where the right side denotes the set of natural transformations from hA to F A similar result holdsin the contravariant case where we get bijections F (A) sim= Nat(hA F ) (There is also a sense inwhich these bijections are themselves natural but wersquoll ignore this bit) The proof of the YonedaLemma was the topic of a student presentation and can be found in Section 22 of the book

Here we will get a bit philosophical and talk about what the point of the Yoneda Lemmaactually is and later the sense in which it says that we can view arbitrary functors as ldquonon-existent

23

objectsrdquo in a category First we mention what we will call the ldquofunctor of pointsrdquo approach tomathematics a mathematical object can be fully characterized by all morphisms into it That isknowledge of Mor(SA) for all S is equivalent to knowledge of A itself Indeed this is an idea wementioned previously in relation to the functors hAmdashwhich we previously said is referred to as beingthe functor of points of Amdashand now we seek to push this idea further In some examples literalpoints of A can be extracted from hA as in the case of the bijection X sim= hX(pt) = Map(ptX) fora topological space X or G sim= hG(Z) = Mor(Z G) for a group G Moreover in these examples therest of the (topological) structure onX or (group) structure on G can also be extracted from hX andhG respectively In general we can view Mor(SA) as in a sense being ldquopoints of A parameterizedby Srdquo which comes from the idea that in say Set Mor(SA) = AS is ldquothe product of S-manycopies of Ardquo so literally points of A ldquoparameterizedrdquo by S In various areas of mathematics anelement of Mor(SA) is thought of as being a ldquogeneralizedrdquo point of A so that the functor of pointshA detects generalized points of A

So an object A is fully encoded by the functor hA Now we claim that morphisms betweenobjects are also encoded by such functors Indeed in the case where F = hB is itself a functor ofpoints the Yoneda result Nat(hA F ) sim= F (A) becomes

Nat(hA hB) sim= hB(A) = Mor(AB)

which says precisely that morphisms between the objects A and B can be obtained solely fromthe functors hA and hB (This fact is one reason why one should prefer the contravariant functorhA to the covariant functor hA in the covariant case Nat(hA F ) sim= F (A) for F = hB becomesNat(hA hB) sim= hB(A) = Mor(BA) so that morphisms between two objects get reversed whencharacterizing them in terms of representable functors)

Pushing our philosophy further we now know that not only objects but also morphisms betweenobjects are fully encoded by the functors hA The final step is to note that we should not only careabout objects and morphisms between them but that an important part of the data of an objectA is also how arbitrary contravariant functors F C rarr Set behave on A and that this too can beobtained from hA precisely via the Yoneda result that F (A) sim= Nat(hA F ) So objects morphismsbetween objects and how arbitrary functors behave on objects are all encoded in knowledge ofthe functors hA so that these functors truly do capture in the most general sense possible allinformation about A

Functors as nonexistent objects The discussion above is all well and good but it still doesnrsquotquite say why we should care about the fact that we can study on object A by studying hA insteadin that we still havenrsquot spoken about what type of insight about A can be gleamed from hA thatwasnrsquot readily apparent in A itself already Now we argue that the true point of the YonedaLemma is that it gives a way to think about arbitrary functors F Cop rarr Set as ldquoobjectsrdquo in C

itself or really as objects in an ldquoenlargementrdquo of C Certainly for those functors F sim= hA which arerepresentable we can indeed think about F as simply being the object A but now we claim thateven non-representable F can be thought of as being ldquononexistentrdquo objects in C The upshot is thatoften in mathematics we come across things which do not actually exist but the Yoneda Lemmagives a way to (indirectly) study such things anyway by interpreting them as functors instead

The starting point of course is that an actual object A is the same as hA as we have explainedabove Thus by analogy we can interpret a non-representable functor F as being an ldquoobjectrdquo in C

as long as we know what it means to ldquomaprdquo into F from other things which are honestly objects in CThat is if F were indeed representable then F (A) sim= Nat(hA F ) sim= Mor(AF ) would characterizethe functor of points of F (here we abuse notation by using the same letter F to denote both arepresentable functor and the object which represents it) so now we take Mor(AF ) = F (A) as

24

the definition of the ldquofunctor of pointsrdquo of the non-existent object F We are thus taking elementsof F (A) as being what it means to ldquomaprdquo from A into F even when F is not representable Thefact that the Yoneda Lemma gives a way to define maps from A into F thus says that F itself isequipped with the same data that a literal object of A would be equipped with so why shouldnrsquotwe interpret F as a being a type of ldquoobjectrdquo in C as well

To phrase this another way note that the Yoneda Lemma gives rise to a functor

C rarr Fun(CopSet)

where Fun(CopSet) denotes the category whose objects are functors from Cop to Set and morphismsare natural transformation defined on objects by A 983041rarr hA and on morphisms by the bijectionMor(AB) sim= Nat(hA hB) arising from the Yoneda Lemma That this is actually bijection saysthat this functor is full and faithful so that we can view C as being a subcategory of the functorcategory Fun(CopSet) (We cay say that C is isomorphic to the subcategory of Fun(CopSet)consisting of the functors hA or that it is equivalent to the subcategory of Fun(CopSet) consistingof representable functors This functor C rarr Fun(CopSet) is usually called the Yoneda embeddingof C into Fun(CopSet)) Thus we can view Fun(CopSet) as an ldquoenlargmentrdquo of C and the ideawe are trying to make sense of is that other objects in this ldquolargerrdquo category should be interpretedas ldquonon-existent objectsrdquo of C

Just how far can we actually push this idea that arbitrary functors Cop rarr Set should be thoughtof as being ldquoobjectsrdquo of C Pretty far in fact For instance often times in geometry one comesacross a type of ldquosmoothrdquo space on which some group acts and one is interested in doing ldquocalculusrdquoon the resulting quotient space which is the space of orbits under the given group action Theproblem is that such quotient spaces are very often (usually in fact) not ldquosmoothrdquo themselves sothat the tools from calculus are not readily available The solution is to nonetheless treat thisquotient as if it were ldquosmoothrdquo and attempt to do calculus anyway To make this precise thisldquonon-smoothrdquo space actually turns out to have a natural interpretation as a ldquosmooth functorrdquo andit is through this functor which we can view this non-smooth space as actually being smooth afterall so we are forced to expand our notion of what we mean by ldquosmooth spacerdquo by allowing thingswhich are not spaces in the usual sense which certainly leads to more abstraction but the benefitis that it actually makes sense to do calculus on such ldquospacesrdquo interpreted as functors

The discussion above is pretty vague so here is one final more concrete example Recall thatequivalence

CHausop rarr ucClowast-Alg

between the category of compact Hausdorff spaces and the category of unital commutative Clowast-algebras mentioned at the start We now claim that a non-commutative (unital) Clowast-algebra shouldbe interpreted as the algebra of functions on some non-existent ldquononcommutative compact Haus-dorff spacerdquo and it is this idea which the subject of noncommutative geometry makes precise Solet B be a non-commutative Clowast-algebra Then the functor of points hB gives a functor

ucClowast-Alg rarr Set

which sends C(X) (where X is some compact Hausdorff space) to Hom(C(X) B) where Hom de-notes homomorphisms of Clowast-algebras whatever that means Composing with the previous functorwe get a functor wersquoll denote by B itself

B CHausop rarr Set

which on objects at least is given by X 983041rarr Hom(C(X) B) This functor is not representable sinceB is not commutative but the philosophy espoused above suggests we should think of it as giving a

25

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

Now in order for this to be category we also need a notion of ldquocompositionrdquo which in this casesince there is only one object amounts to a map of the form

Mor(lowast lowast)timesMor(lowast lowast) rarr Mor(lowast lowast)

which in our case becomesZtimes Z rarr Z

We take this map to be ordinary addition + on Z (So ldquocomposingrdquo the arrow labeled 1 in thepicture above with the arrow labeled 2 results in the arrow labeled 3) The point is that this isNOT ldquocomposition of functionsrdquo in the usual sense but will give a valid ldquocompositionrdquo operationin the category BZ as long as it satisfies the properties required in the definition of a category Onesuch requirement is that composition be associative which is satisfied since + is indeed associativeThe second requirement is that exist a morphism idlowast isin Mor(lowast lowast) which acts as an identity for theldquocompositionrdquo wersquove defined In this case this identity morphism is idlowast = 0 that is the integer0 in Mor(lowast lowast) = Z Saying that this is the ldquoidentity morphismrdquo with respect to ldquocompositionrdquo inthis case simply says that is the identity element for addition

Thus we get a category BZ as claimed albeit one where the notion of ldquomorphismrdquo and ldquocompo-sitionrdquo are more general than simply ldquofunctionrdquo and ordinary composition In fact we can alreadynow give a ldquocategoricalrdquo characterization of a well-known algebraic object a (locally small) cat-egory one with a single object is precisely what is known as a monoid which is a set equippedwith an associative binary operation which admits an identity element The monoid in questionis precisely the set Mor(lowast lowast) and the associative operation which turns this into a monoid is theldquocompositionrdquo of the category in question (A locally small category is one where each collectionof morphisms is actually a set which is needed here in order to guarantee that Mor(lowast lowast) is a set onwhich we can define a binary operation) So for instance any group can be viewed as a categorywith one object where the group elements give the ldquomorphismsrdquo from that one object to itself(Of course a group gives a special type of category where every morphism is in fact ldquoinvertiblerdquoA small category where every morphism is invertible is called a groupoid so a group is nothing buta groupoid with one object Wersquoll talk more about groupoids later on)

A final example Here is one final example wersquoll consider from time to time which is again acategory where the ldquomorphismsrdquo are not simply functions The category Rel of relations has asobjects ordinary sets but as morphisms relations between sets

f isin Mor(AB) is a relation from A to B which is a subset f of AtimesB

The usual graph of a function from A to B gives an example of such a relation so we can say thatRel contains Set as a subcategory but not all relations come from graphs of functions In generalwe can think of a relation as a ldquopartially-defined possibly multivalued functionrdquo if we interpret(a b) being in the relation f as saying that ldquof sends a to brdquo

Composition in Rel comes from ordinary composition of relations given a relation f A rarr Band a relation g B rarr C we define their composition g f A rarr C to be

g f = (a c) isin Atimes C | there exists b isin B such that (a b) isin f and (b c) isin g

So g f consists of all pairs of elements of A and C which are ldquorelatedrdquo to a common elementof B In the case where f and g are graphs of ordinary functions this composition of relationsreproduces the ordinary notion of composing functions if (a b) isin f and (b c) isin g then b = f(a)

3

and c = g(b) (here we abuse notation by denoting the function of which f is graph by the symbolf itself) so

c = g(b) = g(f(a)) = (g f)(a)

meaning that (a c) isin g f precisely when c = (g f)(a) Composition of relations is associativeand you can check that graphs of ordinary identity functions

graph(idA) = (a a) isin AtimesA | a isin A

give the identity morphisms (I spent much time as graduate student trying to do ldquogeometryrdquo witha certain category of relations so this is an example near and dear to my heart)

Special Morphisms Products

Isomorphisms A morphism f isin Mor(AB) between two objects A and B in a category is anisomorphism or is invertible if it has an inverse there exists a morphism g isin Mor(BA) such thatgf = idA and fg = idB where idA isin Mor(AA) and idB isin Mor(BB) are the identity morphismswhich are assumed to exist as part of the definition of a category We say that A and B areisomorphic and we write A sim= B if there exists an isomorphism between them

This general categorical notion of isomorphism reduces to the one you would expect in thestandard examples isomorphisms in Set are bijective functions in Grp they are bijective grouphomomorphisms in Vect they are invertible linear transformations and in Top they are homeo-moprhisms ie continuous bijections with continuous inverses

A category of metric spaces Now consider the category MetC of metric spaces where mor-phisms are given by continuous functions In this category the notion of ldquoisomorphismrdquo againmeans homeomorphism so two metric spaces are isomorphic in this category when they are home-omorphic However often when considering whether or not two metric spaces are the ldquosamerdquo wehave another definition in mind that of two metric spaces being isometric X is isometric to Y ifthere exists a bijective isometry between them which is a bijective map f X rarr Y which preservesdistance in the sense that

dY (f(p) f(q)) = dX(p q) for all p q isin X

After all in a metric space you have more data than simply ldquoopen setsrdquo (which is all that ahomeomorphism detects) so it would be good to have a notion of ldquosamenessrdquo which incorporatedthe extra data of distance as well

The goal is then to construct a category of metric spaces where ldquoisomorphicrdquo indeed meansisometric and we can do so simply by restricting the types of continuous maps we consider Thecategory Met has metric spaces as objects but now a morphism f X rarr Y is taken to be afunction which does not ldquoenlargerdquo distance in the sense that

dY (f(p) f(q)) le dX(p q) for all p q isin X

Maps with this property are often called metric maps Any metric map is actually continuous soMet is a subcategory of MetC (The subscript C used in MetC is meant to denote ldquocontinuousrdquoindicating that we take arbitrary continuous maps as morphisms) In this category now for f X rarr Y to be an isomorphism requires that f and its inverse fminus1 both be metric maps so

dY (f(p) f(q)) le dX(p q) and dX(fminus1(s) fminus1(t)) le dY (s t)

4

for all p q isin X and s t isin Y But writing s and t as s = f(p) and t = f(q) the second conditionbecomes dX(p q) le dY (f(s) f(t)) and the two inequalities together thus imply

dY (f(p) f(q)) = dX(p q) for all p q isin X

so that f X rarr Y is actually an isometry The moral is that certain properties one might wantcan be achieved by changing the morphisms you consider

Monomorphisms and epimorphisms The definitions of monomorphism and epimorphismcan be found in Section 12 of the book The standard examples are those in Set where themonomorphisms are the injective functions and the epimorphisms are the surjective functionsProofs of these facts are left to the homework as well as the interesting problem of determining theepimorphisms in Met or even MetC or Haus (the category of Hausdorff spaces where morphismsare arbitrary continuous maps) where the answer is that epimorphism means more than simplyldquocontinuous and surjectiverdquo (As opposed to all of Top where epimorphism does indeed meancontinuous and surjective)

So monomorphisms and epimorphisms should be treated as categorical analogues of ldquoinjectionsrdquoand ldquosurjectionsrdquo although this analogy only goes so far The thing to note is that we are thusable to give a complete definition of ldquoinjectiverdquo for ordinary functions which makes no referenceto elements and is stated only in terms of the relation between an injective function and otherfunctions certainly proving that ldquomonomorphismrdquo means ldquoinjectionrdquo in Set requires working withelements but stating the definition of an injection as a monomorphism can be done solely usingldquoarrowsrdquo

Products The definition of a product in a category shows up in Section 31 of the book inthe context of the more general notion known of a limit Wersquoll discuss this more general notioneventually but for now we will only focus on products of two objects at a time Here is what themore general notion boils down to in this case which wersquoll take as our definition

A product of two objects A and B in a category is an object P together with morphismspA P rarr A and pB P rarr B (which we call projections) which satisfy the followingproperty given any morphisms f C rarr A and g C rarr B into A and B from acommon object C there exists a unique morphism h C rarr P such that f = pA h andg = pB h We express this by saying that the following diagram commutes

C

A B

Pf g

pA pB

existh

which means that composing any two composable arrows in the diagram results in theother arrow with the same domain and codomain Concretely there are two ways in thisdiagram to get from C to A via f or via the composition pA h and the requirementis that both give the same morphisms (ie f = pA h) and similarly for the two waysto get from C to B The dashed arrow is the one which is required to exist (uniquelyas denoted by the exclamation mark ) in the definition of product

5

So the upshot is that for any C and any f and g in the diagram above there exists a uniqueh which makes the diagram commute The point is that a product gives a way to turn the twopieces of data f C rarr A and g C rarr B into the single piece of data h C rarr P in a uniqueway Being able to turn a possibly large amount of data into a single piece of data in this way isone of the benefits the ldquocategoricalrdquo perspective provides Products in categories are often denotedusing the usual A times B notation even though it is not necessarily true that a product is literallya set-theoretic Cartesian product although it is in the standard examples The map h C rarr Prequired in the definition is then often denoted by h = f times g

Examples In Set the product of two objects A and B always exists and is indeed the ordinaryCartesian product A times B But of course the definition of product also requires that data of theprojection morphisms which in this case are simply the usual ones

pA (a b) 983041rarr a and pB (a b) 983041rarr b

Given f C rarr A and g C rarr B the map h C rarr AtimesB is h(c) = (f(c) g(c)) and you can checkthat these constructions satisfy the properties required in the definition (Note that you would alsohave to argue that h defined in this way is the only map which can make the required diagramcommute)

Products in Grp are given by the usual direct product group structure GtimesH with the groupoperation defined as

(g1 h1) middot (g2 h2) = (g1g2 h1h2)

The projection morphisms are the ordinary ones you would expect as well But let us actuallynow verify that this group structure indeed is among other possible group structures on G times Hthe one required in the definition of a product Suppose f K rarr G and g K rarr H are grouphomomorphisms The key point is that the map h(k) = (f(k) g(k)) which should satisfy therequired product property should itself be a morphism in Grp meaning a group homomorphismand we claim that this is only true for the direct product group structure

Indeed we want it to be true that

h(k1k2) = (f(k1k2) g(k1k2)) equal h(k1)h(k2) = (f(k1) g(k1)) middot (f(k2)g(k2))

for all k1 k2 isin K and some to-be-determined group structure middot on G times H Since f and g arehomomorphisms this requirement becomes

(f(k1)f(k2) g(k1)g(k2)) = (f(k1) g(k1)) middot (f(k2)g(k2))

which says that middot should indeed be componentwise multiplication at least on the image of f andg But recall that the properties required of a product should hold for any K and any f K rarr Gand g K rarr H in particular then it should hold for instances when f and g are surjective whichshows that the componentwise-multiplication property above should in fact hold on all of G timesHso that middot is indeed the ordinary direct product structure

Uniqueness Notice that in the definition of a product in a category we used the word ldquoardquo asopposed to ldquotherdquo we spoke about a product of A and B instead of the product of A and B Sowe now ask are products unique The answer is ldquonordquo if we interpret ldquouniquerdquo in the strictestpossible sense but ldquoyesrdquo if we interpret it correctly For instance in the example of Grp takenow N to be any group which is isomorphic to the direct product G timesH say with isomorphismℓ N rarr G timesH We claim that we can also turn N into a categorical product of G of H as long

6

as we define appropriate ldquoprojectionsrdquo N rarr G and N rarr H take N rarr G to be the compositionpG ℓ and N rarr H to be pH ℓ Given f K rarr G and g K rarr H as in the definition of a productthe map hprime K rarr N defined by

hprime = ℓminus1 h

where h K rarr GtimesH is the usual h = f timesg will satisfy the requirement needed to able to concludethat N is also a product of G and H in Grp A similar trick works in any category so that anyobject isomorphic a product can also itself be turned into a product

The correct answer is that products are unique up to unique isomorphism which means thatgiven a product P of A and B any other product P prime of A and B will in fact be isomorphic to P and in a unique way in the sense that there can exist only one isomorphism P rarr P prime The phraseldquounique up to unique isomorphismrdquo is a common one wersquoll see again and again objects definedby some categorical property are almost never genuinely unique in a strict sense but they willessentially be as close to unique as possible

So suppose that P and P prime are both products of A and B with associated projection morphismspA pB for P and pprimeA p

primeB for P prime Consider the diagram

P

A B

P primepA pB

pprimeA pprimeB

existh

The morphism h P rarr P prime is the unique one guaranteed to exist by the fact that P prime is a productof A and B Commutativity of this diagram says that pA = pprimeA h and pB = pprimeB h A similardiagram with the roles of P and P prime reversed which uses the fact that P is a product results in aunique morphism hprime P prime rarr P satisfying pprimeA = pA hprime and pprimeB = pB hprime

Now consider the following diagram

P

A B

PpA pB

pA pB

hprime h

This commutes which we verify using the properties h and hprime satisfy as follows

pA (hprime h) = (pA hprime) h = pprimeA h = pA

and similar for the morphisms involving B This shows that hprimeh P rarr P satisfies the requirementin the definition of P being a product But of course idP P rarr P satisfies this requirement as wellso since the map satisfying this requirement is assumed to be unique we must have hprime h = idP Similar reasoning considering h hprime P prime rarr P prime shows that h hprime = idprimeP so that h and hprime are indeedinverses and hence that P and P prime are isomorphic The uniqueness of the isomorphism betweenthem comes from the uniqueness of h in the construction above

7

Abstract nonsense It is common to say in the setting above that the fact that products areunique up to unique isomorphism follows from ldquoabstract nonsenserdquo which is a succinct way ofsaying that it follows solely from the definition of a categorical product itself and not any deepermathematical content In particular many of the ldquouniquenessrdquo properties which ldquoproductsrdquo havein the settings we might care aboutmdashCartesian products in Set direct products in Grp producttopologies in Topmdashreally have nothing to do with how those specific products are defined but arejust consequences of the ldquoabstract nonsenserdquo of category theory

But of course we do not use the phrase ldquoabstract nonsenserdquo in a negative or dismissive waysince recognizing what types of properties are consequences of ldquoabstract nonsenserdquo goes a long waytowards understanding well the given problem at hand Soon you too will come to appreciate thisnotion of abstract nonsense

Coproducts Opposite Categories

Initial and terminal objects The notions of initial and terminal objects in a category aredefined in Section 16 of the book where you can also find standard examples Here I want topoint out that initial and terminal objects if they exist are unique up to unique isomorphism asanother example of arguing via ldquoabstract nonsenserdquo

Suppose I and I prime are both initial objects in some category By the fact that I is initial thereexists a unique morphism f I rarr I prime and by the fact that I prime is initial there exists a unique morphismg I prime rarr I But then the composition g f I rarr I must equal idI I rarr I since the latter is theonly morphism from I to itself because I is initial and similarly f g = idIprime Thus I and I prime areunique and there is a unique isomorphism between them

Coproducts The dual concept to that of a product is the notion of a coproduct (The termldquodualrdquo refers to the phenomena you get when you reverse all arrows in a given construction Forinstance the notion of a terminal object is dual to that of an initial object and the notion of anepimorphism is dual to that of a monomorphism) Here is the precise definition a coproduct of anobject A and an object B in a category is an object C together with morphisms iA A rarr C andiB B rarr C such that for any object D and morphisms f A rarr D and g B rarr D there exists aunique morphism h C rarr D which makes the following diagram commute

D

A B

Cf g

iA iB

existh

By abstract nonsense if a coproduct of A and B exists it is unique up to unique isomorphism

Examples In Set the coproduct of two sets A and B is the disjoint union A⊔B (If you havenrsquotseen the notion of a disjoint union before it is essentially the same as the union only that we donrsquotcare about whether A and B have any overlap if they do we treat the common elements as beingldquodifferentrdquo For instance Z ⊔ Z is not Z but consists of two copies of each integer one 0 comesfrom the first copy of Z and another comes from the second but these are different ldquozeroesrdquo)Given f A rarr D and g rarr D the unique map A ⊔B rarr D is defined by applying f to elements of

8

A and g to elements of B (Note that if we simply took the ordinary union AcupB as the candidatefor the coproduct when A and B were not disjoint the required map h A cup B rarr D would notexist since h would have to send any element x isin A cap B to both f(x) and g(x) in order to makethe required diagram commutative)

The coproduct of two objects X and Y in Top is again the disjoint union X ⊔ Y equippedwith the disjoint union topology an open subset of X ⊔ Y is defined to be the (disjoint) union ofan open subset of X with an open subset of Y The coproduct of two groups G and H in Grp isthe free product G lowastH which is further elaborated on in the homework

The coproduct of two vector spaces V and W in Vect is the direct sum V oplusW which is definedto be the set of formal sums

v + w where v isin V and w isin W

where addition and scalar multiplication are defined ldquocomponentwiserdquo

(v1 + w1) + (v2 + w2) = (v1 + v2) + (w1 + w2) and r(v + w) = rv + rw

This is isomorphic to the product V times W with operations also defined componentwise but it iscustomary to speak about the ldquodirect sumrdquo V oplusW when referring to the coproduct as opposed tothe product wersquoll say why in a bit The maps iV V rarr V oplusW and iW W rarr V oplusW required aspart of the data of a coproduct are

iV v 983041rarr v + 0W and iW w 983041rarr 0V + w

To verify that V oplusW is the correct coproduct take any linear transformations S V rarr U andT W rarr U where U is some other vector space We then need a (unique) linear transformationh V oplusW rarr U such that

S = h iV and T = h iW

Let us determine what this map must be thereby not only showing that it exists but also that itis unique Let v + w isin V oplusW which we can write as

v + w = (v + 0W ) + (0V + w)

Since h is required to be linear we have

h(v + w) = h([v + 0W ] + [0V + w]) = h(v + 0W ) + h(0V + w)

The first term at the end is (h iV )v which must thus be Sv and the second term is (h iW )wwhich must be Tw Thus h = S + T is the map

h = S + T v + w 983041rarr Sv + Tw

which you can verify is indeed linear Thus V oplusW does satisfy the property required of a coproductSo the product of two objects in Vect is the same as their coproduct only that we write the

latter using ldquosumrdquo notation Later we will discuss products and coproducts of an arbitrary numberof elements We will see these as special cases of the general notions of limits and colimits butthey are defined via similar requirements as products and coproducts of two objects at a timeIn the category of vector spaces arbitrary products are given by the usual product

983124α Vα with

ldquocomponentwiserdquo addition and scalar multiplication but it turns out that an arbitrary coproductis given by the direct sum

983119α Vα which is defined to be the subspace of

983124α Vα where only finitely

many components are nonzero (This is in a way analogous to how the product topology on an

9

arbitrary number of topological spaces is defined) Wersquoll go over this later but is the underlyingreason why we use oplus instead of times when discussing coproducts

Encoding data via productscoproducts We can nicely summarize the definition of productsand coproducts in the following way the product of A and B is the object with the property thatfor any object D we have

Mor(DA)timesMor(DB) = Mor(D product)

and the coproduct of A and B is the object such that for any object D we have

Mor(AD)timesMor(BD) = Mor(coproductx D)

The first reflects the fact that products give a unique way to turn two morphisms (f and g in thedefinition) mapping into A and B into a single morphism (h in the definition) and the second saysthat coproducts give a unique way to turn two morphisms mapping out of A and B into a singlemorphism

Soon enough we will discuss the idea of universality and the point is that products are ldquouni-versalrdquo for maps into A and B and coproducts are universal for maps from A and B

Opposite categories Finally we give a definition which might seem silly but is actually quiteuseful Given a category C the opposite category is the category Cop is the category with the sameobjects as C but where we defined Mor(AB) in Cop to be Mor(BA) in C To be more concrete weconsider each morphism f A rarr B in C to instead be a morphism fop B rarr A in Cop (Perhaps itwould be better to write this as A larr B fop) The point is that we donrsquot actually care about whatf actually is as say a functioncontinuous maphomomorphismwhatever-we only care about itsbehavior as an ldquoarrowrdquo The opposite category Cop only retains information about how arrowsrelate to one another nothing more

Since arrows are reversed products in C become coproducts in Cop and coproduts in C becomecoproducts in Cop Similarly initialterminal objects and monomorphismsepimorphisms in C be-come terminalinitial objects and epimorphismsmonomorphisms in Cop Again the notion of anopposite category might seem like a strange thing to look at but wersquoll see that it does have uses

Functors Fullness and Faithfulness

One more productcoproduct example Let X be a topological space and define its categoryof open sets Op(X) as follows The objects of Op(X) are simply the open subsets of X and themorphisms are given by inclusions so to be concrete

Mor(U V ) =

983083the unique inclusion U rarr V if U sube V

empty otherwise

The fact that U sube V and V sube W implies U sube W says that the composition of two morphismsin this category is still a morphism in this category The product of U and V in this categoryturns out to be their intersection (with ldquoprojection morphismsrdquo U cap V rarr U and U cap V rarr V beinginclusions) and the coproduct of U and V turns out to be their (ordinary) union Note that weneed X to be a topological space in order to guarantee that U cap V and U cup V are still objects inthis category More generally the product of finitely many objects in Op(X) exists and is theirintersection and the coproduct of an arbitrary number of objects exists and is their union (Wesay that Op(X) has finite products and arbitrary coproducts)

10

Given a set X and a collection T of subsets of X we can define a similar category with elementsof T as objects (I guess we should assume that empty X isin T) and inclusions as morphisms We thenhave that this category has finite products and arbitrary coproducts if and only if T is a topologyon X so that we can rephrase the definition of a topological space itself solely in categorical termsThis perspective on what a ldquotopologyrdquo is makes it simpler to state the definition of whatrsquos calleda sheaf on a topological space which is an important construction in various areas of analysis andgeometry We might discuss this a bit later on

Functors As with most things in mathematics we should care not only about categories asmathematical structures on their own but also about ldquomapsrdquo between these structures whichldquopreserverdquo said structure The notion of a (covariant) functor between categories provides suchldquomapsrdquo in the case at hand where a functor is a way of sending objects to objects and morphismto morphisms in a way which preserves compositions and identities The precise definition is inSection 13 of the book as is the definition of a contravariant functor between categories whichis a functor which reverses the direction of arrows as opposed to ldquocovariantrdquo ones which preservedirections (It is common to use the term ldquofunctorrdquo without qualification to mean one which iscovariant Note that any functor can be made covariant by working with the opposite category acontravariant functor C rarr D is the same as a covariant functor Cop rarr D)

All of the standard examples we looked at are in the book forgetful functors power set functorsthe dual space functor in Vect etc Of particular interest is Example 139 in the book whichshows that various types of group actions in different contexts-the usual notion of a group actionon a set the notion of a continuous action of a group on a topological space and the notion ofa linear action of a group on a vector space (also known as a representation of the group)-are allencoded by a functor with domain category BG

Adjoint functors Let F Top rarr Set be the forgetful functor and let D Set rarr Top be thefunctor which sends a set S to the space S equipped with the discrete topology and a functionS rarr Sprime between sets to the same function only now viewed as a continuous map S rarr Sprime betweendiscrete spaces These two functors satisfy the following property in relation to one another givinga morphism D(S) rarr Y in Top is the same as giving a morphism S rarr F (Y ) in Set or moresymbolically

MorTop(D(S) Y ) = MorSet(S F (Y ))

We say that D is left adjoint to F and that F is right adjoint to D The point is that adjointfunctors can be used to move data back and forth between two categories

Wersquoll talk more about adjoint functors later on but for now here is one more example Nowtake F Grp rarr Set to again be the forgetful functor and take G Grp rarr Set to be the freegroup functor which sends a set S to the free group generated by S G(S) has as elements ldquowordsrdquos1 sn made up of elements of S with the group operation being ldquoconcatenationrdquo A basic factis that giving a group homomorphism from G(S) to some other group H is the same as giving anordinary function from S to F (H) which is the required adjoint property

MorGrp(G(S) H) = MorSet(S F (H))

Faithful and full functors The definitions of what it means for a functor to be faithful and fullare in Section 15 of the book where faithful functors are ones which are injective on morphisms andfull functors are ones which are surjective on morphisms but where in both cases we refer only tomorphisms which occur between the objects F (A) and F (B) in the target category corresponding tofixed objects A and B in the source category (In general a functor could send two objects AB in

11

the source category to the same object F (A) = F (B) in the target category and thus there could betwo morphisms A rarr C and B rarr C which are sent to the same morphism F (A) = F (B) rarr F (C)Such a thing could still possibly be ldquofaithfulrdquo since the non-injectivity here is arising from differentobjects in the source)

Faithfullness and fullness will be nice properties going forward and here is one hint at how theymight be useful Say we have a functor F C rarr D The problem is to determine properties onF which will imply it send products in C to products in D or coproducts in C to coproducts inD So given the former product diagram below we want the second diagram to also be a productdiagram

C

A B

P Fminusminusminusrarr

F (C)

F (A) F (B)

F (P )f g

pA pB

existh

Ff Fg

F (pA) F (pB)

Fh

Of course in order F (P ) to be an actual product of F (A) and F (B) on the right a similarcommutative diagram property would have to hold for all objects in D in place of F (C) and allmorphisms D rarr F (A) and D rarr F (B) not just those of the form Ff and Fg respectively Oneway to guarantee this is to require that all objects in D are of the form F (C) and then for allmorphisms F (C) rarr F (A) and F (C) rarr F (B) to be of the form Ff Fg the former is true whenthe functor F is surjective on objects and the latter is true when F is full We are not saying thatthese are the only conditions under which F might preserve products but it definitely seems togive a sufficient condition at least

Almost there is one more thing to require the uniqueness of the morphism Fh F (C) rarr F (P )making the diagram on the right commute Suppose there were two such morphisms Fh and Fhprimewhere h and hprime are both morphisms C rarr P in C The commutativity of the diagram on the rightgives for instance

Ff = F (pA)Fh and Ff = F (pA)Fhprime

which the functorial properites of F turns into

Ff = F (pA h) and Ff = (pA hprime)

We want h and hprime to make the first diagram commute in order to be able to apply uniqueness ofthe morphism C rarr P But this requires know that f = pA h and f = pA hprime which would followfrom F being faithful In this case it follows that h and hprime both make the first diagram commute(after we go through a similar argument with g) so the fact that P is a product guarantees thath = hprime and hence that Fh = Fhprime so that F (P ) is an actual product on the right

So we conclude that a functor which is full faithful and surjective on objects will send productsto products (Actually we can relax this a bit by requiring not that every object in D be in theliteral image of F but only that it be isomorphic to something in the image A functor with thisproperty is said to be essentially surjective which is a concept will see in a related context soonenough) Again this is not an ldquoif and only ifrdquo statement since functors can preserve products (orcoproducts) without being full faithful and surjective on objects but it at least gives a sufficientcondition Later we will see other ways to guarantee that functors preserve products which arerelated to the existence of adjoints for that given functor

12

Coproduct Examples Concreteness

Products and coproducts for metric spaces The product of two objects in Met the categoryof metric spaces with morphisms given by metric maps is the Cartesian product X times Y equippedwith the box metric and the coproduct of X and Y does not exist if both X and Y are nonemptyThis was the topic of a student presentation and here are some proof sketches

Given metric spaces (X dX) and (Y dY ) the box metric d on X times Y is defined by

d((x1 y1) (x2 y2)) = maxdX(x1 x2) dY (y1 y2))

(The name ldquoboxrdquo metric comes from the fact that in RtimesR = R2 or R3 open balls indeed looks likeldquoboxesrdquo) The key requirement needed in the definition of a product is given morphisms S rarr Xand S rarr Y that the map

h S rarr X times Y defined by h(s) = (f(s) g(s))

actually be a morphism in Met This turns into the requirement that

d((f(s) g(s)) ((f(t) g(t))) = maxdX(f(s) f(t)) dY (g(s) g(t)) le dS(s t) for all s t isin S

given the fact that

dX(f(s) f(t)) le dS(s t) and dY (g(s) g(t)) le dS(s t) for all s t isin S

For the latter inequalities to imply the former d must (at least up to isometry) be given by thebox metric Note that the two other ldquostandardrdquo metrics one might put on a productmdashthe taxicab

dX + dY and Euclidean983156

d2X + d2Y metricsmdashdo not satisfy this required ldquometric maprdquo property

and so do not give the product in MetTo see that coproducts do not exist if XY are nonempty suppose instead (P dP ) was a co-

product for X and Y with inclusion morphisms iX X rarr P and iY Y rarr P For any n isin Ntake 0 n to be a metric space with the absolute value metric d and let f X rarr 0 n andg Y rarr 0 n be the constant function 0 and the constant function n respectively Since P is acoproduct there exists h P rarr 0 n such that

h iX = f and h iY = g

But in order for h to actually be a morphism in Met we must then have

d(h(iX(x) iY (y)) le dP (iX(x) iY (y)) for any x isin X y isin Y

which translates inton = |0minus n| le dP (iX(x) iY (y))

But this should then be true for any n isin N which would imply that the distance in P betweenan element coming from X and one coming from Y should be infinite which is nonsense (Notethat if you altered your definition of ldquometricrdquo to allow for infinite distance then a coproduct wouldexist the disjoint union X ⊔Y where you defined the distance between elements of X and Y to beinfinite)

Coproducts for groups The coproduct of two groups G and H in Grp is the free product GlowastHwhich is defined to be the set of all words consisting of elements of G andH in an alternating fashion

13

with concatenation as the group operation (When we have to elements of G or two elements of Hnext to each other with use their respective group operations to turn these into single elements ofG or H) This was also the topic of a student presentation and here is the basic idea Given grouphomomorphisms g G rarr N and k H rarr N that G lowastH is the coproduct comes down to showingthat the map

h G lowastH rarr N defined by h(g1h1 gnhn) = f(g1)k(h1) middot middot middot f(gn)k(hn)

and similarly on other possible words in GlowastH is a group homomorphism which follows essentiallyfrom the definition of the group operation on G lowastH indeed this group operation arises preciselyfrom this requirement

Now if we instead tried to use GtimesH with its standard inclusions G rarr GtimesH and H rarr GtimesHas the coproduct the issue is that the analogous map

h GtimesH rarr N defined by h((g h)) = f(g)k(h)

would NOT be a homomorphism because h((g1 h1)(g2 h2)) is not equal to h(g1 h1)h(g2 h2) bythe way in which the group operation is defined on the direct product G times H However if westrict our category to only consist of abelian groups so that the N rsquos we consider in the defini-tion of coproduct are also abelian then G times H does serve as a coproduct we can always rewritef(g1)k(g1) middot middot middot f(gn)k(gn) as f(g1) middot middot middot f(gn)k(h1) middot middot middot k(hn) in N Irsquoll leave it to you to flesh out allof these details

Concrete categories The categories Set Top Vect and Grp all have the property thattheir objects are simply sets possibly equipped with extra structure and that their morphisms areordinary set-theoretic functions possibly satisfying some additional requirement Such categoriesare nice since we can often use intuition about sets and functions in their study A concrete categoryis intuitively one where we can think of objects as being ldquosetsrdquo and morphisms as being ldquofunctionsrdquoonly that we have to interpret this in the right way

Here is the definition a concrete category is a category C equipped with a faithful functorF C rarr Set which is part of the data The idea is that such a functor gives us way to turnan object A isin Ob(C) into a set F (A) and a morphism f isin MorC(AB) into a function Ff isinMorSet(F (A) F (B)) where the ldquofaithfulrdquo requirement says that we do not lose information aboutf when doing so so that we can (essentially) learn everything we need to know about f by studyingFf instead In the four examples above the required faithful functor is simply the forgetful functorand indeed in general we think of F as being a ldquoforgetful functorrdquo for the concrete category (C F )

Examples Here are two more examples of concrete categories Let G be a group and considerthe category BG with a single object and elements of G morphisms As written this does notappear to be concrete since morphisms are not honest functions lowast rarr lowast but the point is that thiscategory can indeed by ldquoconcretizedrdquo by specifying an appropriate ldquoforgetful functorrdquo A functorF BG rarr Set is the data of a (left) action of G on a set S = F (lowast) and saying this functor isfaithful is what it means to say that this action is faithful different g h isin G act on S in differentways or equivalently that there exists s isin S such that gs ∕= hs Any group has at least onefaithful left actionmdashnamely the action on itself given by left multiplicationmdashto BG can always beequipped with a faithful functor to Set This is all just a way of phrasing the fact that any groupis isomorphic to a subgroup of a permutation group

The category Rel of relations can also be ldquoconcretizedrdquo even though as written morphismsA rarr B are not honest functions Take F Rel rarr Set to be the power set functor which sends an

14

object in Rel to its power set and a relation f A rarr B to the induced function Ff P (A) rarr P (B)defined by

(Ff)(S) = b isin B | there exists a isin S such that (a b) isin f

In other words if you think of the relation f sube A times B as a ldquopossibly not everywhere-definedpossibly multivalued functionrdquo (Ff)(S) is analogous to taking the image of S That F is faithfulwill be left to the homework

In fact most categories yoursquoll see can be made concrete by specifying an appropriate forgetfulfunctor but not all categories can be made concrete in this way Here is the standard example of anon-concrete category which we mention only to say there is such a thing but which we wonrsquot doanything more with since understanding it better requires knowledge of algebraic topology Thehomotopy category of topological spaces is the category hTop whose objects are topological spacesand whose morphisms are given by equivalence classes of homotopic maps

MorhTop(XY ) = homotopy classes of continuous maps X rarr Y

(Saying that two maps are homotopic means that you can ldquodeformrdquo one into the other but wersquollleave the precise definition to a course in algebraic topology) It turns out that no functor fromhTop to Set can be faithful so that hTop cannot be made concrete but this is actually a deepand difficult thing to show so we make no attempt to do so here

Natural Isomorphisms Representability

Natural transformations The definition of a natural transformation η F rArr G between two(both covariant or both contravariant) functors FG C rarr D is given in Section 14 of the bookThe key idea is that a natural transformation gives a way to turn data about F into data aboutG in a way which is compatible with all possible morphisms as expressed by the commutativity ofthe diagrams

F (A) F (B)

G(A) G(B)

Ff

ηA ηB

Gf

arising from morphisms f A rarr B (or f B rarr A in the contravariant case) in C In the case of anatural isomoprhism η F rArr G so that each ηA F (A) rarr G(A) is actually an isomorphism in Dthe point is not only that F and G produce isomorphic objects but they do so in a way which iscompatible with all possible category-theoretic data

The first standard example of a natural isomorphism is the one between the identity functor onthe category of finite-dimensional and the functor which sends a finite-dimensional vector space Vto V lowastlowast the dual of its dual induced by the isomorphism V rarr V lowastlowast defined by

v 983041rarr (the linear map on V lowast which sends f to f(v))

This was the topic of a student presentation and can be found in Section 14 of the book Theessential reason why this isomorphism is ldquonaturalrdquo is that V rarr V lowastlowast can be defined without referenceto a basis Contrast this with the the functor which sends V to its (single) dual V lowast it is still truethat finite-dimensional vector space is isomorphic to its dual but specifying such an isomorphism

15

requires additional data which prevents the identity functor from being ldquonaturally isomorphicrdquo tothe single dual functor

Some history At this point it makes sense to say a bit about the interesting origins of categorytheory in the 1950rsquos Eilenberg and Mac Lane where studying cohomology (or possible homologyone of those two) in algebraic topology which associates algebraic data (a group ring or similar)to a topological space in a way which encodes some of the topology They had two cohomologicalconstructions which ended up proving isomorphic results but they noticed that not only werethe resulting objects the same but the actual constructions themselves were in some sense theldquosamerdquo They thus then needed a way to formalize this notion and invented the concept of anatural transformation to do so where they interpreted two constructions as being the same asbeing compatible with all morphisms

But this then leads to the question what types of objects should a natural transformationoccur between Eilenberg and Mac Lane then had to invent the notion of a ldquofunctorrdquo to describewhat a natural transformation went from and what it went to In their motivating setting theyinvented the notion of a functor to describe more formally properties which their cohomologicalconstructions were supposed to satisfy But this then leads to the question what types of objectsshould a functor map between In other words what type of data should a functor (cohomologicalconstruction in their example) take as input and should it output They finally had to thus inventthe notion of a ldquocategoryrdquo to describe the source and target of a functor so that their cohomologicalconstructions could be interpreted as functors from a category of topological spaces to a categoryof algebraic objects

Thus historically natural transformations came first then functors and then categories Aswith most things in mathematics categories thus arose from attempts to encode what was beingseen in some concrete examples in a more general and formal setting

Representable functors Representable functors are defined in Section 21 of the book but wersquollgive the required definitions here in a hopefully simpler to digest manner The point is that anyobject A in a (locally small) category C gives two ldquonaturalrdquo functors C rarr Set one covariant andthe other contravariant which as wersquoll see pretty much encode all data about A itself We definehA C rarr Set to be the covariant functor defined by

hA(B) = Mor(AB) on objects and

hA(Bfminusrarr C) = the map Mor(AB)

hAfminusminusrarr Mor(AC) defined by g 983041rarr f g

We define hA C rarr Set to the contravariant functor defined by

hA(B) = Mor(BA) on objects and

hA(Bfminusrarr C) = the map Mor(CA)

hAfminusminusrarr Mor(BA) defined by g 983041rarr g f

(The functor hA is called the functor of points of A and wersquoll usually think of it as being a covariantfunctor hA Cop rarr Set The functor hA does not have a standard name) So hA is defined bymorphisms from A and hA is defined by morphisms into A Wersquoll see later the sense in which suchfunctors encode all information about the object A

A functor F C rarr Set is representable if is naturally isomorphic to such a functor so F sim= hA

for some A in the covariant case and F sim= hA for some A in the contravariant case The idea wersquollbuild towards is that we can essentially treat such a functor as being A itself so that representable

16

functors can be thought as being actual objects in C For now here is a key example wersquove seenbefore which we can now formulate using the language of representability

Given AB isin Ob(C) let F C rarr Set be the contravariant functor defined by

F (C) = Mor(CA)timesMor(CB) and F (Cfminusrarr D) = [(g k) 983041rarr (g f k f)]

where (g k) 983041rarr (g f k f) gives a map Mor(DA) timesMor(DB) rarr Mor(CA) timesMor(CB) ieF (D) rarr F (C) In order for this functor to be representable requires the existence of an object inC satisfying

Mor(CA)timesMor(CB) sim= Mor(C representing object)

in a way which is compatible with all morphisms but we have actually already seen this notionelsewhere it is essentially the definition of a product AtimesB for A and B Indeed we claim that Fis naturally isomorphic to hAtimesB as we now show

First to define a natural isomorphism hAtimesBsim= F requires for each C isin Ob(C) a choice of a

bijection (ie isomorphism in Set)

ηC hAtimesB(C)sim=minusrarr F (C)

which in our case looks like

ηC Mor(CAtimesB))sim=minusrarr Mor(CA)timesMor(CB)

Take this to be the map

(Cfminusrarr AtimesB) 983041rarr (pA f pB f)

where pA A times B rarr A and pB A times B rarr B are the two projection morphisms in the definitionof a product The definition of product implies that this map is invertible since given k C rarr Aand ℓ C rarr B there exists a unique h C rarr A times B such that k = pA h and ℓ = pB h Forthese bijections to define a natural isomorphism hAtimesB

sim= F requires that they be compatible withall morphisms in C which means that given any g C rarr D in C the following diagram shouldcommute

hAtimesB(C) hAtimesB(D)

F (C) F (D)

hAtimesB(g)

ηC ηD

Fg

which in the case at hand looks like

Mor(CAtimesB) Mor(DAtimesB)

Mor(CA)timesMor(CB) Mor(DA)timesMor(DB)

minus g

ηC ηD

(minus gminus g)

Take an element k D rarr AtimesB in the upper right Applying the map on top gives the elementk g in the upper left and then applying ηC on the left side gives

(pA (k g) pB (k g))

17

in the lower left Instead starting with k D rarr AtimesB in the upper right applying ηD on the rightgives (pA k pB k) and applying the map on the bottom gives

((pA k) g (pB k) g)

in the lower left which is the same as the previous element in the lower left by the associativity ofcomposition Hence the isomorphisms ηC do define a natural isomorphism between hAtimesB and F Again the point is that you can then interpret the functor F as being the ldquosamerdquo as the productAtimesB since both encode the same data

More Representable Examples

Coproducts As in the case of products the definition of coproducts can also be phrased in termsof a representable functor Given objects A and B in C let F C rarr Set be the covariant functordefined by

F (C) = Mor(AC)timesMor(BC) and F (Cfminusrarr D) = [(g k) 983041rarr (f g f k)]

where the latter is a function Mor(AC)timesMor(BC) rarr Mor(AD)timesMor(BD) This functor isrepresentable if and only if a coproduct A ⊔ B for A and B exists and the representing object isthat coproduct the key is the requirement that we have bijections

Mor(AC)timesMor(BC) sim= Mor(A ⊔BC)

for all C which is essentially the defining property a coproduct for A and B should have

Power set functor contravariant version Let P Set rarr Set be the contravariant powerset functor defined on objects by sending a set to its power set and on morphisms by sendingf A rarr B to the preimage function Pf P (B) rarr P (A) given by S 983041rarr fminus1(S) In order for this tobe representable requires the existence of a set X with natural bijections

P (A) sim= Mor(AX)

for all sets A that is a set X so that mapping into X is the same data as specifying a subset ofthe domain We take X = 0 1 to be a two-element set and the required bijections

ηA P (A) rarr Mor(A 0 1) to be given by S 983041rarr χS

where XS A rarr 0 1 is the indicator or characteristic function S defined by sending elementsof S to 1 and elements of A minus S to 0 The inverse of ηA sends a function g A rarr 0 1 to thepreimage gminus1(1) sube A

To verify that the bijections ηA define the data of a natural isomorphism η P rArr h01 wecheck the commutativity of some diagrams Given a function f A rarr B consider

P (A) P (B)

h01(A) = Mor(A 0 1) h01(B) = Mor(B 0 1)

Pf = fminus1

ηA ηB

minus f

18

Take S isin P (B) in the upper right Applying the map on top gives fminus1(S) isin P (A) and thenapplying the map on the left gives χfminus1(S) isin Mor(A 0 1) On the other hand applying the mapon the right to S isin P (B) gives χS isin Mor(B 0 1) and then applying the map on the bottomgives χS f isin Mor(A 0 1) The two resulting functions χfminus1(S)χS f in the lower left are

χfminus1(S)(x) =

9830831 x isin fminus1(S)

0 otherwiseand χS(f(x)) =

9830831 f(x) isin S

0 otherwise

so these are the same since x isin fminus1(S) if and only if f(x) isin S Thus the diagram above commutesso P is naturally isomorphic to h01 and hence P is representable

Power set functor covariant version Now consider the covariant power set functor stilldenoted by P Set rarr Set which again sends a set to its power set but now sends a functionf A rarr B to the image function Pf P (A) rarr P (B) defined by S 983041rarr f(S) In order for this tobe representable there would have to exist a set X such that P sim= hX which translates into havingnatural bijections

P (S) sim= Mor(XS)

for all sets S However in this case there is no natural choice for X as there is no natural way todetect subsets of S by mapping into S as opposed to the case of P (S) sim= Mor(S 0 1)

That no such X exists can be made precise using a simple argument when S = empty P (S) hascardinality 1 soX would have to be empty as well since otherwise Mor(XS) would have cardinality0 but if X is empty then Mor(empty S) always cardinality 1 regardless of S which is clearly not trueof P (S) Thus the covariant power set functor is not representable

Forgetful on Top The forgetful functor F Top rarr Set is representable Indeed a representingobject would have to satisfy

X sim= Map(objectX)

as sets for any topological space X (where Map denotes the set of continuous maps) and it iseasy to see that a single point satisfies this property a continuous map f pt rarr X is completelydetermined by the element f(pt) of X and any such element gives a continuous map in this wayOf course checking that F sim= hpt requires checking that various diagrams commute but this issimple to work out in this case so we omit it here

Extracting a topology With an eye towards the idea that the functors hA hA should captureall information about an object A in a category note that the result above shows that in TophX first of all captures all information about X as a set since the set X can be extracted fromhX(pt) = Map(ptX)

We claim that hX actually captures all information about X as a topological space since thetopology on X can be extracted from hX in the following way As we saw in the power set examplewe have a bijection

P (X) sim= Mor(X 0 1)

Now equip 0 1 with the topology in which 1 is open but 0 is not Then a continuous mapf X rarr 0 1 is fully determined by the preimage fminus1(1) since to be continuous only requiresthat this single preimage be open in X Moreoever any open subset of X arises in this way bytaking f to be the indicator function of that open set Thus we get a bijection

Op(X) sim= Map(X 0 1)

19

where Op(X) denotes the set of open subsets of X ie the topology on X Hence the topology onX can be extracted from hX as claimed so hX and hX do encode all information about X

This fact can also be interpreted as saying that the contravariant functor which sends a topo-logical space X to its collection of open subsets (and which sends a continuous map to the mapinduced by taking preimages) is representable with representing object 0 1 equipped with thetopology given above This will be worked out in a homework problem

Forgetful on Grp The forgetful functor Grp rarr Set is also representable In this case arepresenting object should be a group giving set-theoretic bijections

G sim= Hom(objectG)

for all groups G where Hom denotes the set of group homomorphisms The group Z works

G sim= Hom(Z G)

since a homomorphism φ Z rarr G is completely determined by φ(1) since Z is generated by 1 andthere are no restrictions on what φ(1) isin G can actually be The inverse of the desired bijection isthus φ 983041rarr φ(1) and it is simple to check that morphisms behave appropriately so that the forgetfulfunctor in this setting is indeed isomoprhic to hZ

Forgetful on Ring The forgetful functor Ring rarr Set is also representable where by Ring wemean the category of rings with identity elements and where morphisms (ie ring homomorphisms)are required to preserve identities Here we need an object satisfying

R sim= Hom(object R)

for all R isin Ob(Ring) and we can see that Z[x] the ring of polynomials in the variable x withcoefficients in Z works Indeed given a ring homomorphism φ Z[x] rarr R the fact that φ isrequired to preserve identities fully determines φ(n) for any n isin Z (since φ(n) = nφ(1)) and hencesince φ preserves multiplication its behavior is fully determined by φ(x) which can be any elementof R The inverse of the bijection we want is thus φ 983041rarr φ(x) and it will follow that this forgetfulfunctor is isomorphic to hZ[x]

Equivalences between Categories

Definitions iso and equiv Now that we have a notion of a ldquomorphismrdquo between categorieswe can talk about the sense in which two categories are the ldquosamerdquo The first possible definitionis the ldquoobviousrdquo one you might expect given the definition of ldquoisomorphismrdquo wersquove seen in everyother context until now we say that C is isomorphic to D if there exist functors F C rarr D andG D rarr C such that G F = idC and F G = idD where idC and idD are identity functors

However this definition turns out to be too restrictive Indeed the requirement that the givencompositions equal identity functors says that given an object A in C the object G(F (A)) beliterally the same as A itself and similarly for objects in D But we know that two objects in acategory can be thought of as being the ldquosamerdquo without being literally the same as long as theyare isomorphic For instance productscoproducts are not unique in the literal sense but only inthe sense that they are isomorphic in a unique way To take another example the definition ofa functor being representable does not say that F (B) literally be the same as hB(A) only thatthey be isomorphic for sure the power set P (S) of a set S is not literally the same as the set offunctions S rarr 0 1mdashit is only the ldquosamerdquo in the sense that they encode the same data

20

Thus it makes sense to relax the requirement that A andG(F (A)) in the definition of isomorphiccategories be equal to the requirement that they be isomorphic This leads to the following betterdefinition of two categories being the ldquosamerdquo C is equivalent to D if there exist functors F C rarr D

and G D rarr C such that G F sim= idC and F G sim= idD So we require only that for instance thefunctor G F be naturally isomorphic to the identity functor on C which says that for any objectA G(F (A)) be isomorphic to A in a natural way Wersquoll see that this truly is the better notionof what it means for two categories to be the ldquosamerdquo and it reflects the idea that all categoricalproperties of one category can be transported to the other

Alternate characterization of equivalences A functor F C rarr D implementing an equiv-alence between categories is called unsurprisingly and equivalence and the ldquoinverserdquo functorG D rarr C is still referred to as being its inverse even though it is not technically an ldquoinverserdquoin the sense that compositions give literal identities But as in the case of Set where ldquoinvertiblerdquocan be characterized without reference to an inverse as ldquoinjectiverdquo and ldquosurjectiverdquo so too canequivalences between categories be characterized without explicit reference to an inverse

A functor F C rarr D is an equivalence if and only it is full faithful and essentiallysurjective

We have seen the notions of full and faithful before and essentially surjective means that any objectin D is isomorphic to something in the image of F for all D isin Ob(D) there exits C isin Ob(C) suchthat F (C) sim= D Justifying this result was the topic of a student presentation and can be foundin the book in Section 15 (Technically to construct an ldquoinverserdquo of F requires that we work withsmall categories but whatever) The bulk of the real work goes into showing that the ldquoinverserdquofunctor constructed is actually a functor which comes down to some abstract nonsense argumentvia commutative diagrams

Equivalences preserve everything We previously saw the notions of full faithful and es-sentially surjective back when thinking about what properties a functor could have which wouldguarantee it sent products to products so the result we proved back then was that equivalencespreserve products Similarly equivalences preserve coproducts and the proof is basically the sameafter reversing arrows A clean way of deriving this is as follows if F C rarr D is an equivalencethen so is F op Cop rarr Dop so since F op preserves products F preserves coproducts becausecoproducts in a category become products in the opposite category

In general an equivalence between categories will preserve whatever categorical notions youwant monomorphisms epimorphisms initial objects terminal objects etc so that equivalentcategories really do have essentially the ldquosamerdquo properties

Equivalence example Here is a first example (also found in the book) of equivalent categorieswhich are not isomorphic Let FVectR be the category whose objects are finite-dimensional realvector spaces and whose morphisms are linear transformations Let MatR be the category ofmatrices defined as follows

bull objects in MatR are nonnegative integers

bull a morphism n rarr m is an mtimes n matrix with real entries

bull composition in MatR is given by matrix multiplication given B n rarr m and A m rarr kmdashsoB is an mtimes n matrix and A is a k timesm matrixmdashthe composition A B n rarr k is the k times nmatrix AB

bull identity morphisms in MatR are given by identity matrices

21

We claim that FVectR and MatR are equivalent First note that they certainly cannot be isomo-prhic isomorphisms in MatR exist only between two objects which are literally the same (ie anmtimesn matrix can be invertible only when m = n) but isomorphisms in FVectn can exist betweenobjects which are not literally the same So there are in a sense not enough objects in MatR toallow it to be isomorphic to FVectR

To construct an equivalence choose one and for all an isomorphism TV V rarr RdimV forevery object in FVectR or equivalently choose once and for all a basis for every V DefineF FVectR rarr MatR by

V 983041rarr dimV on objects and

(T V rarr W ) 983041rarr (the standard matrix of SW T Sminus1V RdimV rarr RdimW ) on morphisms

Equivalently F sends T to the matrix of T relative to the chosen bases for V and W It isstraightforward to check that this is a functor The inverse functor G MatR rarr VectR is givenby

n 983041rarr Rn on objects and

(mtimes n matrix A) 983041rarr (the linear transformation A Rn rarr Rm induced by A) on morphisms

One can check that G is a functor and that F G and G F are naturally isomorphic to identitiesor instead one can argue that F alone is full faithful and essentially surjective The point is thatG(F (V )) gives back RdimV so not literally V itself but only something isomorphic to V which iswhy this gives an equivalence of categories and not an isomorphism

Yoneda Lemma Functors as Objects

Compact Hausdorff spaces and Clowast-algebras Last time we gave an example of equivalentcategories which were not isomorphic but in many ways that example is unsatisfactory when doinglinear algebra people for sure often work with matrix representations of linear transformations asthe example from last would suggest but no one in their right mind literally thinks about suchthings in terms of the equivalence FVectR rarr MatR itself In other words this equivalence isessentially ldquocooked uprdquo for the sake of giving an example but it does not really give any insightinto the actual mathematics of linear algebra

So with that in mind we will describe another equivalence which actually does have some realmeaning in the sense of producing new ways of thinking about some mathematics Given a compactHausdorff space X we let C(X) denote the space of continuous complex-valued functions on X

C(X) = f X rarr C | f is continuous

We can equip the set C(X) with various additional structures First it is a complex vector spaceunder ordinary addition and scalar multiplication of functions Second we can multiply two func-tions together

(fg)(p) = f(p)g(p)

which turns C(X) into whatrsquos called an algebra over C (We wonrsquot give the definition of anldquoalgebrardquo but the point is that this multiplication is in some sense ldquocompatiblerdquo with the vectorspace structure) Next any element f of C(X) has a conjugate element f obtained by takingcomplex conjugates of the values of f

f(p) = f(p)

22

And finally we can equip C(X) with a norm via

983042f983042 = suppisinX

|f(p)|

where |middot| denotes complex absolute value (This supremum exists sinceX is compact by the ExtremeValue Theorem) All of these structures (vector space algebra multiplication conjugation norm)turn C(X) into whatrsquos called a Clowast-algebra We wonrsquot define this precisely but the definition encodesvarious ways in which these structures are compatible Actually C(X) is a unital commutative Clowast-algebra where ldquounitalrdquo means that the multiplication has an identity element (namely the constantfunction 1) and commutative means that fg = gf There is a natural notion of homomorphismbetween Clowast-algebras which are maps between them which behave nicely with all the structuresinvolved In particular given a continuous map f X rarr Y between compact Hausdorff spaces weget a Clowast-algebra homomorphism flowast C(Y ) rarr C(X) given by pre-composition with f flowast sendsg isin C(Y ) to g f isin C(X)

We thus have a contravariant functor CHaus rarr ucClowast-Alg from the category of compactHausdorff spaces and the category of unital commutative Clowast-algebras It turns out that anyunital commutative Clowast-algebra arises in this way in the sense that given any unital commutativeClowast-algebra B there exists a compact Hausdorff space X such that C(X) sim= B as Clowast-algebrasand that morphisms between these algebra arises from continuous maps in the manner describeabove This says that the functor above is essentially surjective full and it turns out that it isalso faithful so that it is an equivalence Thus all information about compact Hausdorff spaces iscompletely encoded in information about unital commtuative Clowast-algebras and indeed one couldlearn everything there is to know about a compact Hausdorff space X from its Clowast-algebra C(X)of functions The fact that the categories CHaus and ucClowast-Alg are equivalent is the startingpoint in the subject known as noncommutative geometry which wersquoll say something about a bitlater on If we drop the requirement that our algebras be unital it turns out that the categoryof commutative Clowast-algebras in general is equivalent to the category of locally compact Hausdorffspaces via essentially the same functor where we take as morphisms in the category of locallycompact Hausdorff spaces to be continuous functions which ldquovanish at infinityrdquo

We are only introducing Clowast-algebras to give an example of an interesting and actually usefulequivalence between categories but just for the fun of it wersquoll note the following Examples ofnoncommutative Clowast-algebras come from what are known as ldquoalgebras of bounded operators onHilbert spacesrdquo which show up in quantum mechanics as the things which describe physicallyobservable quantities like position momentum and energy (In fact all Clowast-algebras arise fromHilbert spaces in this way) Thus Clowast-algebras show up in mathematical formulations of quantummechanics and the notion of noncommutative geometry mentioned above is at the core of someattempts to understand the elusive concept known as ldquoquantum geometryrdquo

Yoneda Lemma The Yoneda Lemma states that given a (covariant) functor F C rarr Set whereC is locally small for any A isin Ob(C) there exists a bijection

F (A) sim= Nat(hA F )

where the right side denotes the set of natural transformations from hA to F A similar result holdsin the contravariant case where we get bijections F (A) sim= Nat(hA F ) (There is also a sense inwhich these bijections are themselves natural but wersquoll ignore this bit) The proof of the YonedaLemma was the topic of a student presentation and can be found in Section 22 of the book

Here we will get a bit philosophical and talk about what the point of the Yoneda Lemmaactually is and later the sense in which it says that we can view arbitrary functors as ldquonon-existent

23

objectsrdquo in a category First we mention what we will call the ldquofunctor of pointsrdquo approach tomathematics a mathematical object can be fully characterized by all morphisms into it That isknowledge of Mor(SA) for all S is equivalent to knowledge of A itself Indeed this is an idea wementioned previously in relation to the functors hAmdashwhich we previously said is referred to as beingthe functor of points of Amdashand now we seek to push this idea further In some examples literalpoints of A can be extracted from hA as in the case of the bijection X sim= hX(pt) = Map(ptX) fora topological space X or G sim= hG(Z) = Mor(Z G) for a group G Moreover in these examples therest of the (topological) structure onX or (group) structure on G can also be extracted from hX andhG respectively In general we can view Mor(SA) as in a sense being ldquopoints of A parameterizedby Srdquo which comes from the idea that in say Set Mor(SA) = AS is ldquothe product of S-manycopies of Ardquo so literally points of A ldquoparameterizedrdquo by S In various areas of mathematics anelement of Mor(SA) is thought of as being a ldquogeneralizedrdquo point of A so that the functor of pointshA detects generalized points of A

So an object A is fully encoded by the functor hA Now we claim that morphisms betweenobjects are also encoded by such functors Indeed in the case where F = hB is itself a functor ofpoints the Yoneda result Nat(hA F ) sim= F (A) becomes

Nat(hA hB) sim= hB(A) = Mor(AB)

which says precisely that morphisms between the objects A and B can be obtained solely fromthe functors hA and hB (This fact is one reason why one should prefer the contravariant functorhA to the covariant functor hA in the covariant case Nat(hA F ) sim= F (A) for F = hB becomesNat(hA hB) sim= hB(A) = Mor(BA) so that morphisms between two objects get reversed whencharacterizing them in terms of representable functors)

Pushing our philosophy further we now know that not only objects but also morphisms betweenobjects are fully encoded by the functors hA The final step is to note that we should not only careabout objects and morphisms between them but that an important part of the data of an objectA is also how arbitrary contravariant functors F C rarr Set behave on A and that this too can beobtained from hA precisely via the Yoneda result that F (A) sim= Nat(hA F ) So objects morphismsbetween objects and how arbitrary functors behave on objects are all encoded in knowledge ofthe functors hA so that these functors truly do capture in the most general sense possible allinformation about A

Functors as nonexistent objects The discussion above is all well and good but it still doesnrsquotquite say why we should care about the fact that we can study on object A by studying hA insteadin that we still havenrsquot spoken about what type of insight about A can be gleamed from hA thatwasnrsquot readily apparent in A itself already Now we argue that the true point of the YonedaLemma is that it gives a way to think about arbitrary functors F Cop rarr Set as ldquoobjectsrdquo in C

itself or really as objects in an ldquoenlargementrdquo of C Certainly for those functors F sim= hA which arerepresentable we can indeed think about F as simply being the object A but now we claim thateven non-representable F can be thought of as being ldquononexistentrdquo objects in C The upshot is thatoften in mathematics we come across things which do not actually exist but the Yoneda Lemmagives a way to (indirectly) study such things anyway by interpreting them as functors instead

The starting point of course is that an actual object A is the same as hA as we have explainedabove Thus by analogy we can interpret a non-representable functor F as being an ldquoobjectrdquo in C

as long as we know what it means to ldquomaprdquo into F from other things which are honestly objects in CThat is if F were indeed representable then F (A) sim= Nat(hA F ) sim= Mor(AF ) would characterizethe functor of points of F (here we abuse notation by using the same letter F to denote both arepresentable functor and the object which represents it) so now we take Mor(AF ) = F (A) as

24

the definition of the ldquofunctor of pointsrdquo of the non-existent object F We are thus taking elementsof F (A) as being what it means to ldquomaprdquo from A into F even when F is not representable Thefact that the Yoneda Lemma gives a way to define maps from A into F thus says that F itself isequipped with the same data that a literal object of A would be equipped with so why shouldnrsquotwe interpret F as a being a type of ldquoobjectrdquo in C as well

To phrase this another way note that the Yoneda Lemma gives rise to a functor

C rarr Fun(CopSet)

where Fun(CopSet) denotes the category whose objects are functors from Cop to Set and morphismsare natural transformation defined on objects by A 983041rarr hA and on morphisms by the bijectionMor(AB) sim= Nat(hA hB) arising from the Yoneda Lemma That this is actually bijection saysthat this functor is full and faithful so that we can view C as being a subcategory of the functorcategory Fun(CopSet) (We cay say that C is isomorphic to the subcategory of Fun(CopSet)consisting of the functors hA or that it is equivalent to the subcategory of Fun(CopSet) consistingof representable functors This functor C rarr Fun(CopSet) is usually called the Yoneda embeddingof C into Fun(CopSet)) Thus we can view Fun(CopSet) as an ldquoenlargmentrdquo of C and the ideawe are trying to make sense of is that other objects in this ldquolargerrdquo category should be interpretedas ldquonon-existent objectsrdquo of C

Just how far can we actually push this idea that arbitrary functors Cop rarr Set should be thoughtof as being ldquoobjectsrdquo of C Pretty far in fact For instance often times in geometry one comesacross a type of ldquosmoothrdquo space on which some group acts and one is interested in doing ldquocalculusrdquoon the resulting quotient space which is the space of orbits under the given group action Theproblem is that such quotient spaces are very often (usually in fact) not ldquosmoothrdquo themselves sothat the tools from calculus are not readily available The solution is to nonetheless treat thisquotient as if it were ldquosmoothrdquo and attempt to do calculus anyway To make this precise thisldquonon-smoothrdquo space actually turns out to have a natural interpretation as a ldquosmooth functorrdquo andit is through this functor which we can view this non-smooth space as actually being smooth afterall so we are forced to expand our notion of what we mean by ldquosmooth spacerdquo by allowing thingswhich are not spaces in the usual sense which certainly leads to more abstraction but the benefitis that it actually makes sense to do calculus on such ldquospacesrdquo interpreted as functors

The discussion above is pretty vague so here is one final more concrete example Recall thatequivalence

CHausop rarr ucClowast-Alg

between the category of compact Hausdorff spaces and the category of unital commutative Clowast-algebras mentioned at the start We now claim that a non-commutative (unital) Clowast-algebra shouldbe interpreted as the algebra of functions on some non-existent ldquononcommutative compact Haus-dorff spacerdquo and it is this idea which the subject of noncommutative geometry makes precise Solet B be a non-commutative Clowast-algebra Then the functor of points hB gives a functor

ucClowast-Alg rarr Set

which sends C(X) (where X is some compact Hausdorff space) to Hom(C(X) B) where Hom de-notes homomorphisms of Clowast-algebras whatever that means Composing with the previous functorwe get a functor wersquoll denote by B itself

B CHausop rarr Set

which on objects at least is given by X 983041rarr Hom(C(X) B) This functor is not representable sinceB is not commutative but the philosophy espoused above suggests we should think of it as giving a

25

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

and c = g(b) (here we abuse notation by denoting the function of which f is graph by the symbolf itself) so

c = g(b) = g(f(a)) = (g f)(a)

meaning that (a c) isin g f precisely when c = (g f)(a) Composition of relations is associativeand you can check that graphs of ordinary identity functions

graph(idA) = (a a) isin AtimesA | a isin A

give the identity morphisms (I spent much time as graduate student trying to do ldquogeometryrdquo witha certain category of relations so this is an example near and dear to my heart)

Special Morphisms Products

Isomorphisms A morphism f isin Mor(AB) between two objects A and B in a category is anisomorphism or is invertible if it has an inverse there exists a morphism g isin Mor(BA) such thatgf = idA and fg = idB where idA isin Mor(AA) and idB isin Mor(BB) are the identity morphismswhich are assumed to exist as part of the definition of a category We say that A and B areisomorphic and we write A sim= B if there exists an isomorphism between them

This general categorical notion of isomorphism reduces to the one you would expect in thestandard examples isomorphisms in Set are bijective functions in Grp they are bijective grouphomomorphisms in Vect they are invertible linear transformations and in Top they are homeo-moprhisms ie continuous bijections with continuous inverses

A category of metric spaces Now consider the category MetC of metric spaces where mor-phisms are given by continuous functions In this category the notion of ldquoisomorphismrdquo againmeans homeomorphism so two metric spaces are isomorphic in this category when they are home-omorphic However often when considering whether or not two metric spaces are the ldquosamerdquo wehave another definition in mind that of two metric spaces being isometric X is isometric to Y ifthere exists a bijective isometry between them which is a bijective map f X rarr Y which preservesdistance in the sense that

dY (f(p) f(q)) = dX(p q) for all p q isin X

After all in a metric space you have more data than simply ldquoopen setsrdquo (which is all that ahomeomorphism detects) so it would be good to have a notion of ldquosamenessrdquo which incorporatedthe extra data of distance as well

The goal is then to construct a category of metric spaces where ldquoisomorphicrdquo indeed meansisometric and we can do so simply by restricting the types of continuous maps we consider Thecategory Met has metric spaces as objects but now a morphism f X rarr Y is taken to be afunction which does not ldquoenlargerdquo distance in the sense that

dY (f(p) f(q)) le dX(p q) for all p q isin X

Maps with this property are often called metric maps Any metric map is actually continuous soMet is a subcategory of MetC (The subscript C used in MetC is meant to denote ldquocontinuousrdquoindicating that we take arbitrary continuous maps as morphisms) In this category now for f X rarr Y to be an isomorphism requires that f and its inverse fminus1 both be metric maps so

dY (f(p) f(q)) le dX(p q) and dX(fminus1(s) fminus1(t)) le dY (s t)

4

for all p q isin X and s t isin Y But writing s and t as s = f(p) and t = f(q) the second conditionbecomes dX(p q) le dY (f(s) f(t)) and the two inequalities together thus imply

dY (f(p) f(q)) = dX(p q) for all p q isin X

so that f X rarr Y is actually an isometry The moral is that certain properties one might wantcan be achieved by changing the morphisms you consider

Monomorphisms and epimorphisms The definitions of monomorphism and epimorphismcan be found in Section 12 of the book The standard examples are those in Set where themonomorphisms are the injective functions and the epimorphisms are the surjective functionsProofs of these facts are left to the homework as well as the interesting problem of determining theepimorphisms in Met or even MetC or Haus (the category of Hausdorff spaces where morphismsare arbitrary continuous maps) where the answer is that epimorphism means more than simplyldquocontinuous and surjectiverdquo (As opposed to all of Top where epimorphism does indeed meancontinuous and surjective)

So monomorphisms and epimorphisms should be treated as categorical analogues of ldquoinjectionsrdquoand ldquosurjectionsrdquo although this analogy only goes so far The thing to note is that we are thusable to give a complete definition of ldquoinjectiverdquo for ordinary functions which makes no referenceto elements and is stated only in terms of the relation between an injective function and otherfunctions certainly proving that ldquomonomorphismrdquo means ldquoinjectionrdquo in Set requires working withelements but stating the definition of an injection as a monomorphism can be done solely usingldquoarrowsrdquo

Products The definition of a product in a category shows up in Section 31 of the book inthe context of the more general notion known of a limit Wersquoll discuss this more general notioneventually but for now we will only focus on products of two objects at a time Here is what themore general notion boils down to in this case which wersquoll take as our definition

A product of two objects A and B in a category is an object P together with morphismspA P rarr A and pB P rarr B (which we call projections) which satisfy the followingproperty given any morphisms f C rarr A and g C rarr B into A and B from acommon object C there exists a unique morphism h C rarr P such that f = pA h andg = pB h We express this by saying that the following diagram commutes

C

A B

Pf g

pA pB

existh

which means that composing any two composable arrows in the diagram results in theother arrow with the same domain and codomain Concretely there are two ways in thisdiagram to get from C to A via f or via the composition pA h and the requirementis that both give the same morphisms (ie f = pA h) and similarly for the two waysto get from C to B The dashed arrow is the one which is required to exist (uniquelyas denoted by the exclamation mark ) in the definition of product

5

So the upshot is that for any C and any f and g in the diagram above there exists a uniqueh which makes the diagram commute The point is that a product gives a way to turn the twopieces of data f C rarr A and g C rarr B into the single piece of data h C rarr P in a uniqueway Being able to turn a possibly large amount of data into a single piece of data in this way isone of the benefits the ldquocategoricalrdquo perspective provides Products in categories are often denotedusing the usual A times B notation even though it is not necessarily true that a product is literallya set-theoretic Cartesian product although it is in the standard examples The map h C rarr Prequired in the definition is then often denoted by h = f times g

Examples In Set the product of two objects A and B always exists and is indeed the ordinaryCartesian product A times B But of course the definition of product also requires that data of theprojection morphisms which in this case are simply the usual ones

pA (a b) 983041rarr a and pB (a b) 983041rarr b

Given f C rarr A and g C rarr B the map h C rarr AtimesB is h(c) = (f(c) g(c)) and you can checkthat these constructions satisfy the properties required in the definition (Note that you would alsohave to argue that h defined in this way is the only map which can make the required diagramcommute)

Products in Grp are given by the usual direct product group structure GtimesH with the groupoperation defined as

(g1 h1) middot (g2 h2) = (g1g2 h1h2)

The projection morphisms are the ordinary ones you would expect as well But let us actuallynow verify that this group structure indeed is among other possible group structures on G times Hthe one required in the definition of a product Suppose f K rarr G and g K rarr H are grouphomomorphisms The key point is that the map h(k) = (f(k) g(k)) which should satisfy therequired product property should itself be a morphism in Grp meaning a group homomorphismand we claim that this is only true for the direct product group structure

Indeed we want it to be true that

h(k1k2) = (f(k1k2) g(k1k2)) equal h(k1)h(k2) = (f(k1) g(k1)) middot (f(k2)g(k2))

for all k1 k2 isin K and some to-be-determined group structure middot on G times H Since f and g arehomomorphisms this requirement becomes

(f(k1)f(k2) g(k1)g(k2)) = (f(k1) g(k1)) middot (f(k2)g(k2))

which says that middot should indeed be componentwise multiplication at least on the image of f andg But recall that the properties required of a product should hold for any K and any f K rarr Gand g K rarr H in particular then it should hold for instances when f and g are surjective whichshows that the componentwise-multiplication property above should in fact hold on all of G timesHso that middot is indeed the ordinary direct product structure

Uniqueness Notice that in the definition of a product in a category we used the word ldquoardquo asopposed to ldquotherdquo we spoke about a product of A and B instead of the product of A and B Sowe now ask are products unique The answer is ldquonordquo if we interpret ldquouniquerdquo in the strictestpossible sense but ldquoyesrdquo if we interpret it correctly For instance in the example of Grp takenow N to be any group which is isomorphic to the direct product G timesH say with isomorphismℓ N rarr G timesH We claim that we can also turn N into a categorical product of G of H as long

6

as we define appropriate ldquoprojectionsrdquo N rarr G and N rarr H take N rarr G to be the compositionpG ℓ and N rarr H to be pH ℓ Given f K rarr G and g K rarr H as in the definition of a productthe map hprime K rarr N defined by

hprime = ℓminus1 h

where h K rarr GtimesH is the usual h = f timesg will satisfy the requirement needed to able to concludethat N is also a product of G and H in Grp A similar trick works in any category so that anyobject isomorphic a product can also itself be turned into a product

The correct answer is that products are unique up to unique isomorphism which means thatgiven a product P of A and B any other product P prime of A and B will in fact be isomorphic to P and in a unique way in the sense that there can exist only one isomorphism P rarr P prime The phraseldquounique up to unique isomorphismrdquo is a common one wersquoll see again and again objects definedby some categorical property are almost never genuinely unique in a strict sense but they willessentially be as close to unique as possible

So suppose that P and P prime are both products of A and B with associated projection morphismspA pB for P and pprimeA p

primeB for P prime Consider the diagram

P

A B

P primepA pB

pprimeA pprimeB

existh

The morphism h P rarr P prime is the unique one guaranteed to exist by the fact that P prime is a productof A and B Commutativity of this diagram says that pA = pprimeA h and pB = pprimeB h A similardiagram with the roles of P and P prime reversed which uses the fact that P is a product results in aunique morphism hprime P prime rarr P satisfying pprimeA = pA hprime and pprimeB = pB hprime

Now consider the following diagram

P

A B

PpA pB

pA pB

hprime h

This commutes which we verify using the properties h and hprime satisfy as follows

pA (hprime h) = (pA hprime) h = pprimeA h = pA

and similar for the morphisms involving B This shows that hprimeh P rarr P satisfies the requirementin the definition of P being a product But of course idP P rarr P satisfies this requirement as wellso since the map satisfying this requirement is assumed to be unique we must have hprime h = idP Similar reasoning considering h hprime P prime rarr P prime shows that h hprime = idprimeP so that h and hprime are indeedinverses and hence that P and P prime are isomorphic The uniqueness of the isomorphism betweenthem comes from the uniqueness of h in the construction above

7

Abstract nonsense It is common to say in the setting above that the fact that products areunique up to unique isomorphism follows from ldquoabstract nonsenserdquo which is a succinct way ofsaying that it follows solely from the definition of a categorical product itself and not any deepermathematical content In particular many of the ldquouniquenessrdquo properties which ldquoproductsrdquo havein the settings we might care aboutmdashCartesian products in Set direct products in Grp producttopologies in Topmdashreally have nothing to do with how those specific products are defined but arejust consequences of the ldquoabstract nonsenserdquo of category theory

But of course we do not use the phrase ldquoabstract nonsenserdquo in a negative or dismissive waysince recognizing what types of properties are consequences of ldquoabstract nonsenserdquo goes a long waytowards understanding well the given problem at hand Soon you too will come to appreciate thisnotion of abstract nonsense

Coproducts Opposite Categories

Initial and terminal objects The notions of initial and terminal objects in a category aredefined in Section 16 of the book where you can also find standard examples Here I want topoint out that initial and terminal objects if they exist are unique up to unique isomorphism asanother example of arguing via ldquoabstract nonsenserdquo

Suppose I and I prime are both initial objects in some category By the fact that I is initial thereexists a unique morphism f I rarr I prime and by the fact that I prime is initial there exists a unique morphismg I prime rarr I But then the composition g f I rarr I must equal idI I rarr I since the latter is theonly morphism from I to itself because I is initial and similarly f g = idIprime Thus I and I prime areunique and there is a unique isomorphism between them

Coproducts The dual concept to that of a product is the notion of a coproduct (The termldquodualrdquo refers to the phenomena you get when you reverse all arrows in a given construction Forinstance the notion of a terminal object is dual to that of an initial object and the notion of anepimorphism is dual to that of a monomorphism) Here is the precise definition a coproduct of anobject A and an object B in a category is an object C together with morphisms iA A rarr C andiB B rarr C such that for any object D and morphisms f A rarr D and g B rarr D there exists aunique morphism h C rarr D which makes the following diagram commute

D

A B

Cf g

iA iB

existh

By abstract nonsense if a coproduct of A and B exists it is unique up to unique isomorphism

Examples In Set the coproduct of two sets A and B is the disjoint union A⊔B (If you havenrsquotseen the notion of a disjoint union before it is essentially the same as the union only that we donrsquotcare about whether A and B have any overlap if they do we treat the common elements as beingldquodifferentrdquo For instance Z ⊔ Z is not Z but consists of two copies of each integer one 0 comesfrom the first copy of Z and another comes from the second but these are different ldquozeroesrdquo)Given f A rarr D and g rarr D the unique map A ⊔B rarr D is defined by applying f to elements of

8

A and g to elements of B (Note that if we simply took the ordinary union AcupB as the candidatefor the coproduct when A and B were not disjoint the required map h A cup B rarr D would notexist since h would have to send any element x isin A cap B to both f(x) and g(x) in order to makethe required diagram commutative)

The coproduct of two objects X and Y in Top is again the disjoint union X ⊔ Y equippedwith the disjoint union topology an open subset of X ⊔ Y is defined to be the (disjoint) union ofan open subset of X with an open subset of Y The coproduct of two groups G and H in Grp isthe free product G lowastH which is further elaborated on in the homework

The coproduct of two vector spaces V and W in Vect is the direct sum V oplusW which is definedto be the set of formal sums

v + w where v isin V and w isin W

where addition and scalar multiplication are defined ldquocomponentwiserdquo

(v1 + w1) + (v2 + w2) = (v1 + v2) + (w1 + w2) and r(v + w) = rv + rw

This is isomorphic to the product V times W with operations also defined componentwise but it iscustomary to speak about the ldquodirect sumrdquo V oplusW when referring to the coproduct as opposed tothe product wersquoll say why in a bit The maps iV V rarr V oplusW and iW W rarr V oplusW required aspart of the data of a coproduct are

iV v 983041rarr v + 0W and iW w 983041rarr 0V + w

To verify that V oplusW is the correct coproduct take any linear transformations S V rarr U andT W rarr U where U is some other vector space We then need a (unique) linear transformationh V oplusW rarr U such that

S = h iV and T = h iW

Let us determine what this map must be thereby not only showing that it exists but also that itis unique Let v + w isin V oplusW which we can write as

v + w = (v + 0W ) + (0V + w)

Since h is required to be linear we have

h(v + w) = h([v + 0W ] + [0V + w]) = h(v + 0W ) + h(0V + w)

The first term at the end is (h iV )v which must thus be Sv and the second term is (h iW )wwhich must be Tw Thus h = S + T is the map

h = S + T v + w 983041rarr Sv + Tw

which you can verify is indeed linear Thus V oplusW does satisfy the property required of a coproductSo the product of two objects in Vect is the same as their coproduct only that we write the

latter using ldquosumrdquo notation Later we will discuss products and coproducts of an arbitrary numberof elements We will see these as special cases of the general notions of limits and colimits butthey are defined via similar requirements as products and coproducts of two objects at a timeIn the category of vector spaces arbitrary products are given by the usual product

983124α Vα with

ldquocomponentwiserdquo addition and scalar multiplication but it turns out that an arbitrary coproductis given by the direct sum

983119α Vα which is defined to be the subspace of

983124α Vα where only finitely

many components are nonzero (This is in a way analogous to how the product topology on an

9

arbitrary number of topological spaces is defined) Wersquoll go over this later but is the underlyingreason why we use oplus instead of times when discussing coproducts

Encoding data via productscoproducts We can nicely summarize the definition of productsand coproducts in the following way the product of A and B is the object with the property thatfor any object D we have

Mor(DA)timesMor(DB) = Mor(D product)

and the coproduct of A and B is the object such that for any object D we have

Mor(AD)timesMor(BD) = Mor(coproductx D)

The first reflects the fact that products give a unique way to turn two morphisms (f and g in thedefinition) mapping into A and B into a single morphism (h in the definition) and the second saysthat coproducts give a unique way to turn two morphisms mapping out of A and B into a singlemorphism

Soon enough we will discuss the idea of universality and the point is that products are ldquouni-versalrdquo for maps into A and B and coproducts are universal for maps from A and B

Opposite categories Finally we give a definition which might seem silly but is actually quiteuseful Given a category C the opposite category is the category Cop is the category with the sameobjects as C but where we defined Mor(AB) in Cop to be Mor(BA) in C To be more concrete weconsider each morphism f A rarr B in C to instead be a morphism fop B rarr A in Cop (Perhaps itwould be better to write this as A larr B fop) The point is that we donrsquot actually care about whatf actually is as say a functioncontinuous maphomomorphismwhatever-we only care about itsbehavior as an ldquoarrowrdquo The opposite category Cop only retains information about how arrowsrelate to one another nothing more

Since arrows are reversed products in C become coproducts in Cop and coproduts in C becomecoproducts in Cop Similarly initialterminal objects and monomorphismsepimorphisms in C be-come terminalinitial objects and epimorphismsmonomorphisms in Cop Again the notion of anopposite category might seem like a strange thing to look at but wersquoll see that it does have uses

Functors Fullness and Faithfulness

One more productcoproduct example Let X be a topological space and define its categoryof open sets Op(X) as follows The objects of Op(X) are simply the open subsets of X and themorphisms are given by inclusions so to be concrete

Mor(U V ) =

983083the unique inclusion U rarr V if U sube V

empty otherwise

The fact that U sube V and V sube W implies U sube W says that the composition of two morphismsin this category is still a morphism in this category The product of U and V in this categoryturns out to be their intersection (with ldquoprojection morphismsrdquo U cap V rarr U and U cap V rarr V beinginclusions) and the coproduct of U and V turns out to be their (ordinary) union Note that weneed X to be a topological space in order to guarantee that U cap V and U cup V are still objects inthis category More generally the product of finitely many objects in Op(X) exists and is theirintersection and the coproduct of an arbitrary number of objects exists and is their union (Wesay that Op(X) has finite products and arbitrary coproducts)

10

Given a set X and a collection T of subsets of X we can define a similar category with elementsof T as objects (I guess we should assume that empty X isin T) and inclusions as morphisms We thenhave that this category has finite products and arbitrary coproducts if and only if T is a topologyon X so that we can rephrase the definition of a topological space itself solely in categorical termsThis perspective on what a ldquotopologyrdquo is makes it simpler to state the definition of whatrsquos calleda sheaf on a topological space which is an important construction in various areas of analysis andgeometry We might discuss this a bit later on

Functors As with most things in mathematics we should care not only about categories asmathematical structures on their own but also about ldquomapsrdquo between these structures whichldquopreserverdquo said structure The notion of a (covariant) functor between categories provides suchldquomapsrdquo in the case at hand where a functor is a way of sending objects to objects and morphismto morphisms in a way which preserves compositions and identities The precise definition is inSection 13 of the book as is the definition of a contravariant functor between categories whichis a functor which reverses the direction of arrows as opposed to ldquocovariantrdquo ones which preservedirections (It is common to use the term ldquofunctorrdquo without qualification to mean one which iscovariant Note that any functor can be made covariant by working with the opposite category acontravariant functor C rarr D is the same as a covariant functor Cop rarr D)

All of the standard examples we looked at are in the book forgetful functors power set functorsthe dual space functor in Vect etc Of particular interest is Example 139 in the book whichshows that various types of group actions in different contexts-the usual notion of a group actionon a set the notion of a continuous action of a group on a topological space and the notion ofa linear action of a group on a vector space (also known as a representation of the group)-are allencoded by a functor with domain category BG

Adjoint functors Let F Top rarr Set be the forgetful functor and let D Set rarr Top be thefunctor which sends a set S to the space S equipped with the discrete topology and a functionS rarr Sprime between sets to the same function only now viewed as a continuous map S rarr Sprime betweendiscrete spaces These two functors satisfy the following property in relation to one another givinga morphism D(S) rarr Y in Top is the same as giving a morphism S rarr F (Y ) in Set or moresymbolically

MorTop(D(S) Y ) = MorSet(S F (Y ))

We say that D is left adjoint to F and that F is right adjoint to D The point is that adjointfunctors can be used to move data back and forth between two categories

Wersquoll talk more about adjoint functors later on but for now here is one more example Nowtake F Grp rarr Set to again be the forgetful functor and take G Grp rarr Set to be the freegroup functor which sends a set S to the free group generated by S G(S) has as elements ldquowordsrdquos1 sn made up of elements of S with the group operation being ldquoconcatenationrdquo A basic factis that giving a group homomorphism from G(S) to some other group H is the same as giving anordinary function from S to F (H) which is the required adjoint property

MorGrp(G(S) H) = MorSet(S F (H))

Faithful and full functors The definitions of what it means for a functor to be faithful and fullare in Section 15 of the book where faithful functors are ones which are injective on morphisms andfull functors are ones which are surjective on morphisms but where in both cases we refer only tomorphisms which occur between the objects F (A) and F (B) in the target category corresponding tofixed objects A and B in the source category (In general a functor could send two objects AB in

11

the source category to the same object F (A) = F (B) in the target category and thus there could betwo morphisms A rarr C and B rarr C which are sent to the same morphism F (A) = F (B) rarr F (C)Such a thing could still possibly be ldquofaithfulrdquo since the non-injectivity here is arising from differentobjects in the source)

Faithfullness and fullness will be nice properties going forward and here is one hint at how theymight be useful Say we have a functor F C rarr D The problem is to determine properties onF which will imply it send products in C to products in D or coproducts in C to coproducts inD So given the former product diagram below we want the second diagram to also be a productdiagram

C

A B

P Fminusminusminusrarr

F (C)

F (A) F (B)

F (P )f g

pA pB

existh

Ff Fg

F (pA) F (pB)

Fh

Of course in order F (P ) to be an actual product of F (A) and F (B) on the right a similarcommutative diagram property would have to hold for all objects in D in place of F (C) and allmorphisms D rarr F (A) and D rarr F (B) not just those of the form Ff and Fg respectively Oneway to guarantee this is to require that all objects in D are of the form F (C) and then for allmorphisms F (C) rarr F (A) and F (C) rarr F (B) to be of the form Ff Fg the former is true whenthe functor F is surjective on objects and the latter is true when F is full We are not saying thatthese are the only conditions under which F might preserve products but it definitely seems togive a sufficient condition at least

Almost there is one more thing to require the uniqueness of the morphism Fh F (C) rarr F (P )making the diagram on the right commute Suppose there were two such morphisms Fh and Fhprimewhere h and hprime are both morphisms C rarr P in C The commutativity of the diagram on the rightgives for instance

Ff = F (pA)Fh and Ff = F (pA)Fhprime

which the functorial properites of F turns into

Ff = F (pA h) and Ff = (pA hprime)

We want h and hprime to make the first diagram commute in order to be able to apply uniqueness ofthe morphism C rarr P But this requires know that f = pA h and f = pA hprime which would followfrom F being faithful In this case it follows that h and hprime both make the first diagram commute(after we go through a similar argument with g) so the fact that P is a product guarantees thath = hprime and hence that Fh = Fhprime so that F (P ) is an actual product on the right

So we conclude that a functor which is full faithful and surjective on objects will send productsto products (Actually we can relax this a bit by requiring not that every object in D be in theliteral image of F but only that it be isomorphic to something in the image A functor with thisproperty is said to be essentially surjective which is a concept will see in a related context soonenough) Again this is not an ldquoif and only ifrdquo statement since functors can preserve products (orcoproducts) without being full faithful and surjective on objects but it at least gives a sufficientcondition Later we will see other ways to guarantee that functors preserve products which arerelated to the existence of adjoints for that given functor

12

Coproduct Examples Concreteness

Products and coproducts for metric spaces The product of two objects in Met the categoryof metric spaces with morphisms given by metric maps is the Cartesian product X times Y equippedwith the box metric and the coproduct of X and Y does not exist if both X and Y are nonemptyThis was the topic of a student presentation and here are some proof sketches

Given metric spaces (X dX) and (Y dY ) the box metric d on X times Y is defined by

d((x1 y1) (x2 y2)) = maxdX(x1 x2) dY (y1 y2))

(The name ldquoboxrdquo metric comes from the fact that in RtimesR = R2 or R3 open balls indeed looks likeldquoboxesrdquo) The key requirement needed in the definition of a product is given morphisms S rarr Xand S rarr Y that the map

h S rarr X times Y defined by h(s) = (f(s) g(s))

actually be a morphism in Met This turns into the requirement that

d((f(s) g(s)) ((f(t) g(t))) = maxdX(f(s) f(t)) dY (g(s) g(t)) le dS(s t) for all s t isin S

given the fact that

dX(f(s) f(t)) le dS(s t) and dY (g(s) g(t)) le dS(s t) for all s t isin S

For the latter inequalities to imply the former d must (at least up to isometry) be given by thebox metric Note that the two other ldquostandardrdquo metrics one might put on a productmdashthe taxicab

dX + dY and Euclidean983156

d2X + d2Y metricsmdashdo not satisfy this required ldquometric maprdquo property

and so do not give the product in MetTo see that coproducts do not exist if XY are nonempty suppose instead (P dP ) was a co-

product for X and Y with inclusion morphisms iX X rarr P and iY Y rarr P For any n isin Ntake 0 n to be a metric space with the absolute value metric d and let f X rarr 0 n andg Y rarr 0 n be the constant function 0 and the constant function n respectively Since P is acoproduct there exists h P rarr 0 n such that

h iX = f and h iY = g

But in order for h to actually be a morphism in Met we must then have

d(h(iX(x) iY (y)) le dP (iX(x) iY (y)) for any x isin X y isin Y

which translates inton = |0minus n| le dP (iX(x) iY (y))

But this should then be true for any n isin N which would imply that the distance in P betweenan element coming from X and one coming from Y should be infinite which is nonsense (Notethat if you altered your definition of ldquometricrdquo to allow for infinite distance then a coproduct wouldexist the disjoint union X ⊔Y where you defined the distance between elements of X and Y to beinfinite)

Coproducts for groups The coproduct of two groups G and H in Grp is the free product GlowastHwhich is defined to be the set of all words consisting of elements of G andH in an alternating fashion

13

with concatenation as the group operation (When we have to elements of G or two elements of Hnext to each other with use their respective group operations to turn these into single elements ofG or H) This was also the topic of a student presentation and here is the basic idea Given grouphomomorphisms g G rarr N and k H rarr N that G lowastH is the coproduct comes down to showingthat the map

h G lowastH rarr N defined by h(g1h1 gnhn) = f(g1)k(h1) middot middot middot f(gn)k(hn)

and similarly on other possible words in GlowastH is a group homomorphism which follows essentiallyfrom the definition of the group operation on G lowastH indeed this group operation arises preciselyfrom this requirement

Now if we instead tried to use GtimesH with its standard inclusions G rarr GtimesH and H rarr GtimesHas the coproduct the issue is that the analogous map

h GtimesH rarr N defined by h((g h)) = f(g)k(h)

would NOT be a homomorphism because h((g1 h1)(g2 h2)) is not equal to h(g1 h1)h(g2 h2) bythe way in which the group operation is defined on the direct product G times H However if westrict our category to only consist of abelian groups so that the N rsquos we consider in the defini-tion of coproduct are also abelian then G times H does serve as a coproduct we can always rewritef(g1)k(g1) middot middot middot f(gn)k(gn) as f(g1) middot middot middot f(gn)k(h1) middot middot middot k(hn) in N Irsquoll leave it to you to flesh out allof these details

Concrete categories The categories Set Top Vect and Grp all have the property thattheir objects are simply sets possibly equipped with extra structure and that their morphisms areordinary set-theoretic functions possibly satisfying some additional requirement Such categoriesare nice since we can often use intuition about sets and functions in their study A concrete categoryis intuitively one where we can think of objects as being ldquosetsrdquo and morphisms as being ldquofunctionsrdquoonly that we have to interpret this in the right way

Here is the definition a concrete category is a category C equipped with a faithful functorF C rarr Set which is part of the data The idea is that such a functor gives us way to turnan object A isin Ob(C) into a set F (A) and a morphism f isin MorC(AB) into a function Ff isinMorSet(F (A) F (B)) where the ldquofaithfulrdquo requirement says that we do not lose information aboutf when doing so so that we can (essentially) learn everything we need to know about f by studyingFf instead In the four examples above the required faithful functor is simply the forgetful functorand indeed in general we think of F as being a ldquoforgetful functorrdquo for the concrete category (C F )

Examples Here are two more examples of concrete categories Let G be a group and considerthe category BG with a single object and elements of G morphisms As written this does notappear to be concrete since morphisms are not honest functions lowast rarr lowast but the point is that thiscategory can indeed by ldquoconcretizedrdquo by specifying an appropriate ldquoforgetful functorrdquo A functorF BG rarr Set is the data of a (left) action of G on a set S = F (lowast) and saying this functor isfaithful is what it means to say that this action is faithful different g h isin G act on S in differentways or equivalently that there exists s isin S such that gs ∕= hs Any group has at least onefaithful left actionmdashnamely the action on itself given by left multiplicationmdashto BG can always beequipped with a faithful functor to Set This is all just a way of phrasing the fact that any groupis isomorphic to a subgroup of a permutation group

The category Rel of relations can also be ldquoconcretizedrdquo even though as written morphismsA rarr B are not honest functions Take F Rel rarr Set to be the power set functor which sends an

14

object in Rel to its power set and a relation f A rarr B to the induced function Ff P (A) rarr P (B)defined by

(Ff)(S) = b isin B | there exists a isin S such that (a b) isin f

In other words if you think of the relation f sube A times B as a ldquopossibly not everywhere-definedpossibly multivalued functionrdquo (Ff)(S) is analogous to taking the image of S That F is faithfulwill be left to the homework

In fact most categories yoursquoll see can be made concrete by specifying an appropriate forgetfulfunctor but not all categories can be made concrete in this way Here is the standard example of anon-concrete category which we mention only to say there is such a thing but which we wonrsquot doanything more with since understanding it better requires knowledge of algebraic topology Thehomotopy category of topological spaces is the category hTop whose objects are topological spacesand whose morphisms are given by equivalence classes of homotopic maps

MorhTop(XY ) = homotopy classes of continuous maps X rarr Y

(Saying that two maps are homotopic means that you can ldquodeformrdquo one into the other but wersquollleave the precise definition to a course in algebraic topology) It turns out that no functor fromhTop to Set can be faithful so that hTop cannot be made concrete but this is actually a deepand difficult thing to show so we make no attempt to do so here

Natural Isomorphisms Representability

Natural transformations The definition of a natural transformation η F rArr G between two(both covariant or both contravariant) functors FG C rarr D is given in Section 14 of the bookThe key idea is that a natural transformation gives a way to turn data about F into data aboutG in a way which is compatible with all possible morphisms as expressed by the commutativity ofthe diagrams

F (A) F (B)

G(A) G(B)

Ff

ηA ηB

Gf

arising from morphisms f A rarr B (or f B rarr A in the contravariant case) in C In the case of anatural isomoprhism η F rArr G so that each ηA F (A) rarr G(A) is actually an isomorphism in Dthe point is not only that F and G produce isomorphic objects but they do so in a way which iscompatible with all possible category-theoretic data

The first standard example of a natural isomorphism is the one between the identity functor onthe category of finite-dimensional and the functor which sends a finite-dimensional vector space Vto V lowastlowast the dual of its dual induced by the isomorphism V rarr V lowastlowast defined by

v 983041rarr (the linear map on V lowast which sends f to f(v))

This was the topic of a student presentation and can be found in Section 14 of the book Theessential reason why this isomorphism is ldquonaturalrdquo is that V rarr V lowastlowast can be defined without referenceto a basis Contrast this with the the functor which sends V to its (single) dual V lowast it is still truethat finite-dimensional vector space is isomorphic to its dual but specifying such an isomorphism

15

requires additional data which prevents the identity functor from being ldquonaturally isomorphicrdquo tothe single dual functor

Some history At this point it makes sense to say a bit about the interesting origins of categorytheory in the 1950rsquos Eilenberg and Mac Lane where studying cohomology (or possible homologyone of those two) in algebraic topology which associates algebraic data (a group ring or similar)to a topological space in a way which encodes some of the topology They had two cohomologicalconstructions which ended up proving isomorphic results but they noticed that not only werethe resulting objects the same but the actual constructions themselves were in some sense theldquosamerdquo They thus then needed a way to formalize this notion and invented the concept of anatural transformation to do so where they interpreted two constructions as being the same asbeing compatible with all morphisms

But this then leads to the question what types of objects should a natural transformationoccur between Eilenberg and Mac Lane then had to invent the notion of a ldquofunctorrdquo to describewhat a natural transformation went from and what it went to In their motivating setting theyinvented the notion of a functor to describe more formally properties which their cohomologicalconstructions were supposed to satisfy But this then leads to the question what types of objectsshould a functor map between In other words what type of data should a functor (cohomologicalconstruction in their example) take as input and should it output They finally had to thus inventthe notion of a ldquocategoryrdquo to describe the source and target of a functor so that their cohomologicalconstructions could be interpreted as functors from a category of topological spaces to a categoryof algebraic objects

Thus historically natural transformations came first then functors and then categories Aswith most things in mathematics categories thus arose from attempts to encode what was beingseen in some concrete examples in a more general and formal setting

Representable functors Representable functors are defined in Section 21 of the book but wersquollgive the required definitions here in a hopefully simpler to digest manner The point is that anyobject A in a (locally small) category C gives two ldquonaturalrdquo functors C rarr Set one covariant andthe other contravariant which as wersquoll see pretty much encode all data about A itself We definehA C rarr Set to be the covariant functor defined by

hA(B) = Mor(AB) on objects and

hA(Bfminusrarr C) = the map Mor(AB)

hAfminusminusrarr Mor(AC) defined by g 983041rarr f g

We define hA C rarr Set to the contravariant functor defined by

hA(B) = Mor(BA) on objects and

hA(Bfminusrarr C) = the map Mor(CA)

hAfminusminusrarr Mor(BA) defined by g 983041rarr g f

(The functor hA is called the functor of points of A and wersquoll usually think of it as being a covariantfunctor hA Cop rarr Set The functor hA does not have a standard name) So hA is defined bymorphisms from A and hA is defined by morphisms into A Wersquoll see later the sense in which suchfunctors encode all information about the object A

A functor F C rarr Set is representable if is naturally isomorphic to such a functor so F sim= hA

for some A in the covariant case and F sim= hA for some A in the contravariant case The idea wersquollbuild towards is that we can essentially treat such a functor as being A itself so that representable

16

functors can be thought as being actual objects in C For now here is a key example wersquove seenbefore which we can now formulate using the language of representability

Given AB isin Ob(C) let F C rarr Set be the contravariant functor defined by

F (C) = Mor(CA)timesMor(CB) and F (Cfminusrarr D) = [(g k) 983041rarr (g f k f)]

where (g k) 983041rarr (g f k f) gives a map Mor(DA) timesMor(DB) rarr Mor(CA) timesMor(CB) ieF (D) rarr F (C) In order for this functor to be representable requires the existence of an object inC satisfying

Mor(CA)timesMor(CB) sim= Mor(C representing object)

in a way which is compatible with all morphisms but we have actually already seen this notionelsewhere it is essentially the definition of a product AtimesB for A and B Indeed we claim that Fis naturally isomorphic to hAtimesB as we now show

First to define a natural isomorphism hAtimesBsim= F requires for each C isin Ob(C) a choice of a

bijection (ie isomorphism in Set)

ηC hAtimesB(C)sim=minusrarr F (C)

which in our case looks like

ηC Mor(CAtimesB))sim=minusrarr Mor(CA)timesMor(CB)

Take this to be the map

(Cfminusrarr AtimesB) 983041rarr (pA f pB f)

where pA A times B rarr A and pB A times B rarr B are the two projection morphisms in the definitionof a product The definition of product implies that this map is invertible since given k C rarr Aand ℓ C rarr B there exists a unique h C rarr A times B such that k = pA h and ℓ = pB h Forthese bijections to define a natural isomorphism hAtimesB

sim= F requires that they be compatible withall morphisms in C which means that given any g C rarr D in C the following diagram shouldcommute

hAtimesB(C) hAtimesB(D)

F (C) F (D)

hAtimesB(g)

ηC ηD

Fg

which in the case at hand looks like

Mor(CAtimesB) Mor(DAtimesB)

Mor(CA)timesMor(CB) Mor(DA)timesMor(DB)

minus g

ηC ηD

(minus gminus g)

Take an element k D rarr AtimesB in the upper right Applying the map on top gives the elementk g in the upper left and then applying ηC on the left side gives

(pA (k g) pB (k g))

17

in the lower left Instead starting with k D rarr AtimesB in the upper right applying ηD on the rightgives (pA k pB k) and applying the map on the bottom gives

((pA k) g (pB k) g)

in the lower left which is the same as the previous element in the lower left by the associativity ofcomposition Hence the isomorphisms ηC do define a natural isomorphism between hAtimesB and F Again the point is that you can then interpret the functor F as being the ldquosamerdquo as the productAtimesB since both encode the same data

More Representable Examples

Coproducts As in the case of products the definition of coproducts can also be phrased in termsof a representable functor Given objects A and B in C let F C rarr Set be the covariant functordefined by

F (C) = Mor(AC)timesMor(BC) and F (Cfminusrarr D) = [(g k) 983041rarr (f g f k)]

where the latter is a function Mor(AC)timesMor(BC) rarr Mor(AD)timesMor(BD) This functor isrepresentable if and only if a coproduct A ⊔ B for A and B exists and the representing object isthat coproduct the key is the requirement that we have bijections

Mor(AC)timesMor(BC) sim= Mor(A ⊔BC)

for all C which is essentially the defining property a coproduct for A and B should have

Power set functor contravariant version Let P Set rarr Set be the contravariant powerset functor defined on objects by sending a set to its power set and on morphisms by sendingf A rarr B to the preimage function Pf P (B) rarr P (A) given by S 983041rarr fminus1(S) In order for this tobe representable requires the existence of a set X with natural bijections

P (A) sim= Mor(AX)

for all sets A that is a set X so that mapping into X is the same data as specifying a subset ofthe domain We take X = 0 1 to be a two-element set and the required bijections

ηA P (A) rarr Mor(A 0 1) to be given by S 983041rarr χS

where XS A rarr 0 1 is the indicator or characteristic function S defined by sending elementsof S to 1 and elements of A minus S to 0 The inverse of ηA sends a function g A rarr 0 1 to thepreimage gminus1(1) sube A

To verify that the bijections ηA define the data of a natural isomorphism η P rArr h01 wecheck the commutativity of some diagrams Given a function f A rarr B consider

P (A) P (B)

h01(A) = Mor(A 0 1) h01(B) = Mor(B 0 1)

Pf = fminus1

ηA ηB

minus f

18

Take S isin P (B) in the upper right Applying the map on top gives fminus1(S) isin P (A) and thenapplying the map on the left gives χfminus1(S) isin Mor(A 0 1) On the other hand applying the mapon the right to S isin P (B) gives χS isin Mor(B 0 1) and then applying the map on the bottomgives χS f isin Mor(A 0 1) The two resulting functions χfminus1(S)χS f in the lower left are

χfminus1(S)(x) =

9830831 x isin fminus1(S)

0 otherwiseand χS(f(x)) =

9830831 f(x) isin S

0 otherwise

so these are the same since x isin fminus1(S) if and only if f(x) isin S Thus the diagram above commutesso P is naturally isomorphic to h01 and hence P is representable

Power set functor covariant version Now consider the covariant power set functor stilldenoted by P Set rarr Set which again sends a set to its power set but now sends a functionf A rarr B to the image function Pf P (A) rarr P (B) defined by S 983041rarr f(S) In order for this tobe representable there would have to exist a set X such that P sim= hX which translates into havingnatural bijections

P (S) sim= Mor(XS)

for all sets S However in this case there is no natural choice for X as there is no natural way todetect subsets of S by mapping into S as opposed to the case of P (S) sim= Mor(S 0 1)

That no such X exists can be made precise using a simple argument when S = empty P (S) hascardinality 1 soX would have to be empty as well since otherwise Mor(XS) would have cardinality0 but if X is empty then Mor(empty S) always cardinality 1 regardless of S which is clearly not trueof P (S) Thus the covariant power set functor is not representable

Forgetful on Top The forgetful functor F Top rarr Set is representable Indeed a representingobject would have to satisfy

X sim= Map(objectX)

as sets for any topological space X (where Map denotes the set of continuous maps) and it iseasy to see that a single point satisfies this property a continuous map f pt rarr X is completelydetermined by the element f(pt) of X and any such element gives a continuous map in this wayOf course checking that F sim= hpt requires checking that various diagrams commute but this issimple to work out in this case so we omit it here

Extracting a topology With an eye towards the idea that the functors hA hA should captureall information about an object A in a category note that the result above shows that in TophX first of all captures all information about X as a set since the set X can be extracted fromhX(pt) = Map(ptX)

We claim that hX actually captures all information about X as a topological space since thetopology on X can be extracted from hX in the following way As we saw in the power set examplewe have a bijection

P (X) sim= Mor(X 0 1)

Now equip 0 1 with the topology in which 1 is open but 0 is not Then a continuous mapf X rarr 0 1 is fully determined by the preimage fminus1(1) since to be continuous only requiresthat this single preimage be open in X Moreoever any open subset of X arises in this way bytaking f to be the indicator function of that open set Thus we get a bijection

Op(X) sim= Map(X 0 1)

19

where Op(X) denotes the set of open subsets of X ie the topology on X Hence the topology onX can be extracted from hX as claimed so hX and hX do encode all information about X

This fact can also be interpreted as saying that the contravariant functor which sends a topo-logical space X to its collection of open subsets (and which sends a continuous map to the mapinduced by taking preimages) is representable with representing object 0 1 equipped with thetopology given above This will be worked out in a homework problem

Forgetful on Grp The forgetful functor Grp rarr Set is also representable In this case arepresenting object should be a group giving set-theoretic bijections

G sim= Hom(objectG)

for all groups G where Hom denotes the set of group homomorphisms The group Z works

G sim= Hom(Z G)

since a homomorphism φ Z rarr G is completely determined by φ(1) since Z is generated by 1 andthere are no restrictions on what φ(1) isin G can actually be The inverse of the desired bijection isthus φ 983041rarr φ(1) and it is simple to check that morphisms behave appropriately so that the forgetfulfunctor in this setting is indeed isomoprhic to hZ

Forgetful on Ring The forgetful functor Ring rarr Set is also representable where by Ring wemean the category of rings with identity elements and where morphisms (ie ring homomorphisms)are required to preserve identities Here we need an object satisfying

R sim= Hom(object R)

for all R isin Ob(Ring) and we can see that Z[x] the ring of polynomials in the variable x withcoefficients in Z works Indeed given a ring homomorphism φ Z[x] rarr R the fact that φ isrequired to preserve identities fully determines φ(n) for any n isin Z (since φ(n) = nφ(1)) and hencesince φ preserves multiplication its behavior is fully determined by φ(x) which can be any elementof R The inverse of the bijection we want is thus φ 983041rarr φ(x) and it will follow that this forgetfulfunctor is isomorphic to hZ[x]

Equivalences between Categories

Definitions iso and equiv Now that we have a notion of a ldquomorphismrdquo between categorieswe can talk about the sense in which two categories are the ldquosamerdquo The first possible definitionis the ldquoobviousrdquo one you might expect given the definition of ldquoisomorphismrdquo wersquove seen in everyother context until now we say that C is isomorphic to D if there exist functors F C rarr D andG D rarr C such that G F = idC and F G = idD where idC and idD are identity functors

However this definition turns out to be too restrictive Indeed the requirement that the givencompositions equal identity functors says that given an object A in C the object G(F (A)) beliterally the same as A itself and similarly for objects in D But we know that two objects in acategory can be thought of as being the ldquosamerdquo without being literally the same as long as theyare isomorphic For instance productscoproducts are not unique in the literal sense but only inthe sense that they are isomorphic in a unique way To take another example the definition ofa functor being representable does not say that F (B) literally be the same as hB(A) only thatthey be isomorphic for sure the power set P (S) of a set S is not literally the same as the set offunctions S rarr 0 1mdashit is only the ldquosamerdquo in the sense that they encode the same data

20

Thus it makes sense to relax the requirement that A andG(F (A)) in the definition of isomorphiccategories be equal to the requirement that they be isomorphic This leads to the following betterdefinition of two categories being the ldquosamerdquo C is equivalent to D if there exist functors F C rarr D

and G D rarr C such that G F sim= idC and F G sim= idD So we require only that for instance thefunctor G F be naturally isomorphic to the identity functor on C which says that for any objectA G(F (A)) be isomorphic to A in a natural way Wersquoll see that this truly is the better notionof what it means for two categories to be the ldquosamerdquo and it reflects the idea that all categoricalproperties of one category can be transported to the other

Alternate characterization of equivalences A functor F C rarr D implementing an equiv-alence between categories is called unsurprisingly and equivalence and the ldquoinverserdquo functorG D rarr C is still referred to as being its inverse even though it is not technically an ldquoinverserdquoin the sense that compositions give literal identities But as in the case of Set where ldquoinvertiblerdquocan be characterized without reference to an inverse as ldquoinjectiverdquo and ldquosurjectiverdquo so too canequivalences between categories be characterized without explicit reference to an inverse

A functor F C rarr D is an equivalence if and only it is full faithful and essentiallysurjective

We have seen the notions of full and faithful before and essentially surjective means that any objectin D is isomorphic to something in the image of F for all D isin Ob(D) there exits C isin Ob(C) suchthat F (C) sim= D Justifying this result was the topic of a student presentation and can be foundin the book in Section 15 (Technically to construct an ldquoinverserdquo of F requires that we work withsmall categories but whatever) The bulk of the real work goes into showing that the ldquoinverserdquofunctor constructed is actually a functor which comes down to some abstract nonsense argumentvia commutative diagrams

Equivalences preserve everything We previously saw the notions of full faithful and es-sentially surjective back when thinking about what properties a functor could have which wouldguarantee it sent products to products so the result we proved back then was that equivalencespreserve products Similarly equivalences preserve coproducts and the proof is basically the sameafter reversing arrows A clean way of deriving this is as follows if F C rarr D is an equivalencethen so is F op Cop rarr Dop so since F op preserves products F preserves coproducts becausecoproducts in a category become products in the opposite category

In general an equivalence between categories will preserve whatever categorical notions youwant monomorphisms epimorphisms initial objects terminal objects etc so that equivalentcategories really do have essentially the ldquosamerdquo properties

Equivalence example Here is a first example (also found in the book) of equivalent categorieswhich are not isomorphic Let FVectR be the category whose objects are finite-dimensional realvector spaces and whose morphisms are linear transformations Let MatR be the category ofmatrices defined as follows

bull objects in MatR are nonnegative integers

bull a morphism n rarr m is an mtimes n matrix with real entries

bull composition in MatR is given by matrix multiplication given B n rarr m and A m rarr kmdashsoB is an mtimes n matrix and A is a k timesm matrixmdashthe composition A B n rarr k is the k times nmatrix AB

bull identity morphisms in MatR are given by identity matrices

21

We claim that FVectR and MatR are equivalent First note that they certainly cannot be isomo-prhic isomorphisms in MatR exist only between two objects which are literally the same (ie anmtimesn matrix can be invertible only when m = n) but isomorphisms in FVectn can exist betweenobjects which are not literally the same So there are in a sense not enough objects in MatR toallow it to be isomorphic to FVectR

To construct an equivalence choose one and for all an isomorphism TV V rarr RdimV forevery object in FVectR or equivalently choose once and for all a basis for every V DefineF FVectR rarr MatR by

V 983041rarr dimV on objects and

(T V rarr W ) 983041rarr (the standard matrix of SW T Sminus1V RdimV rarr RdimW ) on morphisms

Equivalently F sends T to the matrix of T relative to the chosen bases for V and W It isstraightforward to check that this is a functor The inverse functor G MatR rarr VectR is givenby

n 983041rarr Rn on objects and

(mtimes n matrix A) 983041rarr (the linear transformation A Rn rarr Rm induced by A) on morphisms

One can check that G is a functor and that F G and G F are naturally isomorphic to identitiesor instead one can argue that F alone is full faithful and essentially surjective The point is thatG(F (V )) gives back RdimV so not literally V itself but only something isomorphic to V which iswhy this gives an equivalence of categories and not an isomorphism

Yoneda Lemma Functors as Objects

Compact Hausdorff spaces and Clowast-algebras Last time we gave an example of equivalentcategories which were not isomorphic but in many ways that example is unsatisfactory when doinglinear algebra people for sure often work with matrix representations of linear transformations asthe example from last would suggest but no one in their right mind literally thinks about suchthings in terms of the equivalence FVectR rarr MatR itself In other words this equivalence isessentially ldquocooked uprdquo for the sake of giving an example but it does not really give any insightinto the actual mathematics of linear algebra

So with that in mind we will describe another equivalence which actually does have some realmeaning in the sense of producing new ways of thinking about some mathematics Given a compactHausdorff space X we let C(X) denote the space of continuous complex-valued functions on X

C(X) = f X rarr C | f is continuous

We can equip the set C(X) with various additional structures First it is a complex vector spaceunder ordinary addition and scalar multiplication of functions Second we can multiply two func-tions together

(fg)(p) = f(p)g(p)

which turns C(X) into whatrsquos called an algebra over C (We wonrsquot give the definition of anldquoalgebrardquo but the point is that this multiplication is in some sense ldquocompatiblerdquo with the vectorspace structure) Next any element f of C(X) has a conjugate element f obtained by takingcomplex conjugates of the values of f

f(p) = f(p)

22

And finally we can equip C(X) with a norm via

983042f983042 = suppisinX

|f(p)|

where |middot| denotes complex absolute value (This supremum exists sinceX is compact by the ExtremeValue Theorem) All of these structures (vector space algebra multiplication conjugation norm)turn C(X) into whatrsquos called a Clowast-algebra We wonrsquot define this precisely but the definition encodesvarious ways in which these structures are compatible Actually C(X) is a unital commutative Clowast-algebra where ldquounitalrdquo means that the multiplication has an identity element (namely the constantfunction 1) and commutative means that fg = gf There is a natural notion of homomorphismbetween Clowast-algebras which are maps between them which behave nicely with all the structuresinvolved In particular given a continuous map f X rarr Y between compact Hausdorff spaces weget a Clowast-algebra homomorphism flowast C(Y ) rarr C(X) given by pre-composition with f flowast sendsg isin C(Y ) to g f isin C(X)

We thus have a contravariant functor CHaus rarr ucClowast-Alg from the category of compactHausdorff spaces and the category of unital commutative Clowast-algebras It turns out that anyunital commutative Clowast-algebra arises in this way in the sense that given any unital commutativeClowast-algebra B there exists a compact Hausdorff space X such that C(X) sim= B as Clowast-algebrasand that morphisms between these algebra arises from continuous maps in the manner describeabove This says that the functor above is essentially surjective full and it turns out that it isalso faithful so that it is an equivalence Thus all information about compact Hausdorff spaces iscompletely encoded in information about unital commtuative Clowast-algebras and indeed one couldlearn everything there is to know about a compact Hausdorff space X from its Clowast-algebra C(X)of functions The fact that the categories CHaus and ucClowast-Alg are equivalent is the startingpoint in the subject known as noncommutative geometry which wersquoll say something about a bitlater on If we drop the requirement that our algebras be unital it turns out that the categoryof commutative Clowast-algebras in general is equivalent to the category of locally compact Hausdorffspaces via essentially the same functor where we take as morphisms in the category of locallycompact Hausdorff spaces to be continuous functions which ldquovanish at infinityrdquo

We are only introducing Clowast-algebras to give an example of an interesting and actually usefulequivalence between categories but just for the fun of it wersquoll note the following Examples ofnoncommutative Clowast-algebras come from what are known as ldquoalgebras of bounded operators onHilbert spacesrdquo which show up in quantum mechanics as the things which describe physicallyobservable quantities like position momentum and energy (In fact all Clowast-algebras arise fromHilbert spaces in this way) Thus Clowast-algebras show up in mathematical formulations of quantummechanics and the notion of noncommutative geometry mentioned above is at the core of someattempts to understand the elusive concept known as ldquoquantum geometryrdquo

Yoneda Lemma The Yoneda Lemma states that given a (covariant) functor F C rarr Set whereC is locally small for any A isin Ob(C) there exists a bijection

F (A) sim= Nat(hA F )

where the right side denotes the set of natural transformations from hA to F A similar result holdsin the contravariant case where we get bijections F (A) sim= Nat(hA F ) (There is also a sense inwhich these bijections are themselves natural but wersquoll ignore this bit) The proof of the YonedaLemma was the topic of a student presentation and can be found in Section 22 of the book

Here we will get a bit philosophical and talk about what the point of the Yoneda Lemmaactually is and later the sense in which it says that we can view arbitrary functors as ldquonon-existent

23

objectsrdquo in a category First we mention what we will call the ldquofunctor of pointsrdquo approach tomathematics a mathematical object can be fully characterized by all morphisms into it That isknowledge of Mor(SA) for all S is equivalent to knowledge of A itself Indeed this is an idea wementioned previously in relation to the functors hAmdashwhich we previously said is referred to as beingthe functor of points of Amdashand now we seek to push this idea further In some examples literalpoints of A can be extracted from hA as in the case of the bijection X sim= hX(pt) = Map(ptX) fora topological space X or G sim= hG(Z) = Mor(Z G) for a group G Moreover in these examples therest of the (topological) structure onX or (group) structure on G can also be extracted from hX andhG respectively In general we can view Mor(SA) as in a sense being ldquopoints of A parameterizedby Srdquo which comes from the idea that in say Set Mor(SA) = AS is ldquothe product of S-manycopies of Ardquo so literally points of A ldquoparameterizedrdquo by S In various areas of mathematics anelement of Mor(SA) is thought of as being a ldquogeneralizedrdquo point of A so that the functor of pointshA detects generalized points of A

So an object A is fully encoded by the functor hA Now we claim that morphisms betweenobjects are also encoded by such functors Indeed in the case where F = hB is itself a functor ofpoints the Yoneda result Nat(hA F ) sim= F (A) becomes

Nat(hA hB) sim= hB(A) = Mor(AB)

which says precisely that morphisms between the objects A and B can be obtained solely fromthe functors hA and hB (This fact is one reason why one should prefer the contravariant functorhA to the covariant functor hA in the covariant case Nat(hA F ) sim= F (A) for F = hB becomesNat(hA hB) sim= hB(A) = Mor(BA) so that morphisms between two objects get reversed whencharacterizing them in terms of representable functors)

Pushing our philosophy further we now know that not only objects but also morphisms betweenobjects are fully encoded by the functors hA The final step is to note that we should not only careabout objects and morphisms between them but that an important part of the data of an objectA is also how arbitrary contravariant functors F C rarr Set behave on A and that this too can beobtained from hA precisely via the Yoneda result that F (A) sim= Nat(hA F ) So objects morphismsbetween objects and how arbitrary functors behave on objects are all encoded in knowledge ofthe functors hA so that these functors truly do capture in the most general sense possible allinformation about A

Functors as nonexistent objects The discussion above is all well and good but it still doesnrsquotquite say why we should care about the fact that we can study on object A by studying hA insteadin that we still havenrsquot spoken about what type of insight about A can be gleamed from hA thatwasnrsquot readily apparent in A itself already Now we argue that the true point of the YonedaLemma is that it gives a way to think about arbitrary functors F Cop rarr Set as ldquoobjectsrdquo in C

itself or really as objects in an ldquoenlargementrdquo of C Certainly for those functors F sim= hA which arerepresentable we can indeed think about F as simply being the object A but now we claim thateven non-representable F can be thought of as being ldquononexistentrdquo objects in C The upshot is thatoften in mathematics we come across things which do not actually exist but the Yoneda Lemmagives a way to (indirectly) study such things anyway by interpreting them as functors instead

The starting point of course is that an actual object A is the same as hA as we have explainedabove Thus by analogy we can interpret a non-representable functor F as being an ldquoobjectrdquo in C

as long as we know what it means to ldquomaprdquo into F from other things which are honestly objects in CThat is if F were indeed representable then F (A) sim= Nat(hA F ) sim= Mor(AF ) would characterizethe functor of points of F (here we abuse notation by using the same letter F to denote both arepresentable functor and the object which represents it) so now we take Mor(AF ) = F (A) as

24

the definition of the ldquofunctor of pointsrdquo of the non-existent object F We are thus taking elementsof F (A) as being what it means to ldquomaprdquo from A into F even when F is not representable Thefact that the Yoneda Lemma gives a way to define maps from A into F thus says that F itself isequipped with the same data that a literal object of A would be equipped with so why shouldnrsquotwe interpret F as a being a type of ldquoobjectrdquo in C as well

To phrase this another way note that the Yoneda Lemma gives rise to a functor

C rarr Fun(CopSet)

where Fun(CopSet) denotes the category whose objects are functors from Cop to Set and morphismsare natural transformation defined on objects by A 983041rarr hA and on morphisms by the bijectionMor(AB) sim= Nat(hA hB) arising from the Yoneda Lemma That this is actually bijection saysthat this functor is full and faithful so that we can view C as being a subcategory of the functorcategory Fun(CopSet) (We cay say that C is isomorphic to the subcategory of Fun(CopSet)consisting of the functors hA or that it is equivalent to the subcategory of Fun(CopSet) consistingof representable functors This functor C rarr Fun(CopSet) is usually called the Yoneda embeddingof C into Fun(CopSet)) Thus we can view Fun(CopSet) as an ldquoenlargmentrdquo of C and the ideawe are trying to make sense of is that other objects in this ldquolargerrdquo category should be interpretedas ldquonon-existent objectsrdquo of C

Just how far can we actually push this idea that arbitrary functors Cop rarr Set should be thoughtof as being ldquoobjectsrdquo of C Pretty far in fact For instance often times in geometry one comesacross a type of ldquosmoothrdquo space on which some group acts and one is interested in doing ldquocalculusrdquoon the resulting quotient space which is the space of orbits under the given group action Theproblem is that such quotient spaces are very often (usually in fact) not ldquosmoothrdquo themselves sothat the tools from calculus are not readily available The solution is to nonetheless treat thisquotient as if it were ldquosmoothrdquo and attempt to do calculus anyway To make this precise thisldquonon-smoothrdquo space actually turns out to have a natural interpretation as a ldquosmooth functorrdquo andit is through this functor which we can view this non-smooth space as actually being smooth afterall so we are forced to expand our notion of what we mean by ldquosmooth spacerdquo by allowing thingswhich are not spaces in the usual sense which certainly leads to more abstraction but the benefitis that it actually makes sense to do calculus on such ldquospacesrdquo interpreted as functors

The discussion above is pretty vague so here is one final more concrete example Recall thatequivalence

CHausop rarr ucClowast-Alg

between the category of compact Hausdorff spaces and the category of unital commutative Clowast-algebras mentioned at the start We now claim that a non-commutative (unital) Clowast-algebra shouldbe interpreted as the algebra of functions on some non-existent ldquononcommutative compact Haus-dorff spacerdquo and it is this idea which the subject of noncommutative geometry makes precise Solet B be a non-commutative Clowast-algebra Then the functor of points hB gives a functor

ucClowast-Alg rarr Set

which sends C(X) (where X is some compact Hausdorff space) to Hom(C(X) B) where Hom de-notes homomorphisms of Clowast-algebras whatever that means Composing with the previous functorwe get a functor wersquoll denote by B itself

B CHausop rarr Set

which on objects at least is given by X 983041rarr Hom(C(X) B) This functor is not representable sinceB is not commutative but the philosophy espoused above suggests we should think of it as giving a

25

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

for all p q isin X and s t isin Y But writing s and t as s = f(p) and t = f(q) the second conditionbecomes dX(p q) le dY (f(s) f(t)) and the two inequalities together thus imply

dY (f(p) f(q)) = dX(p q) for all p q isin X

so that f X rarr Y is actually an isometry The moral is that certain properties one might wantcan be achieved by changing the morphisms you consider

Monomorphisms and epimorphisms The definitions of monomorphism and epimorphismcan be found in Section 12 of the book The standard examples are those in Set where themonomorphisms are the injective functions and the epimorphisms are the surjective functionsProofs of these facts are left to the homework as well as the interesting problem of determining theepimorphisms in Met or even MetC or Haus (the category of Hausdorff spaces where morphismsare arbitrary continuous maps) where the answer is that epimorphism means more than simplyldquocontinuous and surjectiverdquo (As opposed to all of Top where epimorphism does indeed meancontinuous and surjective)

So monomorphisms and epimorphisms should be treated as categorical analogues of ldquoinjectionsrdquoand ldquosurjectionsrdquo although this analogy only goes so far The thing to note is that we are thusable to give a complete definition of ldquoinjectiverdquo for ordinary functions which makes no referenceto elements and is stated only in terms of the relation between an injective function and otherfunctions certainly proving that ldquomonomorphismrdquo means ldquoinjectionrdquo in Set requires working withelements but stating the definition of an injection as a monomorphism can be done solely usingldquoarrowsrdquo

Products The definition of a product in a category shows up in Section 31 of the book inthe context of the more general notion known of a limit Wersquoll discuss this more general notioneventually but for now we will only focus on products of two objects at a time Here is what themore general notion boils down to in this case which wersquoll take as our definition

A product of two objects A and B in a category is an object P together with morphismspA P rarr A and pB P rarr B (which we call projections) which satisfy the followingproperty given any morphisms f C rarr A and g C rarr B into A and B from acommon object C there exists a unique morphism h C rarr P such that f = pA h andg = pB h We express this by saying that the following diagram commutes

C

A B

Pf g

pA pB

existh

which means that composing any two composable arrows in the diagram results in theother arrow with the same domain and codomain Concretely there are two ways in thisdiagram to get from C to A via f or via the composition pA h and the requirementis that both give the same morphisms (ie f = pA h) and similarly for the two waysto get from C to B The dashed arrow is the one which is required to exist (uniquelyas denoted by the exclamation mark ) in the definition of product

5

So the upshot is that for any C and any f and g in the diagram above there exists a uniqueh which makes the diagram commute The point is that a product gives a way to turn the twopieces of data f C rarr A and g C rarr B into the single piece of data h C rarr P in a uniqueway Being able to turn a possibly large amount of data into a single piece of data in this way isone of the benefits the ldquocategoricalrdquo perspective provides Products in categories are often denotedusing the usual A times B notation even though it is not necessarily true that a product is literallya set-theoretic Cartesian product although it is in the standard examples The map h C rarr Prequired in the definition is then often denoted by h = f times g

Examples In Set the product of two objects A and B always exists and is indeed the ordinaryCartesian product A times B But of course the definition of product also requires that data of theprojection morphisms which in this case are simply the usual ones

pA (a b) 983041rarr a and pB (a b) 983041rarr b

Given f C rarr A and g C rarr B the map h C rarr AtimesB is h(c) = (f(c) g(c)) and you can checkthat these constructions satisfy the properties required in the definition (Note that you would alsohave to argue that h defined in this way is the only map which can make the required diagramcommute)

Products in Grp are given by the usual direct product group structure GtimesH with the groupoperation defined as

(g1 h1) middot (g2 h2) = (g1g2 h1h2)

The projection morphisms are the ordinary ones you would expect as well But let us actuallynow verify that this group structure indeed is among other possible group structures on G times Hthe one required in the definition of a product Suppose f K rarr G and g K rarr H are grouphomomorphisms The key point is that the map h(k) = (f(k) g(k)) which should satisfy therequired product property should itself be a morphism in Grp meaning a group homomorphismand we claim that this is only true for the direct product group structure

Indeed we want it to be true that

h(k1k2) = (f(k1k2) g(k1k2)) equal h(k1)h(k2) = (f(k1) g(k1)) middot (f(k2)g(k2))

for all k1 k2 isin K and some to-be-determined group structure middot on G times H Since f and g arehomomorphisms this requirement becomes

(f(k1)f(k2) g(k1)g(k2)) = (f(k1) g(k1)) middot (f(k2)g(k2))

which says that middot should indeed be componentwise multiplication at least on the image of f andg But recall that the properties required of a product should hold for any K and any f K rarr Gand g K rarr H in particular then it should hold for instances when f and g are surjective whichshows that the componentwise-multiplication property above should in fact hold on all of G timesHso that middot is indeed the ordinary direct product structure

Uniqueness Notice that in the definition of a product in a category we used the word ldquoardquo asopposed to ldquotherdquo we spoke about a product of A and B instead of the product of A and B Sowe now ask are products unique The answer is ldquonordquo if we interpret ldquouniquerdquo in the strictestpossible sense but ldquoyesrdquo if we interpret it correctly For instance in the example of Grp takenow N to be any group which is isomorphic to the direct product G timesH say with isomorphismℓ N rarr G timesH We claim that we can also turn N into a categorical product of G of H as long

6

as we define appropriate ldquoprojectionsrdquo N rarr G and N rarr H take N rarr G to be the compositionpG ℓ and N rarr H to be pH ℓ Given f K rarr G and g K rarr H as in the definition of a productthe map hprime K rarr N defined by

hprime = ℓminus1 h

where h K rarr GtimesH is the usual h = f timesg will satisfy the requirement needed to able to concludethat N is also a product of G and H in Grp A similar trick works in any category so that anyobject isomorphic a product can also itself be turned into a product

The correct answer is that products are unique up to unique isomorphism which means thatgiven a product P of A and B any other product P prime of A and B will in fact be isomorphic to P and in a unique way in the sense that there can exist only one isomorphism P rarr P prime The phraseldquounique up to unique isomorphismrdquo is a common one wersquoll see again and again objects definedby some categorical property are almost never genuinely unique in a strict sense but they willessentially be as close to unique as possible

So suppose that P and P prime are both products of A and B with associated projection morphismspA pB for P and pprimeA p

primeB for P prime Consider the diagram

P

A B

P primepA pB

pprimeA pprimeB

existh

The morphism h P rarr P prime is the unique one guaranteed to exist by the fact that P prime is a productof A and B Commutativity of this diagram says that pA = pprimeA h and pB = pprimeB h A similardiagram with the roles of P and P prime reversed which uses the fact that P is a product results in aunique morphism hprime P prime rarr P satisfying pprimeA = pA hprime and pprimeB = pB hprime

Now consider the following diagram

P

A B

PpA pB

pA pB

hprime h

This commutes which we verify using the properties h and hprime satisfy as follows

pA (hprime h) = (pA hprime) h = pprimeA h = pA

and similar for the morphisms involving B This shows that hprimeh P rarr P satisfies the requirementin the definition of P being a product But of course idP P rarr P satisfies this requirement as wellso since the map satisfying this requirement is assumed to be unique we must have hprime h = idP Similar reasoning considering h hprime P prime rarr P prime shows that h hprime = idprimeP so that h and hprime are indeedinverses and hence that P and P prime are isomorphic The uniqueness of the isomorphism betweenthem comes from the uniqueness of h in the construction above

7

Abstract nonsense It is common to say in the setting above that the fact that products areunique up to unique isomorphism follows from ldquoabstract nonsenserdquo which is a succinct way ofsaying that it follows solely from the definition of a categorical product itself and not any deepermathematical content In particular many of the ldquouniquenessrdquo properties which ldquoproductsrdquo havein the settings we might care aboutmdashCartesian products in Set direct products in Grp producttopologies in Topmdashreally have nothing to do with how those specific products are defined but arejust consequences of the ldquoabstract nonsenserdquo of category theory

But of course we do not use the phrase ldquoabstract nonsenserdquo in a negative or dismissive waysince recognizing what types of properties are consequences of ldquoabstract nonsenserdquo goes a long waytowards understanding well the given problem at hand Soon you too will come to appreciate thisnotion of abstract nonsense

Coproducts Opposite Categories

Initial and terminal objects The notions of initial and terminal objects in a category aredefined in Section 16 of the book where you can also find standard examples Here I want topoint out that initial and terminal objects if they exist are unique up to unique isomorphism asanother example of arguing via ldquoabstract nonsenserdquo

Suppose I and I prime are both initial objects in some category By the fact that I is initial thereexists a unique morphism f I rarr I prime and by the fact that I prime is initial there exists a unique morphismg I prime rarr I But then the composition g f I rarr I must equal idI I rarr I since the latter is theonly morphism from I to itself because I is initial and similarly f g = idIprime Thus I and I prime areunique and there is a unique isomorphism between them

Coproducts The dual concept to that of a product is the notion of a coproduct (The termldquodualrdquo refers to the phenomena you get when you reverse all arrows in a given construction Forinstance the notion of a terminal object is dual to that of an initial object and the notion of anepimorphism is dual to that of a monomorphism) Here is the precise definition a coproduct of anobject A and an object B in a category is an object C together with morphisms iA A rarr C andiB B rarr C such that for any object D and morphisms f A rarr D and g B rarr D there exists aunique morphism h C rarr D which makes the following diagram commute

D

A B

Cf g

iA iB

existh

By abstract nonsense if a coproduct of A and B exists it is unique up to unique isomorphism

Examples In Set the coproduct of two sets A and B is the disjoint union A⊔B (If you havenrsquotseen the notion of a disjoint union before it is essentially the same as the union only that we donrsquotcare about whether A and B have any overlap if they do we treat the common elements as beingldquodifferentrdquo For instance Z ⊔ Z is not Z but consists of two copies of each integer one 0 comesfrom the first copy of Z and another comes from the second but these are different ldquozeroesrdquo)Given f A rarr D and g rarr D the unique map A ⊔B rarr D is defined by applying f to elements of

8

A and g to elements of B (Note that if we simply took the ordinary union AcupB as the candidatefor the coproduct when A and B were not disjoint the required map h A cup B rarr D would notexist since h would have to send any element x isin A cap B to both f(x) and g(x) in order to makethe required diagram commutative)

The coproduct of two objects X and Y in Top is again the disjoint union X ⊔ Y equippedwith the disjoint union topology an open subset of X ⊔ Y is defined to be the (disjoint) union ofan open subset of X with an open subset of Y The coproduct of two groups G and H in Grp isthe free product G lowastH which is further elaborated on in the homework

The coproduct of two vector spaces V and W in Vect is the direct sum V oplusW which is definedto be the set of formal sums

v + w where v isin V and w isin W

where addition and scalar multiplication are defined ldquocomponentwiserdquo

(v1 + w1) + (v2 + w2) = (v1 + v2) + (w1 + w2) and r(v + w) = rv + rw

This is isomorphic to the product V times W with operations also defined componentwise but it iscustomary to speak about the ldquodirect sumrdquo V oplusW when referring to the coproduct as opposed tothe product wersquoll say why in a bit The maps iV V rarr V oplusW and iW W rarr V oplusW required aspart of the data of a coproduct are

iV v 983041rarr v + 0W and iW w 983041rarr 0V + w

To verify that V oplusW is the correct coproduct take any linear transformations S V rarr U andT W rarr U where U is some other vector space We then need a (unique) linear transformationh V oplusW rarr U such that

S = h iV and T = h iW

Let us determine what this map must be thereby not only showing that it exists but also that itis unique Let v + w isin V oplusW which we can write as

v + w = (v + 0W ) + (0V + w)

Since h is required to be linear we have

h(v + w) = h([v + 0W ] + [0V + w]) = h(v + 0W ) + h(0V + w)

The first term at the end is (h iV )v which must thus be Sv and the second term is (h iW )wwhich must be Tw Thus h = S + T is the map

h = S + T v + w 983041rarr Sv + Tw

which you can verify is indeed linear Thus V oplusW does satisfy the property required of a coproductSo the product of two objects in Vect is the same as their coproduct only that we write the

latter using ldquosumrdquo notation Later we will discuss products and coproducts of an arbitrary numberof elements We will see these as special cases of the general notions of limits and colimits butthey are defined via similar requirements as products and coproducts of two objects at a timeIn the category of vector spaces arbitrary products are given by the usual product

983124α Vα with

ldquocomponentwiserdquo addition and scalar multiplication but it turns out that an arbitrary coproductis given by the direct sum

983119α Vα which is defined to be the subspace of

983124α Vα where only finitely

many components are nonzero (This is in a way analogous to how the product topology on an

9

arbitrary number of topological spaces is defined) Wersquoll go over this later but is the underlyingreason why we use oplus instead of times when discussing coproducts

Encoding data via productscoproducts We can nicely summarize the definition of productsand coproducts in the following way the product of A and B is the object with the property thatfor any object D we have

Mor(DA)timesMor(DB) = Mor(D product)

and the coproduct of A and B is the object such that for any object D we have

Mor(AD)timesMor(BD) = Mor(coproductx D)

The first reflects the fact that products give a unique way to turn two morphisms (f and g in thedefinition) mapping into A and B into a single morphism (h in the definition) and the second saysthat coproducts give a unique way to turn two morphisms mapping out of A and B into a singlemorphism

Soon enough we will discuss the idea of universality and the point is that products are ldquouni-versalrdquo for maps into A and B and coproducts are universal for maps from A and B

Opposite categories Finally we give a definition which might seem silly but is actually quiteuseful Given a category C the opposite category is the category Cop is the category with the sameobjects as C but where we defined Mor(AB) in Cop to be Mor(BA) in C To be more concrete weconsider each morphism f A rarr B in C to instead be a morphism fop B rarr A in Cop (Perhaps itwould be better to write this as A larr B fop) The point is that we donrsquot actually care about whatf actually is as say a functioncontinuous maphomomorphismwhatever-we only care about itsbehavior as an ldquoarrowrdquo The opposite category Cop only retains information about how arrowsrelate to one another nothing more

Since arrows are reversed products in C become coproducts in Cop and coproduts in C becomecoproducts in Cop Similarly initialterminal objects and monomorphismsepimorphisms in C be-come terminalinitial objects and epimorphismsmonomorphisms in Cop Again the notion of anopposite category might seem like a strange thing to look at but wersquoll see that it does have uses

Functors Fullness and Faithfulness

One more productcoproduct example Let X be a topological space and define its categoryof open sets Op(X) as follows The objects of Op(X) are simply the open subsets of X and themorphisms are given by inclusions so to be concrete

Mor(U V ) =

983083the unique inclusion U rarr V if U sube V

empty otherwise

The fact that U sube V and V sube W implies U sube W says that the composition of two morphismsin this category is still a morphism in this category The product of U and V in this categoryturns out to be their intersection (with ldquoprojection morphismsrdquo U cap V rarr U and U cap V rarr V beinginclusions) and the coproduct of U and V turns out to be their (ordinary) union Note that weneed X to be a topological space in order to guarantee that U cap V and U cup V are still objects inthis category More generally the product of finitely many objects in Op(X) exists and is theirintersection and the coproduct of an arbitrary number of objects exists and is their union (Wesay that Op(X) has finite products and arbitrary coproducts)

10

Given a set X and a collection T of subsets of X we can define a similar category with elementsof T as objects (I guess we should assume that empty X isin T) and inclusions as morphisms We thenhave that this category has finite products and arbitrary coproducts if and only if T is a topologyon X so that we can rephrase the definition of a topological space itself solely in categorical termsThis perspective on what a ldquotopologyrdquo is makes it simpler to state the definition of whatrsquos calleda sheaf on a topological space which is an important construction in various areas of analysis andgeometry We might discuss this a bit later on

Functors As with most things in mathematics we should care not only about categories asmathematical structures on their own but also about ldquomapsrdquo between these structures whichldquopreserverdquo said structure The notion of a (covariant) functor between categories provides suchldquomapsrdquo in the case at hand where a functor is a way of sending objects to objects and morphismto morphisms in a way which preserves compositions and identities The precise definition is inSection 13 of the book as is the definition of a contravariant functor between categories whichis a functor which reverses the direction of arrows as opposed to ldquocovariantrdquo ones which preservedirections (It is common to use the term ldquofunctorrdquo without qualification to mean one which iscovariant Note that any functor can be made covariant by working with the opposite category acontravariant functor C rarr D is the same as a covariant functor Cop rarr D)

All of the standard examples we looked at are in the book forgetful functors power set functorsthe dual space functor in Vect etc Of particular interest is Example 139 in the book whichshows that various types of group actions in different contexts-the usual notion of a group actionon a set the notion of a continuous action of a group on a topological space and the notion ofa linear action of a group on a vector space (also known as a representation of the group)-are allencoded by a functor with domain category BG

Adjoint functors Let F Top rarr Set be the forgetful functor and let D Set rarr Top be thefunctor which sends a set S to the space S equipped with the discrete topology and a functionS rarr Sprime between sets to the same function only now viewed as a continuous map S rarr Sprime betweendiscrete spaces These two functors satisfy the following property in relation to one another givinga morphism D(S) rarr Y in Top is the same as giving a morphism S rarr F (Y ) in Set or moresymbolically

MorTop(D(S) Y ) = MorSet(S F (Y ))

We say that D is left adjoint to F and that F is right adjoint to D The point is that adjointfunctors can be used to move data back and forth between two categories

Wersquoll talk more about adjoint functors later on but for now here is one more example Nowtake F Grp rarr Set to again be the forgetful functor and take G Grp rarr Set to be the freegroup functor which sends a set S to the free group generated by S G(S) has as elements ldquowordsrdquos1 sn made up of elements of S with the group operation being ldquoconcatenationrdquo A basic factis that giving a group homomorphism from G(S) to some other group H is the same as giving anordinary function from S to F (H) which is the required adjoint property

MorGrp(G(S) H) = MorSet(S F (H))

Faithful and full functors The definitions of what it means for a functor to be faithful and fullare in Section 15 of the book where faithful functors are ones which are injective on morphisms andfull functors are ones which are surjective on morphisms but where in both cases we refer only tomorphisms which occur between the objects F (A) and F (B) in the target category corresponding tofixed objects A and B in the source category (In general a functor could send two objects AB in

11

the source category to the same object F (A) = F (B) in the target category and thus there could betwo morphisms A rarr C and B rarr C which are sent to the same morphism F (A) = F (B) rarr F (C)Such a thing could still possibly be ldquofaithfulrdquo since the non-injectivity here is arising from differentobjects in the source)

Faithfullness and fullness will be nice properties going forward and here is one hint at how theymight be useful Say we have a functor F C rarr D The problem is to determine properties onF which will imply it send products in C to products in D or coproducts in C to coproducts inD So given the former product diagram below we want the second diagram to also be a productdiagram

C

A B

P Fminusminusminusrarr

F (C)

F (A) F (B)

F (P )f g

pA pB

existh

Ff Fg

F (pA) F (pB)

Fh

Of course in order F (P ) to be an actual product of F (A) and F (B) on the right a similarcommutative diagram property would have to hold for all objects in D in place of F (C) and allmorphisms D rarr F (A) and D rarr F (B) not just those of the form Ff and Fg respectively Oneway to guarantee this is to require that all objects in D are of the form F (C) and then for allmorphisms F (C) rarr F (A) and F (C) rarr F (B) to be of the form Ff Fg the former is true whenthe functor F is surjective on objects and the latter is true when F is full We are not saying thatthese are the only conditions under which F might preserve products but it definitely seems togive a sufficient condition at least

Almost there is one more thing to require the uniqueness of the morphism Fh F (C) rarr F (P )making the diagram on the right commute Suppose there were two such morphisms Fh and Fhprimewhere h and hprime are both morphisms C rarr P in C The commutativity of the diagram on the rightgives for instance

Ff = F (pA)Fh and Ff = F (pA)Fhprime

which the functorial properites of F turns into

Ff = F (pA h) and Ff = (pA hprime)

We want h and hprime to make the first diagram commute in order to be able to apply uniqueness ofthe morphism C rarr P But this requires know that f = pA h and f = pA hprime which would followfrom F being faithful In this case it follows that h and hprime both make the first diagram commute(after we go through a similar argument with g) so the fact that P is a product guarantees thath = hprime and hence that Fh = Fhprime so that F (P ) is an actual product on the right

So we conclude that a functor which is full faithful and surjective on objects will send productsto products (Actually we can relax this a bit by requiring not that every object in D be in theliteral image of F but only that it be isomorphic to something in the image A functor with thisproperty is said to be essentially surjective which is a concept will see in a related context soonenough) Again this is not an ldquoif and only ifrdquo statement since functors can preserve products (orcoproducts) without being full faithful and surjective on objects but it at least gives a sufficientcondition Later we will see other ways to guarantee that functors preserve products which arerelated to the existence of adjoints for that given functor

12

Coproduct Examples Concreteness

Products and coproducts for metric spaces The product of two objects in Met the categoryof metric spaces with morphisms given by metric maps is the Cartesian product X times Y equippedwith the box metric and the coproduct of X and Y does not exist if both X and Y are nonemptyThis was the topic of a student presentation and here are some proof sketches

Given metric spaces (X dX) and (Y dY ) the box metric d on X times Y is defined by

d((x1 y1) (x2 y2)) = maxdX(x1 x2) dY (y1 y2))

(The name ldquoboxrdquo metric comes from the fact that in RtimesR = R2 or R3 open balls indeed looks likeldquoboxesrdquo) The key requirement needed in the definition of a product is given morphisms S rarr Xand S rarr Y that the map

h S rarr X times Y defined by h(s) = (f(s) g(s))

actually be a morphism in Met This turns into the requirement that

d((f(s) g(s)) ((f(t) g(t))) = maxdX(f(s) f(t)) dY (g(s) g(t)) le dS(s t) for all s t isin S

given the fact that

dX(f(s) f(t)) le dS(s t) and dY (g(s) g(t)) le dS(s t) for all s t isin S

For the latter inequalities to imply the former d must (at least up to isometry) be given by thebox metric Note that the two other ldquostandardrdquo metrics one might put on a productmdashthe taxicab

dX + dY and Euclidean983156

d2X + d2Y metricsmdashdo not satisfy this required ldquometric maprdquo property

and so do not give the product in MetTo see that coproducts do not exist if XY are nonempty suppose instead (P dP ) was a co-

product for X and Y with inclusion morphisms iX X rarr P and iY Y rarr P For any n isin Ntake 0 n to be a metric space with the absolute value metric d and let f X rarr 0 n andg Y rarr 0 n be the constant function 0 and the constant function n respectively Since P is acoproduct there exists h P rarr 0 n such that

h iX = f and h iY = g

But in order for h to actually be a morphism in Met we must then have

d(h(iX(x) iY (y)) le dP (iX(x) iY (y)) for any x isin X y isin Y

which translates inton = |0minus n| le dP (iX(x) iY (y))

But this should then be true for any n isin N which would imply that the distance in P betweenan element coming from X and one coming from Y should be infinite which is nonsense (Notethat if you altered your definition of ldquometricrdquo to allow for infinite distance then a coproduct wouldexist the disjoint union X ⊔Y where you defined the distance between elements of X and Y to beinfinite)

Coproducts for groups The coproduct of two groups G and H in Grp is the free product GlowastHwhich is defined to be the set of all words consisting of elements of G andH in an alternating fashion

13

with concatenation as the group operation (When we have to elements of G or two elements of Hnext to each other with use their respective group operations to turn these into single elements ofG or H) This was also the topic of a student presentation and here is the basic idea Given grouphomomorphisms g G rarr N and k H rarr N that G lowastH is the coproduct comes down to showingthat the map

h G lowastH rarr N defined by h(g1h1 gnhn) = f(g1)k(h1) middot middot middot f(gn)k(hn)

and similarly on other possible words in GlowastH is a group homomorphism which follows essentiallyfrom the definition of the group operation on G lowastH indeed this group operation arises preciselyfrom this requirement

Now if we instead tried to use GtimesH with its standard inclusions G rarr GtimesH and H rarr GtimesHas the coproduct the issue is that the analogous map

h GtimesH rarr N defined by h((g h)) = f(g)k(h)

would NOT be a homomorphism because h((g1 h1)(g2 h2)) is not equal to h(g1 h1)h(g2 h2) bythe way in which the group operation is defined on the direct product G times H However if westrict our category to only consist of abelian groups so that the N rsquos we consider in the defini-tion of coproduct are also abelian then G times H does serve as a coproduct we can always rewritef(g1)k(g1) middot middot middot f(gn)k(gn) as f(g1) middot middot middot f(gn)k(h1) middot middot middot k(hn) in N Irsquoll leave it to you to flesh out allof these details

Concrete categories The categories Set Top Vect and Grp all have the property thattheir objects are simply sets possibly equipped with extra structure and that their morphisms areordinary set-theoretic functions possibly satisfying some additional requirement Such categoriesare nice since we can often use intuition about sets and functions in their study A concrete categoryis intuitively one where we can think of objects as being ldquosetsrdquo and morphisms as being ldquofunctionsrdquoonly that we have to interpret this in the right way

Here is the definition a concrete category is a category C equipped with a faithful functorF C rarr Set which is part of the data The idea is that such a functor gives us way to turnan object A isin Ob(C) into a set F (A) and a morphism f isin MorC(AB) into a function Ff isinMorSet(F (A) F (B)) where the ldquofaithfulrdquo requirement says that we do not lose information aboutf when doing so so that we can (essentially) learn everything we need to know about f by studyingFf instead In the four examples above the required faithful functor is simply the forgetful functorand indeed in general we think of F as being a ldquoforgetful functorrdquo for the concrete category (C F )

Examples Here are two more examples of concrete categories Let G be a group and considerthe category BG with a single object and elements of G morphisms As written this does notappear to be concrete since morphisms are not honest functions lowast rarr lowast but the point is that thiscategory can indeed by ldquoconcretizedrdquo by specifying an appropriate ldquoforgetful functorrdquo A functorF BG rarr Set is the data of a (left) action of G on a set S = F (lowast) and saying this functor isfaithful is what it means to say that this action is faithful different g h isin G act on S in differentways or equivalently that there exists s isin S such that gs ∕= hs Any group has at least onefaithful left actionmdashnamely the action on itself given by left multiplicationmdashto BG can always beequipped with a faithful functor to Set This is all just a way of phrasing the fact that any groupis isomorphic to a subgroup of a permutation group

The category Rel of relations can also be ldquoconcretizedrdquo even though as written morphismsA rarr B are not honest functions Take F Rel rarr Set to be the power set functor which sends an

14

object in Rel to its power set and a relation f A rarr B to the induced function Ff P (A) rarr P (B)defined by

(Ff)(S) = b isin B | there exists a isin S such that (a b) isin f

In other words if you think of the relation f sube A times B as a ldquopossibly not everywhere-definedpossibly multivalued functionrdquo (Ff)(S) is analogous to taking the image of S That F is faithfulwill be left to the homework

In fact most categories yoursquoll see can be made concrete by specifying an appropriate forgetfulfunctor but not all categories can be made concrete in this way Here is the standard example of anon-concrete category which we mention only to say there is such a thing but which we wonrsquot doanything more with since understanding it better requires knowledge of algebraic topology Thehomotopy category of topological spaces is the category hTop whose objects are topological spacesand whose morphisms are given by equivalence classes of homotopic maps

MorhTop(XY ) = homotopy classes of continuous maps X rarr Y

(Saying that two maps are homotopic means that you can ldquodeformrdquo one into the other but wersquollleave the precise definition to a course in algebraic topology) It turns out that no functor fromhTop to Set can be faithful so that hTop cannot be made concrete but this is actually a deepand difficult thing to show so we make no attempt to do so here

Natural Isomorphisms Representability

Natural transformations The definition of a natural transformation η F rArr G between two(both covariant or both contravariant) functors FG C rarr D is given in Section 14 of the bookThe key idea is that a natural transformation gives a way to turn data about F into data aboutG in a way which is compatible with all possible morphisms as expressed by the commutativity ofthe diagrams

F (A) F (B)

G(A) G(B)

Ff

ηA ηB

Gf

arising from morphisms f A rarr B (or f B rarr A in the contravariant case) in C In the case of anatural isomoprhism η F rArr G so that each ηA F (A) rarr G(A) is actually an isomorphism in Dthe point is not only that F and G produce isomorphic objects but they do so in a way which iscompatible with all possible category-theoretic data

The first standard example of a natural isomorphism is the one between the identity functor onthe category of finite-dimensional and the functor which sends a finite-dimensional vector space Vto V lowastlowast the dual of its dual induced by the isomorphism V rarr V lowastlowast defined by

v 983041rarr (the linear map on V lowast which sends f to f(v))

This was the topic of a student presentation and can be found in Section 14 of the book Theessential reason why this isomorphism is ldquonaturalrdquo is that V rarr V lowastlowast can be defined without referenceto a basis Contrast this with the the functor which sends V to its (single) dual V lowast it is still truethat finite-dimensional vector space is isomorphic to its dual but specifying such an isomorphism

15

requires additional data which prevents the identity functor from being ldquonaturally isomorphicrdquo tothe single dual functor

Some history At this point it makes sense to say a bit about the interesting origins of categorytheory in the 1950rsquos Eilenberg and Mac Lane where studying cohomology (or possible homologyone of those two) in algebraic topology which associates algebraic data (a group ring or similar)to a topological space in a way which encodes some of the topology They had two cohomologicalconstructions which ended up proving isomorphic results but they noticed that not only werethe resulting objects the same but the actual constructions themselves were in some sense theldquosamerdquo They thus then needed a way to formalize this notion and invented the concept of anatural transformation to do so where they interpreted two constructions as being the same asbeing compatible with all morphisms

But this then leads to the question what types of objects should a natural transformationoccur between Eilenberg and Mac Lane then had to invent the notion of a ldquofunctorrdquo to describewhat a natural transformation went from and what it went to In their motivating setting theyinvented the notion of a functor to describe more formally properties which their cohomologicalconstructions were supposed to satisfy But this then leads to the question what types of objectsshould a functor map between In other words what type of data should a functor (cohomologicalconstruction in their example) take as input and should it output They finally had to thus inventthe notion of a ldquocategoryrdquo to describe the source and target of a functor so that their cohomologicalconstructions could be interpreted as functors from a category of topological spaces to a categoryof algebraic objects

Thus historically natural transformations came first then functors and then categories Aswith most things in mathematics categories thus arose from attempts to encode what was beingseen in some concrete examples in a more general and formal setting

Representable functors Representable functors are defined in Section 21 of the book but wersquollgive the required definitions here in a hopefully simpler to digest manner The point is that anyobject A in a (locally small) category C gives two ldquonaturalrdquo functors C rarr Set one covariant andthe other contravariant which as wersquoll see pretty much encode all data about A itself We definehA C rarr Set to be the covariant functor defined by

hA(B) = Mor(AB) on objects and

hA(Bfminusrarr C) = the map Mor(AB)

hAfminusminusrarr Mor(AC) defined by g 983041rarr f g

We define hA C rarr Set to the contravariant functor defined by

hA(B) = Mor(BA) on objects and

hA(Bfminusrarr C) = the map Mor(CA)

hAfminusminusrarr Mor(BA) defined by g 983041rarr g f

(The functor hA is called the functor of points of A and wersquoll usually think of it as being a covariantfunctor hA Cop rarr Set The functor hA does not have a standard name) So hA is defined bymorphisms from A and hA is defined by morphisms into A Wersquoll see later the sense in which suchfunctors encode all information about the object A

A functor F C rarr Set is representable if is naturally isomorphic to such a functor so F sim= hA

for some A in the covariant case and F sim= hA for some A in the contravariant case The idea wersquollbuild towards is that we can essentially treat such a functor as being A itself so that representable

16

functors can be thought as being actual objects in C For now here is a key example wersquove seenbefore which we can now formulate using the language of representability

Given AB isin Ob(C) let F C rarr Set be the contravariant functor defined by

F (C) = Mor(CA)timesMor(CB) and F (Cfminusrarr D) = [(g k) 983041rarr (g f k f)]

where (g k) 983041rarr (g f k f) gives a map Mor(DA) timesMor(DB) rarr Mor(CA) timesMor(CB) ieF (D) rarr F (C) In order for this functor to be representable requires the existence of an object inC satisfying

Mor(CA)timesMor(CB) sim= Mor(C representing object)

in a way which is compatible with all morphisms but we have actually already seen this notionelsewhere it is essentially the definition of a product AtimesB for A and B Indeed we claim that Fis naturally isomorphic to hAtimesB as we now show

First to define a natural isomorphism hAtimesBsim= F requires for each C isin Ob(C) a choice of a

bijection (ie isomorphism in Set)

ηC hAtimesB(C)sim=minusrarr F (C)

which in our case looks like

ηC Mor(CAtimesB))sim=minusrarr Mor(CA)timesMor(CB)

Take this to be the map

(Cfminusrarr AtimesB) 983041rarr (pA f pB f)

where pA A times B rarr A and pB A times B rarr B are the two projection morphisms in the definitionof a product The definition of product implies that this map is invertible since given k C rarr Aand ℓ C rarr B there exists a unique h C rarr A times B such that k = pA h and ℓ = pB h Forthese bijections to define a natural isomorphism hAtimesB

sim= F requires that they be compatible withall morphisms in C which means that given any g C rarr D in C the following diagram shouldcommute

hAtimesB(C) hAtimesB(D)

F (C) F (D)

hAtimesB(g)

ηC ηD

Fg

which in the case at hand looks like

Mor(CAtimesB) Mor(DAtimesB)

Mor(CA)timesMor(CB) Mor(DA)timesMor(DB)

minus g

ηC ηD

(minus gminus g)

Take an element k D rarr AtimesB in the upper right Applying the map on top gives the elementk g in the upper left and then applying ηC on the left side gives

(pA (k g) pB (k g))

17

in the lower left Instead starting with k D rarr AtimesB in the upper right applying ηD on the rightgives (pA k pB k) and applying the map on the bottom gives

((pA k) g (pB k) g)

in the lower left which is the same as the previous element in the lower left by the associativity ofcomposition Hence the isomorphisms ηC do define a natural isomorphism between hAtimesB and F Again the point is that you can then interpret the functor F as being the ldquosamerdquo as the productAtimesB since both encode the same data

More Representable Examples

Coproducts As in the case of products the definition of coproducts can also be phrased in termsof a representable functor Given objects A and B in C let F C rarr Set be the covariant functordefined by

F (C) = Mor(AC)timesMor(BC) and F (Cfminusrarr D) = [(g k) 983041rarr (f g f k)]

where the latter is a function Mor(AC)timesMor(BC) rarr Mor(AD)timesMor(BD) This functor isrepresentable if and only if a coproduct A ⊔ B for A and B exists and the representing object isthat coproduct the key is the requirement that we have bijections

Mor(AC)timesMor(BC) sim= Mor(A ⊔BC)

for all C which is essentially the defining property a coproduct for A and B should have

Power set functor contravariant version Let P Set rarr Set be the contravariant powerset functor defined on objects by sending a set to its power set and on morphisms by sendingf A rarr B to the preimage function Pf P (B) rarr P (A) given by S 983041rarr fminus1(S) In order for this tobe representable requires the existence of a set X with natural bijections

P (A) sim= Mor(AX)

for all sets A that is a set X so that mapping into X is the same data as specifying a subset ofthe domain We take X = 0 1 to be a two-element set and the required bijections

ηA P (A) rarr Mor(A 0 1) to be given by S 983041rarr χS

where XS A rarr 0 1 is the indicator or characteristic function S defined by sending elementsof S to 1 and elements of A minus S to 0 The inverse of ηA sends a function g A rarr 0 1 to thepreimage gminus1(1) sube A

To verify that the bijections ηA define the data of a natural isomorphism η P rArr h01 wecheck the commutativity of some diagrams Given a function f A rarr B consider

P (A) P (B)

h01(A) = Mor(A 0 1) h01(B) = Mor(B 0 1)

Pf = fminus1

ηA ηB

minus f

18

Take S isin P (B) in the upper right Applying the map on top gives fminus1(S) isin P (A) and thenapplying the map on the left gives χfminus1(S) isin Mor(A 0 1) On the other hand applying the mapon the right to S isin P (B) gives χS isin Mor(B 0 1) and then applying the map on the bottomgives χS f isin Mor(A 0 1) The two resulting functions χfminus1(S)χS f in the lower left are

χfminus1(S)(x) =

9830831 x isin fminus1(S)

0 otherwiseand χS(f(x)) =

9830831 f(x) isin S

0 otherwise

so these are the same since x isin fminus1(S) if and only if f(x) isin S Thus the diagram above commutesso P is naturally isomorphic to h01 and hence P is representable

Power set functor covariant version Now consider the covariant power set functor stilldenoted by P Set rarr Set which again sends a set to its power set but now sends a functionf A rarr B to the image function Pf P (A) rarr P (B) defined by S 983041rarr f(S) In order for this tobe representable there would have to exist a set X such that P sim= hX which translates into havingnatural bijections

P (S) sim= Mor(XS)

for all sets S However in this case there is no natural choice for X as there is no natural way todetect subsets of S by mapping into S as opposed to the case of P (S) sim= Mor(S 0 1)

That no such X exists can be made precise using a simple argument when S = empty P (S) hascardinality 1 soX would have to be empty as well since otherwise Mor(XS) would have cardinality0 but if X is empty then Mor(empty S) always cardinality 1 regardless of S which is clearly not trueof P (S) Thus the covariant power set functor is not representable

Forgetful on Top The forgetful functor F Top rarr Set is representable Indeed a representingobject would have to satisfy

X sim= Map(objectX)

as sets for any topological space X (where Map denotes the set of continuous maps) and it iseasy to see that a single point satisfies this property a continuous map f pt rarr X is completelydetermined by the element f(pt) of X and any such element gives a continuous map in this wayOf course checking that F sim= hpt requires checking that various diagrams commute but this issimple to work out in this case so we omit it here

Extracting a topology With an eye towards the idea that the functors hA hA should captureall information about an object A in a category note that the result above shows that in TophX first of all captures all information about X as a set since the set X can be extracted fromhX(pt) = Map(ptX)

We claim that hX actually captures all information about X as a topological space since thetopology on X can be extracted from hX in the following way As we saw in the power set examplewe have a bijection

P (X) sim= Mor(X 0 1)

Now equip 0 1 with the topology in which 1 is open but 0 is not Then a continuous mapf X rarr 0 1 is fully determined by the preimage fminus1(1) since to be continuous only requiresthat this single preimage be open in X Moreoever any open subset of X arises in this way bytaking f to be the indicator function of that open set Thus we get a bijection

Op(X) sim= Map(X 0 1)

19

where Op(X) denotes the set of open subsets of X ie the topology on X Hence the topology onX can be extracted from hX as claimed so hX and hX do encode all information about X

This fact can also be interpreted as saying that the contravariant functor which sends a topo-logical space X to its collection of open subsets (and which sends a continuous map to the mapinduced by taking preimages) is representable with representing object 0 1 equipped with thetopology given above This will be worked out in a homework problem

Forgetful on Grp The forgetful functor Grp rarr Set is also representable In this case arepresenting object should be a group giving set-theoretic bijections

G sim= Hom(objectG)

for all groups G where Hom denotes the set of group homomorphisms The group Z works

G sim= Hom(Z G)

since a homomorphism φ Z rarr G is completely determined by φ(1) since Z is generated by 1 andthere are no restrictions on what φ(1) isin G can actually be The inverse of the desired bijection isthus φ 983041rarr φ(1) and it is simple to check that morphisms behave appropriately so that the forgetfulfunctor in this setting is indeed isomoprhic to hZ

Forgetful on Ring The forgetful functor Ring rarr Set is also representable where by Ring wemean the category of rings with identity elements and where morphisms (ie ring homomorphisms)are required to preserve identities Here we need an object satisfying

R sim= Hom(object R)

for all R isin Ob(Ring) and we can see that Z[x] the ring of polynomials in the variable x withcoefficients in Z works Indeed given a ring homomorphism φ Z[x] rarr R the fact that φ isrequired to preserve identities fully determines φ(n) for any n isin Z (since φ(n) = nφ(1)) and hencesince φ preserves multiplication its behavior is fully determined by φ(x) which can be any elementof R The inverse of the bijection we want is thus φ 983041rarr φ(x) and it will follow that this forgetfulfunctor is isomorphic to hZ[x]

Equivalences between Categories

Definitions iso and equiv Now that we have a notion of a ldquomorphismrdquo between categorieswe can talk about the sense in which two categories are the ldquosamerdquo The first possible definitionis the ldquoobviousrdquo one you might expect given the definition of ldquoisomorphismrdquo wersquove seen in everyother context until now we say that C is isomorphic to D if there exist functors F C rarr D andG D rarr C such that G F = idC and F G = idD where idC and idD are identity functors

However this definition turns out to be too restrictive Indeed the requirement that the givencompositions equal identity functors says that given an object A in C the object G(F (A)) beliterally the same as A itself and similarly for objects in D But we know that two objects in acategory can be thought of as being the ldquosamerdquo without being literally the same as long as theyare isomorphic For instance productscoproducts are not unique in the literal sense but only inthe sense that they are isomorphic in a unique way To take another example the definition ofa functor being representable does not say that F (B) literally be the same as hB(A) only thatthey be isomorphic for sure the power set P (S) of a set S is not literally the same as the set offunctions S rarr 0 1mdashit is only the ldquosamerdquo in the sense that they encode the same data

20

Thus it makes sense to relax the requirement that A andG(F (A)) in the definition of isomorphiccategories be equal to the requirement that they be isomorphic This leads to the following betterdefinition of two categories being the ldquosamerdquo C is equivalent to D if there exist functors F C rarr D

and G D rarr C such that G F sim= idC and F G sim= idD So we require only that for instance thefunctor G F be naturally isomorphic to the identity functor on C which says that for any objectA G(F (A)) be isomorphic to A in a natural way Wersquoll see that this truly is the better notionof what it means for two categories to be the ldquosamerdquo and it reflects the idea that all categoricalproperties of one category can be transported to the other

Alternate characterization of equivalences A functor F C rarr D implementing an equiv-alence between categories is called unsurprisingly and equivalence and the ldquoinverserdquo functorG D rarr C is still referred to as being its inverse even though it is not technically an ldquoinverserdquoin the sense that compositions give literal identities But as in the case of Set where ldquoinvertiblerdquocan be characterized without reference to an inverse as ldquoinjectiverdquo and ldquosurjectiverdquo so too canequivalences between categories be characterized without explicit reference to an inverse

A functor F C rarr D is an equivalence if and only it is full faithful and essentiallysurjective

We have seen the notions of full and faithful before and essentially surjective means that any objectin D is isomorphic to something in the image of F for all D isin Ob(D) there exits C isin Ob(C) suchthat F (C) sim= D Justifying this result was the topic of a student presentation and can be foundin the book in Section 15 (Technically to construct an ldquoinverserdquo of F requires that we work withsmall categories but whatever) The bulk of the real work goes into showing that the ldquoinverserdquofunctor constructed is actually a functor which comes down to some abstract nonsense argumentvia commutative diagrams

Equivalences preserve everything We previously saw the notions of full faithful and es-sentially surjective back when thinking about what properties a functor could have which wouldguarantee it sent products to products so the result we proved back then was that equivalencespreserve products Similarly equivalences preserve coproducts and the proof is basically the sameafter reversing arrows A clean way of deriving this is as follows if F C rarr D is an equivalencethen so is F op Cop rarr Dop so since F op preserves products F preserves coproducts becausecoproducts in a category become products in the opposite category

In general an equivalence between categories will preserve whatever categorical notions youwant monomorphisms epimorphisms initial objects terminal objects etc so that equivalentcategories really do have essentially the ldquosamerdquo properties

Equivalence example Here is a first example (also found in the book) of equivalent categorieswhich are not isomorphic Let FVectR be the category whose objects are finite-dimensional realvector spaces and whose morphisms are linear transformations Let MatR be the category ofmatrices defined as follows

bull objects in MatR are nonnegative integers

bull a morphism n rarr m is an mtimes n matrix with real entries

bull composition in MatR is given by matrix multiplication given B n rarr m and A m rarr kmdashsoB is an mtimes n matrix and A is a k timesm matrixmdashthe composition A B n rarr k is the k times nmatrix AB

bull identity morphisms in MatR are given by identity matrices

21

We claim that FVectR and MatR are equivalent First note that they certainly cannot be isomo-prhic isomorphisms in MatR exist only between two objects which are literally the same (ie anmtimesn matrix can be invertible only when m = n) but isomorphisms in FVectn can exist betweenobjects which are not literally the same So there are in a sense not enough objects in MatR toallow it to be isomorphic to FVectR

To construct an equivalence choose one and for all an isomorphism TV V rarr RdimV forevery object in FVectR or equivalently choose once and for all a basis for every V DefineF FVectR rarr MatR by

V 983041rarr dimV on objects and

(T V rarr W ) 983041rarr (the standard matrix of SW T Sminus1V RdimV rarr RdimW ) on morphisms

Equivalently F sends T to the matrix of T relative to the chosen bases for V and W It isstraightforward to check that this is a functor The inverse functor G MatR rarr VectR is givenby

n 983041rarr Rn on objects and

(mtimes n matrix A) 983041rarr (the linear transformation A Rn rarr Rm induced by A) on morphisms

One can check that G is a functor and that F G and G F are naturally isomorphic to identitiesor instead one can argue that F alone is full faithful and essentially surjective The point is thatG(F (V )) gives back RdimV so not literally V itself but only something isomorphic to V which iswhy this gives an equivalence of categories and not an isomorphism

Yoneda Lemma Functors as Objects

Compact Hausdorff spaces and Clowast-algebras Last time we gave an example of equivalentcategories which were not isomorphic but in many ways that example is unsatisfactory when doinglinear algebra people for sure often work with matrix representations of linear transformations asthe example from last would suggest but no one in their right mind literally thinks about suchthings in terms of the equivalence FVectR rarr MatR itself In other words this equivalence isessentially ldquocooked uprdquo for the sake of giving an example but it does not really give any insightinto the actual mathematics of linear algebra

So with that in mind we will describe another equivalence which actually does have some realmeaning in the sense of producing new ways of thinking about some mathematics Given a compactHausdorff space X we let C(X) denote the space of continuous complex-valued functions on X

C(X) = f X rarr C | f is continuous

We can equip the set C(X) with various additional structures First it is a complex vector spaceunder ordinary addition and scalar multiplication of functions Second we can multiply two func-tions together

(fg)(p) = f(p)g(p)

which turns C(X) into whatrsquos called an algebra over C (We wonrsquot give the definition of anldquoalgebrardquo but the point is that this multiplication is in some sense ldquocompatiblerdquo with the vectorspace structure) Next any element f of C(X) has a conjugate element f obtained by takingcomplex conjugates of the values of f

f(p) = f(p)

22

And finally we can equip C(X) with a norm via

983042f983042 = suppisinX

|f(p)|

where |middot| denotes complex absolute value (This supremum exists sinceX is compact by the ExtremeValue Theorem) All of these structures (vector space algebra multiplication conjugation norm)turn C(X) into whatrsquos called a Clowast-algebra We wonrsquot define this precisely but the definition encodesvarious ways in which these structures are compatible Actually C(X) is a unital commutative Clowast-algebra where ldquounitalrdquo means that the multiplication has an identity element (namely the constantfunction 1) and commutative means that fg = gf There is a natural notion of homomorphismbetween Clowast-algebras which are maps between them which behave nicely with all the structuresinvolved In particular given a continuous map f X rarr Y between compact Hausdorff spaces weget a Clowast-algebra homomorphism flowast C(Y ) rarr C(X) given by pre-composition with f flowast sendsg isin C(Y ) to g f isin C(X)

We thus have a contravariant functor CHaus rarr ucClowast-Alg from the category of compactHausdorff spaces and the category of unital commutative Clowast-algebras It turns out that anyunital commutative Clowast-algebra arises in this way in the sense that given any unital commutativeClowast-algebra B there exists a compact Hausdorff space X such that C(X) sim= B as Clowast-algebrasand that morphisms between these algebra arises from continuous maps in the manner describeabove This says that the functor above is essentially surjective full and it turns out that it isalso faithful so that it is an equivalence Thus all information about compact Hausdorff spaces iscompletely encoded in information about unital commtuative Clowast-algebras and indeed one couldlearn everything there is to know about a compact Hausdorff space X from its Clowast-algebra C(X)of functions The fact that the categories CHaus and ucClowast-Alg are equivalent is the startingpoint in the subject known as noncommutative geometry which wersquoll say something about a bitlater on If we drop the requirement that our algebras be unital it turns out that the categoryof commutative Clowast-algebras in general is equivalent to the category of locally compact Hausdorffspaces via essentially the same functor where we take as morphisms in the category of locallycompact Hausdorff spaces to be continuous functions which ldquovanish at infinityrdquo

We are only introducing Clowast-algebras to give an example of an interesting and actually usefulequivalence between categories but just for the fun of it wersquoll note the following Examples ofnoncommutative Clowast-algebras come from what are known as ldquoalgebras of bounded operators onHilbert spacesrdquo which show up in quantum mechanics as the things which describe physicallyobservable quantities like position momentum and energy (In fact all Clowast-algebras arise fromHilbert spaces in this way) Thus Clowast-algebras show up in mathematical formulations of quantummechanics and the notion of noncommutative geometry mentioned above is at the core of someattempts to understand the elusive concept known as ldquoquantum geometryrdquo

Yoneda Lemma The Yoneda Lemma states that given a (covariant) functor F C rarr Set whereC is locally small for any A isin Ob(C) there exists a bijection

F (A) sim= Nat(hA F )

where the right side denotes the set of natural transformations from hA to F A similar result holdsin the contravariant case where we get bijections F (A) sim= Nat(hA F ) (There is also a sense inwhich these bijections are themselves natural but wersquoll ignore this bit) The proof of the YonedaLemma was the topic of a student presentation and can be found in Section 22 of the book

Here we will get a bit philosophical and talk about what the point of the Yoneda Lemmaactually is and later the sense in which it says that we can view arbitrary functors as ldquonon-existent

23

objectsrdquo in a category First we mention what we will call the ldquofunctor of pointsrdquo approach tomathematics a mathematical object can be fully characterized by all morphisms into it That isknowledge of Mor(SA) for all S is equivalent to knowledge of A itself Indeed this is an idea wementioned previously in relation to the functors hAmdashwhich we previously said is referred to as beingthe functor of points of Amdashand now we seek to push this idea further In some examples literalpoints of A can be extracted from hA as in the case of the bijection X sim= hX(pt) = Map(ptX) fora topological space X or G sim= hG(Z) = Mor(Z G) for a group G Moreover in these examples therest of the (topological) structure onX or (group) structure on G can also be extracted from hX andhG respectively In general we can view Mor(SA) as in a sense being ldquopoints of A parameterizedby Srdquo which comes from the idea that in say Set Mor(SA) = AS is ldquothe product of S-manycopies of Ardquo so literally points of A ldquoparameterizedrdquo by S In various areas of mathematics anelement of Mor(SA) is thought of as being a ldquogeneralizedrdquo point of A so that the functor of pointshA detects generalized points of A

So an object A is fully encoded by the functor hA Now we claim that morphisms betweenobjects are also encoded by such functors Indeed in the case where F = hB is itself a functor ofpoints the Yoneda result Nat(hA F ) sim= F (A) becomes

Nat(hA hB) sim= hB(A) = Mor(AB)

which says precisely that morphisms between the objects A and B can be obtained solely fromthe functors hA and hB (This fact is one reason why one should prefer the contravariant functorhA to the covariant functor hA in the covariant case Nat(hA F ) sim= F (A) for F = hB becomesNat(hA hB) sim= hB(A) = Mor(BA) so that morphisms between two objects get reversed whencharacterizing them in terms of representable functors)

Pushing our philosophy further we now know that not only objects but also morphisms betweenobjects are fully encoded by the functors hA The final step is to note that we should not only careabout objects and morphisms between them but that an important part of the data of an objectA is also how arbitrary contravariant functors F C rarr Set behave on A and that this too can beobtained from hA precisely via the Yoneda result that F (A) sim= Nat(hA F ) So objects morphismsbetween objects and how arbitrary functors behave on objects are all encoded in knowledge ofthe functors hA so that these functors truly do capture in the most general sense possible allinformation about A

Functors as nonexistent objects The discussion above is all well and good but it still doesnrsquotquite say why we should care about the fact that we can study on object A by studying hA insteadin that we still havenrsquot spoken about what type of insight about A can be gleamed from hA thatwasnrsquot readily apparent in A itself already Now we argue that the true point of the YonedaLemma is that it gives a way to think about arbitrary functors F Cop rarr Set as ldquoobjectsrdquo in C

itself or really as objects in an ldquoenlargementrdquo of C Certainly for those functors F sim= hA which arerepresentable we can indeed think about F as simply being the object A but now we claim thateven non-representable F can be thought of as being ldquononexistentrdquo objects in C The upshot is thatoften in mathematics we come across things which do not actually exist but the Yoneda Lemmagives a way to (indirectly) study such things anyway by interpreting them as functors instead

The starting point of course is that an actual object A is the same as hA as we have explainedabove Thus by analogy we can interpret a non-representable functor F as being an ldquoobjectrdquo in C

as long as we know what it means to ldquomaprdquo into F from other things which are honestly objects in CThat is if F were indeed representable then F (A) sim= Nat(hA F ) sim= Mor(AF ) would characterizethe functor of points of F (here we abuse notation by using the same letter F to denote both arepresentable functor and the object which represents it) so now we take Mor(AF ) = F (A) as

24

the definition of the ldquofunctor of pointsrdquo of the non-existent object F We are thus taking elementsof F (A) as being what it means to ldquomaprdquo from A into F even when F is not representable Thefact that the Yoneda Lemma gives a way to define maps from A into F thus says that F itself isequipped with the same data that a literal object of A would be equipped with so why shouldnrsquotwe interpret F as a being a type of ldquoobjectrdquo in C as well

To phrase this another way note that the Yoneda Lemma gives rise to a functor

C rarr Fun(CopSet)

where Fun(CopSet) denotes the category whose objects are functors from Cop to Set and morphismsare natural transformation defined on objects by A 983041rarr hA and on morphisms by the bijectionMor(AB) sim= Nat(hA hB) arising from the Yoneda Lemma That this is actually bijection saysthat this functor is full and faithful so that we can view C as being a subcategory of the functorcategory Fun(CopSet) (We cay say that C is isomorphic to the subcategory of Fun(CopSet)consisting of the functors hA or that it is equivalent to the subcategory of Fun(CopSet) consistingof representable functors This functor C rarr Fun(CopSet) is usually called the Yoneda embeddingof C into Fun(CopSet)) Thus we can view Fun(CopSet) as an ldquoenlargmentrdquo of C and the ideawe are trying to make sense of is that other objects in this ldquolargerrdquo category should be interpretedas ldquonon-existent objectsrdquo of C

Just how far can we actually push this idea that arbitrary functors Cop rarr Set should be thoughtof as being ldquoobjectsrdquo of C Pretty far in fact For instance often times in geometry one comesacross a type of ldquosmoothrdquo space on which some group acts and one is interested in doing ldquocalculusrdquoon the resulting quotient space which is the space of orbits under the given group action Theproblem is that such quotient spaces are very often (usually in fact) not ldquosmoothrdquo themselves sothat the tools from calculus are not readily available The solution is to nonetheless treat thisquotient as if it were ldquosmoothrdquo and attempt to do calculus anyway To make this precise thisldquonon-smoothrdquo space actually turns out to have a natural interpretation as a ldquosmooth functorrdquo andit is through this functor which we can view this non-smooth space as actually being smooth afterall so we are forced to expand our notion of what we mean by ldquosmooth spacerdquo by allowing thingswhich are not spaces in the usual sense which certainly leads to more abstraction but the benefitis that it actually makes sense to do calculus on such ldquospacesrdquo interpreted as functors

The discussion above is pretty vague so here is one final more concrete example Recall thatequivalence

CHausop rarr ucClowast-Alg

between the category of compact Hausdorff spaces and the category of unital commutative Clowast-algebras mentioned at the start We now claim that a non-commutative (unital) Clowast-algebra shouldbe interpreted as the algebra of functions on some non-existent ldquononcommutative compact Haus-dorff spacerdquo and it is this idea which the subject of noncommutative geometry makes precise Solet B be a non-commutative Clowast-algebra Then the functor of points hB gives a functor

ucClowast-Alg rarr Set

which sends C(X) (where X is some compact Hausdorff space) to Hom(C(X) B) where Hom de-notes homomorphisms of Clowast-algebras whatever that means Composing with the previous functorwe get a functor wersquoll denote by B itself

B CHausop rarr Set

which on objects at least is given by X 983041rarr Hom(C(X) B) This functor is not representable sinceB is not commutative but the philosophy espoused above suggests we should think of it as giving a

25

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

So the upshot is that for any C and any f and g in the diagram above there exists a uniqueh which makes the diagram commute The point is that a product gives a way to turn the twopieces of data f C rarr A and g C rarr B into the single piece of data h C rarr P in a uniqueway Being able to turn a possibly large amount of data into a single piece of data in this way isone of the benefits the ldquocategoricalrdquo perspective provides Products in categories are often denotedusing the usual A times B notation even though it is not necessarily true that a product is literallya set-theoretic Cartesian product although it is in the standard examples The map h C rarr Prequired in the definition is then often denoted by h = f times g

Examples In Set the product of two objects A and B always exists and is indeed the ordinaryCartesian product A times B But of course the definition of product also requires that data of theprojection morphisms which in this case are simply the usual ones

pA (a b) 983041rarr a and pB (a b) 983041rarr b

Given f C rarr A and g C rarr B the map h C rarr AtimesB is h(c) = (f(c) g(c)) and you can checkthat these constructions satisfy the properties required in the definition (Note that you would alsohave to argue that h defined in this way is the only map which can make the required diagramcommute)

Products in Grp are given by the usual direct product group structure GtimesH with the groupoperation defined as

(g1 h1) middot (g2 h2) = (g1g2 h1h2)

The projection morphisms are the ordinary ones you would expect as well But let us actuallynow verify that this group structure indeed is among other possible group structures on G times Hthe one required in the definition of a product Suppose f K rarr G and g K rarr H are grouphomomorphisms The key point is that the map h(k) = (f(k) g(k)) which should satisfy therequired product property should itself be a morphism in Grp meaning a group homomorphismand we claim that this is only true for the direct product group structure

Indeed we want it to be true that

h(k1k2) = (f(k1k2) g(k1k2)) equal h(k1)h(k2) = (f(k1) g(k1)) middot (f(k2)g(k2))

for all k1 k2 isin K and some to-be-determined group structure middot on G times H Since f and g arehomomorphisms this requirement becomes

(f(k1)f(k2) g(k1)g(k2)) = (f(k1) g(k1)) middot (f(k2)g(k2))

which says that middot should indeed be componentwise multiplication at least on the image of f andg But recall that the properties required of a product should hold for any K and any f K rarr Gand g K rarr H in particular then it should hold for instances when f and g are surjective whichshows that the componentwise-multiplication property above should in fact hold on all of G timesHso that middot is indeed the ordinary direct product structure

Uniqueness Notice that in the definition of a product in a category we used the word ldquoardquo asopposed to ldquotherdquo we spoke about a product of A and B instead of the product of A and B Sowe now ask are products unique The answer is ldquonordquo if we interpret ldquouniquerdquo in the strictestpossible sense but ldquoyesrdquo if we interpret it correctly For instance in the example of Grp takenow N to be any group which is isomorphic to the direct product G timesH say with isomorphismℓ N rarr G timesH We claim that we can also turn N into a categorical product of G of H as long

6

as we define appropriate ldquoprojectionsrdquo N rarr G and N rarr H take N rarr G to be the compositionpG ℓ and N rarr H to be pH ℓ Given f K rarr G and g K rarr H as in the definition of a productthe map hprime K rarr N defined by

hprime = ℓminus1 h

where h K rarr GtimesH is the usual h = f timesg will satisfy the requirement needed to able to concludethat N is also a product of G and H in Grp A similar trick works in any category so that anyobject isomorphic a product can also itself be turned into a product

The correct answer is that products are unique up to unique isomorphism which means thatgiven a product P of A and B any other product P prime of A and B will in fact be isomorphic to P and in a unique way in the sense that there can exist only one isomorphism P rarr P prime The phraseldquounique up to unique isomorphismrdquo is a common one wersquoll see again and again objects definedby some categorical property are almost never genuinely unique in a strict sense but they willessentially be as close to unique as possible

So suppose that P and P prime are both products of A and B with associated projection morphismspA pB for P and pprimeA p

primeB for P prime Consider the diagram

P

A B

P primepA pB

pprimeA pprimeB

existh

The morphism h P rarr P prime is the unique one guaranteed to exist by the fact that P prime is a productof A and B Commutativity of this diagram says that pA = pprimeA h and pB = pprimeB h A similardiagram with the roles of P and P prime reversed which uses the fact that P is a product results in aunique morphism hprime P prime rarr P satisfying pprimeA = pA hprime and pprimeB = pB hprime

Now consider the following diagram

P

A B

PpA pB

pA pB

hprime h

This commutes which we verify using the properties h and hprime satisfy as follows

pA (hprime h) = (pA hprime) h = pprimeA h = pA

and similar for the morphisms involving B This shows that hprimeh P rarr P satisfies the requirementin the definition of P being a product But of course idP P rarr P satisfies this requirement as wellso since the map satisfying this requirement is assumed to be unique we must have hprime h = idP Similar reasoning considering h hprime P prime rarr P prime shows that h hprime = idprimeP so that h and hprime are indeedinverses and hence that P and P prime are isomorphic The uniqueness of the isomorphism betweenthem comes from the uniqueness of h in the construction above

7

Abstract nonsense It is common to say in the setting above that the fact that products areunique up to unique isomorphism follows from ldquoabstract nonsenserdquo which is a succinct way ofsaying that it follows solely from the definition of a categorical product itself and not any deepermathematical content In particular many of the ldquouniquenessrdquo properties which ldquoproductsrdquo havein the settings we might care aboutmdashCartesian products in Set direct products in Grp producttopologies in Topmdashreally have nothing to do with how those specific products are defined but arejust consequences of the ldquoabstract nonsenserdquo of category theory

But of course we do not use the phrase ldquoabstract nonsenserdquo in a negative or dismissive waysince recognizing what types of properties are consequences of ldquoabstract nonsenserdquo goes a long waytowards understanding well the given problem at hand Soon you too will come to appreciate thisnotion of abstract nonsense

Coproducts Opposite Categories

Initial and terminal objects The notions of initial and terminal objects in a category aredefined in Section 16 of the book where you can also find standard examples Here I want topoint out that initial and terminal objects if they exist are unique up to unique isomorphism asanother example of arguing via ldquoabstract nonsenserdquo

Suppose I and I prime are both initial objects in some category By the fact that I is initial thereexists a unique morphism f I rarr I prime and by the fact that I prime is initial there exists a unique morphismg I prime rarr I But then the composition g f I rarr I must equal idI I rarr I since the latter is theonly morphism from I to itself because I is initial and similarly f g = idIprime Thus I and I prime areunique and there is a unique isomorphism between them

Coproducts The dual concept to that of a product is the notion of a coproduct (The termldquodualrdquo refers to the phenomena you get when you reverse all arrows in a given construction Forinstance the notion of a terminal object is dual to that of an initial object and the notion of anepimorphism is dual to that of a monomorphism) Here is the precise definition a coproduct of anobject A and an object B in a category is an object C together with morphisms iA A rarr C andiB B rarr C such that for any object D and morphisms f A rarr D and g B rarr D there exists aunique morphism h C rarr D which makes the following diagram commute

D

A B

Cf g

iA iB

existh

By abstract nonsense if a coproduct of A and B exists it is unique up to unique isomorphism

Examples In Set the coproduct of two sets A and B is the disjoint union A⊔B (If you havenrsquotseen the notion of a disjoint union before it is essentially the same as the union only that we donrsquotcare about whether A and B have any overlap if they do we treat the common elements as beingldquodifferentrdquo For instance Z ⊔ Z is not Z but consists of two copies of each integer one 0 comesfrom the first copy of Z and another comes from the second but these are different ldquozeroesrdquo)Given f A rarr D and g rarr D the unique map A ⊔B rarr D is defined by applying f to elements of

8

A and g to elements of B (Note that if we simply took the ordinary union AcupB as the candidatefor the coproduct when A and B were not disjoint the required map h A cup B rarr D would notexist since h would have to send any element x isin A cap B to both f(x) and g(x) in order to makethe required diagram commutative)

The coproduct of two objects X and Y in Top is again the disjoint union X ⊔ Y equippedwith the disjoint union topology an open subset of X ⊔ Y is defined to be the (disjoint) union ofan open subset of X with an open subset of Y The coproduct of two groups G and H in Grp isthe free product G lowastH which is further elaborated on in the homework

The coproduct of two vector spaces V and W in Vect is the direct sum V oplusW which is definedto be the set of formal sums

v + w where v isin V and w isin W

where addition and scalar multiplication are defined ldquocomponentwiserdquo

(v1 + w1) + (v2 + w2) = (v1 + v2) + (w1 + w2) and r(v + w) = rv + rw

This is isomorphic to the product V times W with operations also defined componentwise but it iscustomary to speak about the ldquodirect sumrdquo V oplusW when referring to the coproduct as opposed tothe product wersquoll say why in a bit The maps iV V rarr V oplusW and iW W rarr V oplusW required aspart of the data of a coproduct are

iV v 983041rarr v + 0W and iW w 983041rarr 0V + w

To verify that V oplusW is the correct coproduct take any linear transformations S V rarr U andT W rarr U where U is some other vector space We then need a (unique) linear transformationh V oplusW rarr U such that

S = h iV and T = h iW

Let us determine what this map must be thereby not only showing that it exists but also that itis unique Let v + w isin V oplusW which we can write as

v + w = (v + 0W ) + (0V + w)

Since h is required to be linear we have

h(v + w) = h([v + 0W ] + [0V + w]) = h(v + 0W ) + h(0V + w)

The first term at the end is (h iV )v which must thus be Sv and the second term is (h iW )wwhich must be Tw Thus h = S + T is the map

h = S + T v + w 983041rarr Sv + Tw

which you can verify is indeed linear Thus V oplusW does satisfy the property required of a coproductSo the product of two objects in Vect is the same as their coproduct only that we write the

latter using ldquosumrdquo notation Later we will discuss products and coproducts of an arbitrary numberof elements We will see these as special cases of the general notions of limits and colimits butthey are defined via similar requirements as products and coproducts of two objects at a timeIn the category of vector spaces arbitrary products are given by the usual product

983124α Vα with

ldquocomponentwiserdquo addition and scalar multiplication but it turns out that an arbitrary coproductis given by the direct sum

983119α Vα which is defined to be the subspace of

983124α Vα where only finitely

many components are nonzero (This is in a way analogous to how the product topology on an

9

arbitrary number of topological spaces is defined) Wersquoll go over this later but is the underlyingreason why we use oplus instead of times when discussing coproducts

Encoding data via productscoproducts We can nicely summarize the definition of productsand coproducts in the following way the product of A and B is the object with the property thatfor any object D we have

Mor(DA)timesMor(DB) = Mor(D product)

and the coproduct of A and B is the object such that for any object D we have

Mor(AD)timesMor(BD) = Mor(coproductx D)

The first reflects the fact that products give a unique way to turn two morphisms (f and g in thedefinition) mapping into A and B into a single morphism (h in the definition) and the second saysthat coproducts give a unique way to turn two morphisms mapping out of A and B into a singlemorphism

Soon enough we will discuss the idea of universality and the point is that products are ldquouni-versalrdquo for maps into A and B and coproducts are universal for maps from A and B

Opposite categories Finally we give a definition which might seem silly but is actually quiteuseful Given a category C the opposite category is the category Cop is the category with the sameobjects as C but where we defined Mor(AB) in Cop to be Mor(BA) in C To be more concrete weconsider each morphism f A rarr B in C to instead be a morphism fop B rarr A in Cop (Perhaps itwould be better to write this as A larr B fop) The point is that we donrsquot actually care about whatf actually is as say a functioncontinuous maphomomorphismwhatever-we only care about itsbehavior as an ldquoarrowrdquo The opposite category Cop only retains information about how arrowsrelate to one another nothing more

Since arrows are reversed products in C become coproducts in Cop and coproduts in C becomecoproducts in Cop Similarly initialterminal objects and monomorphismsepimorphisms in C be-come terminalinitial objects and epimorphismsmonomorphisms in Cop Again the notion of anopposite category might seem like a strange thing to look at but wersquoll see that it does have uses

Functors Fullness and Faithfulness

One more productcoproduct example Let X be a topological space and define its categoryof open sets Op(X) as follows The objects of Op(X) are simply the open subsets of X and themorphisms are given by inclusions so to be concrete

Mor(U V ) =

983083the unique inclusion U rarr V if U sube V

empty otherwise

The fact that U sube V and V sube W implies U sube W says that the composition of two morphismsin this category is still a morphism in this category The product of U and V in this categoryturns out to be their intersection (with ldquoprojection morphismsrdquo U cap V rarr U and U cap V rarr V beinginclusions) and the coproduct of U and V turns out to be their (ordinary) union Note that weneed X to be a topological space in order to guarantee that U cap V and U cup V are still objects inthis category More generally the product of finitely many objects in Op(X) exists and is theirintersection and the coproduct of an arbitrary number of objects exists and is their union (Wesay that Op(X) has finite products and arbitrary coproducts)

10

Given a set X and a collection T of subsets of X we can define a similar category with elementsof T as objects (I guess we should assume that empty X isin T) and inclusions as morphisms We thenhave that this category has finite products and arbitrary coproducts if and only if T is a topologyon X so that we can rephrase the definition of a topological space itself solely in categorical termsThis perspective on what a ldquotopologyrdquo is makes it simpler to state the definition of whatrsquos calleda sheaf on a topological space which is an important construction in various areas of analysis andgeometry We might discuss this a bit later on

Functors As with most things in mathematics we should care not only about categories asmathematical structures on their own but also about ldquomapsrdquo between these structures whichldquopreserverdquo said structure The notion of a (covariant) functor between categories provides suchldquomapsrdquo in the case at hand where a functor is a way of sending objects to objects and morphismto morphisms in a way which preserves compositions and identities The precise definition is inSection 13 of the book as is the definition of a contravariant functor between categories whichis a functor which reverses the direction of arrows as opposed to ldquocovariantrdquo ones which preservedirections (It is common to use the term ldquofunctorrdquo without qualification to mean one which iscovariant Note that any functor can be made covariant by working with the opposite category acontravariant functor C rarr D is the same as a covariant functor Cop rarr D)

All of the standard examples we looked at are in the book forgetful functors power set functorsthe dual space functor in Vect etc Of particular interest is Example 139 in the book whichshows that various types of group actions in different contexts-the usual notion of a group actionon a set the notion of a continuous action of a group on a topological space and the notion ofa linear action of a group on a vector space (also known as a representation of the group)-are allencoded by a functor with domain category BG

Adjoint functors Let F Top rarr Set be the forgetful functor and let D Set rarr Top be thefunctor which sends a set S to the space S equipped with the discrete topology and a functionS rarr Sprime between sets to the same function only now viewed as a continuous map S rarr Sprime betweendiscrete spaces These two functors satisfy the following property in relation to one another givinga morphism D(S) rarr Y in Top is the same as giving a morphism S rarr F (Y ) in Set or moresymbolically

MorTop(D(S) Y ) = MorSet(S F (Y ))

We say that D is left adjoint to F and that F is right adjoint to D The point is that adjointfunctors can be used to move data back and forth between two categories

Wersquoll talk more about adjoint functors later on but for now here is one more example Nowtake F Grp rarr Set to again be the forgetful functor and take G Grp rarr Set to be the freegroup functor which sends a set S to the free group generated by S G(S) has as elements ldquowordsrdquos1 sn made up of elements of S with the group operation being ldquoconcatenationrdquo A basic factis that giving a group homomorphism from G(S) to some other group H is the same as giving anordinary function from S to F (H) which is the required adjoint property

MorGrp(G(S) H) = MorSet(S F (H))

Faithful and full functors The definitions of what it means for a functor to be faithful and fullare in Section 15 of the book where faithful functors are ones which are injective on morphisms andfull functors are ones which are surjective on morphisms but where in both cases we refer only tomorphisms which occur between the objects F (A) and F (B) in the target category corresponding tofixed objects A and B in the source category (In general a functor could send two objects AB in

11

the source category to the same object F (A) = F (B) in the target category and thus there could betwo morphisms A rarr C and B rarr C which are sent to the same morphism F (A) = F (B) rarr F (C)Such a thing could still possibly be ldquofaithfulrdquo since the non-injectivity here is arising from differentobjects in the source)

Faithfullness and fullness will be nice properties going forward and here is one hint at how theymight be useful Say we have a functor F C rarr D The problem is to determine properties onF which will imply it send products in C to products in D or coproducts in C to coproducts inD So given the former product diagram below we want the second diagram to also be a productdiagram

C

A B

P Fminusminusminusrarr

F (C)

F (A) F (B)

F (P )f g

pA pB

existh

Ff Fg

F (pA) F (pB)

Fh

Of course in order F (P ) to be an actual product of F (A) and F (B) on the right a similarcommutative diagram property would have to hold for all objects in D in place of F (C) and allmorphisms D rarr F (A) and D rarr F (B) not just those of the form Ff and Fg respectively Oneway to guarantee this is to require that all objects in D are of the form F (C) and then for allmorphisms F (C) rarr F (A) and F (C) rarr F (B) to be of the form Ff Fg the former is true whenthe functor F is surjective on objects and the latter is true when F is full We are not saying thatthese are the only conditions under which F might preserve products but it definitely seems togive a sufficient condition at least

Almost there is one more thing to require the uniqueness of the morphism Fh F (C) rarr F (P )making the diagram on the right commute Suppose there were two such morphisms Fh and Fhprimewhere h and hprime are both morphisms C rarr P in C The commutativity of the diagram on the rightgives for instance

Ff = F (pA)Fh and Ff = F (pA)Fhprime

which the functorial properites of F turns into

Ff = F (pA h) and Ff = (pA hprime)

We want h and hprime to make the first diagram commute in order to be able to apply uniqueness ofthe morphism C rarr P But this requires know that f = pA h and f = pA hprime which would followfrom F being faithful In this case it follows that h and hprime both make the first diagram commute(after we go through a similar argument with g) so the fact that P is a product guarantees thath = hprime and hence that Fh = Fhprime so that F (P ) is an actual product on the right

So we conclude that a functor which is full faithful and surjective on objects will send productsto products (Actually we can relax this a bit by requiring not that every object in D be in theliteral image of F but only that it be isomorphic to something in the image A functor with thisproperty is said to be essentially surjective which is a concept will see in a related context soonenough) Again this is not an ldquoif and only ifrdquo statement since functors can preserve products (orcoproducts) without being full faithful and surjective on objects but it at least gives a sufficientcondition Later we will see other ways to guarantee that functors preserve products which arerelated to the existence of adjoints for that given functor

12

Coproduct Examples Concreteness

Products and coproducts for metric spaces The product of two objects in Met the categoryof metric spaces with morphisms given by metric maps is the Cartesian product X times Y equippedwith the box metric and the coproduct of X and Y does not exist if both X and Y are nonemptyThis was the topic of a student presentation and here are some proof sketches

Given metric spaces (X dX) and (Y dY ) the box metric d on X times Y is defined by

d((x1 y1) (x2 y2)) = maxdX(x1 x2) dY (y1 y2))

(The name ldquoboxrdquo metric comes from the fact that in RtimesR = R2 or R3 open balls indeed looks likeldquoboxesrdquo) The key requirement needed in the definition of a product is given morphisms S rarr Xand S rarr Y that the map

h S rarr X times Y defined by h(s) = (f(s) g(s))

actually be a morphism in Met This turns into the requirement that

d((f(s) g(s)) ((f(t) g(t))) = maxdX(f(s) f(t)) dY (g(s) g(t)) le dS(s t) for all s t isin S

given the fact that

dX(f(s) f(t)) le dS(s t) and dY (g(s) g(t)) le dS(s t) for all s t isin S

For the latter inequalities to imply the former d must (at least up to isometry) be given by thebox metric Note that the two other ldquostandardrdquo metrics one might put on a productmdashthe taxicab

dX + dY and Euclidean983156

d2X + d2Y metricsmdashdo not satisfy this required ldquometric maprdquo property

and so do not give the product in MetTo see that coproducts do not exist if XY are nonempty suppose instead (P dP ) was a co-

product for X and Y with inclusion morphisms iX X rarr P and iY Y rarr P For any n isin Ntake 0 n to be a metric space with the absolute value metric d and let f X rarr 0 n andg Y rarr 0 n be the constant function 0 and the constant function n respectively Since P is acoproduct there exists h P rarr 0 n such that

h iX = f and h iY = g

But in order for h to actually be a morphism in Met we must then have

d(h(iX(x) iY (y)) le dP (iX(x) iY (y)) for any x isin X y isin Y

which translates inton = |0minus n| le dP (iX(x) iY (y))

But this should then be true for any n isin N which would imply that the distance in P betweenan element coming from X and one coming from Y should be infinite which is nonsense (Notethat if you altered your definition of ldquometricrdquo to allow for infinite distance then a coproduct wouldexist the disjoint union X ⊔Y where you defined the distance between elements of X and Y to beinfinite)

Coproducts for groups The coproduct of two groups G and H in Grp is the free product GlowastHwhich is defined to be the set of all words consisting of elements of G andH in an alternating fashion

13

with concatenation as the group operation (When we have to elements of G or two elements of Hnext to each other with use their respective group operations to turn these into single elements ofG or H) This was also the topic of a student presentation and here is the basic idea Given grouphomomorphisms g G rarr N and k H rarr N that G lowastH is the coproduct comes down to showingthat the map

h G lowastH rarr N defined by h(g1h1 gnhn) = f(g1)k(h1) middot middot middot f(gn)k(hn)

and similarly on other possible words in GlowastH is a group homomorphism which follows essentiallyfrom the definition of the group operation on G lowastH indeed this group operation arises preciselyfrom this requirement

Now if we instead tried to use GtimesH with its standard inclusions G rarr GtimesH and H rarr GtimesHas the coproduct the issue is that the analogous map

h GtimesH rarr N defined by h((g h)) = f(g)k(h)

would NOT be a homomorphism because h((g1 h1)(g2 h2)) is not equal to h(g1 h1)h(g2 h2) bythe way in which the group operation is defined on the direct product G times H However if westrict our category to only consist of abelian groups so that the N rsquos we consider in the defini-tion of coproduct are also abelian then G times H does serve as a coproduct we can always rewritef(g1)k(g1) middot middot middot f(gn)k(gn) as f(g1) middot middot middot f(gn)k(h1) middot middot middot k(hn) in N Irsquoll leave it to you to flesh out allof these details

Concrete categories The categories Set Top Vect and Grp all have the property thattheir objects are simply sets possibly equipped with extra structure and that their morphisms areordinary set-theoretic functions possibly satisfying some additional requirement Such categoriesare nice since we can often use intuition about sets and functions in their study A concrete categoryis intuitively one where we can think of objects as being ldquosetsrdquo and morphisms as being ldquofunctionsrdquoonly that we have to interpret this in the right way

Here is the definition a concrete category is a category C equipped with a faithful functorF C rarr Set which is part of the data The idea is that such a functor gives us way to turnan object A isin Ob(C) into a set F (A) and a morphism f isin MorC(AB) into a function Ff isinMorSet(F (A) F (B)) where the ldquofaithfulrdquo requirement says that we do not lose information aboutf when doing so so that we can (essentially) learn everything we need to know about f by studyingFf instead In the four examples above the required faithful functor is simply the forgetful functorand indeed in general we think of F as being a ldquoforgetful functorrdquo for the concrete category (C F )

Examples Here are two more examples of concrete categories Let G be a group and considerthe category BG with a single object and elements of G morphisms As written this does notappear to be concrete since morphisms are not honest functions lowast rarr lowast but the point is that thiscategory can indeed by ldquoconcretizedrdquo by specifying an appropriate ldquoforgetful functorrdquo A functorF BG rarr Set is the data of a (left) action of G on a set S = F (lowast) and saying this functor isfaithful is what it means to say that this action is faithful different g h isin G act on S in differentways or equivalently that there exists s isin S such that gs ∕= hs Any group has at least onefaithful left actionmdashnamely the action on itself given by left multiplicationmdashto BG can always beequipped with a faithful functor to Set This is all just a way of phrasing the fact that any groupis isomorphic to a subgroup of a permutation group

The category Rel of relations can also be ldquoconcretizedrdquo even though as written morphismsA rarr B are not honest functions Take F Rel rarr Set to be the power set functor which sends an

14

object in Rel to its power set and a relation f A rarr B to the induced function Ff P (A) rarr P (B)defined by

(Ff)(S) = b isin B | there exists a isin S such that (a b) isin f

In other words if you think of the relation f sube A times B as a ldquopossibly not everywhere-definedpossibly multivalued functionrdquo (Ff)(S) is analogous to taking the image of S That F is faithfulwill be left to the homework

In fact most categories yoursquoll see can be made concrete by specifying an appropriate forgetfulfunctor but not all categories can be made concrete in this way Here is the standard example of anon-concrete category which we mention only to say there is such a thing but which we wonrsquot doanything more with since understanding it better requires knowledge of algebraic topology Thehomotopy category of topological spaces is the category hTop whose objects are topological spacesand whose morphisms are given by equivalence classes of homotopic maps

MorhTop(XY ) = homotopy classes of continuous maps X rarr Y

(Saying that two maps are homotopic means that you can ldquodeformrdquo one into the other but wersquollleave the precise definition to a course in algebraic topology) It turns out that no functor fromhTop to Set can be faithful so that hTop cannot be made concrete but this is actually a deepand difficult thing to show so we make no attempt to do so here

Natural Isomorphisms Representability

Natural transformations The definition of a natural transformation η F rArr G between two(both covariant or both contravariant) functors FG C rarr D is given in Section 14 of the bookThe key idea is that a natural transformation gives a way to turn data about F into data aboutG in a way which is compatible with all possible morphisms as expressed by the commutativity ofthe diagrams

F (A) F (B)

G(A) G(B)

Ff

ηA ηB

Gf

arising from morphisms f A rarr B (or f B rarr A in the contravariant case) in C In the case of anatural isomoprhism η F rArr G so that each ηA F (A) rarr G(A) is actually an isomorphism in Dthe point is not only that F and G produce isomorphic objects but they do so in a way which iscompatible with all possible category-theoretic data

The first standard example of a natural isomorphism is the one between the identity functor onthe category of finite-dimensional and the functor which sends a finite-dimensional vector space Vto V lowastlowast the dual of its dual induced by the isomorphism V rarr V lowastlowast defined by

v 983041rarr (the linear map on V lowast which sends f to f(v))

This was the topic of a student presentation and can be found in Section 14 of the book Theessential reason why this isomorphism is ldquonaturalrdquo is that V rarr V lowastlowast can be defined without referenceto a basis Contrast this with the the functor which sends V to its (single) dual V lowast it is still truethat finite-dimensional vector space is isomorphic to its dual but specifying such an isomorphism

15

requires additional data which prevents the identity functor from being ldquonaturally isomorphicrdquo tothe single dual functor

Some history At this point it makes sense to say a bit about the interesting origins of categorytheory in the 1950rsquos Eilenberg and Mac Lane where studying cohomology (or possible homologyone of those two) in algebraic topology which associates algebraic data (a group ring or similar)to a topological space in a way which encodes some of the topology They had two cohomologicalconstructions which ended up proving isomorphic results but they noticed that not only werethe resulting objects the same but the actual constructions themselves were in some sense theldquosamerdquo They thus then needed a way to formalize this notion and invented the concept of anatural transformation to do so where they interpreted two constructions as being the same asbeing compatible with all morphisms

But this then leads to the question what types of objects should a natural transformationoccur between Eilenberg and Mac Lane then had to invent the notion of a ldquofunctorrdquo to describewhat a natural transformation went from and what it went to In their motivating setting theyinvented the notion of a functor to describe more formally properties which their cohomologicalconstructions were supposed to satisfy But this then leads to the question what types of objectsshould a functor map between In other words what type of data should a functor (cohomologicalconstruction in their example) take as input and should it output They finally had to thus inventthe notion of a ldquocategoryrdquo to describe the source and target of a functor so that their cohomologicalconstructions could be interpreted as functors from a category of topological spaces to a categoryof algebraic objects

Thus historically natural transformations came first then functors and then categories Aswith most things in mathematics categories thus arose from attempts to encode what was beingseen in some concrete examples in a more general and formal setting

Representable functors Representable functors are defined in Section 21 of the book but wersquollgive the required definitions here in a hopefully simpler to digest manner The point is that anyobject A in a (locally small) category C gives two ldquonaturalrdquo functors C rarr Set one covariant andthe other contravariant which as wersquoll see pretty much encode all data about A itself We definehA C rarr Set to be the covariant functor defined by

hA(B) = Mor(AB) on objects and

hA(Bfminusrarr C) = the map Mor(AB)

hAfminusminusrarr Mor(AC) defined by g 983041rarr f g

We define hA C rarr Set to the contravariant functor defined by

hA(B) = Mor(BA) on objects and

hA(Bfminusrarr C) = the map Mor(CA)

hAfminusminusrarr Mor(BA) defined by g 983041rarr g f

(The functor hA is called the functor of points of A and wersquoll usually think of it as being a covariantfunctor hA Cop rarr Set The functor hA does not have a standard name) So hA is defined bymorphisms from A and hA is defined by morphisms into A Wersquoll see later the sense in which suchfunctors encode all information about the object A

A functor F C rarr Set is representable if is naturally isomorphic to such a functor so F sim= hA

for some A in the covariant case and F sim= hA for some A in the contravariant case The idea wersquollbuild towards is that we can essentially treat such a functor as being A itself so that representable

16

functors can be thought as being actual objects in C For now here is a key example wersquove seenbefore which we can now formulate using the language of representability

Given AB isin Ob(C) let F C rarr Set be the contravariant functor defined by

F (C) = Mor(CA)timesMor(CB) and F (Cfminusrarr D) = [(g k) 983041rarr (g f k f)]

where (g k) 983041rarr (g f k f) gives a map Mor(DA) timesMor(DB) rarr Mor(CA) timesMor(CB) ieF (D) rarr F (C) In order for this functor to be representable requires the existence of an object inC satisfying

Mor(CA)timesMor(CB) sim= Mor(C representing object)

in a way which is compatible with all morphisms but we have actually already seen this notionelsewhere it is essentially the definition of a product AtimesB for A and B Indeed we claim that Fis naturally isomorphic to hAtimesB as we now show

First to define a natural isomorphism hAtimesBsim= F requires for each C isin Ob(C) a choice of a

bijection (ie isomorphism in Set)

ηC hAtimesB(C)sim=minusrarr F (C)

which in our case looks like

ηC Mor(CAtimesB))sim=minusrarr Mor(CA)timesMor(CB)

Take this to be the map

(Cfminusrarr AtimesB) 983041rarr (pA f pB f)

where pA A times B rarr A and pB A times B rarr B are the two projection morphisms in the definitionof a product The definition of product implies that this map is invertible since given k C rarr Aand ℓ C rarr B there exists a unique h C rarr A times B such that k = pA h and ℓ = pB h Forthese bijections to define a natural isomorphism hAtimesB

sim= F requires that they be compatible withall morphisms in C which means that given any g C rarr D in C the following diagram shouldcommute

hAtimesB(C) hAtimesB(D)

F (C) F (D)

hAtimesB(g)

ηC ηD

Fg

which in the case at hand looks like

Mor(CAtimesB) Mor(DAtimesB)

Mor(CA)timesMor(CB) Mor(DA)timesMor(DB)

minus g

ηC ηD

(minus gminus g)

Take an element k D rarr AtimesB in the upper right Applying the map on top gives the elementk g in the upper left and then applying ηC on the left side gives

(pA (k g) pB (k g))

17

in the lower left Instead starting with k D rarr AtimesB in the upper right applying ηD on the rightgives (pA k pB k) and applying the map on the bottom gives

((pA k) g (pB k) g)

in the lower left which is the same as the previous element in the lower left by the associativity ofcomposition Hence the isomorphisms ηC do define a natural isomorphism between hAtimesB and F Again the point is that you can then interpret the functor F as being the ldquosamerdquo as the productAtimesB since both encode the same data

More Representable Examples

Coproducts As in the case of products the definition of coproducts can also be phrased in termsof a representable functor Given objects A and B in C let F C rarr Set be the covariant functordefined by

F (C) = Mor(AC)timesMor(BC) and F (Cfminusrarr D) = [(g k) 983041rarr (f g f k)]

where the latter is a function Mor(AC)timesMor(BC) rarr Mor(AD)timesMor(BD) This functor isrepresentable if and only if a coproduct A ⊔ B for A and B exists and the representing object isthat coproduct the key is the requirement that we have bijections

Mor(AC)timesMor(BC) sim= Mor(A ⊔BC)

for all C which is essentially the defining property a coproduct for A and B should have

Power set functor contravariant version Let P Set rarr Set be the contravariant powerset functor defined on objects by sending a set to its power set and on morphisms by sendingf A rarr B to the preimage function Pf P (B) rarr P (A) given by S 983041rarr fminus1(S) In order for this tobe representable requires the existence of a set X with natural bijections

P (A) sim= Mor(AX)

for all sets A that is a set X so that mapping into X is the same data as specifying a subset ofthe domain We take X = 0 1 to be a two-element set and the required bijections

ηA P (A) rarr Mor(A 0 1) to be given by S 983041rarr χS

where XS A rarr 0 1 is the indicator or characteristic function S defined by sending elementsof S to 1 and elements of A minus S to 0 The inverse of ηA sends a function g A rarr 0 1 to thepreimage gminus1(1) sube A

To verify that the bijections ηA define the data of a natural isomorphism η P rArr h01 wecheck the commutativity of some diagrams Given a function f A rarr B consider

P (A) P (B)

h01(A) = Mor(A 0 1) h01(B) = Mor(B 0 1)

Pf = fminus1

ηA ηB

minus f

18

Take S isin P (B) in the upper right Applying the map on top gives fminus1(S) isin P (A) and thenapplying the map on the left gives χfminus1(S) isin Mor(A 0 1) On the other hand applying the mapon the right to S isin P (B) gives χS isin Mor(B 0 1) and then applying the map on the bottomgives χS f isin Mor(A 0 1) The two resulting functions χfminus1(S)χS f in the lower left are

χfminus1(S)(x) =

9830831 x isin fminus1(S)

0 otherwiseand χS(f(x)) =

9830831 f(x) isin S

0 otherwise

so these are the same since x isin fminus1(S) if and only if f(x) isin S Thus the diagram above commutesso P is naturally isomorphic to h01 and hence P is representable

Power set functor covariant version Now consider the covariant power set functor stilldenoted by P Set rarr Set which again sends a set to its power set but now sends a functionf A rarr B to the image function Pf P (A) rarr P (B) defined by S 983041rarr f(S) In order for this tobe representable there would have to exist a set X such that P sim= hX which translates into havingnatural bijections

P (S) sim= Mor(XS)

for all sets S However in this case there is no natural choice for X as there is no natural way todetect subsets of S by mapping into S as opposed to the case of P (S) sim= Mor(S 0 1)

That no such X exists can be made precise using a simple argument when S = empty P (S) hascardinality 1 soX would have to be empty as well since otherwise Mor(XS) would have cardinality0 but if X is empty then Mor(empty S) always cardinality 1 regardless of S which is clearly not trueof P (S) Thus the covariant power set functor is not representable

Forgetful on Top The forgetful functor F Top rarr Set is representable Indeed a representingobject would have to satisfy

X sim= Map(objectX)

as sets for any topological space X (where Map denotes the set of continuous maps) and it iseasy to see that a single point satisfies this property a continuous map f pt rarr X is completelydetermined by the element f(pt) of X and any such element gives a continuous map in this wayOf course checking that F sim= hpt requires checking that various diagrams commute but this issimple to work out in this case so we omit it here

Extracting a topology With an eye towards the idea that the functors hA hA should captureall information about an object A in a category note that the result above shows that in TophX first of all captures all information about X as a set since the set X can be extracted fromhX(pt) = Map(ptX)

We claim that hX actually captures all information about X as a topological space since thetopology on X can be extracted from hX in the following way As we saw in the power set examplewe have a bijection

P (X) sim= Mor(X 0 1)

Now equip 0 1 with the topology in which 1 is open but 0 is not Then a continuous mapf X rarr 0 1 is fully determined by the preimage fminus1(1) since to be continuous only requiresthat this single preimage be open in X Moreoever any open subset of X arises in this way bytaking f to be the indicator function of that open set Thus we get a bijection

Op(X) sim= Map(X 0 1)

19

where Op(X) denotes the set of open subsets of X ie the topology on X Hence the topology onX can be extracted from hX as claimed so hX and hX do encode all information about X

This fact can also be interpreted as saying that the contravariant functor which sends a topo-logical space X to its collection of open subsets (and which sends a continuous map to the mapinduced by taking preimages) is representable with representing object 0 1 equipped with thetopology given above This will be worked out in a homework problem

Forgetful on Grp The forgetful functor Grp rarr Set is also representable In this case arepresenting object should be a group giving set-theoretic bijections

G sim= Hom(objectG)

for all groups G where Hom denotes the set of group homomorphisms The group Z works

G sim= Hom(Z G)

since a homomorphism φ Z rarr G is completely determined by φ(1) since Z is generated by 1 andthere are no restrictions on what φ(1) isin G can actually be The inverse of the desired bijection isthus φ 983041rarr φ(1) and it is simple to check that morphisms behave appropriately so that the forgetfulfunctor in this setting is indeed isomoprhic to hZ

Forgetful on Ring The forgetful functor Ring rarr Set is also representable where by Ring wemean the category of rings with identity elements and where morphisms (ie ring homomorphisms)are required to preserve identities Here we need an object satisfying

R sim= Hom(object R)

for all R isin Ob(Ring) and we can see that Z[x] the ring of polynomials in the variable x withcoefficients in Z works Indeed given a ring homomorphism φ Z[x] rarr R the fact that φ isrequired to preserve identities fully determines φ(n) for any n isin Z (since φ(n) = nφ(1)) and hencesince φ preserves multiplication its behavior is fully determined by φ(x) which can be any elementof R The inverse of the bijection we want is thus φ 983041rarr φ(x) and it will follow that this forgetfulfunctor is isomorphic to hZ[x]

Equivalences between Categories

Definitions iso and equiv Now that we have a notion of a ldquomorphismrdquo between categorieswe can talk about the sense in which two categories are the ldquosamerdquo The first possible definitionis the ldquoobviousrdquo one you might expect given the definition of ldquoisomorphismrdquo wersquove seen in everyother context until now we say that C is isomorphic to D if there exist functors F C rarr D andG D rarr C such that G F = idC and F G = idD where idC and idD are identity functors

However this definition turns out to be too restrictive Indeed the requirement that the givencompositions equal identity functors says that given an object A in C the object G(F (A)) beliterally the same as A itself and similarly for objects in D But we know that two objects in acategory can be thought of as being the ldquosamerdquo without being literally the same as long as theyare isomorphic For instance productscoproducts are not unique in the literal sense but only inthe sense that they are isomorphic in a unique way To take another example the definition ofa functor being representable does not say that F (B) literally be the same as hB(A) only thatthey be isomorphic for sure the power set P (S) of a set S is not literally the same as the set offunctions S rarr 0 1mdashit is only the ldquosamerdquo in the sense that they encode the same data

20

Thus it makes sense to relax the requirement that A andG(F (A)) in the definition of isomorphiccategories be equal to the requirement that they be isomorphic This leads to the following betterdefinition of two categories being the ldquosamerdquo C is equivalent to D if there exist functors F C rarr D

and G D rarr C such that G F sim= idC and F G sim= idD So we require only that for instance thefunctor G F be naturally isomorphic to the identity functor on C which says that for any objectA G(F (A)) be isomorphic to A in a natural way Wersquoll see that this truly is the better notionof what it means for two categories to be the ldquosamerdquo and it reflects the idea that all categoricalproperties of one category can be transported to the other

Alternate characterization of equivalences A functor F C rarr D implementing an equiv-alence between categories is called unsurprisingly and equivalence and the ldquoinverserdquo functorG D rarr C is still referred to as being its inverse even though it is not technically an ldquoinverserdquoin the sense that compositions give literal identities But as in the case of Set where ldquoinvertiblerdquocan be characterized without reference to an inverse as ldquoinjectiverdquo and ldquosurjectiverdquo so too canequivalences between categories be characterized without explicit reference to an inverse

A functor F C rarr D is an equivalence if and only it is full faithful and essentiallysurjective

We have seen the notions of full and faithful before and essentially surjective means that any objectin D is isomorphic to something in the image of F for all D isin Ob(D) there exits C isin Ob(C) suchthat F (C) sim= D Justifying this result was the topic of a student presentation and can be foundin the book in Section 15 (Technically to construct an ldquoinverserdquo of F requires that we work withsmall categories but whatever) The bulk of the real work goes into showing that the ldquoinverserdquofunctor constructed is actually a functor which comes down to some abstract nonsense argumentvia commutative diagrams

Equivalences preserve everything We previously saw the notions of full faithful and es-sentially surjective back when thinking about what properties a functor could have which wouldguarantee it sent products to products so the result we proved back then was that equivalencespreserve products Similarly equivalences preserve coproducts and the proof is basically the sameafter reversing arrows A clean way of deriving this is as follows if F C rarr D is an equivalencethen so is F op Cop rarr Dop so since F op preserves products F preserves coproducts becausecoproducts in a category become products in the opposite category

In general an equivalence between categories will preserve whatever categorical notions youwant monomorphisms epimorphisms initial objects terminal objects etc so that equivalentcategories really do have essentially the ldquosamerdquo properties

Equivalence example Here is a first example (also found in the book) of equivalent categorieswhich are not isomorphic Let FVectR be the category whose objects are finite-dimensional realvector spaces and whose morphisms are linear transformations Let MatR be the category ofmatrices defined as follows

bull objects in MatR are nonnegative integers

bull a morphism n rarr m is an mtimes n matrix with real entries

bull composition in MatR is given by matrix multiplication given B n rarr m and A m rarr kmdashsoB is an mtimes n matrix and A is a k timesm matrixmdashthe composition A B n rarr k is the k times nmatrix AB

bull identity morphisms in MatR are given by identity matrices

21

We claim that FVectR and MatR are equivalent First note that they certainly cannot be isomo-prhic isomorphisms in MatR exist only between two objects which are literally the same (ie anmtimesn matrix can be invertible only when m = n) but isomorphisms in FVectn can exist betweenobjects which are not literally the same So there are in a sense not enough objects in MatR toallow it to be isomorphic to FVectR

To construct an equivalence choose one and for all an isomorphism TV V rarr RdimV forevery object in FVectR or equivalently choose once and for all a basis for every V DefineF FVectR rarr MatR by

V 983041rarr dimV on objects and

(T V rarr W ) 983041rarr (the standard matrix of SW T Sminus1V RdimV rarr RdimW ) on morphisms

Equivalently F sends T to the matrix of T relative to the chosen bases for V and W It isstraightforward to check that this is a functor The inverse functor G MatR rarr VectR is givenby

n 983041rarr Rn on objects and

(mtimes n matrix A) 983041rarr (the linear transformation A Rn rarr Rm induced by A) on morphisms

One can check that G is a functor and that F G and G F are naturally isomorphic to identitiesor instead one can argue that F alone is full faithful and essentially surjective The point is thatG(F (V )) gives back RdimV so not literally V itself but only something isomorphic to V which iswhy this gives an equivalence of categories and not an isomorphism

Yoneda Lemma Functors as Objects

Compact Hausdorff spaces and Clowast-algebras Last time we gave an example of equivalentcategories which were not isomorphic but in many ways that example is unsatisfactory when doinglinear algebra people for sure often work with matrix representations of linear transformations asthe example from last would suggest but no one in their right mind literally thinks about suchthings in terms of the equivalence FVectR rarr MatR itself In other words this equivalence isessentially ldquocooked uprdquo for the sake of giving an example but it does not really give any insightinto the actual mathematics of linear algebra

So with that in mind we will describe another equivalence which actually does have some realmeaning in the sense of producing new ways of thinking about some mathematics Given a compactHausdorff space X we let C(X) denote the space of continuous complex-valued functions on X

C(X) = f X rarr C | f is continuous

We can equip the set C(X) with various additional structures First it is a complex vector spaceunder ordinary addition and scalar multiplication of functions Second we can multiply two func-tions together

(fg)(p) = f(p)g(p)

which turns C(X) into whatrsquos called an algebra over C (We wonrsquot give the definition of anldquoalgebrardquo but the point is that this multiplication is in some sense ldquocompatiblerdquo with the vectorspace structure) Next any element f of C(X) has a conjugate element f obtained by takingcomplex conjugates of the values of f

f(p) = f(p)

22

And finally we can equip C(X) with a norm via

983042f983042 = suppisinX

|f(p)|

where |middot| denotes complex absolute value (This supremum exists sinceX is compact by the ExtremeValue Theorem) All of these structures (vector space algebra multiplication conjugation norm)turn C(X) into whatrsquos called a Clowast-algebra We wonrsquot define this precisely but the definition encodesvarious ways in which these structures are compatible Actually C(X) is a unital commutative Clowast-algebra where ldquounitalrdquo means that the multiplication has an identity element (namely the constantfunction 1) and commutative means that fg = gf There is a natural notion of homomorphismbetween Clowast-algebras which are maps between them which behave nicely with all the structuresinvolved In particular given a continuous map f X rarr Y between compact Hausdorff spaces weget a Clowast-algebra homomorphism flowast C(Y ) rarr C(X) given by pre-composition with f flowast sendsg isin C(Y ) to g f isin C(X)

We thus have a contravariant functor CHaus rarr ucClowast-Alg from the category of compactHausdorff spaces and the category of unital commutative Clowast-algebras It turns out that anyunital commutative Clowast-algebra arises in this way in the sense that given any unital commutativeClowast-algebra B there exists a compact Hausdorff space X such that C(X) sim= B as Clowast-algebrasand that morphisms between these algebra arises from continuous maps in the manner describeabove This says that the functor above is essentially surjective full and it turns out that it isalso faithful so that it is an equivalence Thus all information about compact Hausdorff spaces iscompletely encoded in information about unital commtuative Clowast-algebras and indeed one couldlearn everything there is to know about a compact Hausdorff space X from its Clowast-algebra C(X)of functions The fact that the categories CHaus and ucClowast-Alg are equivalent is the startingpoint in the subject known as noncommutative geometry which wersquoll say something about a bitlater on If we drop the requirement that our algebras be unital it turns out that the categoryof commutative Clowast-algebras in general is equivalent to the category of locally compact Hausdorffspaces via essentially the same functor where we take as morphisms in the category of locallycompact Hausdorff spaces to be continuous functions which ldquovanish at infinityrdquo

We are only introducing Clowast-algebras to give an example of an interesting and actually usefulequivalence between categories but just for the fun of it wersquoll note the following Examples ofnoncommutative Clowast-algebras come from what are known as ldquoalgebras of bounded operators onHilbert spacesrdquo which show up in quantum mechanics as the things which describe physicallyobservable quantities like position momentum and energy (In fact all Clowast-algebras arise fromHilbert spaces in this way) Thus Clowast-algebras show up in mathematical formulations of quantummechanics and the notion of noncommutative geometry mentioned above is at the core of someattempts to understand the elusive concept known as ldquoquantum geometryrdquo

Yoneda Lemma The Yoneda Lemma states that given a (covariant) functor F C rarr Set whereC is locally small for any A isin Ob(C) there exists a bijection

F (A) sim= Nat(hA F )

where the right side denotes the set of natural transformations from hA to F A similar result holdsin the contravariant case where we get bijections F (A) sim= Nat(hA F ) (There is also a sense inwhich these bijections are themselves natural but wersquoll ignore this bit) The proof of the YonedaLemma was the topic of a student presentation and can be found in Section 22 of the book

Here we will get a bit philosophical and talk about what the point of the Yoneda Lemmaactually is and later the sense in which it says that we can view arbitrary functors as ldquonon-existent

23

objectsrdquo in a category First we mention what we will call the ldquofunctor of pointsrdquo approach tomathematics a mathematical object can be fully characterized by all morphisms into it That isknowledge of Mor(SA) for all S is equivalent to knowledge of A itself Indeed this is an idea wementioned previously in relation to the functors hAmdashwhich we previously said is referred to as beingthe functor of points of Amdashand now we seek to push this idea further In some examples literalpoints of A can be extracted from hA as in the case of the bijection X sim= hX(pt) = Map(ptX) fora topological space X or G sim= hG(Z) = Mor(Z G) for a group G Moreover in these examples therest of the (topological) structure onX or (group) structure on G can also be extracted from hX andhG respectively In general we can view Mor(SA) as in a sense being ldquopoints of A parameterizedby Srdquo which comes from the idea that in say Set Mor(SA) = AS is ldquothe product of S-manycopies of Ardquo so literally points of A ldquoparameterizedrdquo by S In various areas of mathematics anelement of Mor(SA) is thought of as being a ldquogeneralizedrdquo point of A so that the functor of pointshA detects generalized points of A

So an object A is fully encoded by the functor hA Now we claim that morphisms betweenobjects are also encoded by such functors Indeed in the case where F = hB is itself a functor ofpoints the Yoneda result Nat(hA F ) sim= F (A) becomes

Nat(hA hB) sim= hB(A) = Mor(AB)

which says precisely that morphisms between the objects A and B can be obtained solely fromthe functors hA and hB (This fact is one reason why one should prefer the contravariant functorhA to the covariant functor hA in the covariant case Nat(hA F ) sim= F (A) for F = hB becomesNat(hA hB) sim= hB(A) = Mor(BA) so that morphisms between two objects get reversed whencharacterizing them in terms of representable functors)

Pushing our philosophy further we now know that not only objects but also morphisms betweenobjects are fully encoded by the functors hA The final step is to note that we should not only careabout objects and morphisms between them but that an important part of the data of an objectA is also how arbitrary contravariant functors F C rarr Set behave on A and that this too can beobtained from hA precisely via the Yoneda result that F (A) sim= Nat(hA F ) So objects morphismsbetween objects and how arbitrary functors behave on objects are all encoded in knowledge ofthe functors hA so that these functors truly do capture in the most general sense possible allinformation about A

Functors as nonexistent objects The discussion above is all well and good but it still doesnrsquotquite say why we should care about the fact that we can study on object A by studying hA insteadin that we still havenrsquot spoken about what type of insight about A can be gleamed from hA thatwasnrsquot readily apparent in A itself already Now we argue that the true point of the YonedaLemma is that it gives a way to think about arbitrary functors F Cop rarr Set as ldquoobjectsrdquo in C

itself or really as objects in an ldquoenlargementrdquo of C Certainly for those functors F sim= hA which arerepresentable we can indeed think about F as simply being the object A but now we claim thateven non-representable F can be thought of as being ldquononexistentrdquo objects in C The upshot is thatoften in mathematics we come across things which do not actually exist but the Yoneda Lemmagives a way to (indirectly) study such things anyway by interpreting them as functors instead

The starting point of course is that an actual object A is the same as hA as we have explainedabove Thus by analogy we can interpret a non-representable functor F as being an ldquoobjectrdquo in C

as long as we know what it means to ldquomaprdquo into F from other things which are honestly objects in CThat is if F were indeed representable then F (A) sim= Nat(hA F ) sim= Mor(AF ) would characterizethe functor of points of F (here we abuse notation by using the same letter F to denote both arepresentable functor and the object which represents it) so now we take Mor(AF ) = F (A) as

24

the definition of the ldquofunctor of pointsrdquo of the non-existent object F We are thus taking elementsof F (A) as being what it means to ldquomaprdquo from A into F even when F is not representable Thefact that the Yoneda Lemma gives a way to define maps from A into F thus says that F itself isequipped with the same data that a literal object of A would be equipped with so why shouldnrsquotwe interpret F as a being a type of ldquoobjectrdquo in C as well

To phrase this another way note that the Yoneda Lemma gives rise to a functor

C rarr Fun(CopSet)

where Fun(CopSet) denotes the category whose objects are functors from Cop to Set and morphismsare natural transformation defined on objects by A 983041rarr hA and on morphisms by the bijectionMor(AB) sim= Nat(hA hB) arising from the Yoneda Lemma That this is actually bijection saysthat this functor is full and faithful so that we can view C as being a subcategory of the functorcategory Fun(CopSet) (We cay say that C is isomorphic to the subcategory of Fun(CopSet)consisting of the functors hA or that it is equivalent to the subcategory of Fun(CopSet) consistingof representable functors This functor C rarr Fun(CopSet) is usually called the Yoneda embeddingof C into Fun(CopSet)) Thus we can view Fun(CopSet) as an ldquoenlargmentrdquo of C and the ideawe are trying to make sense of is that other objects in this ldquolargerrdquo category should be interpretedas ldquonon-existent objectsrdquo of C

Just how far can we actually push this idea that arbitrary functors Cop rarr Set should be thoughtof as being ldquoobjectsrdquo of C Pretty far in fact For instance often times in geometry one comesacross a type of ldquosmoothrdquo space on which some group acts and one is interested in doing ldquocalculusrdquoon the resulting quotient space which is the space of orbits under the given group action Theproblem is that such quotient spaces are very often (usually in fact) not ldquosmoothrdquo themselves sothat the tools from calculus are not readily available The solution is to nonetheless treat thisquotient as if it were ldquosmoothrdquo and attempt to do calculus anyway To make this precise thisldquonon-smoothrdquo space actually turns out to have a natural interpretation as a ldquosmooth functorrdquo andit is through this functor which we can view this non-smooth space as actually being smooth afterall so we are forced to expand our notion of what we mean by ldquosmooth spacerdquo by allowing thingswhich are not spaces in the usual sense which certainly leads to more abstraction but the benefitis that it actually makes sense to do calculus on such ldquospacesrdquo interpreted as functors

The discussion above is pretty vague so here is one final more concrete example Recall thatequivalence

CHausop rarr ucClowast-Alg

between the category of compact Hausdorff spaces and the category of unital commutative Clowast-algebras mentioned at the start We now claim that a non-commutative (unital) Clowast-algebra shouldbe interpreted as the algebra of functions on some non-existent ldquononcommutative compact Haus-dorff spacerdquo and it is this idea which the subject of noncommutative geometry makes precise Solet B be a non-commutative Clowast-algebra Then the functor of points hB gives a functor

ucClowast-Alg rarr Set

which sends C(X) (where X is some compact Hausdorff space) to Hom(C(X) B) where Hom de-notes homomorphisms of Clowast-algebras whatever that means Composing with the previous functorwe get a functor wersquoll denote by B itself

B CHausop rarr Set

which on objects at least is given by X 983041rarr Hom(C(X) B) This functor is not representable sinceB is not commutative but the philosophy espoused above suggests we should think of it as giving a

25

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

as we define appropriate ldquoprojectionsrdquo N rarr G and N rarr H take N rarr G to be the compositionpG ℓ and N rarr H to be pH ℓ Given f K rarr G and g K rarr H as in the definition of a productthe map hprime K rarr N defined by

hprime = ℓminus1 h

where h K rarr GtimesH is the usual h = f timesg will satisfy the requirement needed to able to concludethat N is also a product of G and H in Grp A similar trick works in any category so that anyobject isomorphic a product can also itself be turned into a product

The correct answer is that products are unique up to unique isomorphism which means thatgiven a product P of A and B any other product P prime of A and B will in fact be isomorphic to P and in a unique way in the sense that there can exist only one isomorphism P rarr P prime The phraseldquounique up to unique isomorphismrdquo is a common one wersquoll see again and again objects definedby some categorical property are almost never genuinely unique in a strict sense but they willessentially be as close to unique as possible

So suppose that P and P prime are both products of A and B with associated projection morphismspA pB for P and pprimeA p

primeB for P prime Consider the diagram

P

A B

P primepA pB

pprimeA pprimeB

existh

The morphism h P rarr P prime is the unique one guaranteed to exist by the fact that P prime is a productof A and B Commutativity of this diagram says that pA = pprimeA h and pB = pprimeB h A similardiagram with the roles of P and P prime reversed which uses the fact that P is a product results in aunique morphism hprime P prime rarr P satisfying pprimeA = pA hprime and pprimeB = pB hprime

Now consider the following diagram

P

A B

PpA pB

pA pB

hprime h

This commutes which we verify using the properties h and hprime satisfy as follows

pA (hprime h) = (pA hprime) h = pprimeA h = pA

and similar for the morphisms involving B This shows that hprimeh P rarr P satisfies the requirementin the definition of P being a product But of course idP P rarr P satisfies this requirement as wellso since the map satisfying this requirement is assumed to be unique we must have hprime h = idP Similar reasoning considering h hprime P prime rarr P prime shows that h hprime = idprimeP so that h and hprime are indeedinverses and hence that P and P prime are isomorphic The uniqueness of the isomorphism betweenthem comes from the uniqueness of h in the construction above

7

Abstract nonsense It is common to say in the setting above that the fact that products areunique up to unique isomorphism follows from ldquoabstract nonsenserdquo which is a succinct way ofsaying that it follows solely from the definition of a categorical product itself and not any deepermathematical content In particular many of the ldquouniquenessrdquo properties which ldquoproductsrdquo havein the settings we might care aboutmdashCartesian products in Set direct products in Grp producttopologies in Topmdashreally have nothing to do with how those specific products are defined but arejust consequences of the ldquoabstract nonsenserdquo of category theory

But of course we do not use the phrase ldquoabstract nonsenserdquo in a negative or dismissive waysince recognizing what types of properties are consequences of ldquoabstract nonsenserdquo goes a long waytowards understanding well the given problem at hand Soon you too will come to appreciate thisnotion of abstract nonsense

Coproducts Opposite Categories

Initial and terminal objects The notions of initial and terminal objects in a category aredefined in Section 16 of the book where you can also find standard examples Here I want topoint out that initial and terminal objects if they exist are unique up to unique isomorphism asanother example of arguing via ldquoabstract nonsenserdquo

Suppose I and I prime are both initial objects in some category By the fact that I is initial thereexists a unique morphism f I rarr I prime and by the fact that I prime is initial there exists a unique morphismg I prime rarr I But then the composition g f I rarr I must equal idI I rarr I since the latter is theonly morphism from I to itself because I is initial and similarly f g = idIprime Thus I and I prime areunique and there is a unique isomorphism between them

Coproducts The dual concept to that of a product is the notion of a coproduct (The termldquodualrdquo refers to the phenomena you get when you reverse all arrows in a given construction Forinstance the notion of a terminal object is dual to that of an initial object and the notion of anepimorphism is dual to that of a monomorphism) Here is the precise definition a coproduct of anobject A and an object B in a category is an object C together with morphisms iA A rarr C andiB B rarr C such that for any object D and morphisms f A rarr D and g B rarr D there exists aunique morphism h C rarr D which makes the following diagram commute

D

A B

Cf g

iA iB

existh

By abstract nonsense if a coproduct of A and B exists it is unique up to unique isomorphism

Examples In Set the coproduct of two sets A and B is the disjoint union A⊔B (If you havenrsquotseen the notion of a disjoint union before it is essentially the same as the union only that we donrsquotcare about whether A and B have any overlap if they do we treat the common elements as beingldquodifferentrdquo For instance Z ⊔ Z is not Z but consists of two copies of each integer one 0 comesfrom the first copy of Z and another comes from the second but these are different ldquozeroesrdquo)Given f A rarr D and g rarr D the unique map A ⊔B rarr D is defined by applying f to elements of

8

A and g to elements of B (Note that if we simply took the ordinary union AcupB as the candidatefor the coproduct when A and B were not disjoint the required map h A cup B rarr D would notexist since h would have to send any element x isin A cap B to both f(x) and g(x) in order to makethe required diagram commutative)

The coproduct of two objects X and Y in Top is again the disjoint union X ⊔ Y equippedwith the disjoint union topology an open subset of X ⊔ Y is defined to be the (disjoint) union ofan open subset of X with an open subset of Y The coproduct of two groups G and H in Grp isthe free product G lowastH which is further elaborated on in the homework

The coproduct of two vector spaces V and W in Vect is the direct sum V oplusW which is definedto be the set of formal sums

v + w where v isin V and w isin W

where addition and scalar multiplication are defined ldquocomponentwiserdquo

(v1 + w1) + (v2 + w2) = (v1 + v2) + (w1 + w2) and r(v + w) = rv + rw

This is isomorphic to the product V times W with operations also defined componentwise but it iscustomary to speak about the ldquodirect sumrdquo V oplusW when referring to the coproduct as opposed tothe product wersquoll say why in a bit The maps iV V rarr V oplusW and iW W rarr V oplusW required aspart of the data of a coproduct are

iV v 983041rarr v + 0W and iW w 983041rarr 0V + w

To verify that V oplusW is the correct coproduct take any linear transformations S V rarr U andT W rarr U where U is some other vector space We then need a (unique) linear transformationh V oplusW rarr U such that

S = h iV and T = h iW

Let us determine what this map must be thereby not only showing that it exists but also that itis unique Let v + w isin V oplusW which we can write as

v + w = (v + 0W ) + (0V + w)

Since h is required to be linear we have

h(v + w) = h([v + 0W ] + [0V + w]) = h(v + 0W ) + h(0V + w)

The first term at the end is (h iV )v which must thus be Sv and the second term is (h iW )wwhich must be Tw Thus h = S + T is the map

h = S + T v + w 983041rarr Sv + Tw

which you can verify is indeed linear Thus V oplusW does satisfy the property required of a coproductSo the product of two objects in Vect is the same as their coproduct only that we write the

latter using ldquosumrdquo notation Later we will discuss products and coproducts of an arbitrary numberof elements We will see these as special cases of the general notions of limits and colimits butthey are defined via similar requirements as products and coproducts of two objects at a timeIn the category of vector spaces arbitrary products are given by the usual product

983124α Vα with

ldquocomponentwiserdquo addition and scalar multiplication but it turns out that an arbitrary coproductis given by the direct sum

983119α Vα which is defined to be the subspace of

983124α Vα where only finitely

many components are nonzero (This is in a way analogous to how the product topology on an

9

arbitrary number of topological spaces is defined) Wersquoll go over this later but is the underlyingreason why we use oplus instead of times when discussing coproducts

Encoding data via productscoproducts We can nicely summarize the definition of productsand coproducts in the following way the product of A and B is the object with the property thatfor any object D we have

Mor(DA)timesMor(DB) = Mor(D product)

and the coproduct of A and B is the object such that for any object D we have

Mor(AD)timesMor(BD) = Mor(coproductx D)

The first reflects the fact that products give a unique way to turn two morphisms (f and g in thedefinition) mapping into A and B into a single morphism (h in the definition) and the second saysthat coproducts give a unique way to turn two morphisms mapping out of A and B into a singlemorphism

Soon enough we will discuss the idea of universality and the point is that products are ldquouni-versalrdquo for maps into A and B and coproducts are universal for maps from A and B

Opposite categories Finally we give a definition which might seem silly but is actually quiteuseful Given a category C the opposite category is the category Cop is the category with the sameobjects as C but where we defined Mor(AB) in Cop to be Mor(BA) in C To be more concrete weconsider each morphism f A rarr B in C to instead be a morphism fop B rarr A in Cop (Perhaps itwould be better to write this as A larr B fop) The point is that we donrsquot actually care about whatf actually is as say a functioncontinuous maphomomorphismwhatever-we only care about itsbehavior as an ldquoarrowrdquo The opposite category Cop only retains information about how arrowsrelate to one another nothing more

Since arrows are reversed products in C become coproducts in Cop and coproduts in C becomecoproducts in Cop Similarly initialterminal objects and monomorphismsepimorphisms in C be-come terminalinitial objects and epimorphismsmonomorphisms in Cop Again the notion of anopposite category might seem like a strange thing to look at but wersquoll see that it does have uses

Functors Fullness and Faithfulness

One more productcoproduct example Let X be a topological space and define its categoryof open sets Op(X) as follows The objects of Op(X) are simply the open subsets of X and themorphisms are given by inclusions so to be concrete

Mor(U V ) =

983083the unique inclusion U rarr V if U sube V

empty otherwise

The fact that U sube V and V sube W implies U sube W says that the composition of two morphismsin this category is still a morphism in this category The product of U and V in this categoryturns out to be their intersection (with ldquoprojection morphismsrdquo U cap V rarr U and U cap V rarr V beinginclusions) and the coproduct of U and V turns out to be their (ordinary) union Note that weneed X to be a topological space in order to guarantee that U cap V and U cup V are still objects inthis category More generally the product of finitely many objects in Op(X) exists and is theirintersection and the coproduct of an arbitrary number of objects exists and is their union (Wesay that Op(X) has finite products and arbitrary coproducts)

10

Given a set X and a collection T of subsets of X we can define a similar category with elementsof T as objects (I guess we should assume that empty X isin T) and inclusions as morphisms We thenhave that this category has finite products and arbitrary coproducts if and only if T is a topologyon X so that we can rephrase the definition of a topological space itself solely in categorical termsThis perspective on what a ldquotopologyrdquo is makes it simpler to state the definition of whatrsquos calleda sheaf on a topological space which is an important construction in various areas of analysis andgeometry We might discuss this a bit later on

Functors As with most things in mathematics we should care not only about categories asmathematical structures on their own but also about ldquomapsrdquo between these structures whichldquopreserverdquo said structure The notion of a (covariant) functor between categories provides suchldquomapsrdquo in the case at hand where a functor is a way of sending objects to objects and morphismto morphisms in a way which preserves compositions and identities The precise definition is inSection 13 of the book as is the definition of a contravariant functor between categories whichis a functor which reverses the direction of arrows as opposed to ldquocovariantrdquo ones which preservedirections (It is common to use the term ldquofunctorrdquo without qualification to mean one which iscovariant Note that any functor can be made covariant by working with the opposite category acontravariant functor C rarr D is the same as a covariant functor Cop rarr D)

All of the standard examples we looked at are in the book forgetful functors power set functorsthe dual space functor in Vect etc Of particular interest is Example 139 in the book whichshows that various types of group actions in different contexts-the usual notion of a group actionon a set the notion of a continuous action of a group on a topological space and the notion ofa linear action of a group on a vector space (also known as a representation of the group)-are allencoded by a functor with domain category BG

Adjoint functors Let F Top rarr Set be the forgetful functor and let D Set rarr Top be thefunctor which sends a set S to the space S equipped with the discrete topology and a functionS rarr Sprime between sets to the same function only now viewed as a continuous map S rarr Sprime betweendiscrete spaces These two functors satisfy the following property in relation to one another givinga morphism D(S) rarr Y in Top is the same as giving a morphism S rarr F (Y ) in Set or moresymbolically

MorTop(D(S) Y ) = MorSet(S F (Y ))

We say that D is left adjoint to F and that F is right adjoint to D The point is that adjointfunctors can be used to move data back and forth between two categories

Wersquoll talk more about adjoint functors later on but for now here is one more example Nowtake F Grp rarr Set to again be the forgetful functor and take G Grp rarr Set to be the freegroup functor which sends a set S to the free group generated by S G(S) has as elements ldquowordsrdquos1 sn made up of elements of S with the group operation being ldquoconcatenationrdquo A basic factis that giving a group homomorphism from G(S) to some other group H is the same as giving anordinary function from S to F (H) which is the required adjoint property

MorGrp(G(S) H) = MorSet(S F (H))

Faithful and full functors The definitions of what it means for a functor to be faithful and fullare in Section 15 of the book where faithful functors are ones which are injective on morphisms andfull functors are ones which are surjective on morphisms but where in both cases we refer only tomorphisms which occur between the objects F (A) and F (B) in the target category corresponding tofixed objects A and B in the source category (In general a functor could send two objects AB in

11

the source category to the same object F (A) = F (B) in the target category and thus there could betwo morphisms A rarr C and B rarr C which are sent to the same morphism F (A) = F (B) rarr F (C)Such a thing could still possibly be ldquofaithfulrdquo since the non-injectivity here is arising from differentobjects in the source)

Faithfullness and fullness will be nice properties going forward and here is one hint at how theymight be useful Say we have a functor F C rarr D The problem is to determine properties onF which will imply it send products in C to products in D or coproducts in C to coproducts inD So given the former product diagram below we want the second diagram to also be a productdiagram

C

A B

P Fminusminusminusrarr

F (C)

F (A) F (B)

F (P )f g

pA pB

existh

Ff Fg

F (pA) F (pB)

Fh

Of course in order F (P ) to be an actual product of F (A) and F (B) on the right a similarcommutative diagram property would have to hold for all objects in D in place of F (C) and allmorphisms D rarr F (A) and D rarr F (B) not just those of the form Ff and Fg respectively Oneway to guarantee this is to require that all objects in D are of the form F (C) and then for allmorphisms F (C) rarr F (A) and F (C) rarr F (B) to be of the form Ff Fg the former is true whenthe functor F is surjective on objects and the latter is true when F is full We are not saying thatthese are the only conditions under which F might preserve products but it definitely seems togive a sufficient condition at least

Almost there is one more thing to require the uniqueness of the morphism Fh F (C) rarr F (P )making the diagram on the right commute Suppose there were two such morphisms Fh and Fhprimewhere h and hprime are both morphisms C rarr P in C The commutativity of the diagram on the rightgives for instance

Ff = F (pA)Fh and Ff = F (pA)Fhprime

which the functorial properites of F turns into

Ff = F (pA h) and Ff = (pA hprime)

We want h and hprime to make the first diagram commute in order to be able to apply uniqueness ofthe morphism C rarr P But this requires know that f = pA h and f = pA hprime which would followfrom F being faithful In this case it follows that h and hprime both make the first diagram commute(after we go through a similar argument with g) so the fact that P is a product guarantees thath = hprime and hence that Fh = Fhprime so that F (P ) is an actual product on the right

So we conclude that a functor which is full faithful and surjective on objects will send productsto products (Actually we can relax this a bit by requiring not that every object in D be in theliteral image of F but only that it be isomorphic to something in the image A functor with thisproperty is said to be essentially surjective which is a concept will see in a related context soonenough) Again this is not an ldquoif and only ifrdquo statement since functors can preserve products (orcoproducts) without being full faithful and surjective on objects but it at least gives a sufficientcondition Later we will see other ways to guarantee that functors preserve products which arerelated to the existence of adjoints for that given functor

12

Coproduct Examples Concreteness

Products and coproducts for metric spaces The product of two objects in Met the categoryof metric spaces with morphisms given by metric maps is the Cartesian product X times Y equippedwith the box metric and the coproduct of X and Y does not exist if both X and Y are nonemptyThis was the topic of a student presentation and here are some proof sketches

Given metric spaces (X dX) and (Y dY ) the box metric d on X times Y is defined by

d((x1 y1) (x2 y2)) = maxdX(x1 x2) dY (y1 y2))

(The name ldquoboxrdquo metric comes from the fact that in RtimesR = R2 or R3 open balls indeed looks likeldquoboxesrdquo) The key requirement needed in the definition of a product is given morphisms S rarr Xand S rarr Y that the map

h S rarr X times Y defined by h(s) = (f(s) g(s))

actually be a morphism in Met This turns into the requirement that

d((f(s) g(s)) ((f(t) g(t))) = maxdX(f(s) f(t)) dY (g(s) g(t)) le dS(s t) for all s t isin S

given the fact that

dX(f(s) f(t)) le dS(s t) and dY (g(s) g(t)) le dS(s t) for all s t isin S

For the latter inequalities to imply the former d must (at least up to isometry) be given by thebox metric Note that the two other ldquostandardrdquo metrics one might put on a productmdashthe taxicab

dX + dY and Euclidean983156

d2X + d2Y metricsmdashdo not satisfy this required ldquometric maprdquo property

and so do not give the product in MetTo see that coproducts do not exist if XY are nonempty suppose instead (P dP ) was a co-

product for X and Y with inclusion morphisms iX X rarr P and iY Y rarr P For any n isin Ntake 0 n to be a metric space with the absolute value metric d and let f X rarr 0 n andg Y rarr 0 n be the constant function 0 and the constant function n respectively Since P is acoproduct there exists h P rarr 0 n such that

h iX = f and h iY = g

But in order for h to actually be a morphism in Met we must then have

d(h(iX(x) iY (y)) le dP (iX(x) iY (y)) for any x isin X y isin Y

which translates inton = |0minus n| le dP (iX(x) iY (y))

But this should then be true for any n isin N which would imply that the distance in P betweenan element coming from X and one coming from Y should be infinite which is nonsense (Notethat if you altered your definition of ldquometricrdquo to allow for infinite distance then a coproduct wouldexist the disjoint union X ⊔Y where you defined the distance between elements of X and Y to beinfinite)

Coproducts for groups The coproduct of two groups G and H in Grp is the free product GlowastHwhich is defined to be the set of all words consisting of elements of G andH in an alternating fashion

13

with concatenation as the group operation (When we have to elements of G or two elements of Hnext to each other with use their respective group operations to turn these into single elements ofG or H) This was also the topic of a student presentation and here is the basic idea Given grouphomomorphisms g G rarr N and k H rarr N that G lowastH is the coproduct comes down to showingthat the map

h G lowastH rarr N defined by h(g1h1 gnhn) = f(g1)k(h1) middot middot middot f(gn)k(hn)

and similarly on other possible words in GlowastH is a group homomorphism which follows essentiallyfrom the definition of the group operation on G lowastH indeed this group operation arises preciselyfrom this requirement

Now if we instead tried to use GtimesH with its standard inclusions G rarr GtimesH and H rarr GtimesHas the coproduct the issue is that the analogous map

h GtimesH rarr N defined by h((g h)) = f(g)k(h)

would NOT be a homomorphism because h((g1 h1)(g2 h2)) is not equal to h(g1 h1)h(g2 h2) bythe way in which the group operation is defined on the direct product G times H However if westrict our category to only consist of abelian groups so that the N rsquos we consider in the defini-tion of coproduct are also abelian then G times H does serve as a coproduct we can always rewritef(g1)k(g1) middot middot middot f(gn)k(gn) as f(g1) middot middot middot f(gn)k(h1) middot middot middot k(hn) in N Irsquoll leave it to you to flesh out allof these details

Concrete categories The categories Set Top Vect and Grp all have the property thattheir objects are simply sets possibly equipped with extra structure and that their morphisms areordinary set-theoretic functions possibly satisfying some additional requirement Such categoriesare nice since we can often use intuition about sets and functions in their study A concrete categoryis intuitively one where we can think of objects as being ldquosetsrdquo and morphisms as being ldquofunctionsrdquoonly that we have to interpret this in the right way

Here is the definition a concrete category is a category C equipped with a faithful functorF C rarr Set which is part of the data The idea is that such a functor gives us way to turnan object A isin Ob(C) into a set F (A) and a morphism f isin MorC(AB) into a function Ff isinMorSet(F (A) F (B)) where the ldquofaithfulrdquo requirement says that we do not lose information aboutf when doing so so that we can (essentially) learn everything we need to know about f by studyingFf instead In the four examples above the required faithful functor is simply the forgetful functorand indeed in general we think of F as being a ldquoforgetful functorrdquo for the concrete category (C F )

Examples Here are two more examples of concrete categories Let G be a group and considerthe category BG with a single object and elements of G morphisms As written this does notappear to be concrete since morphisms are not honest functions lowast rarr lowast but the point is that thiscategory can indeed by ldquoconcretizedrdquo by specifying an appropriate ldquoforgetful functorrdquo A functorF BG rarr Set is the data of a (left) action of G on a set S = F (lowast) and saying this functor isfaithful is what it means to say that this action is faithful different g h isin G act on S in differentways or equivalently that there exists s isin S such that gs ∕= hs Any group has at least onefaithful left actionmdashnamely the action on itself given by left multiplicationmdashto BG can always beequipped with a faithful functor to Set This is all just a way of phrasing the fact that any groupis isomorphic to a subgroup of a permutation group

The category Rel of relations can also be ldquoconcretizedrdquo even though as written morphismsA rarr B are not honest functions Take F Rel rarr Set to be the power set functor which sends an

14

object in Rel to its power set and a relation f A rarr B to the induced function Ff P (A) rarr P (B)defined by

(Ff)(S) = b isin B | there exists a isin S such that (a b) isin f

In other words if you think of the relation f sube A times B as a ldquopossibly not everywhere-definedpossibly multivalued functionrdquo (Ff)(S) is analogous to taking the image of S That F is faithfulwill be left to the homework

In fact most categories yoursquoll see can be made concrete by specifying an appropriate forgetfulfunctor but not all categories can be made concrete in this way Here is the standard example of anon-concrete category which we mention only to say there is such a thing but which we wonrsquot doanything more with since understanding it better requires knowledge of algebraic topology Thehomotopy category of topological spaces is the category hTop whose objects are topological spacesand whose morphisms are given by equivalence classes of homotopic maps

MorhTop(XY ) = homotopy classes of continuous maps X rarr Y

(Saying that two maps are homotopic means that you can ldquodeformrdquo one into the other but wersquollleave the precise definition to a course in algebraic topology) It turns out that no functor fromhTop to Set can be faithful so that hTop cannot be made concrete but this is actually a deepand difficult thing to show so we make no attempt to do so here

Natural Isomorphisms Representability

Natural transformations The definition of a natural transformation η F rArr G between two(both covariant or both contravariant) functors FG C rarr D is given in Section 14 of the bookThe key idea is that a natural transformation gives a way to turn data about F into data aboutG in a way which is compatible with all possible morphisms as expressed by the commutativity ofthe diagrams

F (A) F (B)

G(A) G(B)

Ff

ηA ηB

Gf

arising from morphisms f A rarr B (or f B rarr A in the contravariant case) in C In the case of anatural isomoprhism η F rArr G so that each ηA F (A) rarr G(A) is actually an isomorphism in Dthe point is not only that F and G produce isomorphic objects but they do so in a way which iscompatible with all possible category-theoretic data

The first standard example of a natural isomorphism is the one between the identity functor onthe category of finite-dimensional and the functor which sends a finite-dimensional vector space Vto V lowastlowast the dual of its dual induced by the isomorphism V rarr V lowastlowast defined by

v 983041rarr (the linear map on V lowast which sends f to f(v))

This was the topic of a student presentation and can be found in Section 14 of the book Theessential reason why this isomorphism is ldquonaturalrdquo is that V rarr V lowastlowast can be defined without referenceto a basis Contrast this with the the functor which sends V to its (single) dual V lowast it is still truethat finite-dimensional vector space is isomorphic to its dual but specifying such an isomorphism

15

requires additional data which prevents the identity functor from being ldquonaturally isomorphicrdquo tothe single dual functor

Some history At this point it makes sense to say a bit about the interesting origins of categorytheory in the 1950rsquos Eilenberg and Mac Lane where studying cohomology (or possible homologyone of those two) in algebraic topology which associates algebraic data (a group ring or similar)to a topological space in a way which encodes some of the topology They had two cohomologicalconstructions which ended up proving isomorphic results but they noticed that not only werethe resulting objects the same but the actual constructions themselves were in some sense theldquosamerdquo They thus then needed a way to formalize this notion and invented the concept of anatural transformation to do so where they interpreted two constructions as being the same asbeing compatible with all morphisms

But this then leads to the question what types of objects should a natural transformationoccur between Eilenberg and Mac Lane then had to invent the notion of a ldquofunctorrdquo to describewhat a natural transformation went from and what it went to In their motivating setting theyinvented the notion of a functor to describe more formally properties which their cohomologicalconstructions were supposed to satisfy But this then leads to the question what types of objectsshould a functor map between In other words what type of data should a functor (cohomologicalconstruction in their example) take as input and should it output They finally had to thus inventthe notion of a ldquocategoryrdquo to describe the source and target of a functor so that their cohomologicalconstructions could be interpreted as functors from a category of topological spaces to a categoryof algebraic objects

Thus historically natural transformations came first then functors and then categories Aswith most things in mathematics categories thus arose from attempts to encode what was beingseen in some concrete examples in a more general and formal setting

Representable functors Representable functors are defined in Section 21 of the book but wersquollgive the required definitions here in a hopefully simpler to digest manner The point is that anyobject A in a (locally small) category C gives two ldquonaturalrdquo functors C rarr Set one covariant andthe other contravariant which as wersquoll see pretty much encode all data about A itself We definehA C rarr Set to be the covariant functor defined by

hA(B) = Mor(AB) on objects and

hA(Bfminusrarr C) = the map Mor(AB)

hAfminusminusrarr Mor(AC) defined by g 983041rarr f g

We define hA C rarr Set to the contravariant functor defined by

hA(B) = Mor(BA) on objects and

hA(Bfminusrarr C) = the map Mor(CA)

hAfminusminusrarr Mor(BA) defined by g 983041rarr g f

(The functor hA is called the functor of points of A and wersquoll usually think of it as being a covariantfunctor hA Cop rarr Set The functor hA does not have a standard name) So hA is defined bymorphisms from A and hA is defined by morphisms into A Wersquoll see later the sense in which suchfunctors encode all information about the object A

A functor F C rarr Set is representable if is naturally isomorphic to such a functor so F sim= hA

for some A in the covariant case and F sim= hA for some A in the contravariant case The idea wersquollbuild towards is that we can essentially treat such a functor as being A itself so that representable

16

functors can be thought as being actual objects in C For now here is a key example wersquove seenbefore which we can now formulate using the language of representability

Given AB isin Ob(C) let F C rarr Set be the contravariant functor defined by

F (C) = Mor(CA)timesMor(CB) and F (Cfminusrarr D) = [(g k) 983041rarr (g f k f)]

where (g k) 983041rarr (g f k f) gives a map Mor(DA) timesMor(DB) rarr Mor(CA) timesMor(CB) ieF (D) rarr F (C) In order for this functor to be representable requires the existence of an object inC satisfying

Mor(CA)timesMor(CB) sim= Mor(C representing object)

in a way which is compatible with all morphisms but we have actually already seen this notionelsewhere it is essentially the definition of a product AtimesB for A and B Indeed we claim that Fis naturally isomorphic to hAtimesB as we now show

First to define a natural isomorphism hAtimesBsim= F requires for each C isin Ob(C) a choice of a

bijection (ie isomorphism in Set)

ηC hAtimesB(C)sim=minusrarr F (C)

which in our case looks like

ηC Mor(CAtimesB))sim=minusrarr Mor(CA)timesMor(CB)

Take this to be the map

(Cfminusrarr AtimesB) 983041rarr (pA f pB f)

where pA A times B rarr A and pB A times B rarr B are the two projection morphisms in the definitionof a product The definition of product implies that this map is invertible since given k C rarr Aand ℓ C rarr B there exists a unique h C rarr A times B such that k = pA h and ℓ = pB h Forthese bijections to define a natural isomorphism hAtimesB

sim= F requires that they be compatible withall morphisms in C which means that given any g C rarr D in C the following diagram shouldcommute

hAtimesB(C) hAtimesB(D)

F (C) F (D)

hAtimesB(g)

ηC ηD

Fg

which in the case at hand looks like

Mor(CAtimesB) Mor(DAtimesB)

Mor(CA)timesMor(CB) Mor(DA)timesMor(DB)

minus g

ηC ηD

(minus gminus g)

Take an element k D rarr AtimesB in the upper right Applying the map on top gives the elementk g in the upper left and then applying ηC on the left side gives

(pA (k g) pB (k g))

17

in the lower left Instead starting with k D rarr AtimesB in the upper right applying ηD on the rightgives (pA k pB k) and applying the map on the bottom gives

((pA k) g (pB k) g)

in the lower left which is the same as the previous element in the lower left by the associativity ofcomposition Hence the isomorphisms ηC do define a natural isomorphism between hAtimesB and F Again the point is that you can then interpret the functor F as being the ldquosamerdquo as the productAtimesB since both encode the same data

More Representable Examples

Coproducts As in the case of products the definition of coproducts can also be phrased in termsof a representable functor Given objects A and B in C let F C rarr Set be the covariant functordefined by

F (C) = Mor(AC)timesMor(BC) and F (Cfminusrarr D) = [(g k) 983041rarr (f g f k)]

where the latter is a function Mor(AC)timesMor(BC) rarr Mor(AD)timesMor(BD) This functor isrepresentable if and only if a coproduct A ⊔ B for A and B exists and the representing object isthat coproduct the key is the requirement that we have bijections

Mor(AC)timesMor(BC) sim= Mor(A ⊔BC)

for all C which is essentially the defining property a coproduct for A and B should have

Power set functor contravariant version Let P Set rarr Set be the contravariant powerset functor defined on objects by sending a set to its power set and on morphisms by sendingf A rarr B to the preimage function Pf P (B) rarr P (A) given by S 983041rarr fminus1(S) In order for this tobe representable requires the existence of a set X with natural bijections

P (A) sim= Mor(AX)

for all sets A that is a set X so that mapping into X is the same data as specifying a subset ofthe domain We take X = 0 1 to be a two-element set and the required bijections

ηA P (A) rarr Mor(A 0 1) to be given by S 983041rarr χS

where XS A rarr 0 1 is the indicator or characteristic function S defined by sending elementsof S to 1 and elements of A minus S to 0 The inverse of ηA sends a function g A rarr 0 1 to thepreimage gminus1(1) sube A

To verify that the bijections ηA define the data of a natural isomorphism η P rArr h01 wecheck the commutativity of some diagrams Given a function f A rarr B consider

P (A) P (B)

h01(A) = Mor(A 0 1) h01(B) = Mor(B 0 1)

Pf = fminus1

ηA ηB

minus f

18

Take S isin P (B) in the upper right Applying the map on top gives fminus1(S) isin P (A) and thenapplying the map on the left gives χfminus1(S) isin Mor(A 0 1) On the other hand applying the mapon the right to S isin P (B) gives χS isin Mor(B 0 1) and then applying the map on the bottomgives χS f isin Mor(A 0 1) The two resulting functions χfminus1(S)χS f in the lower left are

χfminus1(S)(x) =

9830831 x isin fminus1(S)

0 otherwiseand χS(f(x)) =

9830831 f(x) isin S

0 otherwise

so these are the same since x isin fminus1(S) if and only if f(x) isin S Thus the diagram above commutesso P is naturally isomorphic to h01 and hence P is representable

Power set functor covariant version Now consider the covariant power set functor stilldenoted by P Set rarr Set which again sends a set to its power set but now sends a functionf A rarr B to the image function Pf P (A) rarr P (B) defined by S 983041rarr f(S) In order for this tobe representable there would have to exist a set X such that P sim= hX which translates into havingnatural bijections

P (S) sim= Mor(XS)

for all sets S However in this case there is no natural choice for X as there is no natural way todetect subsets of S by mapping into S as opposed to the case of P (S) sim= Mor(S 0 1)

That no such X exists can be made precise using a simple argument when S = empty P (S) hascardinality 1 soX would have to be empty as well since otherwise Mor(XS) would have cardinality0 but if X is empty then Mor(empty S) always cardinality 1 regardless of S which is clearly not trueof P (S) Thus the covariant power set functor is not representable

Forgetful on Top The forgetful functor F Top rarr Set is representable Indeed a representingobject would have to satisfy

X sim= Map(objectX)

as sets for any topological space X (where Map denotes the set of continuous maps) and it iseasy to see that a single point satisfies this property a continuous map f pt rarr X is completelydetermined by the element f(pt) of X and any such element gives a continuous map in this wayOf course checking that F sim= hpt requires checking that various diagrams commute but this issimple to work out in this case so we omit it here

Extracting a topology With an eye towards the idea that the functors hA hA should captureall information about an object A in a category note that the result above shows that in TophX first of all captures all information about X as a set since the set X can be extracted fromhX(pt) = Map(ptX)

We claim that hX actually captures all information about X as a topological space since thetopology on X can be extracted from hX in the following way As we saw in the power set examplewe have a bijection

P (X) sim= Mor(X 0 1)

Now equip 0 1 with the topology in which 1 is open but 0 is not Then a continuous mapf X rarr 0 1 is fully determined by the preimage fminus1(1) since to be continuous only requiresthat this single preimage be open in X Moreoever any open subset of X arises in this way bytaking f to be the indicator function of that open set Thus we get a bijection

Op(X) sim= Map(X 0 1)

19

where Op(X) denotes the set of open subsets of X ie the topology on X Hence the topology onX can be extracted from hX as claimed so hX and hX do encode all information about X

This fact can also be interpreted as saying that the contravariant functor which sends a topo-logical space X to its collection of open subsets (and which sends a continuous map to the mapinduced by taking preimages) is representable with representing object 0 1 equipped with thetopology given above This will be worked out in a homework problem

Forgetful on Grp The forgetful functor Grp rarr Set is also representable In this case arepresenting object should be a group giving set-theoretic bijections

G sim= Hom(objectG)

for all groups G where Hom denotes the set of group homomorphisms The group Z works

G sim= Hom(Z G)

since a homomorphism φ Z rarr G is completely determined by φ(1) since Z is generated by 1 andthere are no restrictions on what φ(1) isin G can actually be The inverse of the desired bijection isthus φ 983041rarr φ(1) and it is simple to check that morphisms behave appropriately so that the forgetfulfunctor in this setting is indeed isomoprhic to hZ

Forgetful on Ring The forgetful functor Ring rarr Set is also representable where by Ring wemean the category of rings with identity elements and where morphisms (ie ring homomorphisms)are required to preserve identities Here we need an object satisfying

R sim= Hom(object R)

for all R isin Ob(Ring) and we can see that Z[x] the ring of polynomials in the variable x withcoefficients in Z works Indeed given a ring homomorphism φ Z[x] rarr R the fact that φ isrequired to preserve identities fully determines φ(n) for any n isin Z (since φ(n) = nφ(1)) and hencesince φ preserves multiplication its behavior is fully determined by φ(x) which can be any elementof R The inverse of the bijection we want is thus φ 983041rarr φ(x) and it will follow that this forgetfulfunctor is isomorphic to hZ[x]

Equivalences between Categories

Definitions iso and equiv Now that we have a notion of a ldquomorphismrdquo between categorieswe can talk about the sense in which two categories are the ldquosamerdquo The first possible definitionis the ldquoobviousrdquo one you might expect given the definition of ldquoisomorphismrdquo wersquove seen in everyother context until now we say that C is isomorphic to D if there exist functors F C rarr D andG D rarr C such that G F = idC and F G = idD where idC and idD are identity functors

However this definition turns out to be too restrictive Indeed the requirement that the givencompositions equal identity functors says that given an object A in C the object G(F (A)) beliterally the same as A itself and similarly for objects in D But we know that two objects in acategory can be thought of as being the ldquosamerdquo without being literally the same as long as theyare isomorphic For instance productscoproducts are not unique in the literal sense but only inthe sense that they are isomorphic in a unique way To take another example the definition ofa functor being representable does not say that F (B) literally be the same as hB(A) only thatthey be isomorphic for sure the power set P (S) of a set S is not literally the same as the set offunctions S rarr 0 1mdashit is only the ldquosamerdquo in the sense that they encode the same data

20

Thus it makes sense to relax the requirement that A andG(F (A)) in the definition of isomorphiccategories be equal to the requirement that they be isomorphic This leads to the following betterdefinition of two categories being the ldquosamerdquo C is equivalent to D if there exist functors F C rarr D

and G D rarr C such that G F sim= idC and F G sim= idD So we require only that for instance thefunctor G F be naturally isomorphic to the identity functor on C which says that for any objectA G(F (A)) be isomorphic to A in a natural way Wersquoll see that this truly is the better notionof what it means for two categories to be the ldquosamerdquo and it reflects the idea that all categoricalproperties of one category can be transported to the other

Alternate characterization of equivalences A functor F C rarr D implementing an equiv-alence between categories is called unsurprisingly and equivalence and the ldquoinverserdquo functorG D rarr C is still referred to as being its inverse even though it is not technically an ldquoinverserdquoin the sense that compositions give literal identities But as in the case of Set where ldquoinvertiblerdquocan be characterized without reference to an inverse as ldquoinjectiverdquo and ldquosurjectiverdquo so too canequivalences between categories be characterized without explicit reference to an inverse

A functor F C rarr D is an equivalence if and only it is full faithful and essentiallysurjective

We have seen the notions of full and faithful before and essentially surjective means that any objectin D is isomorphic to something in the image of F for all D isin Ob(D) there exits C isin Ob(C) suchthat F (C) sim= D Justifying this result was the topic of a student presentation and can be foundin the book in Section 15 (Technically to construct an ldquoinverserdquo of F requires that we work withsmall categories but whatever) The bulk of the real work goes into showing that the ldquoinverserdquofunctor constructed is actually a functor which comes down to some abstract nonsense argumentvia commutative diagrams

Equivalences preserve everything We previously saw the notions of full faithful and es-sentially surjective back when thinking about what properties a functor could have which wouldguarantee it sent products to products so the result we proved back then was that equivalencespreserve products Similarly equivalences preserve coproducts and the proof is basically the sameafter reversing arrows A clean way of deriving this is as follows if F C rarr D is an equivalencethen so is F op Cop rarr Dop so since F op preserves products F preserves coproducts becausecoproducts in a category become products in the opposite category

In general an equivalence between categories will preserve whatever categorical notions youwant monomorphisms epimorphisms initial objects terminal objects etc so that equivalentcategories really do have essentially the ldquosamerdquo properties

Equivalence example Here is a first example (also found in the book) of equivalent categorieswhich are not isomorphic Let FVectR be the category whose objects are finite-dimensional realvector spaces and whose morphisms are linear transformations Let MatR be the category ofmatrices defined as follows

bull objects in MatR are nonnegative integers

bull a morphism n rarr m is an mtimes n matrix with real entries

bull composition in MatR is given by matrix multiplication given B n rarr m and A m rarr kmdashsoB is an mtimes n matrix and A is a k timesm matrixmdashthe composition A B n rarr k is the k times nmatrix AB

bull identity morphisms in MatR are given by identity matrices

21

We claim that FVectR and MatR are equivalent First note that they certainly cannot be isomo-prhic isomorphisms in MatR exist only between two objects which are literally the same (ie anmtimesn matrix can be invertible only when m = n) but isomorphisms in FVectn can exist betweenobjects which are not literally the same So there are in a sense not enough objects in MatR toallow it to be isomorphic to FVectR

To construct an equivalence choose one and for all an isomorphism TV V rarr RdimV forevery object in FVectR or equivalently choose once and for all a basis for every V DefineF FVectR rarr MatR by

V 983041rarr dimV on objects and

(T V rarr W ) 983041rarr (the standard matrix of SW T Sminus1V RdimV rarr RdimW ) on morphisms

Equivalently F sends T to the matrix of T relative to the chosen bases for V and W It isstraightforward to check that this is a functor The inverse functor G MatR rarr VectR is givenby

n 983041rarr Rn on objects and

(mtimes n matrix A) 983041rarr (the linear transformation A Rn rarr Rm induced by A) on morphisms

One can check that G is a functor and that F G and G F are naturally isomorphic to identitiesor instead one can argue that F alone is full faithful and essentially surjective The point is thatG(F (V )) gives back RdimV so not literally V itself but only something isomorphic to V which iswhy this gives an equivalence of categories and not an isomorphism

Yoneda Lemma Functors as Objects

Compact Hausdorff spaces and Clowast-algebras Last time we gave an example of equivalentcategories which were not isomorphic but in many ways that example is unsatisfactory when doinglinear algebra people for sure often work with matrix representations of linear transformations asthe example from last would suggest but no one in their right mind literally thinks about suchthings in terms of the equivalence FVectR rarr MatR itself In other words this equivalence isessentially ldquocooked uprdquo for the sake of giving an example but it does not really give any insightinto the actual mathematics of linear algebra

So with that in mind we will describe another equivalence which actually does have some realmeaning in the sense of producing new ways of thinking about some mathematics Given a compactHausdorff space X we let C(X) denote the space of continuous complex-valued functions on X

C(X) = f X rarr C | f is continuous

We can equip the set C(X) with various additional structures First it is a complex vector spaceunder ordinary addition and scalar multiplication of functions Second we can multiply two func-tions together

(fg)(p) = f(p)g(p)

which turns C(X) into whatrsquos called an algebra over C (We wonrsquot give the definition of anldquoalgebrardquo but the point is that this multiplication is in some sense ldquocompatiblerdquo with the vectorspace structure) Next any element f of C(X) has a conjugate element f obtained by takingcomplex conjugates of the values of f

f(p) = f(p)

22

And finally we can equip C(X) with a norm via

983042f983042 = suppisinX

|f(p)|

where |middot| denotes complex absolute value (This supremum exists sinceX is compact by the ExtremeValue Theorem) All of these structures (vector space algebra multiplication conjugation norm)turn C(X) into whatrsquos called a Clowast-algebra We wonrsquot define this precisely but the definition encodesvarious ways in which these structures are compatible Actually C(X) is a unital commutative Clowast-algebra where ldquounitalrdquo means that the multiplication has an identity element (namely the constantfunction 1) and commutative means that fg = gf There is a natural notion of homomorphismbetween Clowast-algebras which are maps between them which behave nicely with all the structuresinvolved In particular given a continuous map f X rarr Y between compact Hausdorff spaces weget a Clowast-algebra homomorphism flowast C(Y ) rarr C(X) given by pre-composition with f flowast sendsg isin C(Y ) to g f isin C(X)

We thus have a contravariant functor CHaus rarr ucClowast-Alg from the category of compactHausdorff spaces and the category of unital commutative Clowast-algebras It turns out that anyunital commutative Clowast-algebra arises in this way in the sense that given any unital commutativeClowast-algebra B there exists a compact Hausdorff space X such that C(X) sim= B as Clowast-algebrasand that morphisms between these algebra arises from continuous maps in the manner describeabove This says that the functor above is essentially surjective full and it turns out that it isalso faithful so that it is an equivalence Thus all information about compact Hausdorff spaces iscompletely encoded in information about unital commtuative Clowast-algebras and indeed one couldlearn everything there is to know about a compact Hausdorff space X from its Clowast-algebra C(X)of functions The fact that the categories CHaus and ucClowast-Alg are equivalent is the startingpoint in the subject known as noncommutative geometry which wersquoll say something about a bitlater on If we drop the requirement that our algebras be unital it turns out that the categoryof commutative Clowast-algebras in general is equivalent to the category of locally compact Hausdorffspaces via essentially the same functor where we take as morphisms in the category of locallycompact Hausdorff spaces to be continuous functions which ldquovanish at infinityrdquo

We are only introducing Clowast-algebras to give an example of an interesting and actually usefulequivalence between categories but just for the fun of it wersquoll note the following Examples ofnoncommutative Clowast-algebras come from what are known as ldquoalgebras of bounded operators onHilbert spacesrdquo which show up in quantum mechanics as the things which describe physicallyobservable quantities like position momentum and energy (In fact all Clowast-algebras arise fromHilbert spaces in this way) Thus Clowast-algebras show up in mathematical formulations of quantummechanics and the notion of noncommutative geometry mentioned above is at the core of someattempts to understand the elusive concept known as ldquoquantum geometryrdquo

Yoneda Lemma The Yoneda Lemma states that given a (covariant) functor F C rarr Set whereC is locally small for any A isin Ob(C) there exists a bijection

F (A) sim= Nat(hA F )

where the right side denotes the set of natural transformations from hA to F A similar result holdsin the contravariant case where we get bijections F (A) sim= Nat(hA F ) (There is also a sense inwhich these bijections are themselves natural but wersquoll ignore this bit) The proof of the YonedaLemma was the topic of a student presentation and can be found in Section 22 of the book

Here we will get a bit philosophical and talk about what the point of the Yoneda Lemmaactually is and later the sense in which it says that we can view arbitrary functors as ldquonon-existent

23

objectsrdquo in a category First we mention what we will call the ldquofunctor of pointsrdquo approach tomathematics a mathematical object can be fully characterized by all morphisms into it That isknowledge of Mor(SA) for all S is equivalent to knowledge of A itself Indeed this is an idea wementioned previously in relation to the functors hAmdashwhich we previously said is referred to as beingthe functor of points of Amdashand now we seek to push this idea further In some examples literalpoints of A can be extracted from hA as in the case of the bijection X sim= hX(pt) = Map(ptX) fora topological space X or G sim= hG(Z) = Mor(Z G) for a group G Moreover in these examples therest of the (topological) structure onX or (group) structure on G can also be extracted from hX andhG respectively In general we can view Mor(SA) as in a sense being ldquopoints of A parameterizedby Srdquo which comes from the idea that in say Set Mor(SA) = AS is ldquothe product of S-manycopies of Ardquo so literally points of A ldquoparameterizedrdquo by S In various areas of mathematics anelement of Mor(SA) is thought of as being a ldquogeneralizedrdquo point of A so that the functor of pointshA detects generalized points of A

So an object A is fully encoded by the functor hA Now we claim that morphisms betweenobjects are also encoded by such functors Indeed in the case where F = hB is itself a functor ofpoints the Yoneda result Nat(hA F ) sim= F (A) becomes

Nat(hA hB) sim= hB(A) = Mor(AB)

which says precisely that morphisms between the objects A and B can be obtained solely fromthe functors hA and hB (This fact is one reason why one should prefer the contravariant functorhA to the covariant functor hA in the covariant case Nat(hA F ) sim= F (A) for F = hB becomesNat(hA hB) sim= hB(A) = Mor(BA) so that morphisms between two objects get reversed whencharacterizing them in terms of representable functors)

Pushing our philosophy further we now know that not only objects but also morphisms betweenobjects are fully encoded by the functors hA The final step is to note that we should not only careabout objects and morphisms between them but that an important part of the data of an objectA is also how arbitrary contravariant functors F C rarr Set behave on A and that this too can beobtained from hA precisely via the Yoneda result that F (A) sim= Nat(hA F ) So objects morphismsbetween objects and how arbitrary functors behave on objects are all encoded in knowledge ofthe functors hA so that these functors truly do capture in the most general sense possible allinformation about A

Functors as nonexistent objects The discussion above is all well and good but it still doesnrsquotquite say why we should care about the fact that we can study on object A by studying hA insteadin that we still havenrsquot spoken about what type of insight about A can be gleamed from hA thatwasnrsquot readily apparent in A itself already Now we argue that the true point of the YonedaLemma is that it gives a way to think about arbitrary functors F Cop rarr Set as ldquoobjectsrdquo in C

itself or really as objects in an ldquoenlargementrdquo of C Certainly for those functors F sim= hA which arerepresentable we can indeed think about F as simply being the object A but now we claim thateven non-representable F can be thought of as being ldquononexistentrdquo objects in C The upshot is thatoften in mathematics we come across things which do not actually exist but the Yoneda Lemmagives a way to (indirectly) study such things anyway by interpreting them as functors instead

The starting point of course is that an actual object A is the same as hA as we have explainedabove Thus by analogy we can interpret a non-representable functor F as being an ldquoobjectrdquo in C

as long as we know what it means to ldquomaprdquo into F from other things which are honestly objects in CThat is if F were indeed representable then F (A) sim= Nat(hA F ) sim= Mor(AF ) would characterizethe functor of points of F (here we abuse notation by using the same letter F to denote both arepresentable functor and the object which represents it) so now we take Mor(AF ) = F (A) as

24

the definition of the ldquofunctor of pointsrdquo of the non-existent object F We are thus taking elementsof F (A) as being what it means to ldquomaprdquo from A into F even when F is not representable Thefact that the Yoneda Lemma gives a way to define maps from A into F thus says that F itself isequipped with the same data that a literal object of A would be equipped with so why shouldnrsquotwe interpret F as a being a type of ldquoobjectrdquo in C as well

To phrase this another way note that the Yoneda Lemma gives rise to a functor

C rarr Fun(CopSet)

where Fun(CopSet) denotes the category whose objects are functors from Cop to Set and morphismsare natural transformation defined on objects by A 983041rarr hA and on morphisms by the bijectionMor(AB) sim= Nat(hA hB) arising from the Yoneda Lemma That this is actually bijection saysthat this functor is full and faithful so that we can view C as being a subcategory of the functorcategory Fun(CopSet) (We cay say that C is isomorphic to the subcategory of Fun(CopSet)consisting of the functors hA or that it is equivalent to the subcategory of Fun(CopSet) consistingof representable functors This functor C rarr Fun(CopSet) is usually called the Yoneda embeddingof C into Fun(CopSet)) Thus we can view Fun(CopSet) as an ldquoenlargmentrdquo of C and the ideawe are trying to make sense of is that other objects in this ldquolargerrdquo category should be interpretedas ldquonon-existent objectsrdquo of C

Just how far can we actually push this idea that arbitrary functors Cop rarr Set should be thoughtof as being ldquoobjectsrdquo of C Pretty far in fact For instance often times in geometry one comesacross a type of ldquosmoothrdquo space on which some group acts and one is interested in doing ldquocalculusrdquoon the resulting quotient space which is the space of orbits under the given group action Theproblem is that such quotient spaces are very often (usually in fact) not ldquosmoothrdquo themselves sothat the tools from calculus are not readily available The solution is to nonetheless treat thisquotient as if it were ldquosmoothrdquo and attempt to do calculus anyway To make this precise thisldquonon-smoothrdquo space actually turns out to have a natural interpretation as a ldquosmooth functorrdquo andit is through this functor which we can view this non-smooth space as actually being smooth afterall so we are forced to expand our notion of what we mean by ldquosmooth spacerdquo by allowing thingswhich are not spaces in the usual sense which certainly leads to more abstraction but the benefitis that it actually makes sense to do calculus on such ldquospacesrdquo interpreted as functors

The discussion above is pretty vague so here is one final more concrete example Recall thatequivalence

CHausop rarr ucClowast-Alg

between the category of compact Hausdorff spaces and the category of unital commutative Clowast-algebras mentioned at the start We now claim that a non-commutative (unital) Clowast-algebra shouldbe interpreted as the algebra of functions on some non-existent ldquononcommutative compact Haus-dorff spacerdquo and it is this idea which the subject of noncommutative geometry makes precise Solet B be a non-commutative Clowast-algebra Then the functor of points hB gives a functor

ucClowast-Alg rarr Set

which sends C(X) (where X is some compact Hausdorff space) to Hom(C(X) B) where Hom de-notes homomorphisms of Clowast-algebras whatever that means Composing with the previous functorwe get a functor wersquoll denote by B itself

B CHausop rarr Set

which on objects at least is given by X 983041rarr Hom(C(X) B) This functor is not representable sinceB is not commutative but the philosophy espoused above suggests we should think of it as giving a

25

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

Abstract nonsense It is common to say in the setting above that the fact that products areunique up to unique isomorphism follows from ldquoabstract nonsenserdquo which is a succinct way ofsaying that it follows solely from the definition of a categorical product itself and not any deepermathematical content In particular many of the ldquouniquenessrdquo properties which ldquoproductsrdquo havein the settings we might care aboutmdashCartesian products in Set direct products in Grp producttopologies in Topmdashreally have nothing to do with how those specific products are defined but arejust consequences of the ldquoabstract nonsenserdquo of category theory

But of course we do not use the phrase ldquoabstract nonsenserdquo in a negative or dismissive waysince recognizing what types of properties are consequences of ldquoabstract nonsenserdquo goes a long waytowards understanding well the given problem at hand Soon you too will come to appreciate thisnotion of abstract nonsense

Coproducts Opposite Categories

Initial and terminal objects The notions of initial and terminal objects in a category aredefined in Section 16 of the book where you can also find standard examples Here I want topoint out that initial and terminal objects if they exist are unique up to unique isomorphism asanother example of arguing via ldquoabstract nonsenserdquo

Suppose I and I prime are both initial objects in some category By the fact that I is initial thereexists a unique morphism f I rarr I prime and by the fact that I prime is initial there exists a unique morphismg I prime rarr I But then the composition g f I rarr I must equal idI I rarr I since the latter is theonly morphism from I to itself because I is initial and similarly f g = idIprime Thus I and I prime areunique and there is a unique isomorphism between them

Coproducts The dual concept to that of a product is the notion of a coproduct (The termldquodualrdquo refers to the phenomena you get when you reverse all arrows in a given construction Forinstance the notion of a terminal object is dual to that of an initial object and the notion of anepimorphism is dual to that of a monomorphism) Here is the precise definition a coproduct of anobject A and an object B in a category is an object C together with morphisms iA A rarr C andiB B rarr C such that for any object D and morphisms f A rarr D and g B rarr D there exists aunique morphism h C rarr D which makes the following diagram commute

D

A B

Cf g

iA iB

existh

By abstract nonsense if a coproduct of A and B exists it is unique up to unique isomorphism

Examples In Set the coproduct of two sets A and B is the disjoint union A⊔B (If you havenrsquotseen the notion of a disjoint union before it is essentially the same as the union only that we donrsquotcare about whether A and B have any overlap if they do we treat the common elements as beingldquodifferentrdquo For instance Z ⊔ Z is not Z but consists of two copies of each integer one 0 comesfrom the first copy of Z and another comes from the second but these are different ldquozeroesrdquo)Given f A rarr D and g rarr D the unique map A ⊔B rarr D is defined by applying f to elements of

8

A and g to elements of B (Note that if we simply took the ordinary union AcupB as the candidatefor the coproduct when A and B were not disjoint the required map h A cup B rarr D would notexist since h would have to send any element x isin A cap B to both f(x) and g(x) in order to makethe required diagram commutative)

The coproduct of two objects X and Y in Top is again the disjoint union X ⊔ Y equippedwith the disjoint union topology an open subset of X ⊔ Y is defined to be the (disjoint) union ofan open subset of X with an open subset of Y The coproduct of two groups G and H in Grp isthe free product G lowastH which is further elaborated on in the homework

The coproduct of two vector spaces V and W in Vect is the direct sum V oplusW which is definedto be the set of formal sums

v + w where v isin V and w isin W

where addition and scalar multiplication are defined ldquocomponentwiserdquo

(v1 + w1) + (v2 + w2) = (v1 + v2) + (w1 + w2) and r(v + w) = rv + rw

This is isomorphic to the product V times W with operations also defined componentwise but it iscustomary to speak about the ldquodirect sumrdquo V oplusW when referring to the coproduct as opposed tothe product wersquoll say why in a bit The maps iV V rarr V oplusW and iW W rarr V oplusW required aspart of the data of a coproduct are

iV v 983041rarr v + 0W and iW w 983041rarr 0V + w

To verify that V oplusW is the correct coproduct take any linear transformations S V rarr U andT W rarr U where U is some other vector space We then need a (unique) linear transformationh V oplusW rarr U such that

S = h iV and T = h iW

Let us determine what this map must be thereby not only showing that it exists but also that itis unique Let v + w isin V oplusW which we can write as

v + w = (v + 0W ) + (0V + w)

Since h is required to be linear we have

h(v + w) = h([v + 0W ] + [0V + w]) = h(v + 0W ) + h(0V + w)

The first term at the end is (h iV )v which must thus be Sv and the second term is (h iW )wwhich must be Tw Thus h = S + T is the map

h = S + T v + w 983041rarr Sv + Tw

which you can verify is indeed linear Thus V oplusW does satisfy the property required of a coproductSo the product of two objects in Vect is the same as their coproduct only that we write the

latter using ldquosumrdquo notation Later we will discuss products and coproducts of an arbitrary numberof elements We will see these as special cases of the general notions of limits and colimits butthey are defined via similar requirements as products and coproducts of two objects at a timeIn the category of vector spaces arbitrary products are given by the usual product

983124α Vα with

ldquocomponentwiserdquo addition and scalar multiplication but it turns out that an arbitrary coproductis given by the direct sum

983119α Vα which is defined to be the subspace of

983124α Vα where only finitely

many components are nonzero (This is in a way analogous to how the product topology on an

9

arbitrary number of topological spaces is defined) Wersquoll go over this later but is the underlyingreason why we use oplus instead of times when discussing coproducts

Encoding data via productscoproducts We can nicely summarize the definition of productsand coproducts in the following way the product of A and B is the object with the property thatfor any object D we have

Mor(DA)timesMor(DB) = Mor(D product)

and the coproduct of A and B is the object such that for any object D we have

Mor(AD)timesMor(BD) = Mor(coproductx D)

The first reflects the fact that products give a unique way to turn two morphisms (f and g in thedefinition) mapping into A and B into a single morphism (h in the definition) and the second saysthat coproducts give a unique way to turn two morphisms mapping out of A and B into a singlemorphism

Soon enough we will discuss the idea of universality and the point is that products are ldquouni-versalrdquo for maps into A and B and coproducts are universal for maps from A and B

Opposite categories Finally we give a definition which might seem silly but is actually quiteuseful Given a category C the opposite category is the category Cop is the category with the sameobjects as C but where we defined Mor(AB) in Cop to be Mor(BA) in C To be more concrete weconsider each morphism f A rarr B in C to instead be a morphism fop B rarr A in Cop (Perhaps itwould be better to write this as A larr B fop) The point is that we donrsquot actually care about whatf actually is as say a functioncontinuous maphomomorphismwhatever-we only care about itsbehavior as an ldquoarrowrdquo The opposite category Cop only retains information about how arrowsrelate to one another nothing more

Since arrows are reversed products in C become coproducts in Cop and coproduts in C becomecoproducts in Cop Similarly initialterminal objects and monomorphismsepimorphisms in C be-come terminalinitial objects and epimorphismsmonomorphisms in Cop Again the notion of anopposite category might seem like a strange thing to look at but wersquoll see that it does have uses

Functors Fullness and Faithfulness

One more productcoproduct example Let X be a topological space and define its categoryof open sets Op(X) as follows The objects of Op(X) are simply the open subsets of X and themorphisms are given by inclusions so to be concrete

Mor(U V ) =

983083the unique inclusion U rarr V if U sube V

empty otherwise

The fact that U sube V and V sube W implies U sube W says that the composition of two morphismsin this category is still a morphism in this category The product of U and V in this categoryturns out to be their intersection (with ldquoprojection morphismsrdquo U cap V rarr U and U cap V rarr V beinginclusions) and the coproduct of U and V turns out to be their (ordinary) union Note that weneed X to be a topological space in order to guarantee that U cap V and U cup V are still objects inthis category More generally the product of finitely many objects in Op(X) exists and is theirintersection and the coproduct of an arbitrary number of objects exists and is their union (Wesay that Op(X) has finite products and arbitrary coproducts)

10

Given a set X and a collection T of subsets of X we can define a similar category with elementsof T as objects (I guess we should assume that empty X isin T) and inclusions as morphisms We thenhave that this category has finite products and arbitrary coproducts if and only if T is a topologyon X so that we can rephrase the definition of a topological space itself solely in categorical termsThis perspective on what a ldquotopologyrdquo is makes it simpler to state the definition of whatrsquos calleda sheaf on a topological space which is an important construction in various areas of analysis andgeometry We might discuss this a bit later on

Functors As with most things in mathematics we should care not only about categories asmathematical structures on their own but also about ldquomapsrdquo between these structures whichldquopreserverdquo said structure The notion of a (covariant) functor between categories provides suchldquomapsrdquo in the case at hand where a functor is a way of sending objects to objects and morphismto morphisms in a way which preserves compositions and identities The precise definition is inSection 13 of the book as is the definition of a contravariant functor between categories whichis a functor which reverses the direction of arrows as opposed to ldquocovariantrdquo ones which preservedirections (It is common to use the term ldquofunctorrdquo without qualification to mean one which iscovariant Note that any functor can be made covariant by working with the opposite category acontravariant functor C rarr D is the same as a covariant functor Cop rarr D)

All of the standard examples we looked at are in the book forgetful functors power set functorsthe dual space functor in Vect etc Of particular interest is Example 139 in the book whichshows that various types of group actions in different contexts-the usual notion of a group actionon a set the notion of a continuous action of a group on a topological space and the notion ofa linear action of a group on a vector space (also known as a representation of the group)-are allencoded by a functor with domain category BG

Adjoint functors Let F Top rarr Set be the forgetful functor and let D Set rarr Top be thefunctor which sends a set S to the space S equipped with the discrete topology and a functionS rarr Sprime between sets to the same function only now viewed as a continuous map S rarr Sprime betweendiscrete spaces These two functors satisfy the following property in relation to one another givinga morphism D(S) rarr Y in Top is the same as giving a morphism S rarr F (Y ) in Set or moresymbolically

MorTop(D(S) Y ) = MorSet(S F (Y ))

We say that D is left adjoint to F and that F is right adjoint to D The point is that adjointfunctors can be used to move data back and forth between two categories

Wersquoll talk more about adjoint functors later on but for now here is one more example Nowtake F Grp rarr Set to again be the forgetful functor and take G Grp rarr Set to be the freegroup functor which sends a set S to the free group generated by S G(S) has as elements ldquowordsrdquos1 sn made up of elements of S with the group operation being ldquoconcatenationrdquo A basic factis that giving a group homomorphism from G(S) to some other group H is the same as giving anordinary function from S to F (H) which is the required adjoint property

MorGrp(G(S) H) = MorSet(S F (H))

Faithful and full functors The definitions of what it means for a functor to be faithful and fullare in Section 15 of the book where faithful functors are ones which are injective on morphisms andfull functors are ones which are surjective on morphisms but where in both cases we refer only tomorphisms which occur between the objects F (A) and F (B) in the target category corresponding tofixed objects A and B in the source category (In general a functor could send two objects AB in

11

the source category to the same object F (A) = F (B) in the target category and thus there could betwo morphisms A rarr C and B rarr C which are sent to the same morphism F (A) = F (B) rarr F (C)Such a thing could still possibly be ldquofaithfulrdquo since the non-injectivity here is arising from differentobjects in the source)

Faithfullness and fullness will be nice properties going forward and here is one hint at how theymight be useful Say we have a functor F C rarr D The problem is to determine properties onF which will imply it send products in C to products in D or coproducts in C to coproducts inD So given the former product diagram below we want the second diagram to also be a productdiagram

C

A B

P Fminusminusminusrarr

F (C)

F (A) F (B)

F (P )f g

pA pB

existh

Ff Fg

F (pA) F (pB)

Fh

Of course in order F (P ) to be an actual product of F (A) and F (B) on the right a similarcommutative diagram property would have to hold for all objects in D in place of F (C) and allmorphisms D rarr F (A) and D rarr F (B) not just those of the form Ff and Fg respectively Oneway to guarantee this is to require that all objects in D are of the form F (C) and then for allmorphisms F (C) rarr F (A) and F (C) rarr F (B) to be of the form Ff Fg the former is true whenthe functor F is surjective on objects and the latter is true when F is full We are not saying thatthese are the only conditions under which F might preserve products but it definitely seems togive a sufficient condition at least

Almost there is one more thing to require the uniqueness of the morphism Fh F (C) rarr F (P )making the diagram on the right commute Suppose there were two such morphisms Fh and Fhprimewhere h and hprime are both morphisms C rarr P in C The commutativity of the diagram on the rightgives for instance

Ff = F (pA)Fh and Ff = F (pA)Fhprime

which the functorial properites of F turns into

Ff = F (pA h) and Ff = (pA hprime)

We want h and hprime to make the first diagram commute in order to be able to apply uniqueness ofthe morphism C rarr P But this requires know that f = pA h and f = pA hprime which would followfrom F being faithful In this case it follows that h and hprime both make the first diagram commute(after we go through a similar argument with g) so the fact that P is a product guarantees thath = hprime and hence that Fh = Fhprime so that F (P ) is an actual product on the right

So we conclude that a functor which is full faithful and surjective on objects will send productsto products (Actually we can relax this a bit by requiring not that every object in D be in theliteral image of F but only that it be isomorphic to something in the image A functor with thisproperty is said to be essentially surjective which is a concept will see in a related context soonenough) Again this is not an ldquoif and only ifrdquo statement since functors can preserve products (orcoproducts) without being full faithful and surjective on objects but it at least gives a sufficientcondition Later we will see other ways to guarantee that functors preserve products which arerelated to the existence of adjoints for that given functor

12

Coproduct Examples Concreteness

Products and coproducts for metric spaces The product of two objects in Met the categoryof metric spaces with morphisms given by metric maps is the Cartesian product X times Y equippedwith the box metric and the coproduct of X and Y does not exist if both X and Y are nonemptyThis was the topic of a student presentation and here are some proof sketches

Given metric spaces (X dX) and (Y dY ) the box metric d on X times Y is defined by

d((x1 y1) (x2 y2)) = maxdX(x1 x2) dY (y1 y2))

(The name ldquoboxrdquo metric comes from the fact that in RtimesR = R2 or R3 open balls indeed looks likeldquoboxesrdquo) The key requirement needed in the definition of a product is given morphisms S rarr Xand S rarr Y that the map

h S rarr X times Y defined by h(s) = (f(s) g(s))

actually be a morphism in Met This turns into the requirement that

d((f(s) g(s)) ((f(t) g(t))) = maxdX(f(s) f(t)) dY (g(s) g(t)) le dS(s t) for all s t isin S

given the fact that

dX(f(s) f(t)) le dS(s t) and dY (g(s) g(t)) le dS(s t) for all s t isin S

For the latter inequalities to imply the former d must (at least up to isometry) be given by thebox metric Note that the two other ldquostandardrdquo metrics one might put on a productmdashthe taxicab

dX + dY and Euclidean983156

d2X + d2Y metricsmdashdo not satisfy this required ldquometric maprdquo property

and so do not give the product in MetTo see that coproducts do not exist if XY are nonempty suppose instead (P dP ) was a co-

product for X and Y with inclusion morphisms iX X rarr P and iY Y rarr P For any n isin Ntake 0 n to be a metric space with the absolute value metric d and let f X rarr 0 n andg Y rarr 0 n be the constant function 0 and the constant function n respectively Since P is acoproduct there exists h P rarr 0 n such that

h iX = f and h iY = g

But in order for h to actually be a morphism in Met we must then have

d(h(iX(x) iY (y)) le dP (iX(x) iY (y)) for any x isin X y isin Y

which translates inton = |0minus n| le dP (iX(x) iY (y))

But this should then be true for any n isin N which would imply that the distance in P betweenan element coming from X and one coming from Y should be infinite which is nonsense (Notethat if you altered your definition of ldquometricrdquo to allow for infinite distance then a coproduct wouldexist the disjoint union X ⊔Y where you defined the distance between elements of X and Y to beinfinite)

Coproducts for groups The coproduct of two groups G and H in Grp is the free product GlowastHwhich is defined to be the set of all words consisting of elements of G andH in an alternating fashion

13

with concatenation as the group operation (When we have to elements of G or two elements of Hnext to each other with use their respective group operations to turn these into single elements ofG or H) This was also the topic of a student presentation and here is the basic idea Given grouphomomorphisms g G rarr N and k H rarr N that G lowastH is the coproduct comes down to showingthat the map

h G lowastH rarr N defined by h(g1h1 gnhn) = f(g1)k(h1) middot middot middot f(gn)k(hn)

and similarly on other possible words in GlowastH is a group homomorphism which follows essentiallyfrom the definition of the group operation on G lowastH indeed this group operation arises preciselyfrom this requirement

Now if we instead tried to use GtimesH with its standard inclusions G rarr GtimesH and H rarr GtimesHas the coproduct the issue is that the analogous map

h GtimesH rarr N defined by h((g h)) = f(g)k(h)

would NOT be a homomorphism because h((g1 h1)(g2 h2)) is not equal to h(g1 h1)h(g2 h2) bythe way in which the group operation is defined on the direct product G times H However if westrict our category to only consist of abelian groups so that the N rsquos we consider in the defini-tion of coproduct are also abelian then G times H does serve as a coproduct we can always rewritef(g1)k(g1) middot middot middot f(gn)k(gn) as f(g1) middot middot middot f(gn)k(h1) middot middot middot k(hn) in N Irsquoll leave it to you to flesh out allof these details

Concrete categories The categories Set Top Vect and Grp all have the property thattheir objects are simply sets possibly equipped with extra structure and that their morphisms areordinary set-theoretic functions possibly satisfying some additional requirement Such categoriesare nice since we can often use intuition about sets and functions in their study A concrete categoryis intuitively one where we can think of objects as being ldquosetsrdquo and morphisms as being ldquofunctionsrdquoonly that we have to interpret this in the right way

Here is the definition a concrete category is a category C equipped with a faithful functorF C rarr Set which is part of the data The idea is that such a functor gives us way to turnan object A isin Ob(C) into a set F (A) and a morphism f isin MorC(AB) into a function Ff isinMorSet(F (A) F (B)) where the ldquofaithfulrdquo requirement says that we do not lose information aboutf when doing so so that we can (essentially) learn everything we need to know about f by studyingFf instead In the four examples above the required faithful functor is simply the forgetful functorand indeed in general we think of F as being a ldquoforgetful functorrdquo for the concrete category (C F )

Examples Here are two more examples of concrete categories Let G be a group and considerthe category BG with a single object and elements of G morphisms As written this does notappear to be concrete since morphisms are not honest functions lowast rarr lowast but the point is that thiscategory can indeed by ldquoconcretizedrdquo by specifying an appropriate ldquoforgetful functorrdquo A functorF BG rarr Set is the data of a (left) action of G on a set S = F (lowast) and saying this functor isfaithful is what it means to say that this action is faithful different g h isin G act on S in differentways or equivalently that there exists s isin S such that gs ∕= hs Any group has at least onefaithful left actionmdashnamely the action on itself given by left multiplicationmdashto BG can always beequipped with a faithful functor to Set This is all just a way of phrasing the fact that any groupis isomorphic to a subgroup of a permutation group

The category Rel of relations can also be ldquoconcretizedrdquo even though as written morphismsA rarr B are not honest functions Take F Rel rarr Set to be the power set functor which sends an

14

object in Rel to its power set and a relation f A rarr B to the induced function Ff P (A) rarr P (B)defined by

(Ff)(S) = b isin B | there exists a isin S such that (a b) isin f

In other words if you think of the relation f sube A times B as a ldquopossibly not everywhere-definedpossibly multivalued functionrdquo (Ff)(S) is analogous to taking the image of S That F is faithfulwill be left to the homework

In fact most categories yoursquoll see can be made concrete by specifying an appropriate forgetfulfunctor but not all categories can be made concrete in this way Here is the standard example of anon-concrete category which we mention only to say there is such a thing but which we wonrsquot doanything more with since understanding it better requires knowledge of algebraic topology Thehomotopy category of topological spaces is the category hTop whose objects are topological spacesand whose morphisms are given by equivalence classes of homotopic maps

MorhTop(XY ) = homotopy classes of continuous maps X rarr Y

(Saying that two maps are homotopic means that you can ldquodeformrdquo one into the other but wersquollleave the precise definition to a course in algebraic topology) It turns out that no functor fromhTop to Set can be faithful so that hTop cannot be made concrete but this is actually a deepand difficult thing to show so we make no attempt to do so here

Natural Isomorphisms Representability

Natural transformations The definition of a natural transformation η F rArr G between two(both covariant or both contravariant) functors FG C rarr D is given in Section 14 of the bookThe key idea is that a natural transformation gives a way to turn data about F into data aboutG in a way which is compatible with all possible morphisms as expressed by the commutativity ofthe diagrams

F (A) F (B)

G(A) G(B)

Ff

ηA ηB

Gf

arising from morphisms f A rarr B (or f B rarr A in the contravariant case) in C In the case of anatural isomoprhism η F rArr G so that each ηA F (A) rarr G(A) is actually an isomorphism in Dthe point is not only that F and G produce isomorphic objects but they do so in a way which iscompatible with all possible category-theoretic data

The first standard example of a natural isomorphism is the one between the identity functor onthe category of finite-dimensional and the functor which sends a finite-dimensional vector space Vto V lowastlowast the dual of its dual induced by the isomorphism V rarr V lowastlowast defined by

v 983041rarr (the linear map on V lowast which sends f to f(v))

This was the topic of a student presentation and can be found in Section 14 of the book Theessential reason why this isomorphism is ldquonaturalrdquo is that V rarr V lowastlowast can be defined without referenceto a basis Contrast this with the the functor which sends V to its (single) dual V lowast it is still truethat finite-dimensional vector space is isomorphic to its dual but specifying such an isomorphism

15

requires additional data which prevents the identity functor from being ldquonaturally isomorphicrdquo tothe single dual functor

Some history At this point it makes sense to say a bit about the interesting origins of categorytheory in the 1950rsquos Eilenberg and Mac Lane where studying cohomology (or possible homologyone of those two) in algebraic topology which associates algebraic data (a group ring or similar)to a topological space in a way which encodes some of the topology They had two cohomologicalconstructions which ended up proving isomorphic results but they noticed that not only werethe resulting objects the same but the actual constructions themselves were in some sense theldquosamerdquo They thus then needed a way to formalize this notion and invented the concept of anatural transformation to do so where they interpreted two constructions as being the same asbeing compatible with all morphisms

But this then leads to the question what types of objects should a natural transformationoccur between Eilenberg and Mac Lane then had to invent the notion of a ldquofunctorrdquo to describewhat a natural transformation went from and what it went to In their motivating setting theyinvented the notion of a functor to describe more formally properties which their cohomologicalconstructions were supposed to satisfy But this then leads to the question what types of objectsshould a functor map between In other words what type of data should a functor (cohomologicalconstruction in their example) take as input and should it output They finally had to thus inventthe notion of a ldquocategoryrdquo to describe the source and target of a functor so that their cohomologicalconstructions could be interpreted as functors from a category of topological spaces to a categoryof algebraic objects

Thus historically natural transformations came first then functors and then categories Aswith most things in mathematics categories thus arose from attempts to encode what was beingseen in some concrete examples in a more general and formal setting

Representable functors Representable functors are defined in Section 21 of the book but wersquollgive the required definitions here in a hopefully simpler to digest manner The point is that anyobject A in a (locally small) category C gives two ldquonaturalrdquo functors C rarr Set one covariant andthe other contravariant which as wersquoll see pretty much encode all data about A itself We definehA C rarr Set to be the covariant functor defined by

hA(B) = Mor(AB) on objects and

hA(Bfminusrarr C) = the map Mor(AB)

hAfminusminusrarr Mor(AC) defined by g 983041rarr f g

We define hA C rarr Set to the contravariant functor defined by

hA(B) = Mor(BA) on objects and

hA(Bfminusrarr C) = the map Mor(CA)

hAfminusminusrarr Mor(BA) defined by g 983041rarr g f

(The functor hA is called the functor of points of A and wersquoll usually think of it as being a covariantfunctor hA Cop rarr Set The functor hA does not have a standard name) So hA is defined bymorphisms from A and hA is defined by morphisms into A Wersquoll see later the sense in which suchfunctors encode all information about the object A

A functor F C rarr Set is representable if is naturally isomorphic to such a functor so F sim= hA

for some A in the covariant case and F sim= hA for some A in the contravariant case The idea wersquollbuild towards is that we can essentially treat such a functor as being A itself so that representable

16

functors can be thought as being actual objects in C For now here is a key example wersquove seenbefore which we can now formulate using the language of representability

Given AB isin Ob(C) let F C rarr Set be the contravariant functor defined by

F (C) = Mor(CA)timesMor(CB) and F (Cfminusrarr D) = [(g k) 983041rarr (g f k f)]

where (g k) 983041rarr (g f k f) gives a map Mor(DA) timesMor(DB) rarr Mor(CA) timesMor(CB) ieF (D) rarr F (C) In order for this functor to be representable requires the existence of an object inC satisfying

Mor(CA)timesMor(CB) sim= Mor(C representing object)

in a way which is compatible with all morphisms but we have actually already seen this notionelsewhere it is essentially the definition of a product AtimesB for A and B Indeed we claim that Fis naturally isomorphic to hAtimesB as we now show

First to define a natural isomorphism hAtimesBsim= F requires for each C isin Ob(C) a choice of a

bijection (ie isomorphism in Set)

ηC hAtimesB(C)sim=minusrarr F (C)

which in our case looks like

ηC Mor(CAtimesB))sim=minusrarr Mor(CA)timesMor(CB)

Take this to be the map

(Cfminusrarr AtimesB) 983041rarr (pA f pB f)

where pA A times B rarr A and pB A times B rarr B are the two projection morphisms in the definitionof a product The definition of product implies that this map is invertible since given k C rarr Aand ℓ C rarr B there exists a unique h C rarr A times B such that k = pA h and ℓ = pB h Forthese bijections to define a natural isomorphism hAtimesB

sim= F requires that they be compatible withall morphisms in C which means that given any g C rarr D in C the following diagram shouldcommute

hAtimesB(C) hAtimesB(D)

F (C) F (D)

hAtimesB(g)

ηC ηD

Fg

which in the case at hand looks like

Mor(CAtimesB) Mor(DAtimesB)

Mor(CA)timesMor(CB) Mor(DA)timesMor(DB)

minus g

ηC ηD

(minus gminus g)

Take an element k D rarr AtimesB in the upper right Applying the map on top gives the elementk g in the upper left and then applying ηC on the left side gives

(pA (k g) pB (k g))

17

in the lower left Instead starting with k D rarr AtimesB in the upper right applying ηD on the rightgives (pA k pB k) and applying the map on the bottom gives

((pA k) g (pB k) g)

in the lower left which is the same as the previous element in the lower left by the associativity ofcomposition Hence the isomorphisms ηC do define a natural isomorphism between hAtimesB and F Again the point is that you can then interpret the functor F as being the ldquosamerdquo as the productAtimesB since both encode the same data

More Representable Examples

Coproducts As in the case of products the definition of coproducts can also be phrased in termsof a representable functor Given objects A and B in C let F C rarr Set be the covariant functordefined by

F (C) = Mor(AC)timesMor(BC) and F (Cfminusrarr D) = [(g k) 983041rarr (f g f k)]

where the latter is a function Mor(AC)timesMor(BC) rarr Mor(AD)timesMor(BD) This functor isrepresentable if and only if a coproduct A ⊔ B for A and B exists and the representing object isthat coproduct the key is the requirement that we have bijections

Mor(AC)timesMor(BC) sim= Mor(A ⊔BC)

for all C which is essentially the defining property a coproduct for A and B should have

Power set functor contravariant version Let P Set rarr Set be the contravariant powerset functor defined on objects by sending a set to its power set and on morphisms by sendingf A rarr B to the preimage function Pf P (B) rarr P (A) given by S 983041rarr fminus1(S) In order for this tobe representable requires the existence of a set X with natural bijections

P (A) sim= Mor(AX)

for all sets A that is a set X so that mapping into X is the same data as specifying a subset ofthe domain We take X = 0 1 to be a two-element set and the required bijections

ηA P (A) rarr Mor(A 0 1) to be given by S 983041rarr χS

where XS A rarr 0 1 is the indicator or characteristic function S defined by sending elementsof S to 1 and elements of A minus S to 0 The inverse of ηA sends a function g A rarr 0 1 to thepreimage gminus1(1) sube A

To verify that the bijections ηA define the data of a natural isomorphism η P rArr h01 wecheck the commutativity of some diagrams Given a function f A rarr B consider

P (A) P (B)

h01(A) = Mor(A 0 1) h01(B) = Mor(B 0 1)

Pf = fminus1

ηA ηB

minus f

18

Take S isin P (B) in the upper right Applying the map on top gives fminus1(S) isin P (A) and thenapplying the map on the left gives χfminus1(S) isin Mor(A 0 1) On the other hand applying the mapon the right to S isin P (B) gives χS isin Mor(B 0 1) and then applying the map on the bottomgives χS f isin Mor(A 0 1) The two resulting functions χfminus1(S)χS f in the lower left are

χfminus1(S)(x) =

9830831 x isin fminus1(S)

0 otherwiseand χS(f(x)) =

9830831 f(x) isin S

0 otherwise

so these are the same since x isin fminus1(S) if and only if f(x) isin S Thus the diagram above commutesso P is naturally isomorphic to h01 and hence P is representable

Power set functor covariant version Now consider the covariant power set functor stilldenoted by P Set rarr Set which again sends a set to its power set but now sends a functionf A rarr B to the image function Pf P (A) rarr P (B) defined by S 983041rarr f(S) In order for this tobe representable there would have to exist a set X such that P sim= hX which translates into havingnatural bijections

P (S) sim= Mor(XS)

for all sets S However in this case there is no natural choice for X as there is no natural way todetect subsets of S by mapping into S as opposed to the case of P (S) sim= Mor(S 0 1)

That no such X exists can be made precise using a simple argument when S = empty P (S) hascardinality 1 soX would have to be empty as well since otherwise Mor(XS) would have cardinality0 but if X is empty then Mor(empty S) always cardinality 1 regardless of S which is clearly not trueof P (S) Thus the covariant power set functor is not representable

Forgetful on Top The forgetful functor F Top rarr Set is representable Indeed a representingobject would have to satisfy

X sim= Map(objectX)

as sets for any topological space X (where Map denotes the set of continuous maps) and it iseasy to see that a single point satisfies this property a continuous map f pt rarr X is completelydetermined by the element f(pt) of X and any such element gives a continuous map in this wayOf course checking that F sim= hpt requires checking that various diagrams commute but this issimple to work out in this case so we omit it here

Extracting a topology With an eye towards the idea that the functors hA hA should captureall information about an object A in a category note that the result above shows that in TophX first of all captures all information about X as a set since the set X can be extracted fromhX(pt) = Map(ptX)

We claim that hX actually captures all information about X as a topological space since thetopology on X can be extracted from hX in the following way As we saw in the power set examplewe have a bijection

P (X) sim= Mor(X 0 1)

Now equip 0 1 with the topology in which 1 is open but 0 is not Then a continuous mapf X rarr 0 1 is fully determined by the preimage fminus1(1) since to be continuous only requiresthat this single preimage be open in X Moreoever any open subset of X arises in this way bytaking f to be the indicator function of that open set Thus we get a bijection

Op(X) sim= Map(X 0 1)

19

where Op(X) denotes the set of open subsets of X ie the topology on X Hence the topology onX can be extracted from hX as claimed so hX and hX do encode all information about X

This fact can also be interpreted as saying that the contravariant functor which sends a topo-logical space X to its collection of open subsets (and which sends a continuous map to the mapinduced by taking preimages) is representable with representing object 0 1 equipped with thetopology given above This will be worked out in a homework problem

Forgetful on Grp The forgetful functor Grp rarr Set is also representable In this case arepresenting object should be a group giving set-theoretic bijections

G sim= Hom(objectG)

for all groups G where Hom denotes the set of group homomorphisms The group Z works

G sim= Hom(Z G)

since a homomorphism φ Z rarr G is completely determined by φ(1) since Z is generated by 1 andthere are no restrictions on what φ(1) isin G can actually be The inverse of the desired bijection isthus φ 983041rarr φ(1) and it is simple to check that morphisms behave appropriately so that the forgetfulfunctor in this setting is indeed isomoprhic to hZ

Forgetful on Ring The forgetful functor Ring rarr Set is also representable where by Ring wemean the category of rings with identity elements and where morphisms (ie ring homomorphisms)are required to preserve identities Here we need an object satisfying

R sim= Hom(object R)

for all R isin Ob(Ring) and we can see that Z[x] the ring of polynomials in the variable x withcoefficients in Z works Indeed given a ring homomorphism φ Z[x] rarr R the fact that φ isrequired to preserve identities fully determines φ(n) for any n isin Z (since φ(n) = nφ(1)) and hencesince φ preserves multiplication its behavior is fully determined by φ(x) which can be any elementof R The inverse of the bijection we want is thus φ 983041rarr φ(x) and it will follow that this forgetfulfunctor is isomorphic to hZ[x]

Equivalences between Categories

Definitions iso and equiv Now that we have a notion of a ldquomorphismrdquo between categorieswe can talk about the sense in which two categories are the ldquosamerdquo The first possible definitionis the ldquoobviousrdquo one you might expect given the definition of ldquoisomorphismrdquo wersquove seen in everyother context until now we say that C is isomorphic to D if there exist functors F C rarr D andG D rarr C such that G F = idC and F G = idD where idC and idD are identity functors

However this definition turns out to be too restrictive Indeed the requirement that the givencompositions equal identity functors says that given an object A in C the object G(F (A)) beliterally the same as A itself and similarly for objects in D But we know that two objects in acategory can be thought of as being the ldquosamerdquo without being literally the same as long as theyare isomorphic For instance productscoproducts are not unique in the literal sense but only inthe sense that they are isomorphic in a unique way To take another example the definition ofa functor being representable does not say that F (B) literally be the same as hB(A) only thatthey be isomorphic for sure the power set P (S) of a set S is not literally the same as the set offunctions S rarr 0 1mdashit is only the ldquosamerdquo in the sense that they encode the same data

20

Thus it makes sense to relax the requirement that A andG(F (A)) in the definition of isomorphiccategories be equal to the requirement that they be isomorphic This leads to the following betterdefinition of two categories being the ldquosamerdquo C is equivalent to D if there exist functors F C rarr D

and G D rarr C such that G F sim= idC and F G sim= idD So we require only that for instance thefunctor G F be naturally isomorphic to the identity functor on C which says that for any objectA G(F (A)) be isomorphic to A in a natural way Wersquoll see that this truly is the better notionof what it means for two categories to be the ldquosamerdquo and it reflects the idea that all categoricalproperties of one category can be transported to the other

Alternate characterization of equivalences A functor F C rarr D implementing an equiv-alence between categories is called unsurprisingly and equivalence and the ldquoinverserdquo functorG D rarr C is still referred to as being its inverse even though it is not technically an ldquoinverserdquoin the sense that compositions give literal identities But as in the case of Set where ldquoinvertiblerdquocan be characterized without reference to an inverse as ldquoinjectiverdquo and ldquosurjectiverdquo so too canequivalences between categories be characterized without explicit reference to an inverse

A functor F C rarr D is an equivalence if and only it is full faithful and essentiallysurjective

We have seen the notions of full and faithful before and essentially surjective means that any objectin D is isomorphic to something in the image of F for all D isin Ob(D) there exits C isin Ob(C) suchthat F (C) sim= D Justifying this result was the topic of a student presentation and can be foundin the book in Section 15 (Technically to construct an ldquoinverserdquo of F requires that we work withsmall categories but whatever) The bulk of the real work goes into showing that the ldquoinverserdquofunctor constructed is actually a functor which comes down to some abstract nonsense argumentvia commutative diagrams

Equivalences preserve everything We previously saw the notions of full faithful and es-sentially surjective back when thinking about what properties a functor could have which wouldguarantee it sent products to products so the result we proved back then was that equivalencespreserve products Similarly equivalences preserve coproducts and the proof is basically the sameafter reversing arrows A clean way of deriving this is as follows if F C rarr D is an equivalencethen so is F op Cop rarr Dop so since F op preserves products F preserves coproducts becausecoproducts in a category become products in the opposite category

In general an equivalence between categories will preserve whatever categorical notions youwant monomorphisms epimorphisms initial objects terminal objects etc so that equivalentcategories really do have essentially the ldquosamerdquo properties

Equivalence example Here is a first example (also found in the book) of equivalent categorieswhich are not isomorphic Let FVectR be the category whose objects are finite-dimensional realvector spaces and whose morphisms are linear transformations Let MatR be the category ofmatrices defined as follows

bull objects in MatR are nonnegative integers

bull a morphism n rarr m is an mtimes n matrix with real entries

bull composition in MatR is given by matrix multiplication given B n rarr m and A m rarr kmdashsoB is an mtimes n matrix and A is a k timesm matrixmdashthe composition A B n rarr k is the k times nmatrix AB

bull identity morphisms in MatR are given by identity matrices

21

We claim that FVectR and MatR are equivalent First note that they certainly cannot be isomo-prhic isomorphisms in MatR exist only between two objects which are literally the same (ie anmtimesn matrix can be invertible only when m = n) but isomorphisms in FVectn can exist betweenobjects which are not literally the same So there are in a sense not enough objects in MatR toallow it to be isomorphic to FVectR

To construct an equivalence choose one and for all an isomorphism TV V rarr RdimV forevery object in FVectR or equivalently choose once and for all a basis for every V DefineF FVectR rarr MatR by

V 983041rarr dimV on objects and

(T V rarr W ) 983041rarr (the standard matrix of SW T Sminus1V RdimV rarr RdimW ) on morphisms

Equivalently F sends T to the matrix of T relative to the chosen bases for V and W It isstraightforward to check that this is a functor The inverse functor G MatR rarr VectR is givenby

n 983041rarr Rn on objects and

(mtimes n matrix A) 983041rarr (the linear transformation A Rn rarr Rm induced by A) on morphisms

One can check that G is a functor and that F G and G F are naturally isomorphic to identitiesor instead one can argue that F alone is full faithful and essentially surjective The point is thatG(F (V )) gives back RdimV so not literally V itself but only something isomorphic to V which iswhy this gives an equivalence of categories and not an isomorphism

Yoneda Lemma Functors as Objects

Compact Hausdorff spaces and Clowast-algebras Last time we gave an example of equivalentcategories which were not isomorphic but in many ways that example is unsatisfactory when doinglinear algebra people for sure often work with matrix representations of linear transformations asthe example from last would suggest but no one in their right mind literally thinks about suchthings in terms of the equivalence FVectR rarr MatR itself In other words this equivalence isessentially ldquocooked uprdquo for the sake of giving an example but it does not really give any insightinto the actual mathematics of linear algebra

So with that in mind we will describe another equivalence which actually does have some realmeaning in the sense of producing new ways of thinking about some mathematics Given a compactHausdorff space X we let C(X) denote the space of continuous complex-valued functions on X

C(X) = f X rarr C | f is continuous

We can equip the set C(X) with various additional structures First it is a complex vector spaceunder ordinary addition and scalar multiplication of functions Second we can multiply two func-tions together

(fg)(p) = f(p)g(p)

which turns C(X) into whatrsquos called an algebra over C (We wonrsquot give the definition of anldquoalgebrardquo but the point is that this multiplication is in some sense ldquocompatiblerdquo with the vectorspace structure) Next any element f of C(X) has a conjugate element f obtained by takingcomplex conjugates of the values of f

f(p) = f(p)

22

And finally we can equip C(X) with a norm via

983042f983042 = suppisinX

|f(p)|

where |middot| denotes complex absolute value (This supremum exists sinceX is compact by the ExtremeValue Theorem) All of these structures (vector space algebra multiplication conjugation norm)turn C(X) into whatrsquos called a Clowast-algebra We wonrsquot define this precisely but the definition encodesvarious ways in which these structures are compatible Actually C(X) is a unital commutative Clowast-algebra where ldquounitalrdquo means that the multiplication has an identity element (namely the constantfunction 1) and commutative means that fg = gf There is a natural notion of homomorphismbetween Clowast-algebras which are maps between them which behave nicely with all the structuresinvolved In particular given a continuous map f X rarr Y between compact Hausdorff spaces weget a Clowast-algebra homomorphism flowast C(Y ) rarr C(X) given by pre-composition with f flowast sendsg isin C(Y ) to g f isin C(X)

We thus have a contravariant functor CHaus rarr ucClowast-Alg from the category of compactHausdorff spaces and the category of unital commutative Clowast-algebras It turns out that anyunital commutative Clowast-algebra arises in this way in the sense that given any unital commutativeClowast-algebra B there exists a compact Hausdorff space X such that C(X) sim= B as Clowast-algebrasand that morphisms between these algebra arises from continuous maps in the manner describeabove This says that the functor above is essentially surjective full and it turns out that it isalso faithful so that it is an equivalence Thus all information about compact Hausdorff spaces iscompletely encoded in information about unital commtuative Clowast-algebras and indeed one couldlearn everything there is to know about a compact Hausdorff space X from its Clowast-algebra C(X)of functions The fact that the categories CHaus and ucClowast-Alg are equivalent is the startingpoint in the subject known as noncommutative geometry which wersquoll say something about a bitlater on If we drop the requirement that our algebras be unital it turns out that the categoryof commutative Clowast-algebras in general is equivalent to the category of locally compact Hausdorffspaces via essentially the same functor where we take as morphisms in the category of locallycompact Hausdorff spaces to be continuous functions which ldquovanish at infinityrdquo

We are only introducing Clowast-algebras to give an example of an interesting and actually usefulequivalence between categories but just for the fun of it wersquoll note the following Examples ofnoncommutative Clowast-algebras come from what are known as ldquoalgebras of bounded operators onHilbert spacesrdquo which show up in quantum mechanics as the things which describe physicallyobservable quantities like position momentum and energy (In fact all Clowast-algebras arise fromHilbert spaces in this way) Thus Clowast-algebras show up in mathematical formulations of quantummechanics and the notion of noncommutative geometry mentioned above is at the core of someattempts to understand the elusive concept known as ldquoquantum geometryrdquo

Yoneda Lemma The Yoneda Lemma states that given a (covariant) functor F C rarr Set whereC is locally small for any A isin Ob(C) there exists a bijection

F (A) sim= Nat(hA F )

where the right side denotes the set of natural transformations from hA to F A similar result holdsin the contravariant case where we get bijections F (A) sim= Nat(hA F ) (There is also a sense inwhich these bijections are themselves natural but wersquoll ignore this bit) The proof of the YonedaLemma was the topic of a student presentation and can be found in Section 22 of the book

Here we will get a bit philosophical and talk about what the point of the Yoneda Lemmaactually is and later the sense in which it says that we can view arbitrary functors as ldquonon-existent

23

objectsrdquo in a category First we mention what we will call the ldquofunctor of pointsrdquo approach tomathematics a mathematical object can be fully characterized by all morphisms into it That isknowledge of Mor(SA) for all S is equivalent to knowledge of A itself Indeed this is an idea wementioned previously in relation to the functors hAmdashwhich we previously said is referred to as beingthe functor of points of Amdashand now we seek to push this idea further In some examples literalpoints of A can be extracted from hA as in the case of the bijection X sim= hX(pt) = Map(ptX) fora topological space X or G sim= hG(Z) = Mor(Z G) for a group G Moreover in these examples therest of the (topological) structure onX or (group) structure on G can also be extracted from hX andhG respectively In general we can view Mor(SA) as in a sense being ldquopoints of A parameterizedby Srdquo which comes from the idea that in say Set Mor(SA) = AS is ldquothe product of S-manycopies of Ardquo so literally points of A ldquoparameterizedrdquo by S In various areas of mathematics anelement of Mor(SA) is thought of as being a ldquogeneralizedrdquo point of A so that the functor of pointshA detects generalized points of A

So an object A is fully encoded by the functor hA Now we claim that morphisms betweenobjects are also encoded by such functors Indeed in the case where F = hB is itself a functor ofpoints the Yoneda result Nat(hA F ) sim= F (A) becomes

Nat(hA hB) sim= hB(A) = Mor(AB)

which says precisely that morphisms between the objects A and B can be obtained solely fromthe functors hA and hB (This fact is one reason why one should prefer the contravariant functorhA to the covariant functor hA in the covariant case Nat(hA F ) sim= F (A) for F = hB becomesNat(hA hB) sim= hB(A) = Mor(BA) so that morphisms between two objects get reversed whencharacterizing them in terms of representable functors)

Pushing our philosophy further we now know that not only objects but also morphisms betweenobjects are fully encoded by the functors hA The final step is to note that we should not only careabout objects and morphisms between them but that an important part of the data of an objectA is also how arbitrary contravariant functors F C rarr Set behave on A and that this too can beobtained from hA precisely via the Yoneda result that F (A) sim= Nat(hA F ) So objects morphismsbetween objects and how arbitrary functors behave on objects are all encoded in knowledge ofthe functors hA so that these functors truly do capture in the most general sense possible allinformation about A

Functors as nonexistent objects The discussion above is all well and good but it still doesnrsquotquite say why we should care about the fact that we can study on object A by studying hA insteadin that we still havenrsquot spoken about what type of insight about A can be gleamed from hA thatwasnrsquot readily apparent in A itself already Now we argue that the true point of the YonedaLemma is that it gives a way to think about arbitrary functors F Cop rarr Set as ldquoobjectsrdquo in C

itself or really as objects in an ldquoenlargementrdquo of C Certainly for those functors F sim= hA which arerepresentable we can indeed think about F as simply being the object A but now we claim thateven non-representable F can be thought of as being ldquononexistentrdquo objects in C The upshot is thatoften in mathematics we come across things which do not actually exist but the Yoneda Lemmagives a way to (indirectly) study such things anyway by interpreting them as functors instead

The starting point of course is that an actual object A is the same as hA as we have explainedabove Thus by analogy we can interpret a non-representable functor F as being an ldquoobjectrdquo in C

as long as we know what it means to ldquomaprdquo into F from other things which are honestly objects in CThat is if F were indeed representable then F (A) sim= Nat(hA F ) sim= Mor(AF ) would characterizethe functor of points of F (here we abuse notation by using the same letter F to denote both arepresentable functor and the object which represents it) so now we take Mor(AF ) = F (A) as

24

the definition of the ldquofunctor of pointsrdquo of the non-existent object F We are thus taking elementsof F (A) as being what it means to ldquomaprdquo from A into F even when F is not representable Thefact that the Yoneda Lemma gives a way to define maps from A into F thus says that F itself isequipped with the same data that a literal object of A would be equipped with so why shouldnrsquotwe interpret F as a being a type of ldquoobjectrdquo in C as well

To phrase this another way note that the Yoneda Lemma gives rise to a functor

C rarr Fun(CopSet)

where Fun(CopSet) denotes the category whose objects are functors from Cop to Set and morphismsare natural transformation defined on objects by A 983041rarr hA and on morphisms by the bijectionMor(AB) sim= Nat(hA hB) arising from the Yoneda Lemma That this is actually bijection saysthat this functor is full and faithful so that we can view C as being a subcategory of the functorcategory Fun(CopSet) (We cay say that C is isomorphic to the subcategory of Fun(CopSet)consisting of the functors hA or that it is equivalent to the subcategory of Fun(CopSet) consistingof representable functors This functor C rarr Fun(CopSet) is usually called the Yoneda embeddingof C into Fun(CopSet)) Thus we can view Fun(CopSet) as an ldquoenlargmentrdquo of C and the ideawe are trying to make sense of is that other objects in this ldquolargerrdquo category should be interpretedas ldquonon-existent objectsrdquo of C

Just how far can we actually push this idea that arbitrary functors Cop rarr Set should be thoughtof as being ldquoobjectsrdquo of C Pretty far in fact For instance often times in geometry one comesacross a type of ldquosmoothrdquo space on which some group acts and one is interested in doing ldquocalculusrdquoon the resulting quotient space which is the space of orbits under the given group action Theproblem is that such quotient spaces are very often (usually in fact) not ldquosmoothrdquo themselves sothat the tools from calculus are not readily available The solution is to nonetheless treat thisquotient as if it were ldquosmoothrdquo and attempt to do calculus anyway To make this precise thisldquonon-smoothrdquo space actually turns out to have a natural interpretation as a ldquosmooth functorrdquo andit is through this functor which we can view this non-smooth space as actually being smooth afterall so we are forced to expand our notion of what we mean by ldquosmooth spacerdquo by allowing thingswhich are not spaces in the usual sense which certainly leads to more abstraction but the benefitis that it actually makes sense to do calculus on such ldquospacesrdquo interpreted as functors

The discussion above is pretty vague so here is one final more concrete example Recall thatequivalence

CHausop rarr ucClowast-Alg

between the category of compact Hausdorff spaces and the category of unital commutative Clowast-algebras mentioned at the start We now claim that a non-commutative (unital) Clowast-algebra shouldbe interpreted as the algebra of functions on some non-existent ldquononcommutative compact Haus-dorff spacerdquo and it is this idea which the subject of noncommutative geometry makes precise Solet B be a non-commutative Clowast-algebra Then the functor of points hB gives a functor

ucClowast-Alg rarr Set

which sends C(X) (where X is some compact Hausdorff space) to Hom(C(X) B) where Hom de-notes homomorphisms of Clowast-algebras whatever that means Composing with the previous functorwe get a functor wersquoll denote by B itself

B CHausop rarr Set

which on objects at least is given by X 983041rarr Hom(C(X) B) This functor is not representable sinceB is not commutative but the philosophy espoused above suggests we should think of it as giving a

25

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

A and g to elements of B (Note that if we simply took the ordinary union AcupB as the candidatefor the coproduct when A and B were not disjoint the required map h A cup B rarr D would notexist since h would have to send any element x isin A cap B to both f(x) and g(x) in order to makethe required diagram commutative)

The coproduct of two objects X and Y in Top is again the disjoint union X ⊔ Y equippedwith the disjoint union topology an open subset of X ⊔ Y is defined to be the (disjoint) union ofan open subset of X with an open subset of Y The coproduct of two groups G and H in Grp isthe free product G lowastH which is further elaborated on in the homework

The coproduct of two vector spaces V and W in Vect is the direct sum V oplusW which is definedto be the set of formal sums

v + w where v isin V and w isin W

where addition and scalar multiplication are defined ldquocomponentwiserdquo

(v1 + w1) + (v2 + w2) = (v1 + v2) + (w1 + w2) and r(v + w) = rv + rw

This is isomorphic to the product V times W with operations also defined componentwise but it iscustomary to speak about the ldquodirect sumrdquo V oplusW when referring to the coproduct as opposed tothe product wersquoll say why in a bit The maps iV V rarr V oplusW and iW W rarr V oplusW required aspart of the data of a coproduct are

iV v 983041rarr v + 0W and iW w 983041rarr 0V + w

To verify that V oplusW is the correct coproduct take any linear transformations S V rarr U andT W rarr U where U is some other vector space We then need a (unique) linear transformationh V oplusW rarr U such that

S = h iV and T = h iW

Let us determine what this map must be thereby not only showing that it exists but also that itis unique Let v + w isin V oplusW which we can write as

v + w = (v + 0W ) + (0V + w)

Since h is required to be linear we have

h(v + w) = h([v + 0W ] + [0V + w]) = h(v + 0W ) + h(0V + w)

The first term at the end is (h iV )v which must thus be Sv and the second term is (h iW )wwhich must be Tw Thus h = S + T is the map

h = S + T v + w 983041rarr Sv + Tw

which you can verify is indeed linear Thus V oplusW does satisfy the property required of a coproductSo the product of two objects in Vect is the same as their coproduct only that we write the

latter using ldquosumrdquo notation Later we will discuss products and coproducts of an arbitrary numberof elements We will see these as special cases of the general notions of limits and colimits butthey are defined via similar requirements as products and coproducts of two objects at a timeIn the category of vector spaces arbitrary products are given by the usual product

983124α Vα with

ldquocomponentwiserdquo addition and scalar multiplication but it turns out that an arbitrary coproductis given by the direct sum

983119α Vα which is defined to be the subspace of

983124α Vα where only finitely

many components are nonzero (This is in a way analogous to how the product topology on an

9

arbitrary number of topological spaces is defined) Wersquoll go over this later but is the underlyingreason why we use oplus instead of times when discussing coproducts

Encoding data via productscoproducts We can nicely summarize the definition of productsand coproducts in the following way the product of A and B is the object with the property thatfor any object D we have

Mor(DA)timesMor(DB) = Mor(D product)

and the coproduct of A and B is the object such that for any object D we have

Mor(AD)timesMor(BD) = Mor(coproductx D)

The first reflects the fact that products give a unique way to turn two morphisms (f and g in thedefinition) mapping into A and B into a single morphism (h in the definition) and the second saysthat coproducts give a unique way to turn two morphisms mapping out of A and B into a singlemorphism

Soon enough we will discuss the idea of universality and the point is that products are ldquouni-versalrdquo for maps into A and B and coproducts are universal for maps from A and B

Opposite categories Finally we give a definition which might seem silly but is actually quiteuseful Given a category C the opposite category is the category Cop is the category with the sameobjects as C but where we defined Mor(AB) in Cop to be Mor(BA) in C To be more concrete weconsider each morphism f A rarr B in C to instead be a morphism fop B rarr A in Cop (Perhaps itwould be better to write this as A larr B fop) The point is that we donrsquot actually care about whatf actually is as say a functioncontinuous maphomomorphismwhatever-we only care about itsbehavior as an ldquoarrowrdquo The opposite category Cop only retains information about how arrowsrelate to one another nothing more

Since arrows are reversed products in C become coproducts in Cop and coproduts in C becomecoproducts in Cop Similarly initialterminal objects and monomorphismsepimorphisms in C be-come terminalinitial objects and epimorphismsmonomorphisms in Cop Again the notion of anopposite category might seem like a strange thing to look at but wersquoll see that it does have uses

Functors Fullness and Faithfulness

One more productcoproduct example Let X be a topological space and define its categoryof open sets Op(X) as follows The objects of Op(X) are simply the open subsets of X and themorphisms are given by inclusions so to be concrete

Mor(U V ) =

983083the unique inclusion U rarr V if U sube V

empty otherwise

The fact that U sube V and V sube W implies U sube W says that the composition of two morphismsin this category is still a morphism in this category The product of U and V in this categoryturns out to be their intersection (with ldquoprojection morphismsrdquo U cap V rarr U and U cap V rarr V beinginclusions) and the coproduct of U and V turns out to be their (ordinary) union Note that weneed X to be a topological space in order to guarantee that U cap V and U cup V are still objects inthis category More generally the product of finitely many objects in Op(X) exists and is theirintersection and the coproduct of an arbitrary number of objects exists and is their union (Wesay that Op(X) has finite products and arbitrary coproducts)

10

Given a set X and a collection T of subsets of X we can define a similar category with elementsof T as objects (I guess we should assume that empty X isin T) and inclusions as morphisms We thenhave that this category has finite products and arbitrary coproducts if and only if T is a topologyon X so that we can rephrase the definition of a topological space itself solely in categorical termsThis perspective on what a ldquotopologyrdquo is makes it simpler to state the definition of whatrsquos calleda sheaf on a topological space which is an important construction in various areas of analysis andgeometry We might discuss this a bit later on

Functors As with most things in mathematics we should care not only about categories asmathematical structures on their own but also about ldquomapsrdquo between these structures whichldquopreserverdquo said structure The notion of a (covariant) functor between categories provides suchldquomapsrdquo in the case at hand where a functor is a way of sending objects to objects and morphismto morphisms in a way which preserves compositions and identities The precise definition is inSection 13 of the book as is the definition of a contravariant functor between categories whichis a functor which reverses the direction of arrows as opposed to ldquocovariantrdquo ones which preservedirections (It is common to use the term ldquofunctorrdquo without qualification to mean one which iscovariant Note that any functor can be made covariant by working with the opposite category acontravariant functor C rarr D is the same as a covariant functor Cop rarr D)

All of the standard examples we looked at are in the book forgetful functors power set functorsthe dual space functor in Vect etc Of particular interest is Example 139 in the book whichshows that various types of group actions in different contexts-the usual notion of a group actionon a set the notion of a continuous action of a group on a topological space and the notion ofa linear action of a group on a vector space (also known as a representation of the group)-are allencoded by a functor with domain category BG

Adjoint functors Let F Top rarr Set be the forgetful functor and let D Set rarr Top be thefunctor which sends a set S to the space S equipped with the discrete topology and a functionS rarr Sprime between sets to the same function only now viewed as a continuous map S rarr Sprime betweendiscrete spaces These two functors satisfy the following property in relation to one another givinga morphism D(S) rarr Y in Top is the same as giving a morphism S rarr F (Y ) in Set or moresymbolically

MorTop(D(S) Y ) = MorSet(S F (Y ))

We say that D is left adjoint to F and that F is right adjoint to D The point is that adjointfunctors can be used to move data back and forth between two categories

Wersquoll talk more about adjoint functors later on but for now here is one more example Nowtake F Grp rarr Set to again be the forgetful functor and take G Grp rarr Set to be the freegroup functor which sends a set S to the free group generated by S G(S) has as elements ldquowordsrdquos1 sn made up of elements of S with the group operation being ldquoconcatenationrdquo A basic factis that giving a group homomorphism from G(S) to some other group H is the same as giving anordinary function from S to F (H) which is the required adjoint property

MorGrp(G(S) H) = MorSet(S F (H))

Faithful and full functors The definitions of what it means for a functor to be faithful and fullare in Section 15 of the book where faithful functors are ones which are injective on morphisms andfull functors are ones which are surjective on morphisms but where in both cases we refer only tomorphisms which occur between the objects F (A) and F (B) in the target category corresponding tofixed objects A and B in the source category (In general a functor could send two objects AB in

11

the source category to the same object F (A) = F (B) in the target category and thus there could betwo morphisms A rarr C and B rarr C which are sent to the same morphism F (A) = F (B) rarr F (C)Such a thing could still possibly be ldquofaithfulrdquo since the non-injectivity here is arising from differentobjects in the source)

Faithfullness and fullness will be nice properties going forward and here is one hint at how theymight be useful Say we have a functor F C rarr D The problem is to determine properties onF which will imply it send products in C to products in D or coproducts in C to coproducts inD So given the former product diagram below we want the second diagram to also be a productdiagram

C

A B

P Fminusminusminusrarr

F (C)

F (A) F (B)

F (P )f g

pA pB

existh

Ff Fg

F (pA) F (pB)

Fh

Of course in order F (P ) to be an actual product of F (A) and F (B) on the right a similarcommutative diagram property would have to hold for all objects in D in place of F (C) and allmorphisms D rarr F (A) and D rarr F (B) not just those of the form Ff and Fg respectively Oneway to guarantee this is to require that all objects in D are of the form F (C) and then for allmorphisms F (C) rarr F (A) and F (C) rarr F (B) to be of the form Ff Fg the former is true whenthe functor F is surjective on objects and the latter is true when F is full We are not saying thatthese are the only conditions under which F might preserve products but it definitely seems togive a sufficient condition at least

Almost there is one more thing to require the uniqueness of the morphism Fh F (C) rarr F (P )making the diagram on the right commute Suppose there were two such morphisms Fh and Fhprimewhere h and hprime are both morphisms C rarr P in C The commutativity of the diagram on the rightgives for instance

Ff = F (pA)Fh and Ff = F (pA)Fhprime

which the functorial properites of F turns into

Ff = F (pA h) and Ff = (pA hprime)

We want h and hprime to make the first diagram commute in order to be able to apply uniqueness ofthe morphism C rarr P But this requires know that f = pA h and f = pA hprime which would followfrom F being faithful In this case it follows that h and hprime both make the first diagram commute(after we go through a similar argument with g) so the fact that P is a product guarantees thath = hprime and hence that Fh = Fhprime so that F (P ) is an actual product on the right

So we conclude that a functor which is full faithful and surjective on objects will send productsto products (Actually we can relax this a bit by requiring not that every object in D be in theliteral image of F but only that it be isomorphic to something in the image A functor with thisproperty is said to be essentially surjective which is a concept will see in a related context soonenough) Again this is not an ldquoif and only ifrdquo statement since functors can preserve products (orcoproducts) without being full faithful and surjective on objects but it at least gives a sufficientcondition Later we will see other ways to guarantee that functors preserve products which arerelated to the existence of adjoints for that given functor

12

Coproduct Examples Concreteness

Products and coproducts for metric spaces The product of two objects in Met the categoryof metric spaces with morphisms given by metric maps is the Cartesian product X times Y equippedwith the box metric and the coproduct of X and Y does not exist if both X and Y are nonemptyThis was the topic of a student presentation and here are some proof sketches

Given metric spaces (X dX) and (Y dY ) the box metric d on X times Y is defined by

d((x1 y1) (x2 y2)) = maxdX(x1 x2) dY (y1 y2))

(The name ldquoboxrdquo metric comes from the fact that in RtimesR = R2 or R3 open balls indeed looks likeldquoboxesrdquo) The key requirement needed in the definition of a product is given morphisms S rarr Xand S rarr Y that the map

h S rarr X times Y defined by h(s) = (f(s) g(s))

actually be a morphism in Met This turns into the requirement that

d((f(s) g(s)) ((f(t) g(t))) = maxdX(f(s) f(t)) dY (g(s) g(t)) le dS(s t) for all s t isin S

given the fact that

dX(f(s) f(t)) le dS(s t) and dY (g(s) g(t)) le dS(s t) for all s t isin S

For the latter inequalities to imply the former d must (at least up to isometry) be given by thebox metric Note that the two other ldquostandardrdquo metrics one might put on a productmdashthe taxicab

dX + dY and Euclidean983156

d2X + d2Y metricsmdashdo not satisfy this required ldquometric maprdquo property

and so do not give the product in MetTo see that coproducts do not exist if XY are nonempty suppose instead (P dP ) was a co-

product for X and Y with inclusion morphisms iX X rarr P and iY Y rarr P For any n isin Ntake 0 n to be a metric space with the absolute value metric d and let f X rarr 0 n andg Y rarr 0 n be the constant function 0 and the constant function n respectively Since P is acoproduct there exists h P rarr 0 n such that

h iX = f and h iY = g

But in order for h to actually be a morphism in Met we must then have

d(h(iX(x) iY (y)) le dP (iX(x) iY (y)) for any x isin X y isin Y

which translates inton = |0minus n| le dP (iX(x) iY (y))

But this should then be true for any n isin N which would imply that the distance in P betweenan element coming from X and one coming from Y should be infinite which is nonsense (Notethat if you altered your definition of ldquometricrdquo to allow for infinite distance then a coproduct wouldexist the disjoint union X ⊔Y where you defined the distance between elements of X and Y to beinfinite)

Coproducts for groups The coproduct of two groups G and H in Grp is the free product GlowastHwhich is defined to be the set of all words consisting of elements of G andH in an alternating fashion

13

with concatenation as the group operation (When we have to elements of G or two elements of Hnext to each other with use their respective group operations to turn these into single elements ofG or H) This was also the topic of a student presentation and here is the basic idea Given grouphomomorphisms g G rarr N and k H rarr N that G lowastH is the coproduct comes down to showingthat the map

h G lowastH rarr N defined by h(g1h1 gnhn) = f(g1)k(h1) middot middot middot f(gn)k(hn)

and similarly on other possible words in GlowastH is a group homomorphism which follows essentiallyfrom the definition of the group operation on G lowastH indeed this group operation arises preciselyfrom this requirement

Now if we instead tried to use GtimesH with its standard inclusions G rarr GtimesH and H rarr GtimesHas the coproduct the issue is that the analogous map

h GtimesH rarr N defined by h((g h)) = f(g)k(h)

would NOT be a homomorphism because h((g1 h1)(g2 h2)) is not equal to h(g1 h1)h(g2 h2) bythe way in which the group operation is defined on the direct product G times H However if westrict our category to only consist of abelian groups so that the N rsquos we consider in the defini-tion of coproduct are also abelian then G times H does serve as a coproduct we can always rewritef(g1)k(g1) middot middot middot f(gn)k(gn) as f(g1) middot middot middot f(gn)k(h1) middot middot middot k(hn) in N Irsquoll leave it to you to flesh out allof these details

Concrete categories The categories Set Top Vect and Grp all have the property thattheir objects are simply sets possibly equipped with extra structure and that their morphisms areordinary set-theoretic functions possibly satisfying some additional requirement Such categoriesare nice since we can often use intuition about sets and functions in their study A concrete categoryis intuitively one where we can think of objects as being ldquosetsrdquo and morphisms as being ldquofunctionsrdquoonly that we have to interpret this in the right way

Here is the definition a concrete category is a category C equipped with a faithful functorF C rarr Set which is part of the data The idea is that such a functor gives us way to turnan object A isin Ob(C) into a set F (A) and a morphism f isin MorC(AB) into a function Ff isinMorSet(F (A) F (B)) where the ldquofaithfulrdquo requirement says that we do not lose information aboutf when doing so so that we can (essentially) learn everything we need to know about f by studyingFf instead In the four examples above the required faithful functor is simply the forgetful functorand indeed in general we think of F as being a ldquoforgetful functorrdquo for the concrete category (C F )

Examples Here are two more examples of concrete categories Let G be a group and considerthe category BG with a single object and elements of G morphisms As written this does notappear to be concrete since morphisms are not honest functions lowast rarr lowast but the point is that thiscategory can indeed by ldquoconcretizedrdquo by specifying an appropriate ldquoforgetful functorrdquo A functorF BG rarr Set is the data of a (left) action of G on a set S = F (lowast) and saying this functor isfaithful is what it means to say that this action is faithful different g h isin G act on S in differentways or equivalently that there exists s isin S such that gs ∕= hs Any group has at least onefaithful left actionmdashnamely the action on itself given by left multiplicationmdashto BG can always beequipped with a faithful functor to Set This is all just a way of phrasing the fact that any groupis isomorphic to a subgroup of a permutation group

The category Rel of relations can also be ldquoconcretizedrdquo even though as written morphismsA rarr B are not honest functions Take F Rel rarr Set to be the power set functor which sends an

14

object in Rel to its power set and a relation f A rarr B to the induced function Ff P (A) rarr P (B)defined by

(Ff)(S) = b isin B | there exists a isin S such that (a b) isin f

In other words if you think of the relation f sube A times B as a ldquopossibly not everywhere-definedpossibly multivalued functionrdquo (Ff)(S) is analogous to taking the image of S That F is faithfulwill be left to the homework

In fact most categories yoursquoll see can be made concrete by specifying an appropriate forgetfulfunctor but not all categories can be made concrete in this way Here is the standard example of anon-concrete category which we mention only to say there is such a thing but which we wonrsquot doanything more with since understanding it better requires knowledge of algebraic topology Thehomotopy category of topological spaces is the category hTop whose objects are topological spacesand whose morphisms are given by equivalence classes of homotopic maps

MorhTop(XY ) = homotopy classes of continuous maps X rarr Y

(Saying that two maps are homotopic means that you can ldquodeformrdquo one into the other but wersquollleave the precise definition to a course in algebraic topology) It turns out that no functor fromhTop to Set can be faithful so that hTop cannot be made concrete but this is actually a deepand difficult thing to show so we make no attempt to do so here

Natural Isomorphisms Representability

Natural transformations The definition of a natural transformation η F rArr G between two(both covariant or both contravariant) functors FG C rarr D is given in Section 14 of the bookThe key idea is that a natural transformation gives a way to turn data about F into data aboutG in a way which is compatible with all possible morphisms as expressed by the commutativity ofthe diagrams

F (A) F (B)

G(A) G(B)

Ff

ηA ηB

Gf

arising from morphisms f A rarr B (or f B rarr A in the contravariant case) in C In the case of anatural isomoprhism η F rArr G so that each ηA F (A) rarr G(A) is actually an isomorphism in Dthe point is not only that F and G produce isomorphic objects but they do so in a way which iscompatible with all possible category-theoretic data

The first standard example of a natural isomorphism is the one between the identity functor onthe category of finite-dimensional and the functor which sends a finite-dimensional vector space Vto V lowastlowast the dual of its dual induced by the isomorphism V rarr V lowastlowast defined by

v 983041rarr (the linear map on V lowast which sends f to f(v))

This was the topic of a student presentation and can be found in Section 14 of the book Theessential reason why this isomorphism is ldquonaturalrdquo is that V rarr V lowastlowast can be defined without referenceto a basis Contrast this with the the functor which sends V to its (single) dual V lowast it is still truethat finite-dimensional vector space is isomorphic to its dual but specifying such an isomorphism

15

requires additional data which prevents the identity functor from being ldquonaturally isomorphicrdquo tothe single dual functor

Some history At this point it makes sense to say a bit about the interesting origins of categorytheory in the 1950rsquos Eilenberg and Mac Lane where studying cohomology (or possible homologyone of those two) in algebraic topology which associates algebraic data (a group ring or similar)to a topological space in a way which encodes some of the topology They had two cohomologicalconstructions which ended up proving isomorphic results but they noticed that not only werethe resulting objects the same but the actual constructions themselves were in some sense theldquosamerdquo They thus then needed a way to formalize this notion and invented the concept of anatural transformation to do so where they interpreted two constructions as being the same asbeing compatible with all morphisms

But this then leads to the question what types of objects should a natural transformationoccur between Eilenberg and Mac Lane then had to invent the notion of a ldquofunctorrdquo to describewhat a natural transformation went from and what it went to In their motivating setting theyinvented the notion of a functor to describe more formally properties which their cohomologicalconstructions were supposed to satisfy But this then leads to the question what types of objectsshould a functor map between In other words what type of data should a functor (cohomologicalconstruction in their example) take as input and should it output They finally had to thus inventthe notion of a ldquocategoryrdquo to describe the source and target of a functor so that their cohomologicalconstructions could be interpreted as functors from a category of topological spaces to a categoryof algebraic objects

Thus historically natural transformations came first then functors and then categories Aswith most things in mathematics categories thus arose from attempts to encode what was beingseen in some concrete examples in a more general and formal setting

Representable functors Representable functors are defined in Section 21 of the book but wersquollgive the required definitions here in a hopefully simpler to digest manner The point is that anyobject A in a (locally small) category C gives two ldquonaturalrdquo functors C rarr Set one covariant andthe other contravariant which as wersquoll see pretty much encode all data about A itself We definehA C rarr Set to be the covariant functor defined by

hA(B) = Mor(AB) on objects and

hA(Bfminusrarr C) = the map Mor(AB)

hAfminusminusrarr Mor(AC) defined by g 983041rarr f g

We define hA C rarr Set to the contravariant functor defined by

hA(B) = Mor(BA) on objects and

hA(Bfminusrarr C) = the map Mor(CA)

hAfminusminusrarr Mor(BA) defined by g 983041rarr g f

(The functor hA is called the functor of points of A and wersquoll usually think of it as being a covariantfunctor hA Cop rarr Set The functor hA does not have a standard name) So hA is defined bymorphisms from A and hA is defined by morphisms into A Wersquoll see later the sense in which suchfunctors encode all information about the object A

A functor F C rarr Set is representable if is naturally isomorphic to such a functor so F sim= hA

for some A in the covariant case and F sim= hA for some A in the contravariant case The idea wersquollbuild towards is that we can essentially treat such a functor as being A itself so that representable

16

functors can be thought as being actual objects in C For now here is a key example wersquove seenbefore which we can now formulate using the language of representability

Given AB isin Ob(C) let F C rarr Set be the contravariant functor defined by

F (C) = Mor(CA)timesMor(CB) and F (Cfminusrarr D) = [(g k) 983041rarr (g f k f)]

where (g k) 983041rarr (g f k f) gives a map Mor(DA) timesMor(DB) rarr Mor(CA) timesMor(CB) ieF (D) rarr F (C) In order for this functor to be representable requires the existence of an object inC satisfying

Mor(CA)timesMor(CB) sim= Mor(C representing object)

in a way which is compatible with all morphisms but we have actually already seen this notionelsewhere it is essentially the definition of a product AtimesB for A and B Indeed we claim that Fis naturally isomorphic to hAtimesB as we now show

First to define a natural isomorphism hAtimesBsim= F requires for each C isin Ob(C) a choice of a

bijection (ie isomorphism in Set)

ηC hAtimesB(C)sim=minusrarr F (C)

which in our case looks like

ηC Mor(CAtimesB))sim=minusrarr Mor(CA)timesMor(CB)

Take this to be the map

(Cfminusrarr AtimesB) 983041rarr (pA f pB f)

where pA A times B rarr A and pB A times B rarr B are the two projection morphisms in the definitionof a product The definition of product implies that this map is invertible since given k C rarr Aand ℓ C rarr B there exists a unique h C rarr A times B such that k = pA h and ℓ = pB h Forthese bijections to define a natural isomorphism hAtimesB

sim= F requires that they be compatible withall morphisms in C which means that given any g C rarr D in C the following diagram shouldcommute

hAtimesB(C) hAtimesB(D)

F (C) F (D)

hAtimesB(g)

ηC ηD

Fg

which in the case at hand looks like

Mor(CAtimesB) Mor(DAtimesB)

Mor(CA)timesMor(CB) Mor(DA)timesMor(DB)

minus g

ηC ηD

(minus gminus g)

Take an element k D rarr AtimesB in the upper right Applying the map on top gives the elementk g in the upper left and then applying ηC on the left side gives

(pA (k g) pB (k g))

17

in the lower left Instead starting with k D rarr AtimesB in the upper right applying ηD on the rightgives (pA k pB k) and applying the map on the bottom gives

((pA k) g (pB k) g)

in the lower left which is the same as the previous element in the lower left by the associativity ofcomposition Hence the isomorphisms ηC do define a natural isomorphism between hAtimesB and F Again the point is that you can then interpret the functor F as being the ldquosamerdquo as the productAtimesB since both encode the same data

More Representable Examples

Coproducts As in the case of products the definition of coproducts can also be phrased in termsof a representable functor Given objects A and B in C let F C rarr Set be the covariant functordefined by

F (C) = Mor(AC)timesMor(BC) and F (Cfminusrarr D) = [(g k) 983041rarr (f g f k)]

where the latter is a function Mor(AC)timesMor(BC) rarr Mor(AD)timesMor(BD) This functor isrepresentable if and only if a coproduct A ⊔ B for A and B exists and the representing object isthat coproduct the key is the requirement that we have bijections

Mor(AC)timesMor(BC) sim= Mor(A ⊔BC)

for all C which is essentially the defining property a coproduct for A and B should have

Power set functor contravariant version Let P Set rarr Set be the contravariant powerset functor defined on objects by sending a set to its power set and on morphisms by sendingf A rarr B to the preimage function Pf P (B) rarr P (A) given by S 983041rarr fminus1(S) In order for this tobe representable requires the existence of a set X with natural bijections

P (A) sim= Mor(AX)

for all sets A that is a set X so that mapping into X is the same data as specifying a subset ofthe domain We take X = 0 1 to be a two-element set and the required bijections

ηA P (A) rarr Mor(A 0 1) to be given by S 983041rarr χS

where XS A rarr 0 1 is the indicator or characteristic function S defined by sending elementsof S to 1 and elements of A minus S to 0 The inverse of ηA sends a function g A rarr 0 1 to thepreimage gminus1(1) sube A

To verify that the bijections ηA define the data of a natural isomorphism η P rArr h01 wecheck the commutativity of some diagrams Given a function f A rarr B consider

P (A) P (B)

h01(A) = Mor(A 0 1) h01(B) = Mor(B 0 1)

Pf = fminus1

ηA ηB

minus f

18

Take S isin P (B) in the upper right Applying the map on top gives fminus1(S) isin P (A) and thenapplying the map on the left gives χfminus1(S) isin Mor(A 0 1) On the other hand applying the mapon the right to S isin P (B) gives χS isin Mor(B 0 1) and then applying the map on the bottomgives χS f isin Mor(A 0 1) The two resulting functions χfminus1(S)χS f in the lower left are

χfminus1(S)(x) =

9830831 x isin fminus1(S)

0 otherwiseand χS(f(x)) =

9830831 f(x) isin S

0 otherwise

so these are the same since x isin fminus1(S) if and only if f(x) isin S Thus the diagram above commutesso P is naturally isomorphic to h01 and hence P is representable

Power set functor covariant version Now consider the covariant power set functor stilldenoted by P Set rarr Set which again sends a set to its power set but now sends a functionf A rarr B to the image function Pf P (A) rarr P (B) defined by S 983041rarr f(S) In order for this tobe representable there would have to exist a set X such that P sim= hX which translates into havingnatural bijections

P (S) sim= Mor(XS)

for all sets S However in this case there is no natural choice for X as there is no natural way todetect subsets of S by mapping into S as opposed to the case of P (S) sim= Mor(S 0 1)

That no such X exists can be made precise using a simple argument when S = empty P (S) hascardinality 1 soX would have to be empty as well since otherwise Mor(XS) would have cardinality0 but if X is empty then Mor(empty S) always cardinality 1 regardless of S which is clearly not trueof P (S) Thus the covariant power set functor is not representable

Forgetful on Top The forgetful functor F Top rarr Set is representable Indeed a representingobject would have to satisfy

X sim= Map(objectX)

as sets for any topological space X (where Map denotes the set of continuous maps) and it iseasy to see that a single point satisfies this property a continuous map f pt rarr X is completelydetermined by the element f(pt) of X and any such element gives a continuous map in this wayOf course checking that F sim= hpt requires checking that various diagrams commute but this issimple to work out in this case so we omit it here

Extracting a topology With an eye towards the idea that the functors hA hA should captureall information about an object A in a category note that the result above shows that in TophX first of all captures all information about X as a set since the set X can be extracted fromhX(pt) = Map(ptX)

We claim that hX actually captures all information about X as a topological space since thetopology on X can be extracted from hX in the following way As we saw in the power set examplewe have a bijection

P (X) sim= Mor(X 0 1)

Now equip 0 1 with the topology in which 1 is open but 0 is not Then a continuous mapf X rarr 0 1 is fully determined by the preimage fminus1(1) since to be continuous only requiresthat this single preimage be open in X Moreoever any open subset of X arises in this way bytaking f to be the indicator function of that open set Thus we get a bijection

Op(X) sim= Map(X 0 1)

19

where Op(X) denotes the set of open subsets of X ie the topology on X Hence the topology onX can be extracted from hX as claimed so hX and hX do encode all information about X

This fact can also be interpreted as saying that the contravariant functor which sends a topo-logical space X to its collection of open subsets (and which sends a continuous map to the mapinduced by taking preimages) is representable with representing object 0 1 equipped with thetopology given above This will be worked out in a homework problem

Forgetful on Grp The forgetful functor Grp rarr Set is also representable In this case arepresenting object should be a group giving set-theoretic bijections

G sim= Hom(objectG)

for all groups G where Hom denotes the set of group homomorphisms The group Z works

G sim= Hom(Z G)

since a homomorphism φ Z rarr G is completely determined by φ(1) since Z is generated by 1 andthere are no restrictions on what φ(1) isin G can actually be The inverse of the desired bijection isthus φ 983041rarr φ(1) and it is simple to check that morphisms behave appropriately so that the forgetfulfunctor in this setting is indeed isomoprhic to hZ

Forgetful on Ring The forgetful functor Ring rarr Set is also representable where by Ring wemean the category of rings with identity elements and where morphisms (ie ring homomorphisms)are required to preserve identities Here we need an object satisfying

R sim= Hom(object R)

for all R isin Ob(Ring) and we can see that Z[x] the ring of polynomials in the variable x withcoefficients in Z works Indeed given a ring homomorphism φ Z[x] rarr R the fact that φ isrequired to preserve identities fully determines φ(n) for any n isin Z (since φ(n) = nφ(1)) and hencesince φ preserves multiplication its behavior is fully determined by φ(x) which can be any elementof R The inverse of the bijection we want is thus φ 983041rarr φ(x) and it will follow that this forgetfulfunctor is isomorphic to hZ[x]

Equivalences between Categories

Definitions iso and equiv Now that we have a notion of a ldquomorphismrdquo between categorieswe can talk about the sense in which two categories are the ldquosamerdquo The first possible definitionis the ldquoobviousrdquo one you might expect given the definition of ldquoisomorphismrdquo wersquove seen in everyother context until now we say that C is isomorphic to D if there exist functors F C rarr D andG D rarr C such that G F = idC and F G = idD where idC and idD are identity functors

However this definition turns out to be too restrictive Indeed the requirement that the givencompositions equal identity functors says that given an object A in C the object G(F (A)) beliterally the same as A itself and similarly for objects in D But we know that two objects in acategory can be thought of as being the ldquosamerdquo without being literally the same as long as theyare isomorphic For instance productscoproducts are not unique in the literal sense but only inthe sense that they are isomorphic in a unique way To take another example the definition ofa functor being representable does not say that F (B) literally be the same as hB(A) only thatthey be isomorphic for sure the power set P (S) of a set S is not literally the same as the set offunctions S rarr 0 1mdashit is only the ldquosamerdquo in the sense that they encode the same data

20

Thus it makes sense to relax the requirement that A andG(F (A)) in the definition of isomorphiccategories be equal to the requirement that they be isomorphic This leads to the following betterdefinition of two categories being the ldquosamerdquo C is equivalent to D if there exist functors F C rarr D

and G D rarr C such that G F sim= idC and F G sim= idD So we require only that for instance thefunctor G F be naturally isomorphic to the identity functor on C which says that for any objectA G(F (A)) be isomorphic to A in a natural way Wersquoll see that this truly is the better notionof what it means for two categories to be the ldquosamerdquo and it reflects the idea that all categoricalproperties of one category can be transported to the other

Alternate characterization of equivalences A functor F C rarr D implementing an equiv-alence between categories is called unsurprisingly and equivalence and the ldquoinverserdquo functorG D rarr C is still referred to as being its inverse even though it is not technically an ldquoinverserdquoin the sense that compositions give literal identities But as in the case of Set where ldquoinvertiblerdquocan be characterized without reference to an inverse as ldquoinjectiverdquo and ldquosurjectiverdquo so too canequivalences between categories be characterized without explicit reference to an inverse

A functor F C rarr D is an equivalence if and only it is full faithful and essentiallysurjective

We have seen the notions of full and faithful before and essentially surjective means that any objectin D is isomorphic to something in the image of F for all D isin Ob(D) there exits C isin Ob(C) suchthat F (C) sim= D Justifying this result was the topic of a student presentation and can be foundin the book in Section 15 (Technically to construct an ldquoinverserdquo of F requires that we work withsmall categories but whatever) The bulk of the real work goes into showing that the ldquoinverserdquofunctor constructed is actually a functor which comes down to some abstract nonsense argumentvia commutative diagrams

Equivalences preserve everything We previously saw the notions of full faithful and es-sentially surjective back when thinking about what properties a functor could have which wouldguarantee it sent products to products so the result we proved back then was that equivalencespreserve products Similarly equivalences preserve coproducts and the proof is basically the sameafter reversing arrows A clean way of deriving this is as follows if F C rarr D is an equivalencethen so is F op Cop rarr Dop so since F op preserves products F preserves coproducts becausecoproducts in a category become products in the opposite category

In general an equivalence between categories will preserve whatever categorical notions youwant monomorphisms epimorphisms initial objects terminal objects etc so that equivalentcategories really do have essentially the ldquosamerdquo properties

Equivalence example Here is a first example (also found in the book) of equivalent categorieswhich are not isomorphic Let FVectR be the category whose objects are finite-dimensional realvector spaces and whose morphisms are linear transformations Let MatR be the category ofmatrices defined as follows

bull objects in MatR are nonnegative integers

bull a morphism n rarr m is an mtimes n matrix with real entries

bull composition in MatR is given by matrix multiplication given B n rarr m and A m rarr kmdashsoB is an mtimes n matrix and A is a k timesm matrixmdashthe composition A B n rarr k is the k times nmatrix AB

bull identity morphisms in MatR are given by identity matrices

21

We claim that FVectR and MatR are equivalent First note that they certainly cannot be isomo-prhic isomorphisms in MatR exist only between two objects which are literally the same (ie anmtimesn matrix can be invertible only when m = n) but isomorphisms in FVectn can exist betweenobjects which are not literally the same So there are in a sense not enough objects in MatR toallow it to be isomorphic to FVectR

To construct an equivalence choose one and for all an isomorphism TV V rarr RdimV forevery object in FVectR or equivalently choose once and for all a basis for every V DefineF FVectR rarr MatR by

V 983041rarr dimV on objects and

(T V rarr W ) 983041rarr (the standard matrix of SW T Sminus1V RdimV rarr RdimW ) on morphisms

Equivalently F sends T to the matrix of T relative to the chosen bases for V and W It isstraightforward to check that this is a functor The inverse functor G MatR rarr VectR is givenby

n 983041rarr Rn on objects and

(mtimes n matrix A) 983041rarr (the linear transformation A Rn rarr Rm induced by A) on morphisms

One can check that G is a functor and that F G and G F are naturally isomorphic to identitiesor instead one can argue that F alone is full faithful and essentially surjective The point is thatG(F (V )) gives back RdimV so not literally V itself but only something isomorphic to V which iswhy this gives an equivalence of categories and not an isomorphism

Yoneda Lemma Functors as Objects

Compact Hausdorff spaces and Clowast-algebras Last time we gave an example of equivalentcategories which were not isomorphic but in many ways that example is unsatisfactory when doinglinear algebra people for sure often work with matrix representations of linear transformations asthe example from last would suggest but no one in their right mind literally thinks about suchthings in terms of the equivalence FVectR rarr MatR itself In other words this equivalence isessentially ldquocooked uprdquo for the sake of giving an example but it does not really give any insightinto the actual mathematics of linear algebra

So with that in mind we will describe another equivalence which actually does have some realmeaning in the sense of producing new ways of thinking about some mathematics Given a compactHausdorff space X we let C(X) denote the space of continuous complex-valued functions on X

C(X) = f X rarr C | f is continuous

We can equip the set C(X) with various additional structures First it is a complex vector spaceunder ordinary addition and scalar multiplication of functions Second we can multiply two func-tions together

(fg)(p) = f(p)g(p)

which turns C(X) into whatrsquos called an algebra over C (We wonrsquot give the definition of anldquoalgebrardquo but the point is that this multiplication is in some sense ldquocompatiblerdquo with the vectorspace structure) Next any element f of C(X) has a conjugate element f obtained by takingcomplex conjugates of the values of f

f(p) = f(p)

22

And finally we can equip C(X) with a norm via

983042f983042 = suppisinX

|f(p)|

where |middot| denotes complex absolute value (This supremum exists sinceX is compact by the ExtremeValue Theorem) All of these structures (vector space algebra multiplication conjugation norm)turn C(X) into whatrsquos called a Clowast-algebra We wonrsquot define this precisely but the definition encodesvarious ways in which these structures are compatible Actually C(X) is a unital commutative Clowast-algebra where ldquounitalrdquo means that the multiplication has an identity element (namely the constantfunction 1) and commutative means that fg = gf There is a natural notion of homomorphismbetween Clowast-algebras which are maps between them which behave nicely with all the structuresinvolved In particular given a continuous map f X rarr Y between compact Hausdorff spaces weget a Clowast-algebra homomorphism flowast C(Y ) rarr C(X) given by pre-composition with f flowast sendsg isin C(Y ) to g f isin C(X)

We thus have a contravariant functor CHaus rarr ucClowast-Alg from the category of compactHausdorff spaces and the category of unital commutative Clowast-algebras It turns out that anyunital commutative Clowast-algebra arises in this way in the sense that given any unital commutativeClowast-algebra B there exists a compact Hausdorff space X such that C(X) sim= B as Clowast-algebrasand that morphisms between these algebra arises from continuous maps in the manner describeabove This says that the functor above is essentially surjective full and it turns out that it isalso faithful so that it is an equivalence Thus all information about compact Hausdorff spaces iscompletely encoded in information about unital commtuative Clowast-algebras and indeed one couldlearn everything there is to know about a compact Hausdorff space X from its Clowast-algebra C(X)of functions The fact that the categories CHaus and ucClowast-Alg are equivalent is the startingpoint in the subject known as noncommutative geometry which wersquoll say something about a bitlater on If we drop the requirement that our algebras be unital it turns out that the categoryof commutative Clowast-algebras in general is equivalent to the category of locally compact Hausdorffspaces via essentially the same functor where we take as morphisms in the category of locallycompact Hausdorff spaces to be continuous functions which ldquovanish at infinityrdquo

We are only introducing Clowast-algebras to give an example of an interesting and actually usefulequivalence between categories but just for the fun of it wersquoll note the following Examples ofnoncommutative Clowast-algebras come from what are known as ldquoalgebras of bounded operators onHilbert spacesrdquo which show up in quantum mechanics as the things which describe physicallyobservable quantities like position momentum and energy (In fact all Clowast-algebras arise fromHilbert spaces in this way) Thus Clowast-algebras show up in mathematical formulations of quantummechanics and the notion of noncommutative geometry mentioned above is at the core of someattempts to understand the elusive concept known as ldquoquantum geometryrdquo

Yoneda Lemma The Yoneda Lemma states that given a (covariant) functor F C rarr Set whereC is locally small for any A isin Ob(C) there exists a bijection

F (A) sim= Nat(hA F )

where the right side denotes the set of natural transformations from hA to F A similar result holdsin the contravariant case where we get bijections F (A) sim= Nat(hA F ) (There is also a sense inwhich these bijections are themselves natural but wersquoll ignore this bit) The proof of the YonedaLemma was the topic of a student presentation and can be found in Section 22 of the book

Here we will get a bit philosophical and talk about what the point of the Yoneda Lemmaactually is and later the sense in which it says that we can view arbitrary functors as ldquonon-existent

23

objectsrdquo in a category First we mention what we will call the ldquofunctor of pointsrdquo approach tomathematics a mathematical object can be fully characterized by all morphisms into it That isknowledge of Mor(SA) for all S is equivalent to knowledge of A itself Indeed this is an idea wementioned previously in relation to the functors hAmdashwhich we previously said is referred to as beingthe functor of points of Amdashand now we seek to push this idea further In some examples literalpoints of A can be extracted from hA as in the case of the bijection X sim= hX(pt) = Map(ptX) fora topological space X or G sim= hG(Z) = Mor(Z G) for a group G Moreover in these examples therest of the (topological) structure onX or (group) structure on G can also be extracted from hX andhG respectively In general we can view Mor(SA) as in a sense being ldquopoints of A parameterizedby Srdquo which comes from the idea that in say Set Mor(SA) = AS is ldquothe product of S-manycopies of Ardquo so literally points of A ldquoparameterizedrdquo by S In various areas of mathematics anelement of Mor(SA) is thought of as being a ldquogeneralizedrdquo point of A so that the functor of pointshA detects generalized points of A

So an object A is fully encoded by the functor hA Now we claim that morphisms betweenobjects are also encoded by such functors Indeed in the case where F = hB is itself a functor ofpoints the Yoneda result Nat(hA F ) sim= F (A) becomes

Nat(hA hB) sim= hB(A) = Mor(AB)

which says precisely that morphisms between the objects A and B can be obtained solely fromthe functors hA and hB (This fact is one reason why one should prefer the contravariant functorhA to the covariant functor hA in the covariant case Nat(hA F ) sim= F (A) for F = hB becomesNat(hA hB) sim= hB(A) = Mor(BA) so that morphisms between two objects get reversed whencharacterizing them in terms of representable functors)

Pushing our philosophy further we now know that not only objects but also morphisms betweenobjects are fully encoded by the functors hA The final step is to note that we should not only careabout objects and morphisms between them but that an important part of the data of an objectA is also how arbitrary contravariant functors F C rarr Set behave on A and that this too can beobtained from hA precisely via the Yoneda result that F (A) sim= Nat(hA F ) So objects morphismsbetween objects and how arbitrary functors behave on objects are all encoded in knowledge ofthe functors hA so that these functors truly do capture in the most general sense possible allinformation about A

Functors as nonexistent objects The discussion above is all well and good but it still doesnrsquotquite say why we should care about the fact that we can study on object A by studying hA insteadin that we still havenrsquot spoken about what type of insight about A can be gleamed from hA thatwasnrsquot readily apparent in A itself already Now we argue that the true point of the YonedaLemma is that it gives a way to think about arbitrary functors F Cop rarr Set as ldquoobjectsrdquo in C

itself or really as objects in an ldquoenlargementrdquo of C Certainly for those functors F sim= hA which arerepresentable we can indeed think about F as simply being the object A but now we claim thateven non-representable F can be thought of as being ldquononexistentrdquo objects in C The upshot is thatoften in mathematics we come across things which do not actually exist but the Yoneda Lemmagives a way to (indirectly) study such things anyway by interpreting them as functors instead

The starting point of course is that an actual object A is the same as hA as we have explainedabove Thus by analogy we can interpret a non-representable functor F as being an ldquoobjectrdquo in C

as long as we know what it means to ldquomaprdquo into F from other things which are honestly objects in CThat is if F were indeed representable then F (A) sim= Nat(hA F ) sim= Mor(AF ) would characterizethe functor of points of F (here we abuse notation by using the same letter F to denote both arepresentable functor and the object which represents it) so now we take Mor(AF ) = F (A) as

24

the definition of the ldquofunctor of pointsrdquo of the non-existent object F We are thus taking elementsof F (A) as being what it means to ldquomaprdquo from A into F even when F is not representable Thefact that the Yoneda Lemma gives a way to define maps from A into F thus says that F itself isequipped with the same data that a literal object of A would be equipped with so why shouldnrsquotwe interpret F as a being a type of ldquoobjectrdquo in C as well

To phrase this another way note that the Yoneda Lemma gives rise to a functor

C rarr Fun(CopSet)

where Fun(CopSet) denotes the category whose objects are functors from Cop to Set and morphismsare natural transformation defined on objects by A 983041rarr hA and on morphisms by the bijectionMor(AB) sim= Nat(hA hB) arising from the Yoneda Lemma That this is actually bijection saysthat this functor is full and faithful so that we can view C as being a subcategory of the functorcategory Fun(CopSet) (We cay say that C is isomorphic to the subcategory of Fun(CopSet)consisting of the functors hA or that it is equivalent to the subcategory of Fun(CopSet) consistingof representable functors This functor C rarr Fun(CopSet) is usually called the Yoneda embeddingof C into Fun(CopSet)) Thus we can view Fun(CopSet) as an ldquoenlargmentrdquo of C and the ideawe are trying to make sense of is that other objects in this ldquolargerrdquo category should be interpretedas ldquonon-existent objectsrdquo of C

Just how far can we actually push this idea that arbitrary functors Cop rarr Set should be thoughtof as being ldquoobjectsrdquo of C Pretty far in fact For instance often times in geometry one comesacross a type of ldquosmoothrdquo space on which some group acts and one is interested in doing ldquocalculusrdquoon the resulting quotient space which is the space of orbits under the given group action Theproblem is that such quotient spaces are very often (usually in fact) not ldquosmoothrdquo themselves sothat the tools from calculus are not readily available The solution is to nonetheless treat thisquotient as if it were ldquosmoothrdquo and attempt to do calculus anyway To make this precise thisldquonon-smoothrdquo space actually turns out to have a natural interpretation as a ldquosmooth functorrdquo andit is through this functor which we can view this non-smooth space as actually being smooth afterall so we are forced to expand our notion of what we mean by ldquosmooth spacerdquo by allowing thingswhich are not spaces in the usual sense which certainly leads to more abstraction but the benefitis that it actually makes sense to do calculus on such ldquospacesrdquo interpreted as functors

The discussion above is pretty vague so here is one final more concrete example Recall thatequivalence

CHausop rarr ucClowast-Alg

between the category of compact Hausdorff spaces and the category of unital commutative Clowast-algebras mentioned at the start We now claim that a non-commutative (unital) Clowast-algebra shouldbe interpreted as the algebra of functions on some non-existent ldquononcommutative compact Haus-dorff spacerdquo and it is this idea which the subject of noncommutative geometry makes precise Solet B be a non-commutative Clowast-algebra Then the functor of points hB gives a functor

ucClowast-Alg rarr Set

which sends C(X) (where X is some compact Hausdorff space) to Hom(C(X) B) where Hom de-notes homomorphisms of Clowast-algebras whatever that means Composing with the previous functorwe get a functor wersquoll denote by B itself

B CHausop rarr Set

which on objects at least is given by X 983041rarr Hom(C(X) B) This functor is not representable sinceB is not commutative but the philosophy espoused above suggests we should think of it as giving a

25

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

arbitrary number of topological spaces is defined) Wersquoll go over this later but is the underlyingreason why we use oplus instead of times when discussing coproducts

Encoding data via productscoproducts We can nicely summarize the definition of productsand coproducts in the following way the product of A and B is the object with the property thatfor any object D we have

Mor(DA)timesMor(DB) = Mor(D product)

and the coproduct of A and B is the object such that for any object D we have

Mor(AD)timesMor(BD) = Mor(coproductx D)

The first reflects the fact that products give a unique way to turn two morphisms (f and g in thedefinition) mapping into A and B into a single morphism (h in the definition) and the second saysthat coproducts give a unique way to turn two morphisms mapping out of A and B into a singlemorphism

Soon enough we will discuss the idea of universality and the point is that products are ldquouni-versalrdquo for maps into A and B and coproducts are universal for maps from A and B

Opposite categories Finally we give a definition which might seem silly but is actually quiteuseful Given a category C the opposite category is the category Cop is the category with the sameobjects as C but where we defined Mor(AB) in Cop to be Mor(BA) in C To be more concrete weconsider each morphism f A rarr B in C to instead be a morphism fop B rarr A in Cop (Perhaps itwould be better to write this as A larr B fop) The point is that we donrsquot actually care about whatf actually is as say a functioncontinuous maphomomorphismwhatever-we only care about itsbehavior as an ldquoarrowrdquo The opposite category Cop only retains information about how arrowsrelate to one another nothing more

Since arrows are reversed products in C become coproducts in Cop and coproduts in C becomecoproducts in Cop Similarly initialterminal objects and monomorphismsepimorphisms in C be-come terminalinitial objects and epimorphismsmonomorphisms in Cop Again the notion of anopposite category might seem like a strange thing to look at but wersquoll see that it does have uses

Functors Fullness and Faithfulness

One more productcoproduct example Let X be a topological space and define its categoryof open sets Op(X) as follows The objects of Op(X) are simply the open subsets of X and themorphisms are given by inclusions so to be concrete

Mor(U V ) =

983083the unique inclusion U rarr V if U sube V

empty otherwise

The fact that U sube V and V sube W implies U sube W says that the composition of two morphismsin this category is still a morphism in this category The product of U and V in this categoryturns out to be their intersection (with ldquoprojection morphismsrdquo U cap V rarr U and U cap V rarr V beinginclusions) and the coproduct of U and V turns out to be their (ordinary) union Note that weneed X to be a topological space in order to guarantee that U cap V and U cup V are still objects inthis category More generally the product of finitely many objects in Op(X) exists and is theirintersection and the coproduct of an arbitrary number of objects exists and is their union (Wesay that Op(X) has finite products and arbitrary coproducts)

10

Given a set X and a collection T of subsets of X we can define a similar category with elementsof T as objects (I guess we should assume that empty X isin T) and inclusions as morphisms We thenhave that this category has finite products and arbitrary coproducts if and only if T is a topologyon X so that we can rephrase the definition of a topological space itself solely in categorical termsThis perspective on what a ldquotopologyrdquo is makes it simpler to state the definition of whatrsquos calleda sheaf on a topological space which is an important construction in various areas of analysis andgeometry We might discuss this a bit later on

Functors As with most things in mathematics we should care not only about categories asmathematical structures on their own but also about ldquomapsrdquo between these structures whichldquopreserverdquo said structure The notion of a (covariant) functor between categories provides suchldquomapsrdquo in the case at hand where a functor is a way of sending objects to objects and morphismto morphisms in a way which preserves compositions and identities The precise definition is inSection 13 of the book as is the definition of a contravariant functor between categories whichis a functor which reverses the direction of arrows as opposed to ldquocovariantrdquo ones which preservedirections (It is common to use the term ldquofunctorrdquo without qualification to mean one which iscovariant Note that any functor can be made covariant by working with the opposite category acontravariant functor C rarr D is the same as a covariant functor Cop rarr D)

All of the standard examples we looked at are in the book forgetful functors power set functorsthe dual space functor in Vect etc Of particular interest is Example 139 in the book whichshows that various types of group actions in different contexts-the usual notion of a group actionon a set the notion of a continuous action of a group on a topological space and the notion ofa linear action of a group on a vector space (also known as a representation of the group)-are allencoded by a functor with domain category BG

Adjoint functors Let F Top rarr Set be the forgetful functor and let D Set rarr Top be thefunctor which sends a set S to the space S equipped with the discrete topology and a functionS rarr Sprime between sets to the same function only now viewed as a continuous map S rarr Sprime betweendiscrete spaces These two functors satisfy the following property in relation to one another givinga morphism D(S) rarr Y in Top is the same as giving a morphism S rarr F (Y ) in Set or moresymbolically

MorTop(D(S) Y ) = MorSet(S F (Y ))

We say that D is left adjoint to F and that F is right adjoint to D The point is that adjointfunctors can be used to move data back and forth between two categories

Wersquoll talk more about adjoint functors later on but for now here is one more example Nowtake F Grp rarr Set to again be the forgetful functor and take G Grp rarr Set to be the freegroup functor which sends a set S to the free group generated by S G(S) has as elements ldquowordsrdquos1 sn made up of elements of S with the group operation being ldquoconcatenationrdquo A basic factis that giving a group homomorphism from G(S) to some other group H is the same as giving anordinary function from S to F (H) which is the required adjoint property

MorGrp(G(S) H) = MorSet(S F (H))

Faithful and full functors The definitions of what it means for a functor to be faithful and fullare in Section 15 of the book where faithful functors are ones which are injective on morphisms andfull functors are ones which are surjective on morphisms but where in both cases we refer only tomorphisms which occur between the objects F (A) and F (B) in the target category corresponding tofixed objects A and B in the source category (In general a functor could send two objects AB in

11

the source category to the same object F (A) = F (B) in the target category and thus there could betwo morphisms A rarr C and B rarr C which are sent to the same morphism F (A) = F (B) rarr F (C)Such a thing could still possibly be ldquofaithfulrdquo since the non-injectivity here is arising from differentobjects in the source)

Faithfullness and fullness will be nice properties going forward and here is one hint at how theymight be useful Say we have a functor F C rarr D The problem is to determine properties onF which will imply it send products in C to products in D or coproducts in C to coproducts inD So given the former product diagram below we want the second diagram to also be a productdiagram

C

A B

P Fminusminusminusrarr

F (C)

F (A) F (B)

F (P )f g

pA pB

existh

Ff Fg

F (pA) F (pB)

Fh

Of course in order F (P ) to be an actual product of F (A) and F (B) on the right a similarcommutative diagram property would have to hold for all objects in D in place of F (C) and allmorphisms D rarr F (A) and D rarr F (B) not just those of the form Ff and Fg respectively Oneway to guarantee this is to require that all objects in D are of the form F (C) and then for allmorphisms F (C) rarr F (A) and F (C) rarr F (B) to be of the form Ff Fg the former is true whenthe functor F is surjective on objects and the latter is true when F is full We are not saying thatthese are the only conditions under which F might preserve products but it definitely seems togive a sufficient condition at least

Almost there is one more thing to require the uniqueness of the morphism Fh F (C) rarr F (P )making the diagram on the right commute Suppose there were two such morphisms Fh and Fhprimewhere h and hprime are both morphisms C rarr P in C The commutativity of the diagram on the rightgives for instance

Ff = F (pA)Fh and Ff = F (pA)Fhprime

which the functorial properites of F turns into

Ff = F (pA h) and Ff = (pA hprime)

We want h and hprime to make the first diagram commute in order to be able to apply uniqueness ofthe morphism C rarr P But this requires know that f = pA h and f = pA hprime which would followfrom F being faithful In this case it follows that h and hprime both make the first diagram commute(after we go through a similar argument with g) so the fact that P is a product guarantees thath = hprime and hence that Fh = Fhprime so that F (P ) is an actual product on the right

So we conclude that a functor which is full faithful and surjective on objects will send productsto products (Actually we can relax this a bit by requiring not that every object in D be in theliteral image of F but only that it be isomorphic to something in the image A functor with thisproperty is said to be essentially surjective which is a concept will see in a related context soonenough) Again this is not an ldquoif and only ifrdquo statement since functors can preserve products (orcoproducts) without being full faithful and surjective on objects but it at least gives a sufficientcondition Later we will see other ways to guarantee that functors preserve products which arerelated to the existence of adjoints for that given functor

12

Coproduct Examples Concreteness

Products and coproducts for metric spaces The product of two objects in Met the categoryof metric spaces with morphisms given by metric maps is the Cartesian product X times Y equippedwith the box metric and the coproduct of X and Y does not exist if both X and Y are nonemptyThis was the topic of a student presentation and here are some proof sketches

Given metric spaces (X dX) and (Y dY ) the box metric d on X times Y is defined by

d((x1 y1) (x2 y2)) = maxdX(x1 x2) dY (y1 y2))

(The name ldquoboxrdquo metric comes from the fact that in RtimesR = R2 or R3 open balls indeed looks likeldquoboxesrdquo) The key requirement needed in the definition of a product is given morphisms S rarr Xand S rarr Y that the map

h S rarr X times Y defined by h(s) = (f(s) g(s))

actually be a morphism in Met This turns into the requirement that

d((f(s) g(s)) ((f(t) g(t))) = maxdX(f(s) f(t)) dY (g(s) g(t)) le dS(s t) for all s t isin S

given the fact that

dX(f(s) f(t)) le dS(s t) and dY (g(s) g(t)) le dS(s t) for all s t isin S

For the latter inequalities to imply the former d must (at least up to isometry) be given by thebox metric Note that the two other ldquostandardrdquo metrics one might put on a productmdashthe taxicab

dX + dY and Euclidean983156

d2X + d2Y metricsmdashdo not satisfy this required ldquometric maprdquo property

and so do not give the product in MetTo see that coproducts do not exist if XY are nonempty suppose instead (P dP ) was a co-

product for X and Y with inclusion morphisms iX X rarr P and iY Y rarr P For any n isin Ntake 0 n to be a metric space with the absolute value metric d and let f X rarr 0 n andg Y rarr 0 n be the constant function 0 and the constant function n respectively Since P is acoproduct there exists h P rarr 0 n such that

h iX = f and h iY = g

But in order for h to actually be a morphism in Met we must then have

d(h(iX(x) iY (y)) le dP (iX(x) iY (y)) for any x isin X y isin Y

which translates inton = |0minus n| le dP (iX(x) iY (y))

But this should then be true for any n isin N which would imply that the distance in P betweenan element coming from X and one coming from Y should be infinite which is nonsense (Notethat if you altered your definition of ldquometricrdquo to allow for infinite distance then a coproduct wouldexist the disjoint union X ⊔Y where you defined the distance between elements of X and Y to beinfinite)

Coproducts for groups The coproduct of two groups G and H in Grp is the free product GlowastHwhich is defined to be the set of all words consisting of elements of G andH in an alternating fashion

13

with concatenation as the group operation (When we have to elements of G or two elements of Hnext to each other with use their respective group operations to turn these into single elements ofG or H) This was also the topic of a student presentation and here is the basic idea Given grouphomomorphisms g G rarr N and k H rarr N that G lowastH is the coproduct comes down to showingthat the map

h G lowastH rarr N defined by h(g1h1 gnhn) = f(g1)k(h1) middot middot middot f(gn)k(hn)

and similarly on other possible words in GlowastH is a group homomorphism which follows essentiallyfrom the definition of the group operation on G lowastH indeed this group operation arises preciselyfrom this requirement

Now if we instead tried to use GtimesH with its standard inclusions G rarr GtimesH and H rarr GtimesHas the coproduct the issue is that the analogous map

h GtimesH rarr N defined by h((g h)) = f(g)k(h)

would NOT be a homomorphism because h((g1 h1)(g2 h2)) is not equal to h(g1 h1)h(g2 h2) bythe way in which the group operation is defined on the direct product G times H However if westrict our category to only consist of abelian groups so that the N rsquos we consider in the defini-tion of coproduct are also abelian then G times H does serve as a coproduct we can always rewritef(g1)k(g1) middot middot middot f(gn)k(gn) as f(g1) middot middot middot f(gn)k(h1) middot middot middot k(hn) in N Irsquoll leave it to you to flesh out allof these details

Concrete categories The categories Set Top Vect and Grp all have the property thattheir objects are simply sets possibly equipped with extra structure and that their morphisms areordinary set-theoretic functions possibly satisfying some additional requirement Such categoriesare nice since we can often use intuition about sets and functions in their study A concrete categoryis intuitively one where we can think of objects as being ldquosetsrdquo and morphisms as being ldquofunctionsrdquoonly that we have to interpret this in the right way

Here is the definition a concrete category is a category C equipped with a faithful functorF C rarr Set which is part of the data The idea is that such a functor gives us way to turnan object A isin Ob(C) into a set F (A) and a morphism f isin MorC(AB) into a function Ff isinMorSet(F (A) F (B)) where the ldquofaithfulrdquo requirement says that we do not lose information aboutf when doing so so that we can (essentially) learn everything we need to know about f by studyingFf instead In the four examples above the required faithful functor is simply the forgetful functorand indeed in general we think of F as being a ldquoforgetful functorrdquo for the concrete category (C F )

Examples Here are two more examples of concrete categories Let G be a group and considerthe category BG with a single object and elements of G morphisms As written this does notappear to be concrete since morphisms are not honest functions lowast rarr lowast but the point is that thiscategory can indeed by ldquoconcretizedrdquo by specifying an appropriate ldquoforgetful functorrdquo A functorF BG rarr Set is the data of a (left) action of G on a set S = F (lowast) and saying this functor isfaithful is what it means to say that this action is faithful different g h isin G act on S in differentways or equivalently that there exists s isin S such that gs ∕= hs Any group has at least onefaithful left actionmdashnamely the action on itself given by left multiplicationmdashto BG can always beequipped with a faithful functor to Set This is all just a way of phrasing the fact that any groupis isomorphic to a subgroup of a permutation group

The category Rel of relations can also be ldquoconcretizedrdquo even though as written morphismsA rarr B are not honest functions Take F Rel rarr Set to be the power set functor which sends an

14

object in Rel to its power set and a relation f A rarr B to the induced function Ff P (A) rarr P (B)defined by

(Ff)(S) = b isin B | there exists a isin S such that (a b) isin f

In other words if you think of the relation f sube A times B as a ldquopossibly not everywhere-definedpossibly multivalued functionrdquo (Ff)(S) is analogous to taking the image of S That F is faithfulwill be left to the homework

In fact most categories yoursquoll see can be made concrete by specifying an appropriate forgetfulfunctor but not all categories can be made concrete in this way Here is the standard example of anon-concrete category which we mention only to say there is such a thing but which we wonrsquot doanything more with since understanding it better requires knowledge of algebraic topology Thehomotopy category of topological spaces is the category hTop whose objects are topological spacesand whose morphisms are given by equivalence classes of homotopic maps

MorhTop(XY ) = homotopy classes of continuous maps X rarr Y

(Saying that two maps are homotopic means that you can ldquodeformrdquo one into the other but wersquollleave the precise definition to a course in algebraic topology) It turns out that no functor fromhTop to Set can be faithful so that hTop cannot be made concrete but this is actually a deepand difficult thing to show so we make no attempt to do so here

Natural Isomorphisms Representability

Natural transformations The definition of a natural transformation η F rArr G between two(both covariant or both contravariant) functors FG C rarr D is given in Section 14 of the bookThe key idea is that a natural transformation gives a way to turn data about F into data aboutG in a way which is compatible with all possible morphisms as expressed by the commutativity ofthe diagrams

F (A) F (B)

G(A) G(B)

Ff

ηA ηB

Gf

arising from morphisms f A rarr B (or f B rarr A in the contravariant case) in C In the case of anatural isomoprhism η F rArr G so that each ηA F (A) rarr G(A) is actually an isomorphism in Dthe point is not only that F and G produce isomorphic objects but they do so in a way which iscompatible with all possible category-theoretic data

The first standard example of a natural isomorphism is the one between the identity functor onthe category of finite-dimensional and the functor which sends a finite-dimensional vector space Vto V lowastlowast the dual of its dual induced by the isomorphism V rarr V lowastlowast defined by

v 983041rarr (the linear map on V lowast which sends f to f(v))

This was the topic of a student presentation and can be found in Section 14 of the book Theessential reason why this isomorphism is ldquonaturalrdquo is that V rarr V lowastlowast can be defined without referenceto a basis Contrast this with the the functor which sends V to its (single) dual V lowast it is still truethat finite-dimensional vector space is isomorphic to its dual but specifying such an isomorphism

15

requires additional data which prevents the identity functor from being ldquonaturally isomorphicrdquo tothe single dual functor

Some history At this point it makes sense to say a bit about the interesting origins of categorytheory in the 1950rsquos Eilenberg and Mac Lane where studying cohomology (or possible homologyone of those two) in algebraic topology which associates algebraic data (a group ring or similar)to a topological space in a way which encodes some of the topology They had two cohomologicalconstructions which ended up proving isomorphic results but they noticed that not only werethe resulting objects the same but the actual constructions themselves were in some sense theldquosamerdquo They thus then needed a way to formalize this notion and invented the concept of anatural transformation to do so where they interpreted two constructions as being the same asbeing compatible with all morphisms

But this then leads to the question what types of objects should a natural transformationoccur between Eilenberg and Mac Lane then had to invent the notion of a ldquofunctorrdquo to describewhat a natural transformation went from and what it went to In their motivating setting theyinvented the notion of a functor to describe more formally properties which their cohomologicalconstructions were supposed to satisfy But this then leads to the question what types of objectsshould a functor map between In other words what type of data should a functor (cohomologicalconstruction in their example) take as input and should it output They finally had to thus inventthe notion of a ldquocategoryrdquo to describe the source and target of a functor so that their cohomologicalconstructions could be interpreted as functors from a category of topological spaces to a categoryof algebraic objects

Thus historically natural transformations came first then functors and then categories Aswith most things in mathematics categories thus arose from attempts to encode what was beingseen in some concrete examples in a more general and formal setting

Representable functors Representable functors are defined in Section 21 of the book but wersquollgive the required definitions here in a hopefully simpler to digest manner The point is that anyobject A in a (locally small) category C gives two ldquonaturalrdquo functors C rarr Set one covariant andthe other contravariant which as wersquoll see pretty much encode all data about A itself We definehA C rarr Set to be the covariant functor defined by

hA(B) = Mor(AB) on objects and

hA(Bfminusrarr C) = the map Mor(AB)

hAfminusminusrarr Mor(AC) defined by g 983041rarr f g

We define hA C rarr Set to the contravariant functor defined by

hA(B) = Mor(BA) on objects and

hA(Bfminusrarr C) = the map Mor(CA)

hAfminusminusrarr Mor(BA) defined by g 983041rarr g f

(The functor hA is called the functor of points of A and wersquoll usually think of it as being a covariantfunctor hA Cop rarr Set The functor hA does not have a standard name) So hA is defined bymorphisms from A and hA is defined by morphisms into A Wersquoll see later the sense in which suchfunctors encode all information about the object A

A functor F C rarr Set is representable if is naturally isomorphic to such a functor so F sim= hA

for some A in the covariant case and F sim= hA for some A in the contravariant case The idea wersquollbuild towards is that we can essentially treat such a functor as being A itself so that representable

16

functors can be thought as being actual objects in C For now here is a key example wersquove seenbefore which we can now formulate using the language of representability

Given AB isin Ob(C) let F C rarr Set be the contravariant functor defined by

F (C) = Mor(CA)timesMor(CB) and F (Cfminusrarr D) = [(g k) 983041rarr (g f k f)]

where (g k) 983041rarr (g f k f) gives a map Mor(DA) timesMor(DB) rarr Mor(CA) timesMor(CB) ieF (D) rarr F (C) In order for this functor to be representable requires the existence of an object inC satisfying

Mor(CA)timesMor(CB) sim= Mor(C representing object)

in a way which is compatible with all morphisms but we have actually already seen this notionelsewhere it is essentially the definition of a product AtimesB for A and B Indeed we claim that Fis naturally isomorphic to hAtimesB as we now show

First to define a natural isomorphism hAtimesBsim= F requires for each C isin Ob(C) a choice of a

bijection (ie isomorphism in Set)

ηC hAtimesB(C)sim=minusrarr F (C)

which in our case looks like

ηC Mor(CAtimesB))sim=minusrarr Mor(CA)timesMor(CB)

Take this to be the map

(Cfminusrarr AtimesB) 983041rarr (pA f pB f)

where pA A times B rarr A and pB A times B rarr B are the two projection morphisms in the definitionof a product The definition of product implies that this map is invertible since given k C rarr Aand ℓ C rarr B there exists a unique h C rarr A times B such that k = pA h and ℓ = pB h Forthese bijections to define a natural isomorphism hAtimesB

sim= F requires that they be compatible withall morphisms in C which means that given any g C rarr D in C the following diagram shouldcommute

hAtimesB(C) hAtimesB(D)

F (C) F (D)

hAtimesB(g)

ηC ηD

Fg

which in the case at hand looks like

Mor(CAtimesB) Mor(DAtimesB)

Mor(CA)timesMor(CB) Mor(DA)timesMor(DB)

minus g

ηC ηD

(minus gminus g)

Take an element k D rarr AtimesB in the upper right Applying the map on top gives the elementk g in the upper left and then applying ηC on the left side gives

(pA (k g) pB (k g))

17

in the lower left Instead starting with k D rarr AtimesB in the upper right applying ηD on the rightgives (pA k pB k) and applying the map on the bottom gives

((pA k) g (pB k) g)

in the lower left which is the same as the previous element in the lower left by the associativity ofcomposition Hence the isomorphisms ηC do define a natural isomorphism between hAtimesB and F Again the point is that you can then interpret the functor F as being the ldquosamerdquo as the productAtimesB since both encode the same data

More Representable Examples

Coproducts As in the case of products the definition of coproducts can also be phrased in termsof a representable functor Given objects A and B in C let F C rarr Set be the covariant functordefined by

F (C) = Mor(AC)timesMor(BC) and F (Cfminusrarr D) = [(g k) 983041rarr (f g f k)]

where the latter is a function Mor(AC)timesMor(BC) rarr Mor(AD)timesMor(BD) This functor isrepresentable if and only if a coproduct A ⊔ B for A and B exists and the representing object isthat coproduct the key is the requirement that we have bijections

Mor(AC)timesMor(BC) sim= Mor(A ⊔BC)

for all C which is essentially the defining property a coproduct for A and B should have

Power set functor contravariant version Let P Set rarr Set be the contravariant powerset functor defined on objects by sending a set to its power set and on morphisms by sendingf A rarr B to the preimage function Pf P (B) rarr P (A) given by S 983041rarr fminus1(S) In order for this tobe representable requires the existence of a set X with natural bijections

P (A) sim= Mor(AX)

for all sets A that is a set X so that mapping into X is the same data as specifying a subset ofthe domain We take X = 0 1 to be a two-element set and the required bijections

ηA P (A) rarr Mor(A 0 1) to be given by S 983041rarr χS

where XS A rarr 0 1 is the indicator or characteristic function S defined by sending elementsof S to 1 and elements of A minus S to 0 The inverse of ηA sends a function g A rarr 0 1 to thepreimage gminus1(1) sube A

To verify that the bijections ηA define the data of a natural isomorphism η P rArr h01 wecheck the commutativity of some diagrams Given a function f A rarr B consider

P (A) P (B)

h01(A) = Mor(A 0 1) h01(B) = Mor(B 0 1)

Pf = fminus1

ηA ηB

minus f

18

Take S isin P (B) in the upper right Applying the map on top gives fminus1(S) isin P (A) and thenapplying the map on the left gives χfminus1(S) isin Mor(A 0 1) On the other hand applying the mapon the right to S isin P (B) gives χS isin Mor(B 0 1) and then applying the map on the bottomgives χS f isin Mor(A 0 1) The two resulting functions χfminus1(S)χS f in the lower left are

χfminus1(S)(x) =

9830831 x isin fminus1(S)

0 otherwiseand χS(f(x)) =

9830831 f(x) isin S

0 otherwise

so these are the same since x isin fminus1(S) if and only if f(x) isin S Thus the diagram above commutesso P is naturally isomorphic to h01 and hence P is representable

Power set functor covariant version Now consider the covariant power set functor stilldenoted by P Set rarr Set which again sends a set to its power set but now sends a functionf A rarr B to the image function Pf P (A) rarr P (B) defined by S 983041rarr f(S) In order for this tobe representable there would have to exist a set X such that P sim= hX which translates into havingnatural bijections

P (S) sim= Mor(XS)

for all sets S However in this case there is no natural choice for X as there is no natural way todetect subsets of S by mapping into S as opposed to the case of P (S) sim= Mor(S 0 1)

That no such X exists can be made precise using a simple argument when S = empty P (S) hascardinality 1 soX would have to be empty as well since otherwise Mor(XS) would have cardinality0 but if X is empty then Mor(empty S) always cardinality 1 regardless of S which is clearly not trueof P (S) Thus the covariant power set functor is not representable

Forgetful on Top The forgetful functor F Top rarr Set is representable Indeed a representingobject would have to satisfy

X sim= Map(objectX)

as sets for any topological space X (where Map denotes the set of continuous maps) and it iseasy to see that a single point satisfies this property a continuous map f pt rarr X is completelydetermined by the element f(pt) of X and any such element gives a continuous map in this wayOf course checking that F sim= hpt requires checking that various diagrams commute but this issimple to work out in this case so we omit it here

Extracting a topology With an eye towards the idea that the functors hA hA should captureall information about an object A in a category note that the result above shows that in TophX first of all captures all information about X as a set since the set X can be extracted fromhX(pt) = Map(ptX)

We claim that hX actually captures all information about X as a topological space since thetopology on X can be extracted from hX in the following way As we saw in the power set examplewe have a bijection

P (X) sim= Mor(X 0 1)

Now equip 0 1 with the topology in which 1 is open but 0 is not Then a continuous mapf X rarr 0 1 is fully determined by the preimage fminus1(1) since to be continuous only requiresthat this single preimage be open in X Moreoever any open subset of X arises in this way bytaking f to be the indicator function of that open set Thus we get a bijection

Op(X) sim= Map(X 0 1)

19

where Op(X) denotes the set of open subsets of X ie the topology on X Hence the topology onX can be extracted from hX as claimed so hX and hX do encode all information about X

This fact can also be interpreted as saying that the contravariant functor which sends a topo-logical space X to its collection of open subsets (and which sends a continuous map to the mapinduced by taking preimages) is representable with representing object 0 1 equipped with thetopology given above This will be worked out in a homework problem

Forgetful on Grp The forgetful functor Grp rarr Set is also representable In this case arepresenting object should be a group giving set-theoretic bijections

G sim= Hom(objectG)

for all groups G where Hom denotes the set of group homomorphisms The group Z works

G sim= Hom(Z G)

since a homomorphism φ Z rarr G is completely determined by φ(1) since Z is generated by 1 andthere are no restrictions on what φ(1) isin G can actually be The inverse of the desired bijection isthus φ 983041rarr φ(1) and it is simple to check that morphisms behave appropriately so that the forgetfulfunctor in this setting is indeed isomoprhic to hZ

Forgetful on Ring The forgetful functor Ring rarr Set is also representable where by Ring wemean the category of rings with identity elements and where morphisms (ie ring homomorphisms)are required to preserve identities Here we need an object satisfying

R sim= Hom(object R)

for all R isin Ob(Ring) and we can see that Z[x] the ring of polynomials in the variable x withcoefficients in Z works Indeed given a ring homomorphism φ Z[x] rarr R the fact that φ isrequired to preserve identities fully determines φ(n) for any n isin Z (since φ(n) = nφ(1)) and hencesince φ preserves multiplication its behavior is fully determined by φ(x) which can be any elementof R The inverse of the bijection we want is thus φ 983041rarr φ(x) and it will follow that this forgetfulfunctor is isomorphic to hZ[x]

Equivalences between Categories

Definitions iso and equiv Now that we have a notion of a ldquomorphismrdquo between categorieswe can talk about the sense in which two categories are the ldquosamerdquo The first possible definitionis the ldquoobviousrdquo one you might expect given the definition of ldquoisomorphismrdquo wersquove seen in everyother context until now we say that C is isomorphic to D if there exist functors F C rarr D andG D rarr C such that G F = idC and F G = idD where idC and idD are identity functors

However this definition turns out to be too restrictive Indeed the requirement that the givencompositions equal identity functors says that given an object A in C the object G(F (A)) beliterally the same as A itself and similarly for objects in D But we know that two objects in acategory can be thought of as being the ldquosamerdquo without being literally the same as long as theyare isomorphic For instance productscoproducts are not unique in the literal sense but only inthe sense that they are isomorphic in a unique way To take another example the definition ofa functor being representable does not say that F (B) literally be the same as hB(A) only thatthey be isomorphic for sure the power set P (S) of a set S is not literally the same as the set offunctions S rarr 0 1mdashit is only the ldquosamerdquo in the sense that they encode the same data

20

Thus it makes sense to relax the requirement that A andG(F (A)) in the definition of isomorphiccategories be equal to the requirement that they be isomorphic This leads to the following betterdefinition of two categories being the ldquosamerdquo C is equivalent to D if there exist functors F C rarr D

and G D rarr C such that G F sim= idC and F G sim= idD So we require only that for instance thefunctor G F be naturally isomorphic to the identity functor on C which says that for any objectA G(F (A)) be isomorphic to A in a natural way Wersquoll see that this truly is the better notionof what it means for two categories to be the ldquosamerdquo and it reflects the idea that all categoricalproperties of one category can be transported to the other

Alternate characterization of equivalences A functor F C rarr D implementing an equiv-alence between categories is called unsurprisingly and equivalence and the ldquoinverserdquo functorG D rarr C is still referred to as being its inverse even though it is not technically an ldquoinverserdquoin the sense that compositions give literal identities But as in the case of Set where ldquoinvertiblerdquocan be characterized without reference to an inverse as ldquoinjectiverdquo and ldquosurjectiverdquo so too canequivalences between categories be characterized without explicit reference to an inverse

A functor F C rarr D is an equivalence if and only it is full faithful and essentiallysurjective

We have seen the notions of full and faithful before and essentially surjective means that any objectin D is isomorphic to something in the image of F for all D isin Ob(D) there exits C isin Ob(C) suchthat F (C) sim= D Justifying this result was the topic of a student presentation and can be foundin the book in Section 15 (Technically to construct an ldquoinverserdquo of F requires that we work withsmall categories but whatever) The bulk of the real work goes into showing that the ldquoinverserdquofunctor constructed is actually a functor which comes down to some abstract nonsense argumentvia commutative diagrams

Equivalences preserve everything We previously saw the notions of full faithful and es-sentially surjective back when thinking about what properties a functor could have which wouldguarantee it sent products to products so the result we proved back then was that equivalencespreserve products Similarly equivalences preserve coproducts and the proof is basically the sameafter reversing arrows A clean way of deriving this is as follows if F C rarr D is an equivalencethen so is F op Cop rarr Dop so since F op preserves products F preserves coproducts becausecoproducts in a category become products in the opposite category

In general an equivalence between categories will preserve whatever categorical notions youwant monomorphisms epimorphisms initial objects terminal objects etc so that equivalentcategories really do have essentially the ldquosamerdquo properties

Equivalence example Here is a first example (also found in the book) of equivalent categorieswhich are not isomorphic Let FVectR be the category whose objects are finite-dimensional realvector spaces and whose morphisms are linear transformations Let MatR be the category ofmatrices defined as follows

bull objects in MatR are nonnegative integers

bull a morphism n rarr m is an mtimes n matrix with real entries

bull composition in MatR is given by matrix multiplication given B n rarr m and A m rarr kmdashsoB is an mtimes n matrix and A is a k timesm matrixmdashthe composition A B n rarr k is the k times nmatrix AB

bull identity morphisms in MatR are given by identity matrices

21

We claim that FVectR and MatR are equivalent First note that they certainly cannot be isomo-prhic isomorphisms in MatR exist only between two objects which are literally the same (ie anmtimesn matrix can be invertible only when m = n) but isomorphisms in FVectn can exist betweenobjects which are not literally the same So there are in a sense not enough objects in MatR toallow it to be isomorphic to FVectR

To construct an equivalence choose one and for all an isomorphism TV V rarr RdimV forevery object in FVectR or equivalently choose once and for all a basis for every V DefineF FVectR rarr MatR by

V 983041rarr dimV on objects and

(T V rarr W ) 983041rarr (the standard matrix of SW T Sminus1V RdimV rarr RdimW ) on morphisms

Equivalently F sends T to the matrix of T relative to the chosen bases for V and W It isstraightforward to check that this is a functor The inverse functor G MatR rarr VectR is givenby

n 983041rarr Rn on objects and

(mtimes n matrix A) 983041rarr (the linear transformation A Rn rarr Rm induced by A) on morphisms

One can check that G is a functor and that F G and G F are naturally isomorphic to identitiesor instead one can argue that F alone is full faithful and essentially surjective The point is thatG(F (V )) gives back RdimV so not literally V itself but only something isomorphic to V which iswhy this gives an equivalence of categories and not an isomorphism

Yoneda Lemma Functors as Objects

Compact Hausdorff spaces and Clowast-algebras Last time we gave an example of equivalentcategories which were not isomorphic but in many ways that example is unsatisfactory when doinglinear algebra people for sure often work with matrix representations of linear transformations asthe example from last would suggest but no one in their right mind literally thinks about suchthings in terms of the equivalence FVectR rarr MatR itself In other words this equivalence isessentially ldquocooked uprdquo for the sake of giving an example but it does not really give any insightinto the actual mathematics of linear algebra

So with that in mind we will describe another equivalence which actually does have some realmeaning in the sense of producing new ways of thinking about some mathematics Given a compactHausdorff space X we let C(X) denote the space of continuous complex-valued functions on X

C(X) = f X rarr C | f is continuous

We can equip the set C(X) with various additional structures First it is a complex vector spaceunder ordinary addition and scalar multiplication of functions Second we can multiply two func-tions together

(fg)(p) = f(p)g(p)

which turns C(X) into whatrsquos called an algebra over C (We wonrsquot give the definition of anldquoalgebrardquo but the point is that this multiplication is in some sense ldquocompatiblerdquo with the vectorspace structure) Next any element f of C(X) has a conjugate element f obtained by takingcomplex conjugates of the values of f

f(p) = f(p)

22

And finally we can equip C(X) with a norm via

983042f983042 = suppisinX

|f(p)|

where |middot| denotes complex absolute value (This supremum exists sinceX is compact by the ExtremeValue Theorem) All of these structures (vector space algebra multiplication conjugation norm)turn C(X) into whatrsquos called a Clowast-algebra We wonrsquot define this precisely but the definition encodesvarious ways in which these structures are compatible Actually C(X) is a unital commutative Clowast-algebra where ldquounitalrdquo means that the multiplication has an identity element (namely the constantfunction 1) and commutative means that fg = gf There is a natural notion of homomorphismbetween Clowast-algebras which are maps between them which behave nicely with all the structuresinvolved In particular given a continuous map f X rarr Y between compact Hausdorff spaces weget a Clowast-algebra homomorphism flowast C(Y ) rarr C(X) given by pre-composition with f flowast sendsg isin C(Y ) to g f isin C(X)

We thus have a contravariant functor CHaus rarr ucClowast-Alg from the category of compactHausdorff spaces and the category of unital commutative Clowast-algebras It turns out that anyunital commutative Clowast-algebra arises in this way in the sense that given any unital commutativeClowast-algebra B there exists a compact Hausdorff space X such that C(X) sim= B as Clowast-algebrasand that morphisms between these algebra arises from continuous maps in the manner describeabove This says that the functor above is essentially surjective full and it turns out that it isalso faithful so that it is an equivalence Thus all information about compact Hausdorff spaces iscompletely encoded in information about unital commtuative Clowast-algebras and indeed one couldlearn everything there is to know about a compact Hausdorff space X from its Clowast-algebra C(X)of functions The fact that the categories CHaus and ucClowast-Alg are equivalent is the startingpoint in the subject known as noncommutative geometry which wersquoll say something about a bitlater on If we drop the requirement that our algebras be unital it turns out that the categoryof commutative Clowast-algebras in general is equivalent to the category of locally compact Hausdorffspaces via essentially the same functor where we take as morphisms in the category of locallycompact Hausdorff spaces to be continuous functions which ldquovanish at infinityrdquo

We are only introducing Clowast-algebras to give an example of an interesting and actually usefulequivalence between categories but just for the fun of it wersquoll note the following Examples ofnoncommutative Clowast-algebras come from what are known as ldquoalgebras of bounded operators onHilbert spacesrdquo which show up in quantum mechanics as the things which describe physicallyobservable quantities like position momentum and energy (In fact all Clowast-algebras arise fromHilbert spaces in this way) Thus Clowast-algebras show up in mathematical formulations of quantummechanics and the notion of noncommutative geometry mentioned above is at the core of someattempts to understand the elusive concept known as ldquoquantum geometryrdquo

Yoneda Lemma The Yoneda Lemma states that given a (covariant) functor F C rarr Set whereC is locally small for any A isin Ob(C) there exists a bijection

F (A) sim= Nat(hA F )

where the right side denotes the set of natural transformations from hA to F A similar result holdsin the contravariant case where we get bijections F (A) sim= Nat(hA F ) (There is also a sense inwhich these bijections are themselves natural but wersquoll ignore this bit) The proof of the YonedaLemma was the topic of a student presentation and can be found in Section 22 of the book

Here we will get a bit philosophical and talk about what the point of the Yoneda Lemmaactually is and later the sense in which it says that we can view arbitrary functors as ldquonon-existent

23

objectsrdquo in a category First we mention what we will call the ldquofunctor of pointsrdquo approach tomathematics a mathematical object can be fully characterized by all morphisms into it That isknowledge of Mor(SA) for all S is equivalent to knowledge of A itself Indeed this is an idea wementioned previously in relation to the functors hAmdashwhich we previously said is referred to as beingthe functor of points of Amdashand now we seek to push this idea further In some examples literalpoints of A can be extracted from hA as in the case of the bijection X sim= hX(pt) = Map(ptX) fora topological space X or G sim= hG(Z) = Mor(Z G) for a group G Moreover in these examples therest of the (topological) structure onX or (group) structure on G can also be extracted from hX andhG respectively In general we can view Mor(SA) as in a sense being ldquopoints of A parameterizedby Srdquo which comes from the idea that in say Set Mor(SA) = AS is ldquothe product of S-manycopies of Ardquo so literally points of A ldquoparameterizedrdquo by S In various areas of mathematics anelement of Mor(SA) is thought of as being a ldquogeneralizedrdquo point of A so that the functor of pointshA detects generalized points of A

So an object A is fully encoded by the functor hA Now we claim that morphisms betweenobjects are also encoded by such functors Indeed in the case where F = hB is itself a functor ofpoints the Yoneda result Nat(hA F ) sim= F (A) becomes

Nat(hA hB) sim= hB(A) = Mor(AB)

which says precisely that morphisms between the objects A and B can be obtained solely fromthe functors hA and hB (This fact is one reason why one should prefer the contravariant functorhA to the covariant functor hA in the covariant case Nat(hA F ) sim= F (A) for F = hB becomesNat(hA hB) sim= hB(A) = Mor(BA) so that morphisms between two objects get reversed whencharacterizing them in terms of representable functors)

Pushing our philosophy further we now know that not only objects but also morphisms betweenobjects are fully encoded by the functors hA The final step is to note that we should not only careabout objects and morphisms between them but that an important part of the data of an objectA is also how arbitrary contravariant functors F C rarr Set behave on A and that this too can beobtained from hA precisely via the Yoneda result that F (A) sim= Nat(hA F ) So objects morphismsbetween objects and how arbitrary functors behave on objects are all encoded in knowledge ofthe functors hA so that these functors truly do capture in the most general sense possible allinformation about A

Functors as nonexistent objects The discussion above is all well and good but it still doesnrsquotquite say why we should care about the fact that we can study on object A by studying hA insteadin that we still havenrsquot spoken about what type of insight about A can be gleamed from hA thatwasnrsquot readily apparent in A itself already Now we argue that the true point of the YonedaLemma is that it gives a way to think about arbitrary functors F Cop rarr Set as ldquoobjectsrdquo in C

itself or really as objects in an ldquoenlargementrdquo of C Certainly for those functors F sim= hA which arerepresentable we can indeed think about F as simply being the object A but now we claim thateven non-representable F can be thought of as being ldquononexistentrdquo objects in C The upshot is thatoften in mathematics we come across things which do not actually exist but the Yoneda Lemmagives a way to (indirectly) study such things anyway by interpreting them as functors instead

The starting point of course is that an actual object A is the same as hA as we have explainedabove Thus by analogy we can interpret a non-representable functor F as being an ldquoobjectrdquo in C

as long as we know what it means to ldquomaprdquo into F from other things which are honestly objects in CThat is if F were indeed representable then F (A) sim= Nat(hA F ) sim= Mor(AF ) would characterizethe functor of points of F (here we abuse notation by using the same letter F to denote both arepresentable functor and the object which represents it) so now we take Mor(AF ) = F (A) as

24

the definition of the ldquofunctor of pointsrdquo of the non-existent object F We are thus taking elementsof F (A) as being what it means to ldquomaprdquo from A into F even when F is not representable Thefact that the Yoneda Lemma gives a way to define maps from A into F thus says that F itself isequipped with the same data that a literal object of A would be equipped with so why shouldnrsquotwe interpret F as a being a type of ldquoobjectrdquo in C as well

To phrase this another way note that the Yoneda Lemma gives rise to a functor

C rarr Fun(CopSet)

where Fun(CopSet) denotes the category whose objects are functors from Cop to Set and morphismsare natural transformation defined on objects by A 983041rarr hA and on morphisms by the bijectionMor(AB) sim= Nat(hA hB) arising from the Yoneda Lemma That this is actually bijection saysthat this functor is full and faithful so that we can view C as being a subcategory of the functorcategory Fun(CopSet) (We cay say that C is isomorphic to the subcategory of Fun(CopSet)consisting of the functors hA or that it is equivalent to the subcategory of Fun(CopSet) consistingof representable functors This functor C rarr Fun(CopSet) is usually called the Yoneda embeddingof C into Fun(CopSet)) Thus we can view Fun(CopSet) as an ldquoenlargmentrdquo of C and the ideawe are trying to make sense of is that other objects in this ldquolargerrdquo category should be interpretedas ldquonon-existent objectsrdquo of C

Just how far can we actually push this idea that arbitrary functors Cop rarr Set should be thoughtof as being ldquoobjectsrdquo of C Pretty far in fact For instance often times in geometry one comesacross a type of ldquosmoothrdquo space on which some group acts and one is interested in doing ldquocalculusrdquoon the resulting quotient space which is the space of orbits under the given group action Theproblem is that such quotient spaces are very often (usually in fact) not ldquosmoothrdquo themselves sothat the tools from calculus are not readily available The solution is to nonetheless treat thisquotient as if it were ldquosmoothrdquo and attempt to do calculus anyway To make this precise thisldquonon-smoothrdquo space actually turns out to have a natural interpretation as a ldquosmooth functorrdquo andit is through this functor which we can view this non-smooth space as actually being smooth afterall so we are forced to expand our notion of what we mean by ldquosmooth spacerdquo by allowing thingswhich are not spaces in the usual sense which certainly leads to more abstraction but the benefitis that it actually makes sense to do calculus on such ldquospacesrdquo interpreted as functors

The discussion above is pretty vague so here is one final more concrete example Recall thatequivalence

CHausop rarr ucClowast-Alg

between the category of compact Hausdorff spaces and the category of unital commutative Clowast-algebras mentioned at the start We now claim that a non-commutative (unital) Clowast-algebra shouldbe interpreted as the algebra of functions on some non-existent ldquononcommutative compact Haus-dorff spacerdquo and it is this idea which the subject of noncommutative geometry makes precise Solet B be a non-commutative Clowast-algebra Then the functor of points hB gives a functor

ucClowast-Alg rarr Set

which sends C(X) (where X is some compact Hausdorff space) to Hom(C(X) B) where Hom de-notes homomorphisms of Clowast-algebras whatever that means Composing with the previous functorwe get a functor wersquoll denote by B itself

B CHausop rarr Set

which on objects at least is given by X 983041rarr Hom(C(X) B) This functor is not representable sinceB is not commutative but the philosophy espoused above suggests we should think of it as giving a

25

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

Given a set X and a collection T of subsets of X we can define a similar category with elementsof T as objects (I guess we should assume that empty X isin T) and inclusions as morphisms We thenhave that this category has finite products and arbitrary coproducts if and only if T is a topologyon X so that we can rephrase the definition of a topological space itself solely in categorical termsThis perspective on what a ldquotopologyrdquo is makes it simpler to state the definition of whatrsquos calleda sheaf on a topological space which is an important construction in various areas of analysis andgeometry We might discuss this a bit later on

Functors As with most things in mathematics we should care not only about categories asmathematical structures on their own but also about ldquomapsrdquo between these structures whichldquopreserverdquo said structure The notion of a (covariant) functor between categories provides suchldquomapsrdquo in the case at hand where a functor is a way of sending objects to objects and morphismto morphisms in a way which preserves compositions and identities The precise definition is inSection 13 of the book as is the definition of a contravariant functor between categories whichis a functor which reverses the direction of arrows as opposed to ldquocovariantrdquo ones which preservedirections (It is common to use the term ldquofunctorrdquo without qualification to mean one which iscovariant Note that any functor can be made covariant by working with the opposite category acontravariant functor C rarr D is the same as a covariant functor Cop rarr D)

All of the standard examples we looked at are in the book forgetful functors power set functorsthe dual space functor in Vect etc Of particular interest is Example 139 in the book whichshows that various types of group actions in different contexts-the usual notion of a group actionon a set the notion of a continuous action of a group on a topological space and the notion ofa linear action of a group on a vector space (also known as a representation of the group)-are allencoded by a functor with domain category BG

Adjoint functors Let F Top rarr Set be the forgetful functor and let D Set rarr Top be thefunctor which sends a set S to the space S equipped with the discrete topology and a functionS rarr Sprime between sets to the same function only now viewed as a continuous map S rarr Sprime betweendiscrete spaces These two functors satisfy the following property in relation to one another givinga morphism D(S) rarr Y in Top is the same as giving a morphism S rarr F (Y ) in Set or moresymbolically

MorTop(D(S) Y ) = MorSet(S F (Y ))

We say that D is left adjoint to F and that F is right adjoint to D The point is that adjointfunctors can be used to move data back and forth between two categories

Wersquoll talk more about adjoint functors later on but for now here is one more example Nowtake F Grp rarr Set to again be the forgetful functor and take G Grp rarr Set to be the freegroup functor which sends a set S to the free group generated by S G(S) has as elements ldquowordsrdquos1 sn made up of elements of S with the group operation being ldquoconcatenationrdquo A basic factis that giving a group homomorphism from G(S) to some other group H is the same as giving anordinary function from S to F (H) which is the required adjoint property

MorGrp(G(S) H) = MorSet(S F (H))

Faithful and full functors The definitions of what it means for a functor to be faithful and fullare in Section 15 of the book where faithful functors are ones which are injective on morphisms andfull functors are ones which are surjective on morphisms but where in both cases we refer only tomorphisms which occur between the objects F (A) and F (B) in the target category corresponding tofixed objects A and B in the source category (In general a functor could send two objects AB in

11

the source category to the same object F (A) = F (B) in the target category and thus there could betwo morphisms A rarr C and B rarr C which are sent to the same morphism F (A) = F (B) rarr F (C)Such a thing could still possibly be ldquofaithfulrdquo since the non-injectivity here is arising from differentobjects in the source)

Faithfullness and fullness will be nice properties going forward and here is one hint at how theymight be useful Say we have a functor F C rarr D The problem is to determine properties onF which will imply it send products in C to products in D or coproducts in C to coproducts inD So given the former product diagram below we want the second diagram to also be a productdiagram

C

A B

P Fminusminusminusrarr

F (C)

F (A) F (B)

F (P )f g

pA pB

existh

Ff Fg

F (pA) F (pB)

Fh

Of course in order F (P ) to be an actual product of F (A) and F (B) on the right a similarcommutative diagram property would have to hold for all objects in D in place of F (C) and allmorphisms D rarr F (A) and D rarr F (B) not just those of the form Ff and Fg respectively Oneway to guarantee this is to require that all objects in D are of the form F (C) and then for allmorphisms F (C) rarr F (A) and F (C) rarr F (B) to be of the form Ff Fg the former is true whenthe functor F is surjective on objects and the latter is true when F is full We are not saying thatthese are the only conditions under which F might preserve products but it definitely seems togive a sufficient condition at least

Almost there is one more thing to require the uniqueness of the morphism Fh F (C) rarr F (P )making the diagram on the right commute Suppose there were two such morphisms Fh and Fhprimewhere h and hprime are both morphisms C rarr P in C The commutativity of the diagram on the rightgives for instance

Ff = F (pA)Fh and Ff = F (pA)Fhprime

which the functorial properites of F turns into

Ff = F (pA h) and Ff = (pA hprime)

We want h and hprime to make the first diagram commute in order to be able to apply uniqueness ofthe morphism C rarr P But this requires know that f = pA h and f = pA hprime which would followfrom F being faithful In this case it follows that h and hprime both make the first diagram commute(after we go through a similar argument with g) so the fact that P is a product guarantees thath = hprime and hence that Fh = Fhprime so that F (P ) is an actual product on the right

So we conclude that a functor which is full faithful and surjective on objects will send productsto products (Actually we can relax this a bit by requiring not that every object in D be in theliteral image of F but only that it be isomorphic to something in the image A functor with thisproperty is said to be essentially surjective which is a concept will see in a related context soonenough) Again this is not an ldquoif and only ifrdquo statement since functors can preserve products (orcoproducts) without being full faithful and surjective on objects but it at least gives a sufficientcondition Later we will see other ways to guarantee that functors preserve products which arerelated to the existence of adjoints for that given functor

12

Coproduct Examples Concreteness

Products and coproducts for metric spaces The product of two objects in Met the categoryof metric spaces with morphisms given by metric maps is the Cartesian product X times Y equippedwith the box metric and the coproduct of X and Y does not exist if both X and Y are nonemptyThis was the topic of a student presentation and here are some proof sketches

Given metric spaces (X dX) and (Y dY ) the box metric d on X times Y is defined by

d((x1 y1) (x2 y2)) = maxdX(x1 x2) dY (y1 y2))

(The name ldquoboxrdquo metric comes from the fact that in RtimesR = R2 or R3 open balls indeed looks likeldquoboxesrdquo) The key requirement needed in the definition of a product is given morphisms S rarr Xand S rarr Y that the map

h S rarr X times Y defined by h(s) = (f(s) g(s))

actually be a morphism in Met This turns into the requirement that

d((f(s) g(s)) ((f(t) g(t))) = maxdX(f(s) f(t)) dY (g(s) g(t)) le dS(s t) for all s t isin S

given the fact that

dX(f(s) f(t)) le dS(s t) and dY (g(s) g(t)) le dS(s t) for all s t isin S

For the latter inequalities to imply the former d must (at least up to isometry) be given by thebox metric Note that the two other ldquostandardrdquo metrics one might put on a productmdashthe taxicab

dX + dY and Euclidean983156

d2X + d2Y metricsmdashdo not satisfy this required ldquometric maprdquo property

and so do not give the product in MetTo see that coproducts do not exist if XY are nonempty suppose instead (P dP ) was a co-

product for X and Y with inclusion morphisms iX X rarr P and iY Y rarr P For any n isin Ntake 0 n to be a metric space with the absolute value metric d and let f X rarr 0 n andg Y rarr 0 n be the constant function 0 and the constant function n respectively Since P is acoproduct there exists h P rarr 0 n such that

h iX = f and h iY = g

But in order for h to actually be a morphism in Met we must then have

d(h(iX(x) iY (y)) le dP (iX(x) iY (y)) for any x isin X y isin Y

which translates inton = |0minus n| le dP (iX(x) iY (y))

But this should then be true for any n isin N which would imply that the distance in P betweenan element coming from X and one coming from Y should be infinite which is nonsense (Notethat if you altered your definition of ldquometricrdquo to allow for infinite distance then a coproduct wouldexist the disjoint union X ⊔Y where you defined the distance between elements of X and Y to beinfinite)

Coproducts for groups The coproduct of two groups G and H in Grp is the free product GlowastHwhich is defined to be the set of all words consisting of elements of G andH in an alternating fashion

13

with concatenation as the group operation (When we have to elements of G or two elements of Hnext to each other with use their respective group operations to turn these into single elements ofG or H) This was also the topic of a student presentation and here is the basic idea Given grouphomomorphisms g G rarr N and k H rarr N that G lowastH is the coproduct comes down to showingthat the map

h G lowastH rarr N defined by h(g1h1 gnhn) = f(g1)k(h1) middot middot middot f(gn)k(hn)

and similarly on other possible words in GlowastH is a group homomorphism which follows essentiallyfrom the definition of the group operation on G lowastH indeed this group operation arises preciselyfrom this requirement

Now if we instead tried to use GtimesH with its standard inclusions G rarr GtimesH and H rarr GtimesHas the coproduct the issue is that the analogous map

h GtimesH rarr N defined by h((g h)) = f(g)k(h)

would NOT be a homomorphism because h((g1 h1)(g2 h2)) is not equal to h(g1 h1)h(g2 h2) bythe way in which the group operation is defined on the direct product G times H However if westrict our category to only consist of abelian groups so that the N rsquos we consider in the defini-tion of coproduct are also abelian then G times H does serve as a coproduct we can always rewritef(g1)k(g1) middot middot middot f(gn)k(gn) as f(g1) middot middot middot f(gn)k(h1) middot middot middot k(hn) in N Irsquoll leave it to you to flesh out allof these details

Concrete categories The categories Set Top Vect and Grp all have the property thattheir objects are simply sets possibly equipped with extra structure and that their morphisms areordinary set-theoretic functions possibly satisfying some additional requirement Such categoriesare nice since we can often use intuition about sets and functions in their study A concrete categoryis intuitively one where we can think of objects as being ldquosetsrdquo and morphisms as being ldquofunctionsrdquoonly that we have to interpret this in the right way

Here is the definition a concrete category is a category C equipped with a faithful functorF C rarr Set which is part of the data The idea is that such a functor gives us way to turnan object A isin Ob(C) into a set F (A) and a morphism f isin MorC(AB) into a function Ff isinMorSet(F (A) F (B)) where the ldquofaithfulrdquo requirement says that we do not lose information aboutf when doing so so that we can (essentially) learn everything we need to know about f by studyingFf instead In the four examples above the required faithful functor is simply the forgetful functorand indeed in general we think of F as being a ldquoforgetful functorrdquo for the concrete category (C F )

Examples Here are two more examples of concrete categories Let G be a group and considerthe category BG with a single object and elements of G morphisms As written this does notappear to be concrete since morphisms are not honest functions lowast rarr lowast but the point is that thiscategory can indeed by ldquoconcretizedrdquo by specifying an appropriate ldquoforgetful functorrdquo A functorF BG rarr Set is the data of a (left) action of G on a set S = F (lowast) and saying this functor isfaithful is what it means to say that this action is faithful different g h isin G act on S in differentways or equivalently that there exists s isin S such that gs ∕= hs Any group has at least onefaithful left actionmdashnamely the action on itself given by left multiplicationmdashto BG can always beequipped with a faithful functor to Set This is all just a way of phrasing the fact that any groupis isomorphic to a subgroup of a permutation group

The category Rel of relations can also be ldquoconcretizedrdquo even though as written morphismsA rarr B are not honest functions Take F Rel rarr Set to be the power set functor which sends an

14

object in Rel to its power set and a relation f A rarr B to the induced function Ff P (A) rarr P (B)defined by

(Ff)(S) = b isin B | there exists a isin S such that (a b) isin f

In other words if you think of the relation f sube A times B as a ldquopossibly not everywhere-definedpossibly multivalued functionrdquo (Ff)(S) is analogous to taking the image of S That F is faithfulwill be left to the homework

In fact most categories yoursquoll see can be made concrete by specifying an appropriate forgetfulfunctor but not all categories can be made concrete in this way Here is the standard example of anon-concrete category which we mention only to say there is such a thing but which we wonrsquot doanything more with since understanding it better requires knowledge of algebraic topology Thehomotopy category of topological spaces is the category hTop whose objects are topological spacesand whose morphisms are given by equivalence classes of homotopic maps

MorhTop(XY ) = homotopy classes of continuous maps X rarr Y

(Saying that two maps are homotopic means that you can ldquodeformrdquo one into the other but wersquollleave the precise definition to a course in algebraic topology) It turns out that no functor fromhTop to Set can be faithful so that hTop cannot be made concrete but this is actually a deepand difficult thing to show so we make no attempt to do so here

Natural Isomorphisms Representability

Natural transformations The definition of a natural transformation η F rArr G between two(both covariant or both contravariant) functors FG C rarr D is given in Section 14 of the bookThe key idea is that a natural transformation gives a way to turn data about F into data aboutG in a way which is compatible with all possible morphisms as expressed by the commutativity ofthe diagrams

F (A) F (B)

G(A) G(B)

Ff

ηA ηB

Gf

arising from morphisms f A rarr B (or f B rarr A in the contravariant case) in C In the case of anatural isomoprhism η F rArr G so that each ηA F (A) rarr G(A) is actually an isomorphism in Dthe point is not only that F and G produce isomorphic objects but they do so in a way which iscompatible with all possible category-theoretic data

The first standard example of a natural isomorphism is the one between the identity functor onthe category of finite-dimensional and the functor which sends a finite-dimensional vector space Vto V lowastlowast the dual of its dual induced by the isomorphism V rarr V lowastlowast defined by

v 983041rarr (the linear map on V lowast which sends f to f(v))

This was the topic of a student presentation and can be found in Section 14 of the book Theessential reason why this isomorphism is ldquonaturalrdquo is that V rarr V lowastlowast can be defined without referenceto a basis Contrast this with the the functor which sends V to its (single) dual V lowast it is still truethat finite-dimensional vector space is isomorphic to its dual but specifying such an isomorphism

15

requires additional data which prevents the identity functor from being ldquonaturally isomorphicrdquo tothe single dual functor

Some history At this point it makes sense to say a bit about the interesting origins of categorytheory in the 1950rsquos Eilenberg and Mac Lane where studying cohomology (or possible homologyone of those two) in algebraic topology which associates algebraic data (a group ring or similar)to a topological space in a way which encodes some of the topology They had two cohomologicalconstructions which ended up proving isomorphic results but they noticed that not only werethe resulting objects the same but the actual constructions themselves were in some sense theldquosamerdquo They thus then needed a way to formalize this notion and invented the concept of anatural transformation to do so where they interpreted two constructions as being the same asbeing compatible with all morphisms

But this then leads to the question what types of objects should a natural transformationoccur between Eilenberg and Mac Lane then had to invent the notion of a ldquofunctorrdquo to describewhat a natural transformation went from and what it went to In their motivating setting theyinvented the notion of a functor to describe more formally properties which their cohomologicalconstructions were supposed to satisfy But this then leads to the question what types of objectsshould a functor map between In other words what type of data should a functor (cohomologicalconstruction in their example) take as input and should it output They finally had to thus inventthe notion of a ldquocategoryrdquo to describe the source and target of a functor so that their cohomologicalconstructions could be interpreted as functors from a category of topological spaces to a categoryof algebraic objects

Thus historically natural transformations came first then functors and then categories Aswith most things in mathematics categories thus arose from attempts to encode what was beingseen in some concrete examples in a more general and formal setting

Representable functors Representable functors are defined in Section 21 of the book but wersquollgive the required definitions here in a hopefully simpler to digest manner The point is that anyobject A in a (locally small) category C gives two ldquonaturalrdquo functors C rarr Set one covariant andthe other contravariant which as wersquoll see pretty much encode all data about A itself We definehA C rarr Set to be the covariant functor defined by

hA(B) = Mor(AB) on objects and

hA(Bfminusrarr C) = the map Mor(AB)

hAfminusminusrarr Mor(AC) defined by g 983041rarr f g

We define hA C rarr Set to the contravariant functor defined by

hA(B) = Mor(BA) on objects and

hA(Bfminusrarr C) = the map Mor(CA)

hAfminusminusrarr Mor(BA) defined by g 983041rarr g f

(The functor hA is called the functor of points of A and wersquoll usually think of it as being a covariantfunctor hA Cop rarr Set The functor hA does not have a standard name) So hA is defined bymorphisms from A and hA is defined by morphisms into A Wersquoll see later the sense in which suchfunctors encode all information about the object A

A functor F C rarr Set is representable if is naturally isomorphic to such a functor so F sim= hA

for some A in the covariant case and F sim= hA for some A in the contravariant case The idea wersquollbuild towards is that we can essentially treat such a functor as being A itself so that representable

16

functors can be thought as being actual objects in C For now here is a key example wersquove seenbefore which we can now formulate using the language of representability

Given AB isin Ob(C) let F C rarr Set be the contravariant functor defined by

F (C) = Mor(CA)timesMor(CB) and F (Cfminusrarr D) = [(g k) 983041rarr (g f k f)]

where (g k) 983041rarr (g f k f) gives a map Mor(DA) timesMor(DB) rarr Mor(CA) timesMor(CB) ieF (D) rarr F (C) In order for this functor to be representable requires the existence of an object inC satisfying

Mor(CA)timesMor(CB) sim= Mor(C representing object)

in a way which is compatible with all morphisms but we have actually already seen this notionelsewhere it is essentially the definition of a product AtimesB for A and B Indeed we claim that Fis naturally isomorphic to hAtimesB as we now show

First to define a natural isomorphism hAtimesBsim= F requires for each C isin Ob(C) a choice of a

bijection (ie isomorphism in Set)

ηC hAtimesB(C)sim=minusrarr F (C)

which in our case looks like

ηC Mor(CAtimesB))sim=minusrarr Mor(CA)timesMor(CB)

Take this to be the map

(Cfminusrarr AtimesB) 983041rarr (pA f pB f)

where pA A times B rarr A and pB A times B rarr B are the two projection morphisms in the definitionof a product The definition of product implies that this map is invertible since given k C rarr Aand ℓ C rarr B there exists a unique h C rarr A times B such that k = pA h and ℓ = pB h Forthese bijections to define a natural isomorphism hAtimesB

sim= F requires that they be compatible withall morphisms in C which means that given any g C rarr D in C the following diagram shouldcommute

hAtimesB(C) hAtimesB(D)

F (C) F (D)

hAtimesB(g)

ηC ηD

Fg

which in the case at hand looks like

Mor(CAtimesB) Mor(DAtimesB)

Mor(CA)timesMor(CB) Mor(DA)timesMor(DB)

minus g

ηC ηD

(minus gminus g)

Take an element k D rarr AtimesB in the upper right Applying the map on top gives the elementk g in the upper left and then applying ηC on the left side gives

(pA (k g) pB (k g))

17

in the lower left Instead starting with k D rarr AtimesB in the upper right applying ηD on the rightgives (pA k pB k) and applying the map on the bottom gives

((pA k) g (pB k) g)

in the lower left which is the same as the previous element in the lower left by the associativity ofcomposition Hence the isomorphisms ηC do define a natural isomorphism between hAtimesB and F Again the point is that you can then interpret the functor F as being the ldquosamerdquo as the productAtimesB since both encode the same data

More Representable Examples

Coproducts As in the case of products the definition of coproducts can also be phrased in termsof a representable functor Given objects A and B in C let F C rarr Set be the covariant functordefined by

F (C) = Mor(AC)timesMor(BC) and F (Cfminusrarr D) = [(g k) 983041rarr (f g f k)]

where the latter is a function Mor(AC)timesMor(BC) rarr Mor(AD)timesMor(BD) This functor isrepresentable if and only if a coproduct A ⊔ B for A and B exists and the representing object isthat coproduct the key is the requirement that we have bijections

Mor(AC)timesMor(BC) sim= Mor(A ⊔BC)

for all C which is essentially the defining property a coproduct for A and B should have

Power set functor contravariant version Let P Set rarr Set be the contravariant powerset functor defined on objects by sending a set to its power set and on morphisms by sendingf A rarr B to the preimage function Pf P (B) rarr P (A) given by S 983041rarr fminus1(S) In order for this tobe representable requires the existence of a set X with natural bijections

P (A) sim= Mor(AX)

for all sets A that is a set X so that mapping into X is the same data as specifying a subset ofthe domain We take X = 0 1 to be a two-element set and the required bijections

ηA P (A) rarr Mor(A 0 1) to be given by S 983041rarr χS

where XS A rarr 0 1 is the indicator or characteristic function S defined by sending elementsof S to 1 and elements of A minus S to 0 The inverse of ηA sends a function g A rarr 0 1 to thepreimage gminus1(1) sube A

To verify that the bijections ηA define the data of a natural isomorphism η P rArr h01 wecheck the commutativity of some diagrams Given a function f A rarr B consider

P (A) P (B)

h01(A) = Mor(A 0 1) h01(B) = Mor(B 0 1)

Pf = fminus1

ηA ηB

minus f

18

Take S isin P (B) in the upper right Applying the map on top gives fminus1(S) isin P (A) and thenapplying the map on the left gives χfminus1(S) isin Mor(A 0 1) On the other hand applying the mapon the right to S isin P (B) gives χS isin Mor(B 0 1) and then applying the map on the bottomgives χS f isin Mor(A 0 1) The two resulting functions χfminus1(S)χS f in the lower left are

χfminus1(S)(x) =

9830831 x isin fminus1(S)

0 otherwiseand χS(f(x)) =

9830831 f(x) isin S

0 otherwise

so these are the same since x isin fminus1(S) if and only if f(x) isin S Thus the diagram above commutesso P is naturally isomorphic to h01 and hence P is representable

Power set functor covariant version Now consider the covariant power set functor stilldenoted by P Set rarr Set which again sends a set to its power set but now sends a functionf A rarr B to the image function Pf P (A) rarr P (B) defined by S 983041rarr f(S) In order for this tobe representable there would have to exist a set X such that P sim= hX which translates into havingnatural bijections

P (S) sim= Mor(XS)

for all sets S However in this case there is no natural choice for X as there is no natural way todetect subsets of S by mapping into S as opposed to the case of P (S) sim= Mor(S 0 1)

That no such X exists can be made precise using a simple argument when S = empty P (S) hascardinality 1 soX would have to be empty as well since otherwise Mor(XS) would have cardinality0 but if X is empty then Mor(empty S) always cardinality 1 regardless of S which is clearly not trueof P (S) Thus the covariant power set functor is not representable

Forgetful on Top The forgetful functor F Top rarr Set is representable Indeed a representingobject would have to satisfy

X sim= Map(objectX)

as sets for any topological space X (where Map denotes the set of continuous maps) and it iseasy to see that a single point satisfies this property a continuous map f pt rarr X is completelydetermined by the element f(pt) of X and any such element gives a continuous map in this wayOf course checking that F sim= hpt requires checking that various diagrams commute but this issimple to work out in this case so we omit it here

Extracting a topology With an eye towards the idea that the functors hA hA should captureall information about an object A in a category note that the result above shows that in TophX first of all captures all information about X as a set since the set X can be extracted fromhX(pt) = Map(ptX)

We claim that hX actually captures all information about X as a topological space since thetopology on X can be extracted from hX in the following way As we saw in the power set examplewe have a bijection

P (X) sim= Mor(X 0 1)

Now equip 0 1 with the topology in which 1 is open but 0 is not Then a continuous mapf X rarr 0 1 is fully determined by the preimage fminus1(1) since to be continuous only requiresthat this single preimage be open in X Moreoever any open subset of X arises in this way bytaking f to be the indicator function of that open set Thus we get a bijection

Op(X) sim= Map(X 0 1)

19

where Op(X) denotes the set of open subsets of X ie the topology on X Hence the topology onX can be extracted from hX as claimed so hX and hX do encode all information about X

This fact can also be interpreted as saying that the contravariant functor which sends a topo-logical space X to its collection of open subsets (and which sends a continuous map to the mapinduced by taking preimages) is representable with representing object 0 1 equipped with thetopology given above This will be worked out in a homework problem

Forgetful on Grp The forgetful functor Grp rarr Set is also representable In this case arepresenting object should be a group giving set-theoretic bijections

G sim= Hom(objectG)

for all groups G where Hom denotes the set of group homomorphisms The group Z works

G sim= Hom(Z G)

since a homomorphism φ Z rarr G is completely determined by φ(1) since Z is generated by 1 andthere are no restrictions on what φ(1) isin G can actually be The inverse of the desired bijection isthus φ 983041rarr φ(1) and it is simple to check that morphisms behave appropriately so that the forgetfulfunctor in this setting is indeed isomoprhic to hZ

Forgetful on Ring The forgetful functor Ring rarr Set is also representable where by Ring wemean the category of rings with identity elements and where morphisms (ie ring homomorphisms)are required to preserve identities Here we need an object satisfying

R sim= Hom(object R)

for all R isin Ob(Ring) and we can see that Z[x] the ring of polynomials in the variable x withcoefficients in Z works Indeed given a ring homomorphism φ Z[x] rarr R the fact that φ isrequired to preserve identities fully determines φ(n) for any n isin Z (since φ(n) = nφ(1)) and hencesince φ preserves multiplication its behavior is fully determined by φ(x) which can be any elementof R The inverse of the bijection we want is thus φ 983041rarr φ(x) and it will follow that this forgetfulfunctor is isomorphic to hZ[x]

Equivalences between Categories

Definitions iso and equiv Now that we have a notion of a ldquomorphismrdquo between categorieswe can talk about the sense in which two categories are the ldquosamerdquo The first possible definitionis the ldquoobviousrdquo one you might expect given the definition of ldquoisomorphismrdquo wersquove seen in everyother context until now we say that C is isomorphic to D if there exist functors F C rarr D andG D rarr C such that G F = idC and F G = idD where idC and idD are identity functors

However this definition turns out to be too restrictive Indeed the requirement that the givencompositions equal identity functors says that given an object A in C the object G(F (A)) beliterally the same as A itself and similarly for objects in D But we know that two objects in acategory can be thought of as being the ldquosamerdquo without being literally the same as long as theyare isomorphic For instance productscoproducts are not unique in the literal sense but only inthe sense that they are isomorphic in a unique way To take another example the definition ofa functor being representable does not say that F (B) literally be the same as hB(A) only thatthey be isomorphic for sure the power set P (S) of a set S is not literally the same as the set offunctions S rarr 0 1mdashit is only the ldquosamerdquo in the sense that they encode the same data

20

Thus it makes sense to relax the requirement that A andG(F (A)) in the definition of isomorphiccategories be equal to the requirement that they be isomorphic This leads to the following betterdefinition of two categories being the ldquosamerdquo C is equivalent to D if there exist functors F C rarr D

and G D rarr C such that G F sim= idC and F G sim= idD So we require only that for instance thefunctor G F be naturally isomorphic to the identity functor on C which says that for any objectA G(F (A)) be isomorphic to A in a natural way Wersquoll see that this truly is the better notionof what it means for two categories to be the ldquosamerdquo and it reflects the idea that all categoricalproperties of one category can be transported to the other

Alternate characterization of equivalences A functor F C rarr D implementing an equiv-alence between categories is called unsurprisingly and equivalence and the ldquoinverserdquo functorG D rarr C is still referred to as being its inverse even though it is not technically an ldquoinverserdquoin the sense that compositions give literal identities But as in the case of Set where ldquoinvertiblerdquocan be characterized without reference to an inverse as ldquoinjectiverdquo and ldquosurjectiverdquo so too canequivalences between categories be characterized without explicit reference to an inverse

A functor F C rarr D is an equivalence if and only it is full faithful and essentiallysurjective

We have seen the notions of full and faithful before and essentially surjective means that any objectin D is isomorphic to something in the image of F for all D isin Ob(D) there exits C isin Ob(C) suchthat F (C) sim= D Justifying this result was the topic of a student presentation and can be foundin the book in Section 15 (Technically to construct an ldquoinverserdquo of F requires that we work withsmall categories but whatever) The bulk of the real work goes into showing that the ldquoinverserdquofunctor constructed is actually a functor which comes down to some abstract nonsense argumentvia commutative diagrams

Equivalences preserve everything We previously saw the notions of full faithful and es-sentially surjective back when thinking about what properties a functor could have which wouldguarantee it sent products to products so the result we proved back then was that equivalencespreserve products Similarly equivalences preserve coproducts and the proof is basically the sameafter reversing arrows A clean way of deriving this is as follows if F C rarr D is an equivalencethen so is F op Cop rarr Dop so since F op preserves products F preserves coproducts becausecoproducts in a category become products in the opposite category

In general an equivalence between categories will preserve whatever categorical notions youwant monomorphisms epimorphisms initial objects terminal objects etc so that equivalentcategories really do have essentially the ldquosamerdquo properties

Equivalence example Here is a first example (also found in the book) of equivalent categorieswhich are not isomorphic Let FVectR be the category whose objects are finite-dimensional realvector spaces and whose morphisms are linear transformations Let MatR be the category ofmatrices defined as follows

bull objects in MatR are nonnegative integers

bull a morphism n rarr m is an mtimes n matrix with real entries

bull composition in MatR is given by matrix multiplication given B n rarr m and A m rarr kmdashsoB is an mtimes n matrix and A is a k timesm matrixmdashthe composition A B n rarr k is the k times nmatrix AB

bull identity morphisms in MatR are given by identity matrices

21

We claim that FVectR and MatR are equivalent First note that they certainly cannot be isomo-prhic isomorphisms in MatR exist only between two objects which are literally the same (ie anmtimesn matrix can be invertible only when m = n) but isomorphisms in FVectn can exist betweenobjects which are not literally the same So there are in a sense not enough objects in MatR toallow it to be isomorphic to FVectR

To construct an equivalence choose one and for all an isomorphism TV V rarr RdimV forevery object in FVectR or equivalently choose once and for all a basis for every V DefineF FVectR rarr MatR by

V 983041rarr dimV on objects and

(T V rarr W ) 983041rarr (the standard matrix of SW T Sminus1V RdimV rarr RdimW ) on morphisms

Equivalently F sends T to the matrix of T relative to the chosen bases for V and W It isstraightforward to check that this is a functor The inverse functor G MatR rarr VectR is givenby

n 983041rarr Rn on objects and

(mtimes n matrix A) 983041rarr (the linear transformation A Rn rarr Rm induced by A) on morphisms

One can check that G is a functor and that F G and G F are naturally isomorphic to identitiesor instead one can argue that F alone is full faithful and essentially surjective The point is thatG(F (V )) gives back RdimV so not literally V itself but only something isomorphic to V which iswhy this gives an equivalence of categories and not an isomorphism

Yoneda Lemma Functors as Objects

Compact Hausdorff spaces and Clowast-algebras Last time we gave an example of equivalentcategories which were not isomorphic but in many ways that example is unsatisfactory when doinglinear algebra people for sure often work with matrix representations of linear transformations asthe example from last would suggest but no one in their right mind literally thinks about suchthings in terms of the equivalence FVectR rarr MatR itself In other words this equivalence isessentially ldquocooked uprdquo for the sake of giving an example but it does not really give any insightinto the actual mathematics of linear algebra

So with that in mind we will describe another equivalence which actually does have some realmeaning in the sense of producing new ways of thinking about some mathematics Given a compactHausdorff space X we let C(X) denote the space of continuous complex-valued functions on X

C(X) = f X rarr C | f is continuous

We can equip the set C(X) with various additional structures First it is a complex vector spaceunder ordinary addition and scalar multiplication of functions Second we can multiply two func-tions together

(fg)(p) = f(p)g(p)

which turns C(X) into whatrsquos called an algebra over C (We wonrsquot give the definition of anldquoalgebrardquo but the point is that this multiplication is in some sense ldquocompatiblerdquo with the vectorspace structure) Next any element f of C(X) has a conjugate element f obtained by takingcomplex conjugates of the values of f

f(p) = f(p)

22

And finally we can equip C(X) with a norm via

983042f983042 = suppisinX

|f(p)|

where |middot| denotes complex absolute value (This supremum exists sinceX is compact by the ExtremeValue Theorem) All of these structures (vector space algebra multiplication conjugation norm)turn C(X) into whatrsquos called a Clowast-algebra We wonrsquot define this precisely but the definition encodesvarious ways in which these structures are compatible Actually C(X) is a unital commutative Clowast-algebra where ldquounitalrdquo means that the multiplication has an identity element (namely the constantfunction 1) and commutative means that fg = gf There is a natural notion of homomorphismbetween Clowast-algebras which are maps between them which behave nicely with all the structuresinvolved In particular given a continuous map f X rarr Y between compact Hausdorff spaces weget a Clowast-algebra homomorphism flowast C(Y ) rarr C(X) given by pre-composition with f flowast sendsg isin C(Y ) to g f isin C(X)

We thus have a contravariant functor CHaus rarr ucClowast-Alg from the category of compactHausdorff spaces and the category of unital commutative Clowast-algebras It turns out that anyunital commutative Clowast-algebra arises in this way in the sense that given any unital commutativeClowast-algebra B there exists a compact Hausdorff space X such that C(X) sim= B as Clowast-algebrasand that morphisms between these algebra arises from continuous maps in the manner describeabove This says that the functor above is essentially surjective full and it turns out that it isalso faithful so that it is an equivalence Thus all information about compact Hausdorff spaces iscompletely encoded in information about unital commtuative Clowast-algebras and indeed one couldlearn everything there is to know about a compact Hausdorff space X from its Clowast-algebra C(X)of functions The fact that the categories CHaus and ucClowast-Alg are equivalent is the startingpoint in the subject known as noncommutative geometry which wersquoll say something about a bitlater on If we drop the requirement that our algebras be unital it turns out that the categoryof commutative Clowast-algebras in general is equivalent to the category of locally compact Hausdorffspaces via essentially the same functor where we take as morphisms in the category of locallycompact Hausdorff spaces to be continuous functions which ldquovanish at infinityrdquo

We are only introducing Clowast-algebras to give an example of an interesting and actually usefulequivalence between categories but just for the fun of it wersquoll note the following Examples ofnoncommutative Clowast-algebras come from what are known as ldquoalgebras of bounded operators onHilbert spacesrdquo which show up in quantum mechanics as the things which describe physicallyobservable quantities like position momentum and energy (In fact all Clowast-algebras arise fromHilbert spaces in this way) Thus Clowast-algebras show up in mathematical formulations of quantummechanics and the notion of noncommutative geometry mentioned above is at the core of someattempts to understand the elusive concept known as ldquoquantum geometryrdquo

Yoneda Lemma The Yoneda Lemma states that given a (covariant) functor F C rarr Set whereC is locally small for any A isin Ob(C) there exists a bijection

F (A) sim= Nat(hA F )

where the right side denotes the set of natural transformations from hA to F A similar result holdsin the contravariant case where we get bijections F (A) sim= Nat(hA F ) (There is also a sense inwhich these bijections are themselves natural but wersquoll ignore this bit) The proof of the YonedaLemma was the topic of a student presentation and can be found in Section 22 of the book

Here we will get a bit philosophical and talk about what the point of the Yoneda Lemmaactually is and later the sense in which it says that we can view arbitrary functors as ldquonon-existent

23

objectsrdquo in a category First we mention what we will call the ldquofunctor of pointsrdquo approach tomathematics a mathematical object can be fully characterized by all morphisms into it That isknowledge of Mor(SA) for all S is equivalent to knowledge of A itself Indeed this is an idea wementioned previously in relation to the functors hAmdashwhich we previously said is referred to as beingthe functor of points of Amdashand now we seek to push this idea further In some examples literalpoints of A can be extracted from hA as in the case of the bijection X sim= hX(pt) = Map(ptX) fora topological space X or G sim= hG(Z) = Mor(Z G) for a group G Moreover in these examples therest of the (topological) structure onX or (group) structure on G can also be extracted from hX andhG respectively In general we can view Mor(SA) as in a sense being ldquopoints of A parameterizedby Srdquo which comes from the idea that in say Set Mor(SA) = AS is ldquothe product of S-manycopies of Ardquo so literally points of A ldquoparameterizedrdquo by S In various areas of mathematics anelement of Mor(SA) is thought of as being a ldquogeneralizedrdquo point of A so that the functor of pointshA detects generalized points of A

So an object A is fully encoded by the functor hA Now we claim that morphisms betweenobjects are also encoded by such functors Indeed in the case where F = hB is itself a functor ofpoints the Yoneda result Nat(hA F ) sim= F (A) becomes

Nat(hA hB) sim= hB(A) = Mor(AB)

which says precisely that morphisms between the objects A and B can be obtained solely fromthe functors hA and hB (This fact is one reason why one should prefer the contravariant functorhA to the covariant functor hA in the covariant case Nat(hA F ) sim= F (A) for F = hB becomesNat(hA hB) sim= hB(A) = Mor(BA) so that morphisms between two objects get reversed whencharacterizing them in terms of representable functors)

Pushing our philosophy further we now know that not only objects but also morphisms betweenobjects are fully encoded by the functors hA The final step is to note that we should not only careabout objects and morphisms between them but that an important part of the data of an objectA is also how arbitrary contravariant functors F C rarr Set behave on A and that this too can beobtained from hA precisely via the Yoneda result that F (A) sim= Nat(hA F ) So objects morphismsbetween objects and how arbitrary functors behave on objects are all encoded in knowledge ofthe functors hA so that these functors truly do capture in the most general sense possible allinformation about A

Functors as nonexistent objects The discussion above is all well and good but it still doesnrsquotquite say why we should care about the fact that we can study on object A by studying hA insteadin that we still havenrsquot spoken about what type of insight about A can be gleamed from hA thatwasnrsquot readily apparent in A itself already Now we argue that the true point of the YonedaLemma is that it gives a way to think about arbitrary functors F Cop rarr Set as ldquoobjectsrdquo in C

itself or really as objects in an ldquoenlargementrdquo of C Certainly for those functors F sim= hA which arerepresentable we can indeed think about F as simply being the object A but now we claim thateven non-representable F can be thought of as being ldquononexistentrdquo objects in C The upshot is thatoften in mathematics we come across things which do not actually exist but the Yoneda Lemmagives a way to (indirectly) study such things anyway by interpreting them as functors instead

The starting point of course is that an actual object A is the same as hA as we have explainedabove Thus by analogy we can interpret a non-representable functor F as being an ldquoobjectrdquo in C

as long as we know what it means to ldquomaprdquo into F from other things which are honestly objects in CThat is if F were indeed representable then F (A) sim= Nat(hA F ) sim= Mor(AF ) would characterizethe functor of points of F (here we abuse notation by using the same letter F to denote both arepresentable functor and the object which represents it) so now we take Mor(AF ) = F (A) as

24

the definition of the ldquofunctor of pointsrdquo of the non-existent object F We are thus taking elementsof F (A) as being what it means to ldquomaprdquo from A into F even when F is not representable Thefact that the Yoneda Lemma gives a way to define maps from A into F thus says that F itself isequipped with the same data that a literal object of A would be equipped with so why shouldnrsquotwe interpret F as a being a type of ldquoobjectrdquo in C as well

To phrase this another way note that the Yoneda Lemma gives rise to a functor

C rarr Fun(CopSet)

where Fun(CopSet) denotes the category whose objects are functors from Cop to Set and morphismsare natural transformation defined on objects by A 983041rarr hA and on morphisms by the bijectionMor(AB) sim= Nat(hA hB) arising from the Yoneda Lemma That this is actually bijection saysthat this functor is full and faithful so that we can view C as being a subcategory of the functorcategory Fun(CopSet) (We cay say that C is isomorphic to the subcategory of Fun(CopSet)consisting of the functors hA or that it is equivalent to the subcategory of Fun(CopSet) consistingof representable functors This functor C rarr Fun(CopSet) is usually called the Yoneda embeddingof C into Fun(CopSet)) Thus we can view Fun(CopSet) as an ldquoenlargmentrdquo of C and the ideawe are trying to make sense of is that other objects in this ldquolargerrdquo category should be interpretedas ldquonon-existent objectsrdquo of C

Just how far can we actually push this idea that arbitrary functors Cop rarr Set should be thoughtof as being ldquoobjectsrdquo of C Pretty far in fact For instance often times in geometry one comesacross a type of ldquosmoothrdquo space on which some group acts and one is interested in doing ldquocalculusrdquoon the resulting quotient space which is the space of orbits under the given group action Theproblem is that such quotient spaces are very often (usually in fact) not ldquosmoothrdquo themselves sothat the tools from calculus are not readily available The solution is to nonetheless treat thisquotient as if it were ldquosmoothrdquo and attempt to do calculus anyway To make this precise thisldquonon-smoothrdquo space actually turns out to have a natural interpretation as a ldquosmooth functorrdquo andit is through this functor which we can view this non-smooth space as actually being smooth afterall so we are forced to expand our notion of what we mean by ldquosmooth spacerdquo by allowing thingswhich are not spaces in the usual sense which certainly leads to more abstraction but the benefitis that it actually makes sense to do calculus on such ldquospacesrdquo interpreted as functors

The discussion above is pretty vague so here is one final more concrete example Recall thatequivalence

CHausop rarr ucClowast-Alg

between the category of compact Hausdorff spaces and the category of unital commutative Clowast-algebras mentioned at the start We now claim that a non-commutative (unital) Clowast-algebra shouldbe interpreted as the algebra of functions on some non-existent ldquononcommutative compact Haus-dorff spacerdquo and it is this idea which the subject of noncommutative geometry makes precise Solet B be a non-commutative Clowast-algebra Then the functor of points hB gives a functor

ucClowast-Alg rarr Set

which sends C(X) (where X is some compact Hausdorff space) to Hom(C(X) B) where Hom de-notes homomorphisms of Clowast-algebras whatever that means Composing with the previous functorwe get a functor wersquoll denote by B itself

B CHausop rarr Set

which on objects at least is given by X 983041rarr Hom(C(X) B) This functor is not representable sinceB is not commutative but the philosophy espoused above suggests we should think of it as giving a

25

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

the source category to the same object F (A) = F (B) in the target category and thus there could betwo morphisms A rarr C and B rarr C which are sent to the same morphism F (A) = F (B) rarr F (C)Such a thing could still possibly be ldquofaithfulrdquo since the non-injectivity here is arising from differentobjects in the source)

Faithfullness and fullness will be nice properties going forward and here is one hint at how theymight be useful Say we have a functor F C rarr D The problem is to determine properties onF which will imply it send products in C to products in D or coproducts in C to coproducts inD So given the former product diagram below we want the second diagram to also be a productdiagram

C

A B

P Fminusminusminusrarr

F (C)

F (A) F (B)

F (P )f g

pA pB

existh

Ff Fg

F (pA) F (pB)

Fh

Of course in order F (P ) to be an actual product of F (A) and F (B) on the right a similarcommutative diagram property would have to hold for all objects in D in place of F (C) and allmorphisms D rarr F (A) and D rarr F (B) not just those of the form Ff and Fg respectively Oneway to guarantee this is to require that all objects in D are of the form F (C) and then for allmorphisms F (C) rarr F (A) and F (C) rarr F (B) to be of the form Ff Fg the former is true whenthe functor F is surjective on objects and the latter is true when F is full We are not saying thatthese are the only conditions under which F might preserve products but it definitely seems togive a sufficient condition at least

Almost there is one more thing to require the uniqueness of the morphism Fh F (C) rarr F (P )making the diagram on the right commute Suppose there were two such morphisms Fh and Fhprimewhere h and hprime are both morphisms C rarr P in C The commutativity of the diagram on the rightgives for instance

Ff = F (pA)Fh and Ff = F (pA)Fhprime

which the functorial properites of F turns into

Ff = F (pA h) and Ff = (pA hprime)

We want h and hprime to make the first diagram commute in order to be able to apply uniqueness ofthe morphism C rarr P But this requires know that f = pA h and f = pA hprime which would followfrom F being faithful In this case it follows that h and hprime both make the first diagram commute(after we go through a similar argument with g) so the fact that P is a product guarantees thath = hprime and hence that Fh = Fhprime so that F (P ) is an actual product on the right

So we conclude that a functor which is full faithful and surjective on objects will send productsto products (Actually we can relax this a bit by requiring not that every object in D be in theliteral image of F but only that it be isomorphic to something in the image A functor with thisproperty is said to be essentially surjective which is a concept will see in a related context soonenough) Again this is not an ldquoif and only ifrdquo statement since functors can preserve products (orcoproducts) without being full faithful and surjective on objects but it at least gives a sufficientcondition Later we will see other ways to guarantee that functors preserve products which arerelated to the existence of adjoints for that given functor

12

Coproduct Examples Concreteness

Products and coproducts for metric spaces The product of two objects in Met the categoryof metric spaces with morphisms given by metric maps is the Cartesian product X times Y equippedwith the box metric and the coproduct of X and Y does not exist if both X and Y are nonemptyThis was the topic of a student presentation and here are some proof sketches

Given metric spaces (X dX) and (Y dY ) the box metric d on X times Y is defined by

d((x1 y1) (x2 y2)) = maxdX(x1 x2) dY (y1 y2))

(The name ldquoboxrdquo metric comes from the fact that in RtimesR = R2 or R3 open balls indeed looks likeldquoboxesrdquo) The key requirement needed in the definition of a product is given morphisms S rarr Xand S rarr Y that the map

h S rarr X times Y defined by h(s) = (f(s) g(s))

actually be a morphism in Met This turns into the requirement that

d((f(s) g(s)) ((f(t) g(t))) = maxdX(f(s) f(t)) dY (g(s) g(t)) le dS(s t) for all s t isin S

given the fact that

dX(f(s) f(t)) le dS(s t) and dY (g(s) g(t)) le dS(s t) for all s t isin S

For the latter inequalities to imply the former d must (at least up to isometry) be given by thebox metric Note that the two other ldquostandardrdquo metrics one might put on a productmdashthe taxicab

dX + dY and Euclidean983156

d2X + d2Y metricsmdashdo not satisfy this required ldquometric maprdquo property

and so do not give the product in MetTo see that coproducts do not exist if XY are nonempty suppose instead (P dP ) was a co-

product for X and Y with inclusion morphisms iX X rarr P and iY Y rarr P For any n isin Ntake 0 n to be a metric space with the absolute value metric d and let f X rarr 0 n andg Y rarr 0 n be the constant function 0 and the constant function n respectively Since P is acoproduct there exists h P rarr 0 n such that

h iX = f and h iY = g

But in order for h to actually be a morphism in Met we must then have

d(h(iX(x) iY (y)) le dP (iX(x) iY (y)) for any x isin X y isin Y

which translates inton = |0minus n| le dP (iX(x) iY (y))

But this should then be true for any n isin N which would imply that the distance in P betweenan element coming from X and one coming from Y should be infinite which is nonsense (Notethat if you altered your definition of ldquometricrdquo to allow for infinite distance then a coproduct wouldexist the disjoint union X ⊔Y where you defined the distance between elements of X and Y to beinfinite)

Coproducts for groups The coproduct of two groups G and H in Grp is the free product GlowastHwhich is defined to be the set of all words consisting of elements of G andH in an alternating fashion

13

with concatenation as the group operation (When we have to elements of G or two elements of Hnext to each other with use their respective group operations to turn these into single elements ofG or H) This was also the topic of a student presentation and here is the basic idea Given grouphomomorphisms g G rarr N and k H rarr N that G lowastH is the coproduct comes down to showingthat the map

h G lowastH rarr N defined by h(g1h1 gnhn) = f(g1)k(h1) middot middot middot f(gn)k(hn)

and similarly on other possible words in GlowastH is a group homomorphism which follows essentiallyfrom the definition of the group operation on G lowastH indeed this group operation arises preciselyfrom this requirement

Now if we instead tried to use GtimesH with its standard inclusions G rarr GtimesH and H rarr GtimesHas the coproduct the issue is that the analogous map

h GtimesH rarr N defined by h((g h)) = f(g)k(h)

would NOT be a homomorphism because h((g1 h1)(g2 h2)) is not equal to h(g1 h1)h(g2 h2) bythe way in which the group operation is defined on the direct product G times H However if westrict our category to only consist of abelian groups so that the N rsquos we consider in the defini-tion of coproduct are also abelian then G times H does serve as a coproduct we can always rewritef(g1)k(g1) middot middot middot f(gn)k(gn) as f(g1) middot middot middot f(gn)k(h1) middot middot middot k(hn) in N Irsquoll leave it to you to flesh out allof these details

Concrete categories The categories Set Top Vect and Grp all have the property thattheir objects are simply sets possibly equipped with extra structure and that their morphisms areordinary set-theoretic functions possibly satisfying some additional requirement Such categoriesare nice since we can often use intuition about sets and functions in their study A concrete categoryis intuitively one where we can think of objects as being ldquosetsrdquo and morphisms as being ldquofunctionsrdquoonly that we have to interpret this in the right way

Here is the definition a concrete category is a category C equipped with a faithful functorF C rarr Set which is part of the data The idea is that such a functor gives us way to turnan object A isin Ob(C) into a set F (A) and a morphism f isin MorC(AB) into a function Ff isinMorSet(F (A) F (B)) where the ldquofaithfulrdquo requirement says that we do not lose information aboutf when doing so so that we can (essentially) learn everything we need to know about f by studyingFf instead In the four examples above the required faithful functor is simply the forgetful functorand indeed in general we think of F as being a ldquoforgetful functorrdquo for the concrete category (C F )

Examples Here are two more examples of concrete categories Let G be a group and considerthe category BG with a single object and elements of G morphisms As written this does notappear to be concrete since morphisms are not honest functions lowast rarr lowast but the point is that thiscategory can indeed by ldquoconcretizedrdquo by specifying an appropriate ldquoforgetful functorrdquo A functorF BG rarr Set is the data of a (left) action of G on a set S = F (lowast) and saying this functor isfaithful is what it means to say that this action is faithful different g h isin G act on S in differentways or equivalently that there exists s isin S such that gs ∕= hs Any group has at least onefaithful left actionmdashnamely the action on itself given by left multiplicationmdashto BG can always beequipped with a faithful functor to Set This is all just a way of phrasing the fact that any groupis isomorphic to a subgroup of a permutation group

The category Rel of relations can also be ldquoconcretizedrdquo even though as written morphismsA rarr B are not honest functions Take F Rel rarr Set to be the power set functor which sends an

14

object in Rel to its power set and a relation f A rarr B to the induced function Ff P (A) rarr P (B)defined by

(Ff)(S) = b isin B | there exists a isin S such that (a b) isin f

In other words if you think of the relation f sube A times B as a ldquopossibly not everywhere-definedpossibly multivalued functionrdquo (Ff)(S) is analogous to taking the image of S That F is faithfulwill be left to the homework

In fact most categories yoursquoll see can be made concrete by specifying an appropriate forgetfulfunctor but not all categories can be made concrete in this way Here is the standard example of anon-concrete category which we mention only to say there is such a thing but which we wonrsquot doanything more with since understanding it better requires knowledge of algebraic topology Thehomotopy category of topological spaces is the category hTop whose objects are topological spacesand whose morphisms are given by equivalence classes of homotopic maps

MorhTop(XY ) = homotopy classes of continuous maps X rarr Y

(Saying that two maps are homotopic means that you can ldquodeformrdquo one into the other but wersquollleave the precise definition to a course in algebraic topology) It turns out that no functor fromhTop to Set can be faithful so that hTop cannot be made concrete but this is actually a deepand difficult thing to show so we make no attempt to do so here

Natural Isomorphisms Representability

Natural transformations The definition of a natural transformation η F rArr G between two(both covariant or both contravariant) functors FG C rarr D is given in Section 14 of the bookThe key idea is that a natural transformation gives a way to turn data about F into data aboutG in a way which is compatible with all possible morphisms as expressed by the commutativity ofthe diagrams

F (A) F (B)

G(A) G(B)

Ff

ηA ηB

Gf

arising from morphisms f A rarr B (or f B rarr A in the contravariant case) in C In the case of anatural isomoprhism η F rArr G so that each ηA F (A) rarr G(A) is actually an isomorphism in Dthe point is not only that F and G produce isomorphic objects but they do so in a way which iscompatible with all possible category-theoretic data

The first standard example of a natural isomorphism is the one between the identity functor onthe category of finite-dimensional and the functor which sends a finite-dimensional vector space Vto V lowastlowast the dual of its dual induced by the isomorphism V rarr V lowastlowast defined by

v 983041rarr (the linear map on V lowast which sends f to f(v))

This was the topic of a student presentation and can be found in Section 14 of the book Theessential reason why this isomorphism is ldquonaturalrdquo is that V rarr V lowastlowast can be defined without referenceto a basis Contrast this with the the functor which sends V to its (single) dual V lowast it is still truethat finite-dimensional vector space is isomorphic to its dual but specifying such an isomorphism

15

requires additional data which prevents the identity functor from being ldquonaturally isomorphicrdquo tothe single dual functor

Some history At this point it makes sense to say a bit about the interesting origins of categorytheory in the 1950rsquos Eilenberg and Mac Lane where studying cohomology (or possible homologyone of those two) in algebraic topology which associates algebraic data (a group ring or similar)to a topological space in a way which encodes some of the topology They had two cohomologicalconstructions which ended up proving isomorphic results but they noticed that not only werethe resulting objects the same but the actual constructions themselves were in some sense theldquosamerdquo They thus then needed a way to formalize this notion and invented the concept of anatural transformation to do so where they interpreted two constructions as being the same asbeing compatible with all morphisms

But this then leads to the question what types of objects should a natural transformationoccur between Eilenberg and Mac Lane then had to invent the notion of a ldquofunctorrdquo to describewhat a natural transformation went from and what it went to In their motivating setting theyinvented the notion of a functor to describe more formally properties which their cohomologicalconstructions were supposed to satisfy But this then leads to the question what types of objectsshould a functor map between In other words what type of data should a functor (cohomologicalconstruction in their example) take as input and should it output They finally had to thus inventthe notion of a ldquocategoryrdquo to describe the source and target of a functor so that their cohomologicalconstructions could be interpreted as functors from a category of topological spaces to a categoryof algebraic objects

Thus historically natural transformations came first then functors and then categories Aswith most things in mathematics categories thus arose from attempts to encode what was beingseen in some concrete examples in a more general and formal setting

Representable functors Representable functors are defined in Section 21 of the book but wersquollgive the required definitions here in a hopefully simpler to digest manner The point is that anyobject A in a (locally small) category C gives two ldquonaturalrdquo functors C rarr Set one covariant andthe other contravariant which as wersquoll see pretty much encode all data about A itself We definehA C rarr Set to be the covariant functor defined by

hA(B) = Mor(AB) on objects and

hA(Bfminusrarr C) = the map Mor(AB)

hAfminusminusrarr Mor(AC) defined by g 983041rarr f g

We define hA C rarr Set to the contravariant functor defined by

hA(B) = Mor(BA) on objects and

hA(Bfminusrarr C) = the map Mor(CA)

hAfminusminusrarr Mor(BA) defined by g 983041rarr g f

(The functor hA is called the functor of points of A and wersquoll usually think of it as being a covariantfunctor hA Cop rarr Set The functor hA does not have a standard name) So hA is defined bymorphisms from A and hA is defined by morphisms into A Wersquoll see later the sense in which suchfunctors encode all information about the object A

A functor F C rarr Set is representable if is naturally isomorphic to such a functor so F sim= hA

for some A in the covariant case and F sim= hA for some A in the contravariant case The idea wersquollbuild towards is that we can essentially treat such a functor as being A itself so that representable

16

functors can be thought as being actual objects in C For now here is a key example wersquove seenbefore which we can now formulate using the language of representability

Given AB isin Ob(C) let F C rarr Set be the contravariant functor defined by

F (C) = Mor(CA)timesMor(CB) and F (Cfminusrarr D) = [(g k) 983041rarr (g f k f)]

where (g k) 983041rarr (g f k f) gives a map Mor(DA) timesMor(DB) rarr Mor(CA) timesMor(CB) ieF (D) rarr F (C) In order for this functor to be representable requires the existence of an object inC satisfying

Mor(CA)timesMor(CB) sim= Mor(C representing object)

in a way which is compatible with all morphisms but we have actually already seen this notionelsewhere it is essentially the definition of a product AtimesB for A and B Indeed we claim that Fis naturally isomorphic to hAtimesB as we now show

First to define a natural isomorphism hAtimesBsim= F requires for each C isin Ob(C) a choice of a

bijection (ie isomorphism in Set)

ηC hAtimesB(C)sim=minusrarr F (C)

which in our case looks like

ηC Mor(CAtimesB))sim=minusrarr Mor(CA)timesMor(CB)

Take this to be the map

(Cfminusrarr AtimesB) 983041rarr (pA f pB f)

where pA A times B rarr A and pB A times B rarr B are the two projection morphisms in the definitionof a product The definition of product implies that this map is invertible since given k C rarr Aand ℓ C rarr B there exists a unique h C rarr A times B such that k = pA h and ℓ = pB h Forthese bijections to define a natural isomorphism hAtimesB

sim= F requires that they be compatible withall morphisms in C which means that given any g C rarr D in C the following diagram shouldcommute

hAtimesB(C) hAtimesB(D)

F (C) F (D)

hAtimesB(g)

ηC ηD

Fg

which in the case at hand looks like

Mor(CAtimesB) Mor(DAtimesB)

Mor(CA)timesMor(CB) Mor(DA)timesMor(DB)

minus g

ηC ηD

(minus gminus g)

Take an element k D rarr AtimesB in the upper right Applying the map on top gives the elementk g in the upper left and then applying ηC on the left side gives

(pA (k g) pB (k g))

17

in the lower left Instead starting with k D rarr AtimesB in the upper right applying ηD on the rightgives (pA k pB k) and applying the map on the bottom gives

((pA k) g (pB k) g)

in the lower left which is the same as the previous element in the lower left by the associativity ofcomposition Hence the isomorphisms ηC do define a natural isomorphism between hAtimesB and F Again the point is that you can then interpret the functor F as being the ldquosamerdquo as the productAtimesB since both encode the same data

More Representable Examples

Coproducts As in the case of products the definition of coproducts can also be phrased in termsof a representable functor Given objects A and B in C let F C rarr Set be the covariant functordefined by

F (C) = Mor(AC)timesMor(BC) and F (Cfminusrarr D) = [(g k) 983041rarr (f g f k)]

where the latter is a function Mor(AC)timesMor(BC) rarr Mor(AD)timesMor(BD) This functor isrepresentable if and only if a coproduct A ⊔ B for A and B exists and the representing object isthat coproduct the key is the requirement that we have bijections

Mor(AC)timesMor(BC) sim= Mor(A ⊔BC)

for all C which is essentially the defining property a coproduct for A and B should have

Power set functor contravariant version Let P Set rarr Set be the contravariant powerset functor defined on objects by sending a set to its power set and on morphisms by sendingf A rarr B to the preimage function Pf P (B) rarr P (A) given by S 983041rarr fminus1(S) In order for this tobe representable requires the existence of a set X with natural bijections

P (A) sim= Mor(AX)

for all sets A that is a set X so that mapping into X is the same data as specifying a subset ofthe domain We take X = 0 1 to be a two-element set and the required bijections

ηA P (A) rarr Mor(A 0 1) to be given by S 983041rarr χS

where XS A rarr 0 1 is the indicator or characteristic function S defined by sending elementsof S to 1 and elements of A minus S to 0 The inverse of ηA sends a function g A rarr 0 1 to thepreimage gminus1(1) sube A

To verify that the bijections ηA define the data of a natural isomorphism η P rArr h01 wecheck the commutativity of some diagrams Given a function f A rarr B consider

P (A) P (B)

h01(A) = Mor(A 0 1) h01(B) = Mor(B 0 1)

Pf = fminus1

ηA ηB

minus f

18

Take S isin P (B) in the upper right Applying the map on top gives fminus1(S) isin P (A) and thenapplying the map on the left gives χfminus1(S) isin Mor(A 0 1) On the other hand applying the mapon the right to S isin P (B) gives χS isin Mor(B 0 1) and then applying the map on the bottomgives χS f isin Mor(A 0 1) The two resulting functions χfminus1(S)χS f in the lower left are

χfminus1(S)(x) =

9830831 x isin fminus1(S)

0 otherwiseand χS(f(x)) =

9830831 f(x) isin S

0 otherwise

so these are the same since x isin fminus1(S) if and only if f(x) isin S Thus the diagram above commutesso P is naturally isomorphic to h01 and hence P is representable

Power set functor covariant version Now consider the covariant power set functor stilldenoted by P Set rarr Set which again sends a set to its power set but now sends a functionf A rarr B to the image function Pf P (A) rarr P (B) defined by S 983041rarr f(S) In order for this tobe representable there would have to exist a set X such that P sim= hX which translates into havingnatural bijections

P (S) sim= Mor(XS)

for all sets S However in this case there is no natural choice for X as there is no natural way todetect subsets of S by mapping into S as opposed to the case of P (S) sim= Mor(S 0 1)

That no such X exists can be made precise using a simple argument when S = empty P (S) hascardinality 1 soX would have to be empty as well since otherwise Mor(XS) would have cardinality0 but if X is empty then Mor(empty S) always cardinality 1 regardless of S which is clearly not trueof P (S) Thus the covariant power set functor is not representable

Forgetful on Top The forgetful functor F Top rarr Set is representable Indeed a representingobject would have to satisfy

X sim= Map(objectX)

as sets for any topological space X (where Map denotes the set of continuous maps) and it iseasy to see that a single point satisfies this property a continuous map f pt rarr X is completelydetermined by the element f(pt) of X and any such element gives a continuous map in this wayOf course checking that F sim= hpt requires checking that various diagrams commute but this issimple to work out in this case so we omit it here

Extracting a topology With an eye towards the idea that the functors hA hA should captureall information about an object A in a category note that the result above shows that in TophX first of all captures all information about X as a set since the set X can be extracted fromhX(pt) = Map(ptX)

We claim that hX actually captures all information about X as a topological space since thetopology on X can be extracted from hX in the following way As we saw in the power set examplewe have a bijection

P (X) sim= Mor(X 0 1)

Now equip 0 1 with the topology in which 1 is open but 0 is not Then a continuous mapf X rarr 0 1 is fully determined by the preimage fminus1(1) since to be continuous only requiresthat this single preimage be open in X Moreoever any open subset of X arises in this way bytaking f to be the indicator function of that open set Thus we get a bijection

Op(X) sim= Map(X 0 1)

19

where Op(X) denotes the set of open subsets of X ie the topology on X Hence the topology onX can be extracted from hX as claimed so hX and hX do encode all information about X

This fact can also be interpreted as saying that the contravariant functor which sends a topo-logical space X to its collection of open subsets (and which sends a continuous map to the mapinduced by taking preimages) is representable with representing object 0 1 equipped with thetopology given above This will be worked out in a homework problem

Forgetful on Grp The forgetful functor Grp rarr Set is also representable In this case arepresenting object should be a group giving set-theoretic bijections

G sim= Hom(objectG)

for all groups G where Hom denotes the set of group homomorphisms The group Z works

G sim= Hom(Z G)

since a homomorphism φ Z rarr G is completely determined by φ(1) since Z is generated by 1 andthere are no restrictions on what φ(1) isin G can actually be The inverse of the desired bijection isthus φ 983041rarr φ(1) and it is simple to check that morphisms behave appropriately so that the forgetfulfunctor in this setting is indeed isomoprhic to hZ

Forgetful on Ring The forgetful functor Ring rarr Set is also representable where by Ring wemean the category of rings with identity elements and where morphisms (ie ring homomorphisms)are required to preserve identities Here we need an object satisfying

R sim= Hom(object R)

for all R isin Ob(Ring) and we can see that Z[x] the ring of polynomials in the variable x withcoefficients in Z works Indeed given a ring homomorphism φ Z[x] rarr R the fact that φ isrequired to preserve identities fully determines φ(n) for any n isin Z (since φ(n) = nφ(1)) and hencesince φ preserves multiplication its behavior is fully determined by φ(x) which can be any elementof R The inverse of the bijection we want is thus φ 983041rarr φ(x) and it will follow that this forgetfulfunctor is isomorphic to hZ[x]

Equivalences between Categories

Definitions iso and equiv Now that we have a notion of a ldquomorphismrdquo between categorieswe can talk about the sense in which two categories are the ldquosamerdquo The first possible definitionis the ldquoobviousrdquo one you might expect given the definition of ldquoisomorphismrdquo wersquove seen in everyother context until now we say that C is isomorphic to D if there exist functors F C rarr D andG D rarr C such that G F = idC and F G = idD where idC and idD are identity functors

However this definition turns out to be too restrictive Indeed the requirement that the givencompositions equal identity functors says that given an object A in C the object G(F (A)) beliterally the same as A itself and similarly for objects in D But we know that two objects in acategory can be thought of as being the ldquosamerdquo without being literally the same as long as theyare isomorphic For instance productscoproducts are not unique in the literal sense but only inthe sense that they are isomorphic in a unique way To take another example the definition ofa functor being representable does not say that F (B) literally be the same as hB(A) only thatthey be isomorphic for sure the power set P (S) of a set S is not literally the same as the set offunctions S rarr 0 1mdashit is only the ldquosamerdquo in the sense that they encode the same data

20

Thus it makes sense to relax the requirement that A andG(F (A)) in the definition of isomorphiccategories be equal to the requirement that they be isomorphic This leads to the following betterdefinition of two categories being the ldquosamerdquo C is equivalent to D if there exist functors F C rarr D

and G D rarr C such that G F sim= idC and F G sim= idD So we require only that for instance thefunctor G F be naturally isomorphic to the identity functor on C which says that for any objectA G(F (A)) be isomorphic to A in a natural way Wersquoll see that this truly is the better notionof what it means for two categories to be the ldquosamerdquo and it reflects the idea that all categoricalproperties of one category can be transported to the other

Alternate characterization of equivalences A functor F C rarr D implementing an equiv-alence between categories is called unsurprisingly and equivalence and the ldquoinverserdquo functorG D rarr C is still referred to as being its inverse even though it is not technically an ldquoinverserdquoin the sense that compositions give literal identities But as in the case of Set where ldquoinvertiblerdquocan be characterized without reference to an inverse as ldquoinjectiverdquo and ldquosurjectiverdquo so too canequivalences between categories be characterized without explicit reference to an inverse

A functor F C rarr D is an equivalence if and only it is full faithful and essentiallysurjective

We have seen the notions of full and faithful before and essentially surjective means that any objectin D is isomorphic to something in the image of F for all D isin Ob(D) there exits C isin Ob(C) suchthat F (C) sim= D Justifying this result was the topic of a student presentation and can be foundin the book in Section 15 (Technically to construct an ldquoinverserdquo of F requires that we work withsmall categories but whatever) The bulk of the real work goes into showing that the ldquoinverserdquofunctor constructed is actually a functor which comes down to some abstract nonsense argumentvia commutative diagrams

Equivalences preserve everything We previously saw the notions of full faithful and es-sentially surjective back when thinking about what properties a functor could have which wouldguarantee it sent products to products so the result we proved back then was that equivalencespreserve products Similarly equivalences preserve coproducts and the proof is basically the sameafter reversing arrows A clean way of deriving this is as follows if F C rarr D is an equivalencethen so is F op Cop rarr Dop so since F op preserves products F preserves coproducts becausecoproducts in a category become products in the opposite category

In general an equivalence between categories will preserve whatever categorical notions youwant monomorphisms epimorphisms initial objects terminal objects etc so that equivalentcategories really do have essentially the ldquosamerdquo properties

Equivalence example Here is a first example (also found in the book) of equivalent categorieswhich are not isomorphic Let FVectR be the category whose objects are finite-dimensional realvector spaces and whose morphisms are linear transformations Let MatR be the category ofmatrices defined as follows

bull objects in MatR are nonnegative integers

bull a morphism n rarr m is an mtimes n matrix with real entries

bull composition in MatR is given by matrix multiplication given B n rarr m and A m rarr kmdashsoB is an mtimes n matrix and A is a k timesm matrixmdashthe composition A B n rarr k is the k times nmatrix AB

bull identity morphisms in MatR are given by identity matrices

21

We claim that FVectR and MatR are equivalent First note that they certainly cannot be isomo-prhic isomorphisms in MatR exist only between two objects which are literally the same (ie anmtimesn matrix can be invertible only when m = n) but isomorphisms in FVectn can exist betweenobjects which are not literally the same So there are in a sense not enough objects in MatR toallow it to be isomorphic to FVectR

To construct an equivalence choose one and for all an isomorphism TV V rarr RdimV forevery object in FVectR or equivalently choose once and for all a basis for every V DefineF FVectR rarr MatR by

V 983041rarr dimV on objects and

(T V rarr W ) 983041rarr (the standard matrix of SW T Sminus1V RdimV rarr RdimW ) on morphisms

Equivalently F sends T to the matrix of T relative to the chosen bases for V and W It isstraightforward to check that this is a functor The inverse functor G MatR rarr VectR is givenby

n 983041rarr Rn on objects and

(mtimes n matrix A) 983041rarr (the linear transformation A Rn rarr Rm induced by A) on morphisms

One can check that G is a functor and that F G and G F are naturally isomorphic to identitiesor instead one can argue that F alone is full faithful and essentially surjective The point is thatG(F (V )) gives back RdimV so not literally V itself but only something isomorphic to V which iswhy this gives an equivalence of categories and not an isomorphism

Yoneda Lemma Functors as Objects

Compact Hausdorff spaces and Clowast-algebras Last time we gave an example of equivalentcategories which were not isomorphic but in many ways that example is unsatisfactory when doinglinear algebra people for sure often work with matrix representations of linear transformations asthe example from last would suggest but no one in their right mind literally thinks about suchthings in terms of the equivalence FVectR rarr MatR itself In other words this equivalence isessentially ldquocooked uprdquo for the sake of giving an example but it does not really give any insightinto the actual mathematics of linear algebra

So with that in mind we will describe another equivalence which actually does have some realmeaning in the sense of producing new ways of thinking about some mathematics Given a compactHausdorff space X we let C(X) denote the space of continuous complex-valued functions on X

C(X) = f X rarr C | f is continuous

We can equip the set C(X) with various additional structures First it is a complex vector spaceunder ordinary addition and scalar multiplication of functions Second we can multiply two func-tions together

(fg)(p) = f(p)g(p)

which turns C(X) into whatrsquos called an algebra over C (We wonrsquot give the definition of anldquoalgebrardquo but the point is that this multiplication is in some sense ldquocompatiblerdquo with the vectorspace structure) Next any element f of C(X) has a conjugate element f obtained by takingcomplex conjugates of the values of f

f(p) = f(p)

22

And finally we can equip C(X) with a norm via

983042f983042 = suppisinX

|f(p)|

where |middot| denotes complex absolute value (This supremum exists sinceX is compact by the ExtremeValue Theorem) All of these structures (vector space algebra multiplication conjugation norm)turn C(X) into whatrsquos called a Clowast-algebra We wonrsquot define this precisely but the definition encodesvarious ways in which these structures are compatible Actually C(X) is a unital commutative Clowast-algebra where ldquounitalrdquo means that the multiplication has an identity element (namely the constantfunction 1) and commutative means that fg = gf There is a natural notion of homomorphismbetween Clowast-algebras which are maps between them which behave nicely with all the structuresinvolved In particular given a continuous map f X rarr Y between compact Hausdorff spaces weget a Clowast-algebra homomorphism flowast C(Y ) rarr C(X) given by pre-composition with f flowast sendsg isin C(Y ) to g f isin C(X)

We thus have a contravariant functor CHaus rarr ucClowast-Alg from the category of compactHausdorff spaces and the category of unital commutative Clowast-algebras It turns out that anyunital commutative Clowast-algebra arises in this way in the sense that given any unital commutativeClowast-algebra B there exists a compact Hausdorff space X such that C(X) sim= B as Clowast-algebrasand that morphisms between these algebra arises from continuous maps in the manner describeabove This says that the functor above is essentially surjective full and it turns out that it isalso faithful so that it is an equivalence Thus all information about compact Hausdorff spaces iscompletely encoded in information about unital commtuative Clowast-algebras and indeed one couldlearn everything there is to know about a compact Hausdorff space X from its Clowast-algebra C(X)of functions The fact that the categories CHaus and ucClowast-Alg are equivalent is the startingpoint in the subject known as noncommutative geometry which wersquoll say something about a bitlater on If we drop the requirement that our algebras be unital it turns out that the categoryof commutative Clowast-algebras in general is equivalent to the category of locally compact Hausdorffspaces via essentially the same functor where we take as morphisms in the category of locallycompact Hausdorff spaces to be continuous functions which ldquovanish at infinityrdquo

We are only introducing Clowast-algebras to give an example of an interesting and actually usefulequivalence between categories but just for the fun of it wersquoll note the following Examples ofnoncommutative Clowast-algebras come from what are known as ldquoalgebras of bounded operators onHilbert spacesrdquo which show up in quantum mechanics as the things which describe physicallyobservable quantities like position momentum and energy (In fact all Clowast-algebras arise fromHilbert spaces in this way) Thus Clowast-algebras show up in mathematical formulations of quantummechanics and the notion of noncommutative geometry mentioned above is at the core of someattempts to understand the elusive concept known as ldquoquantum geometryrdquo

Yoneda Lemma The Yoneda Lemma states that given a (covariant) functor F C rarr Set whereC is locally small for any A isin Ob(C) there exists a bijection

F (A) sim= Nat(hA F )

where the right side denotes the set of natural transformations from hA to F A similar result holdsin the contravariant case where we get bijections F (A) sim= Nat(hA F ) (There is also a sense inwhich these bijections are themselves natural but wersquoll ignore this bit) The proof of the YonedaLemma was the topic of a student presentation and can be found in Section 22 of the book

Here we will get a bit philosophical and talk about what the point of the Yoneda Lemmaactually is and later the sense in which it says that we can view arbitrary functors as ldquonon-existent

23

objectsrdquo in a category First we mention what we will call the ldquofunctor of pointsrdquo approach tomathematics a mathematical object can be fully characterized by all morphisms into it That isknowledge of Mor(SA) for all S is equivalent to knowledge of A itself Indeed this is an idea wementioned previously in relation to the functors hAmdashwhich we previously said is referred to as beingthe functor of points of Amdashand now we seek to push this idea further In some examples literalpoints of A can be extracted from hA as in the case of the bijection X sim= hX(pt) = Map(ptX) fora topological space X or G sim= hG(Z) = Mor(Z G) for a group G Moreover in these examples therest of the (topological) structure onX or (group) structure on G can also be extracted from hX andhG respectively In general we can view Mor(SA) as in a sense being ldquopoints of A parameterizedby Srdquo which comes from the idea that in say Set Mor(SA) = AS is ldquothe product of S-manycopies of Ardquo so literally points of A ldquoparameterizedrdquo by S In various areas of mathematics anelement of Mor(SA) is thought of as being a ldquogeneralizedrdquo point of A so that the functor of pointshA detects generalized points of A

So an object A is fully encoded by the functor hA Now we claim that morphisms betweenobjects are also encoded by such functors Indeed in the case where F = hB is itself a functor ofpoints the Yoneda result Nat(hA F ) sim= F (A) becomes

Nat(hA hB) sim= hB(A) = Mor(AB)

which says precisely that morphisms between the objects A and B can be obtained solely fromthe functors hA and hB (This fact is one reason why one should prefer the contravariant functorhA to the covariant functor hA in the covariant case Nat(hA F ) sim= F (A) for F = hB becomesNat(hA hB) sim= hB(A) = Mor(BA) so that morphisms between two objects get reversed whencharacterizing them in terms of representable functors)

Pushing our philosophy further we now know that not only objects but also morphisms betweenobjects are fully encoded by the functors hA The final step is to note that we should not only careabout objects and morphisms between them but that an important part of the data of an objectA is also how arbitrary contravariant functors F C rarr Set behave on A and that this too can beobtained from hA precisely via the Yoneda result that F (A) sim= Nat(hA F ) So objects morphismsbetween objects and how arbitrary functors behave on objects are all encoded in knowledge ofthe functors hA so that these functors truly do capture in the most general sense possible allinformation about A

Functors as nonexistent objects The discussion above is all well and good but it still doesnrsquotquite say why we should care about the fact that we can study on object A by studying hA insteadin that we still havenrsquot spoken about what type of insight about A can be gleamed from hA thatwasnrsquot readily apparent in A itself already Now we argue that the true point of the YonedaLemma is that it gives a way to think about arbitrary functors F Cop rarr Set as ldquoobjectsrdquo in C

itself or really as objects in an ldquoenlargementrdquo of C Certainly for those functors F sim= hA which arerepresentable we can indeed think about F as simply being the object A but now we claim thateven non-representable F can be thought of as being ldquononexistentrdquo objects in C The upshot is thatoften in mathematics we come across things which do not actually exist but the Yoneda Lemmagives a way to (indirectly) study such things anyway by interpreting them as functors instead

The starting point of course is that an actual object A is the same as hA as we have explainedabove Thus by analogy we can interpret a non-representable functor F as being an ldquoobjectrdquo in C

as long as we know what it means to ldquomaprdquo into F from other things which are honestly objects in CThat is if F were indeed representable then F (A) sim= Nat(hA F ) sim= Mor(AF ) would characterizethe functor of points of F (here we abuse notation by using the same letter F to denote both arepresentable functor and the object which represents it) so now we take Mor(AF ) = F (A) as

24

the definition of the ldquofunctor of pointsrdquo of the non-existent object F We are thus taking elementsof F (A) as being what it means to ldquomaprdquo from A into F even when F is not representable Thefact that the Yoneda Lemma gives a way to define maps from A into F thus says that F itself isequipped with the same data that a literal object of A would be equipped with so why shouldnrsquotwe interpret F as a being a type of ldquoobjectrdquo in C as well

To phrase this another way note that the Yoneda Lemma gives rise to a functor

C rarr Fun(CopSet)

where Fun(CopSet) denotes the category whose objects are functors from Cop to Set and morphismsare natural transformation defined on objects by A 983041rarr hA and on morphisms by the bijectionMor(AB) sim= Nat(hA hB) arising from the Yoneda Lemma That this is actually bijection saysthat this functor is full and faithful so that we can view C as being a subcategory of the functorcategory Fun(CopSet) (We cay say that C is isomorphic to the subcategory of Fun(CopSet)consisting of the functors hA or that it is equivalent to the subcategory of Fun(CopSet) consistingof representable functors This functor C rarr Fun(CopSet) is usually called the Yoneda embeddingof C into Fun(CopSet)) Thus we can view Fun(CopSet) as an ldquoenlargmentrdquo of C and the ideawe are trying to make sense of is that other objects in this ldquolargerrdquo category should be interpretedas ldquonon-existent objectsrdquo of C

Just how far can we actually push this idea that arbitrary functors Cop rarr Set should be thoughtof as being ldquoobjectsrdquo of C Pretty far in fact For instance often times in geometry one comesacross a type of ldquosmoothrdquo space on which some group acts and one is interested in doing ldquocalculusrdquoon the resulting quotient space which is the space of orbits under the given group action Theproblem is that such quotient spaces are very often (usually in fact) not ldquosmoothrdquo themselves sothat the tools from calculus are not readily available The solution is to nonetheless treat thisquotient as if it were ldquosmoothrdquo and attempt to do calculus anyway To make this precise thisldquonon-smoothrdquo space actually turns out to have a natural interpretation as a ldquosmooth functorrdquo andit is through this functor which we can view this non-smooth space as actually being smooth afterall so we are forced to expand our notion of what we mean by ldquosmooth spacerdquo by allowing thingswhich are not spaces in the usual sense which certainly leads to more abstraction but the benefitis that it actually makes sense to do calculus on such ldquospacesrdquo interpreted as functors

The discussion above is pretty vague so here is one final more concrete example Recall thatequivalence

CHausop rarr ucClowast-Alg

between the category of compact Hausdorff spaces and the category of unital commutative Clowast-algebras mentioned at the start We now claim that a non-commutative (unital) Clowast-algebra shouldbe interpreted as the algebra of functions on some non-existent ldquononcommutative compact Haus-dorff spacerdquo and it is this idea which the subject of noncommutative geometry makes precise Solet B be a non-commutative Clowast-algebra Then the functor of points hB gives a functor

ucClowast-Alg rarr Set

which sends C(X) (where X is some compact Hausdorff space) to Hom(C(X) B) where Hom de-notes homomorphisms of Clowast-algebras whatever that means Composing with the previous functorwe get a functor wersquoll denote by B itself

B CHausop rarr Set

which on objects at least is given by X 983041rarr Hom(C(X) B) This functor is not representable sinceB is not commutative but the philosophy espoused above suggests we should think of it as giving a

25

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

Coproduct Examples Concreteness

Products and coproducts for metric spaces The product of two objects in Met the categoryof metric spaces with morphisms given by metric maps is the Cartesian product X times Y equippedwith the box metric and the coproduct of X and Y does not exist if both X and Y are nonemptyThis was the topic of a student presentation and here are some proof sketches

Given metric spaces (X dX) and (Y dY ) the box metric d on X times Y is defined by

d((x1 y1) (x2 y2)) = maxdX(x1 x2) dY (y1 y2))

(The name ldquoboxrdquo metric comes from the fact that in RtimesR = R2 or R3 open balls indeed looks likeldquoboxesrdquo) The key requirement needed in the definition of a product is given morphisms S rarr Xand S rarr Y that the map

h S rarr X times Y defined by h(s) = (f(s) g(s))

actually be a morphism in Met This turns into the requirement that

d((f(s) g(s)) ((f(t) g(t))) = maxdX(f(s) f(t)) dY (g(s) g(t)) le dS(s t) for all s t isin S

given the fact that

dX(f(s) f(t)) le dS(s t) and dY (g(s) g(t)) le dS(s t) for all s t isin S

For the latter inequalities to imply the former d must (at least up to isometry) be given by thebox metric Note that the two other ldquostandardrdquo metrics one might put on a productmdashthe taxicab

dX + dY and Euclidean983156

d2X + d2Y metricsmdashdo not satisfy this required ldquometric maprdquo property

and so do not give the product in MetTo see that coproducts do not exist if XY are nonempty suppose instead (P dP ) was a co-

product for X and Y with inclusion morphisms iX X rarr P and iY Y rarr P For any n isin Ntake 0 n to be a metric space with the absolute value metric d and let f X rarr 0 n andg Y rarr 0 n be the constant function 0 and the constant function n respectively Since P is acoproduct there exists h P rarr 0 n such that

h iX = f and h iY = g

But in order for h to actually be a morphism in Met we must then have

d(h(iX(x) iY (y)) le dP (iX(x) iY (y)) for any x isin X y isin Y

which translates inton = |0minus n| le dP (iX(x) iY (y))

But this should then be true for any n isin N which would imply that the distance in P betweenan element coming from X and one coming from Y should be infinite which is nonsense (Notethat if you altered your definition of ldquometricrdquo to allow for infinite distance then a coproduct wouldexist the disjoint union X ⊔Y where you defined the distance between elements of X and Y to beinfinite)

Coproducts for groups The coproduct of two groups G and H in Grp is the free product GlowastHwhich is defined to be the set of all words consisting of elements of G andH in an alternating fashion

13

with concatenation as the group operation (When we have to elements of G or two elements of Hnext to each other with use their respective group operations to turn these into single elements ofG or H) This was also the topic of a student presentation and here is the basic idea Given grouphomomorphisms g G rarr N and k H rarr N that G lowastH is the coproduct comes down to showingthat the map

h G lowastH rarr N defined by h(g1h1 gnhn) = f(g1)k(h1) middot middot middot f(gn)k(hn)

and similarly on other possible words in GlowastH is a group homomorphism which follows essentiallyfrom the definition of the group operation on G lowastH indeed this group operation arises preciselyfrom this requirement

Now if we instead tried to use GtimesH with its standard inclusions G rarr GtimesH and H rarr GtimesHas the coproduct the issue is that the analogous map

h GtimesH rarr N defined by h((g h)) = f(g)k(h)

would NOT be a homomorphism because h((g1 h1)(g2 h2)) is not equal to h(g1 h1)h(g2 h2) bythe way in which the group operation is defined on the direct product G times H However if westrict our category to only consist of abelian groups so that the N rsquos we consider in the defini-tion of coproduct are also abelian then G times H does serve as a coproduct we can always rewritef(g1)k(g1) middot middot middot f(gn)k(gn) as f(g1) middot middot middot f(gn)k(h1) middot middot middot k(hn) in N Irsquoll leave it to you to flesh out allof these details

Concrete categories The categories Set Top Vect and Grp all have the property thattheir objects are simply sets possibly equipped with extra structure and that their morphisms areordinary set-theoretic functions possibly satisfying some additional requirement Such categoriesare nice since we can often use intuition about sets and functions in their study A concrete categoryis intuitively one where we can think of objects as being ldquosetsrdquo and morphisms as being ldquofunctionsrdquoonly that we have to interpret this in the right way

Here is the definition a concrete category is a category C equipped with a faithful functorF C rarr Set which is part of the data The idea is that such a functor gives us way to turnan object A isin Ob(C) into a set F (A) and a morphism f isin MorC(AB) into a function Ff isinMorSet(F (A) F (B)) where the ldquofaithfulrdquo requirement says that we do not lose information aboutf when doing so so that we can (essentially) learn everything we need to know about f by studyingFf instead In the four examples above the required faithful functor is simply the forgetful functorand indeed in general we think of F as being a ldquoforgetful functorrdquo for the concrete category (C F )

Examples Here are two more examples of concrete categories Let G be a group and considerthe category BG with a single object and elements of G morphisms As written this does notappear to be concrete since morphisms are not honest functions lowast rarr lowast but the point is that thiscategory can indeed by ldquoconcretizedrdquo by specifying an appropriate ldquoforgetful functorrdquo A functorF BG rarr Set is the data of a (left) action of G on a set S = F (lowast) and saying this functor isfaithful is what it means to say that this action is faithful different g h isin G act on S in differentways or equivalently that there exists s isin S such that gs ∕= hs Any group has at least onefaithful left actionmdashnamely the action on itself given by left multiplicationmdashto BG can always beequipped with a faithful functor to Set This is all just a way of phrasing the fact that any groupis isomorphic to a subgroup of a permutation group

The category Rel of relations can also be ldquoconcretizedrdquo even though as written morphismsA rarr B are not honest functions Take F Rel rarr Set to be the power set functor which sends an

14

object in Rel to its power set and a relation f A rarr B to the induced function Ff P (A) rarr P (B)defined by

(Ff)(S) = b isin B | there exists a isin S such that (a b) isin f

In other words if you think of the relation f sube A times B as a ldquopossibly not everywhere-definedpossibly multivalued functionrdquo (Ff)(S) is analogous to taking the image of S That F is faithfulwill be left to the homework

In fact most categories yoursquoll see can be made concrete by specifying an appropriate forgetfulfunctor but not all categories can be made concrete in this way Here is the standard example of anon-concrete category which we mention only to say there is such a thing but which we wonrsquot doanything more with since understanding it better requires knowledge of algebraic topology Thehomotopy category of topological spaces is the category hTop whose objects are topological spacesand whose morphisms are given by equivalence classes of homotopic maps

MorhTop(XY ) = homotopy classes of continuous maps X rarr Y

(Saying that two maps are homotopic means that you can ldquodeformrdquo one into the other but wersquollleave the precise definition to a course in algebraic topology) It turns out that no functor fromhTop to Set can be faithful so that hTop cannot be made concrete but this is actually a deepand difficult thing to show so we make no attempt to do so here

Natural Isomorphisms Representability

Natural transformations The definition of a natural transformation η F rArr G between two(both covariant or both contravariant) functors FG C rarr D is given in Section 14 of the bookThe key idea is that a natural transformation gives a way to turn data about F into data aboutG in a way which is compatible with all possible morphisms as expressed by the commutativity ofthe diagrams

F (A) F (B)

G(A) G(B)

Ff

ηA ηB

Gf

arising from morphisms f A rarr B (or f B rarr A in the contravariant case) in C In the case of anatural isomoprhism η F rArr G so that each ηA F (A) rarr G(A) is actually an isomorphism in Dthe point is not only that F and G produce isomorphic objects but they do so in a way which iscompatible with all possible category-theoretic data

The first standard example of a natural isomorphism is the one between the identity functor onthe category of finite-dimensional and the functor which sends a finite-dimensional vector space Vto V lowastlowast the dual of its dual induced by the isomorphism V rarr V lowastlowast defined by

v 983041rarr (the linear map on V lowast which sends f to f(v))

This was the topic of a student presentation and can be found in Section 14 of the book Theessential reason why this isomorphism is ldquonaturalrdquo is that V rarr V lowastlowast can be defined without referenceto a basis Contrast this with the the functor which sends V to its (single) dual V lowast it is still truethat finite-dimensional vector space is isomorphic to its dual but specifying such an isomorphism

15

requires additional data which prevents the identity functor from being ldquonaturally isomorphicrdquo tothe single dual functor

Some history At this point it makes sense to say a bit about the interesting origins of categorytheory in the 1950rsquos Eilenberg and Mac Lane where studying cohomology (or possible homologyone of those two) in algebraic topology which associates algebraic data (a group ring or similar)to a topological space in a way which encodes some of the topology They had two cohomologicalconstructions which ended up proving isomorphic results but they noticed that not only werethe resulting objects the same but the actual constructions themselves were in some sense theldquosamerdquo They thus then needed a way to formalize this notion and invented the concept of anatural transformation to do so where they interpreted two constructions as being the same asbeing compatible with all morphisms

But this then leads to the question what types of objects should a natural transformationoccur between Eilenberg and Mac Lane then had to invent the notion of a ldquofunctorrdquo to describewhat a natural transformation went from and what it went to In their motivating setting theyinvented the notion of a functor to describe more formally properties which their cohomologicalconstructions were supposed to satisfy But this then leads to the question what types of objectsshould a functor map between In other words what type of data should a functor (cohomologicalconstruction in their example) take as input and should it output They finally had to thus inventthe notion of a ldquocategoryrdquo to describe the source and target of a functor so that their cohomologicalconstructions could be interpreted as functors from a category of topological spaces to a categoryof algebraic objects

Thus historically natural transformations came first then functors and then categories Aswith most things in mathematics categories thus arose from attempts to encode what was beingseen in some concrete examples in a more general and formal setting

Representable functors Representable functors are defined in Section 21 of the book but wersquollgive the required definitions here in a hopefully simpler to digest manner The point is that anyobject A in a (locally small) category C gives two ldquonaturalrdquo functors C rarr Set one covariant andthe other contravariant which as wersquoll see pretty much encode all data about A itself We definehA C rarr Set to be the covariant functor defined by

hA(B) = Mor(AB) on objects and

hA(Bfminusrarr C) = the map Mor(AB)

hAfminusminusrarr Mor(AC) defined by g 983041rarr f g

We define hA C rarr Set to the contravariant functor defined by

hA(B) = Mor(BA) on objects and

hA(Bfminusrarr C) = the map Mor(CA)

hAfminusminusrarr Mor(BA) defined by g 983041rarr g f

(The functor hA is called the functor of points of A and wersquoll usually think of it as being a covariantfunctor hA Cop rarr Set The functor hA does not have a standard name) So hA is defined bymorphisms from A and hA is defined by morphisms into A Wersquoll see later the sense in which suchfunctors encode all information about the object A

A functor F C rarr Set is representable if is naturally isomorphic to such a functor so F sim= hA

for some A in the covariant case and F sim= hA for some A in the contravariant case The idea wersquollbuild towards is that we can essentially treat such a functor as being A itself so that representable

16

functors can be thought as being actual objects in C For now here is a key example wersquove seenbefore which we can now formulate using the language of representability

Given AB isin Ob(C) let F C rarr Set be the contravariant functor defined by

F (C) = Mor(CA)timesMor(CB) and F (Cfminusrarr D) = [(g k) 983041rarr (g f k f)]

where (g k) 983041rarr (g f k f) gives a map Mor(DA) timesMor(DB) rarr Mor(CA) timesMor(CB) ieF (D) rarr F (C) In order for this functor to be representable requires the existence of an object inC satisfying

Mor(CA)timesMor(CB) sim= Mor(C representing object)

in a way which is compatible with all morphisms but we have actually already seen this notionelsewhere it is essentially the definition of a product AtimesB for A and B Indeed we claim that Fis naturally isomorphic to hAtimesB as we now show

First to define a natural isomorphism hAtimesBsim= F requires for each C isin Ob(C) a choice of a

bijection (ie isomorphism in Set)

ηC hAtimesB(C)sim=minusrarr F (C)

which in our case looks like

ηC Mor(CAtimesB))sim=minusrarr Mor(CA)timesMor(CB)

Take this to be the map

(Cfminusrarr AtimesB) 983041rarr (pA f pB f)

where pA A times B rarr A and pB A times B rarr B are the two projection morphisms in the definitionof a product The definition of product implies that this map is invertible since given k C rarr Aand ℓ C rarr B there exists a unique h C rarr A times B such that k = pA h and ℓ = pB h Forthese bijections to define a natural isomorphism hAtimesB

sim= F requires that they be compatible withall morphisms in C which means that given any g C rarr D in C the following diagram shouldcommute

hAtimesB(C) hAtimesB(D)

F (C) F (D)

hAtimesB(g)

ηC ηD

Fg

which in the case at hand looks like

Mor(CAtimesB) Mor(DAtimesB)

Mor(CA)timesMor(CB) Mor(DA)timesMor(DB)

minus g

ηC ηD

(minus gminus g)

Take an element k D rarr AtimesB in the upper right Applying the map on top gives the elementk g in the upper left and then applying ηC on the left side gives

(pA (k g) pB (k g))

17

in the lower left Instead starting with k D rarr AtimesB in the upper right applying ηD on the rightgives (pA k pB k) and applying the map on the bottom gives

((pA k) g (pB k) g)

in the lower left which is the same as the previous element in the lower left by the associativity ofcomposition Hence the isomorphisms ηC do define a natural isomorphism between hAtimesB and F Again the point is that you can then interpret the functor F as being the ldquosamerdquo as the productAtimesB since both encode the same data

More Representable Examples

Coproducts As in the case of products the definition of coproducts can also be phrased in termsof a representable functor Given objects A and B in C let F C rarr Set be the covariant functordefined by

F (C) = Mor(AC)timesMor(BC) and F (Cfminusrarr D) = [(g k) 983041rarr (f g f k)]

where the latter is a function Mor(AC)timesMor(BC) rarr Mor(AD)timesMor(BD) This functor isrepresentable if and only if a coproduct A ⊔ B for A and B exists and the representing object isthat coproduct the key is the requirement that we have bijections

Mor(AC)timesMor(BC) sim= Mor(A ⊔BC)

for all C which is essentially the defining property a coproduct for A and B should have

Power set functor contravariant version Let P Set rarr Set be the contravariant powerset functor defined on objects by sending a set to its power set and on morphisms by sendingf A rarr B to the preimage function Pf P (B) rarr P (A) given by S 983041rarr fminus1(S) In order for this tobe representable requires the existence of a set X with natural bijections

P (A) sim= Mor(AX)

for all sets A that is a set X so that mapping into X is the same data as specifying a subset ofthe domain We take X = 0 1 to be a two-element set and the required bijections

ηA P (A) rarr Mor(A 0 1) to be given by S 983041rarr χS

where XS A rarr 0 1 is the indicator or characteristic function S defined by sending elementsof S to 1 and elements of A minus S to 0 The inverse of ηA sends a function g A rarr 0 1 to thepreimage gminus1(1) sube A

To verify that the bijections ηA define the data of a natural isomorphism η P rArr h01 wecheck the commutativity of some diagrams Given a function f A rarr B consider

P (A) P (B)

h01(A) = Mor(A 0 1) h01(B) = Mor(B 0 1)

Pf = fminus1

ηA ηB

minus f

18

Take S isin P (B) in the upper right Applying the map on top gives fminus1(S) isin P (A) and thenapplying the map on the left gives χfminus1(S) isin Mor(A 0 1) On the other hand applying the mapon the right to S isin P (B) gives χS isin Mor(B 0 1) and then applying the map on the bottomgives χS f isin Mor(A 0 1) The two resulting functions χfminus1(S)χS f in the lower left are

χfminus1(S)(x) =

9830831 x isin fminus1(S)

0 otherwiseand χS(f(x)) =

9830831 f(x) isin S

0 otherwise

so these are the same since x isin fminus1(S) if and only if f(x) isin S Thus the diagram above commutesso P is naturally isomorphic to h01 and hence P is representable

Power set functor covariant version Now consider the covariant power set functor stilldenoted by P Set rarr Set which again sends a set to its power set but now sends a functionf A rarr B to the image function Pf P (A) rarr P (B) defined by S 983041rarr f(S) In order for this tobe representable there would have to exist a set X such that P sim= hX which translates into havingnatural bijections

P (S) sim= Mor(XS)

for all sets S However in this case there is no natural choice for X as there is no natural way todetect subsets of S by mapping into S as opposed to the case of P (S) sim= Mor(S 0 1)

That no such X exists can be made precise using a simple argument when S = empty P (S) hascardinality 1 soX would have to be empty as well since otherwise Mor(XS) would have cardinality0 but if X is empty then Mor(empty S) always cardinality 1 regardless of S which is clearly not trueof P (S) Thus the covariant power set functor is not representable

Forgetful on Top The forgetful functor F Top rarr Set is representable Indeed a representingobject would have to satisfy

X sim= Map(objectX)

as sets for any topological space X (where Map denotes the set of continuous maps) and it iseasy to see that a single point satisfies this property a continuous map f pt rarr X is completelydetermined by the element f(pt) of X and any such element gives a continuous map in this wayOf course checking that F sim= hpt requires checking that various diagrams commute but this issimple to work out in this case so we omit it here

Extracting a topology With an eye towards the idea that the functors hA hA should captureall information about an object A in a category note that the result above shows that in TophX first of all captures all information about X as a set since the set X can be extracted fromhX(pt) = Map(ptX)

We claim that hX actually captures all information about X as a topological space since thetopology on X can be extracted from hX in the following way As we saw in the power set examplewe have a bijection

P (X) sim= Mor(X 0 1)

Now equip 0 1 with the topology in which 1 is open but 0 is not Then a continuous mapf X rarr 0 1 is fully determined by the preimage fminus1(1) since to be continuous only requiresthat this single preimage be open in X Moreoever any open subset of X arises in this way bytaking f to be the indicator function of that open set Thus we get a bijection

Op(X) sim= Map(X 0 1)

19

where Op(X) denotes the set of open subsets of X ie the topology on X Hence the topology onX can be extracted from hX as claimed so hX and hX do encode all information about X

This fact can also be interpreted as saying that the contravariant functor which sends a topo-logical space X to its collection of open subsets (and which sends a continuous map to the mapinduced by taking preimages) is representable with representing object 0 1 equipped with thetopology given above This will be worked out in a homework problem

Forgetful on Grp The forgetful functor Grp rarr Set is also representable In this case arepresenting object should be a group giving set-theoretic bijections

G sim= Hom(objectG)

for all groups G where Hom denotes the set of group homomorphisms The group Z works

G sim= Hom(Z G)

since a homomorphism φ Z rarr G is completely determined by φ(1) since Z is generated by 1 andthere are no restrictions on what φ(1) isin G can actually be The inverse of the desired bijection isthus φ 983041rarr φ(1) and it is simple to check that morphisms behave appropriately so that the forgetfulfunctor in this setting is indeed isomoprhic to hZ

Forgetful on Ring The forgetful functor Ring rarr Set is also representable where by Ring wemean the category of rings with identity elements and where morphisms (ie ring homomorphisms)are required to preserve identities Here we need an object satisfying

R sim= Hom(object R)

for all R isin Ob(Ring) and we can see that Z[x] the ring of polynomials in the variable x withcoefficients in Z works Indeed given a ring homomorphism φ Z[x] rarr R the fact that φ isrequired to preserve identities fully determines φ(n) for any n isin Z (since φ(n) = nφ(1)) and hencesince φ preserves multiplication its behavior is fully determined by φ(x) which can be any elementof R The inverse of the bijection we want is thus φ 983041rarr φ(x) and it will follow that this forgetfulfunctor is isomorphic to hZ[x]

Equivalences between Categories

Definitions iso and equiv Now that we have a notion of a ldquomorphismrdquo between categorieswe can talk about the sense in which two categories are the ldquosamerdquo The first possible definitionis the ldquoobviousrdquo one you might expect given the definition of ldquoisomorphismrdquo wersquove seen in everyother context until now we say that C is isomorphic to D if there exist functors F C rarr D andG D rarr C such that G F = idC and F G = idD where idC and idD are identity functors

However this definition turns out to be too restrictive Indeed the requirement that the givencompositions equal identity functors says that given an object A in C the object G(F (A)) beliterally the same as A itself and similarly for objects in D But we know that two objects in acategory can be thought of as being the ldquosamerdquo without being literally the same as long as theyare isomorphic For instance productscoproducts are not unique in the literal sense but only inthe sense that they are isomorphic in a unique way To take another example the definition ofa functor being representable does not say that F (B) literally be the same as hB(A) only thatthey be isomorphic for sure the power set P (S) of a set S is not literally the same as the set offunctions S rarr 0 1mdashit is only the ldquosamerdquo in the sense that they encode the same data

20

Thus it makes sense to relax the requirement that A andG(F (A)) in the definition of isomorphiccategories be equal to the requirement that they be isomorphic This leads to the following betterdefinition of two categories being the ldquosamerdquo C is equivalent to D if there exist functors F C rarr D

and G D rarr C such that G F sim= idC and F G sim= idD So we require only that for instance thefunctor G F be naturally isomorphic to the identity functor on C which says that for any objectA G(F (A)) be isomorphic to A in a natural way Wersquoll see that this truly is the better notionof what it means for two categories to be the ldquosamerdquo and it reflects the idea that all categoricalproperties of one category can be transported to the other

Alternate characterization of equivalences A functor F C rarr D implementing an equiv-alence between categories is called unsurprisingly and equivalence and the ldquoinverserdquo functorG D rarr C is still referred to as being its inverse even though it is not technically an ldquoinverserdquoin the sense that compositions give literal identities But as in the case of Set where ldquoinvertiblerdquocan be characterized without reference to an inverse as ldquoinjectiverdquo and ldquosurjectiverdquo so too canequivalences between categories be characterized without explicit reference to an inverse

A functor F C rarr D is an equivalence if and only it is full faithful and essentiallysurjective

We have seen the notions of full and faithful before and essentially surjective means that any objectin D is isomorphic to something in the image of F for all D isin Ob(D) there exits C isin Ob(C) suchthat F (C) sim= D Justifying this result was the topic of a student presentation and can be foundin the book in Section 15 (Technically to construct an ldquoinverserdquo of F requires that we work withsmall categories but whatever) The bulk of the real work goes into showing that the ldquoinverserdquofunctor constructed is actually a functor which comes down to some abstract nonsense argumentvia commutative diagrams

Equivalences preserve everything We previously saw the notions of full faithful and es-sentially surjective back when thinking about what properties a functor could have which wouldguarantee it sent products to products so the result we proved back then was that equivalencespreserve products Similarly equivalences preserve coproducts and the proof is basically the sameafter reversing arrows A clean way of deriving this is as follows if F C rarr D is an equivalencethen so is F op Cop rarr Dop so since F op preserves products F preserves coproducts becausecoproducts in a category become products in the opposite category

In general an equivalence between categories will preserve whatever categorical notions youwant monomorphisms epimorphisms initial objects terminal objects etc so that equivalentcategories really do have essentially the ldquosamerdquo properties

Equivalence example Here is a first example (also found in the book) of equivalent categorieswhich are not isomorphic Let FVectR be the category whose objects are finite-dimensional realvector spaces and whose morphisms are linear transformations Let MatR be the category ofmatrices defined as follows

bull objects in MatR are nonnegative integers

bull a morphism n rarr m is an mtimes n matrix with real entries

bull composition in MatR is given by matrix multiplication given B n rarr m and A m rarr kmdashsoB is an mtimes n matrix and A is a k timesm matrixmdashthe composition A B n rarr k is the k times nmatrix AB

bull identity morphisms in MatR are given by identity matrices

21

We claim that FVectR and MatR are equivalent First note that they certainly cannot be isomo-prhic isomorphisms in MatR exist only between two objects which are literally the same (ie anmtimesn matrix can be invertible only when m = n) but isomorphisms in FVectn can exist betweenobjects which are not literally the same So there are in a sense not enough objects in MatR toallow it to be isomorphic to FVectR

To construct an equivalence choose one and for all an isomorphism TV V rarr RdimV forevery object in FVectR or equivalently choose once and for all a basis for every V DefineF FVectR rarr MatR by

V 983041rarr dimV on objects and

(T V rarr W ) 983041rarr (the standard matrix of SW T Sminus1V RdimV rarr RdimW ) on morphisms

Equivalently F sends T to the matrix of T relative to the chosen bases for V and W It isstraightforward to check that this is a functor The inverse functor G MatR rarr VectR is givenby

n 983041rarr Rn on objects and

(mtimes n matrix A) 983041rarr (the linear transformation A Rn rarr Rm induced by A) on morphisms

One can check that G is a functor and that F G and G F are naturally isomorphic to identitiesor instead one can argue that F alone is full faithful and essentially surjective The point is thatG(F (V )) gives back RdimV so not literally V itself but only something isomorphic to V which iswhy this gives an equivalence of categories and not an isomorphism

Yoneda Lemma Functors as Objects

Compact Hausdorff spaces and Clowast-algebras Last time we gave an example of equivalentcategories which were not isomorphic but in many ways that example is unsatisfactory when doinglinear algebra people for sure often work with matrix representations of linear transformations asthe example from last would suggest but no one in their right mind literally thinks about suchthings in terms of the equivalence FVectR rarr MatR itself In other words this equivalence isessentially ldquocooked uprdquo for the sake of giving an example but it does not really give any insightinto the actual mathematics of linear algebra

So with that in mind we will describe another equivalence which actually does have some realmeaning in the sense of producing new ways of thinking about some mathematics Given a compactHausdorff space X we let C(X) denote the space of continuous complex-valued functions on X

C(X) = f X rarr C | f is continuous

We can equip the set C(X) with various additional structures First it is a complex vector spaceunder ordinary addition and scalar multiplication of functions Second we can multiply two func-tions together

(fg)(p) = f(p)g(p)

which turns C(X) into whatrsquos called an algebra over C (We wonrsquot give the definition of anldquoalgebrardquo but the point is that this multiplication is in some sense ldquocompatiblerdquo with the vectorspace structure) Next any element f of C(X) has a conjugate element f obtained by takingcomplex conjugates of the values of f

f(p) = f(p)

22

And finally we can equip C(X) with a norm via

983042f983042 = suppisinX

|f(p)|

where |middot| denotes complex absolute value (This supremum exists sinceX is compact by the ExtremeValue Theorem) All of these structures (vector space algebra multiplication conjugation norm)turn C(X) into whatrsquos called a Clowast-algebra We wonrsquot define this precisely but the definition encodesvarious ways in which these structures are compatible Actually C(X) is a unital commutative Clowast-algebra where ldquounitalrdquo means that the multiplication has an identity element (namely the constantfunction 1) and commutative means that fg = gf There is a natural notion of homomorphismbetween Clowast-algebras which are maps between them which behave nicely with all the structuresinvolved In particular given a continuous map f X rarr Y between compact Hausdorff spaces weget a Clowast-algebra homomorphism flowast C(Y ) rarr C(X) given by pre-composition with f flowast sendsg isin C(Y ) to g f isin C(X)

We thus have a contravariant functor CHaus rarr ucClowast-Alg from the category of compactHausdorff spaces and the category of unital commutative Clowast-algebras It turns out that anyunital commutative Clowast-algebra arises in this way in the sense that given any unital commutativeClowast-algebra B there exists a compact Hausdorff space X such that C(X) sim= B as Clowast-algebrasand that morphisms between these algebra arises from continuous maps in the manner describeabove This says that the functor above is essentially surjective full and it turns out that it isalso faithful so that it is an equivalence Thus all information about compact Hausdorff spaces iscompletely encoded in information about unital commtuative Clowast-algebras and indeed one couldlearn everything there is to know about a compact Hausdorff space X from its Clowast-algebra C(X)of functions The fact that the categories CHaus and ucClowast-Alg are equivalent is the startingpoint in the subject known as noncommutative geometry which wersquoll say something about a bitlater on If we drop the requirement that our algebras be unital it turns out that the categoryof commutative Clowast-algebras in general is equivalent to the category of locally compact Hausdorffspaces via essentially the same functor where we take as morphisms in the category of locallycompact Hausdorff spaces to be continuous functions which ldquovanish at infinityrdquo

We are only introducing Clowast-algebras to give an example of an interesting and actually usefulequivalence between categories but just for the fun of it wersquoll note the following Examples ofnoncommutative Clowast-algebras come from what are known as ldquoalgebras of bounded operators onHilbert spacesrdquo which show up in quantum mechanics as the things which describe physicallyobservable quantities like position momentum and energy (In fact all Clowast-algebras arise fromHilbert spaces in this way) Thus Clowast-algebras show up in mathematical formulations of quantummechanics and the notion of noncommutative geometry mentioned above is at the core of someattempts to understand the elusive concept known as ldquoquantum geometryrdquo

Yoneda Lemma The Yoneda Lemma states that given a (covariant) functor F C rarr Set whereC is locally small for any A isin Ob(C) there exists a bijection

F (A) sim= Nat(hA F )

where the right side denotes the set of natural transformations from hA to F A similar result holdsin the contravariant case where we get bijections F (A) sim= Nat(hA F ) (There is also a sense inwhich these bijections are themselves natural but wersquoll ignore this bit) The proof of the YonedaLemma was the topic of a student presentation and can be found in Section 22 of the book

Here we will get a bit philosophical and talk about what the point of the Yoneda Lemmaactually is and later the sense in which it says that we can view arbitrary functors as ldquonon-existent

23

objectsrdquo in a category First we mention what we will call the ldquofunctor of pointsrdquo approach tomathematics a mathematical object can be fully characterized by all morphisms into it That isknowledge of Mor(SA) for all S is equivalent to knowledge of A itself Indeed this is an idea wementioned previously in relation to the functors hAmdashwhich we previously said is referred to as beingthe functor of points of Amdashand now we seek to push this idea further In some examples literalpoints of A can be extracted from hA as in the case of the bijection X sim= hX(pt) = Map(ptX) fora topological space X or G sim= hG(Z) = Mor(Z G) for a group G Moreover in these examples therest of the (topological) structure onX or (group) structure on G can also be extracted from hX andhG respectively In general we can view Mor(SA) as in a sense being ldquopoints of A parameterizedby Srdquo which comes from the idea that in say Set Mor(SA) = AS is ldquothe product of S-manycopies of Ardquo so literally points of A ldquoparameterizedrdquo by S In various areas of mathematics anelement of Mor(SA) is thought of as being a ldquogeneralizedrdquo point of A so that the functor of pointshA detects generalized points of A

So an object A is fully encoded by the functor hA Now we claim that morphisms betweenobjects are also encoded by such functors Indeed in the case where F = hB is itself a functor ofpoints the Yoneda result Nat(hA F ) sim= F (A) becomes

Nat(hA hB) sim= hB(A) = Mor(AB)

which says precisely that morphisms between the objects A and B can be obtained solely fromthe functors hA and hB (This fact is one reason why one should prefer the contravariant functorhA to the covariant functor hA in the covariant case Nat(hA F ) sim= F (A) for F = hB becomesNat(hA hB) sim= hB(A) = Mor(BA) so that morphisms between two objects get reversed whencharacterizing them in terms of representable functors)

Pushing our philosophy further we now know that not only objects but also morphisms betweenobjects are fully encoded by the functors hA The final step is to note that we should not only careabout objects and morphisms between them but that an important part of the data of an objectA is also how arbitrary contravariant functors F C rarr Set behave on A and that this too can beobtained from hA precisely via the Yoneda result that F (A) sim= Nat(hA F ) So objects morphismsbetween objects and how arbitrary functors behave on objects are all encoded in knowledge ofthe functors hA so that these functors truly do capture in the most general sense possible allinformation about A

Functors as nonexistent objects The discussion above is all well and good but it still doesnrsquotquite say why we should care about the fact that we can study on object A by studying hA insteadin that we still havenrsquot spoken about what type of insight about A can be gleamed from hA thatwasnrsquot readily apparent in A itself already Now we argue that the true point of the YonedaLemma is that it gives a way to think about arbitrary functors F Cop rarr Set as ldquoobjectsrdquo in C

itself or really as objects in an ldquoenlargementrdquo of C Certainly for those functors F sim= hA which arerepresentable we can indeed think about F as simply being the object A but now we claim thateven non-representable F can be thought of as being ldquononexistentrdquo objects in C The upshot is thatoften in mathematics we come across things which do not actually exist but the Yoneda Lemmagives a way to (indirectly) study such things anyway by interpreting them as functors instead

The starting point of course is that an actual object A is the same as hA as we have explainedabove Thus by analogy we can interpret a non-representable functor F as being an ldquoobjectrdquo in C

as long as we know what it means to ldquomaprdquo into F from other things which are honestly objects in CThat is if F were indeed representable then F (A) sim= Nat(hA F ) sim= Mor(AF ) would characterizethe functor of points of F (here we abuse notation by using the same letter F to denote both arepresentable functor and the object which represents it) so now we take Mor(AF ) = F (A) as

24

the definition of the ldquofunctor of pointsrdquo of the non-existent object F We are thus taking elementsof F (A) as being what it means to ldquomaprdquo from A into F even when F is not representable Thefact that the Yoneda Lemma gives a way to define maps from A into F thus says that F itself isequipped with the same data that a literal object of A would be equipped with so why shouldnrsquotwe interpret F as a being a type of ldquoobjectrdquo in C as well

To phrase this another way note that the Yoneda Lemma gives rise to a functor

C rarr Fun(CopSet)

where Fun(CopSet) denotes the category whose objects are functors from Cop to Set and morphismsare natural transformation defined on objects by A 983041rarr hA and on morphisms by the bijectionMor(AB) sim= Nat(hA hB) arising from the Yoneda Lemma That this is actually bijection saysthat this functor is full and faithful so that we can view C as being a subcategory of the functorcategory Fun(CopSet) (We cay say that C is isomorphic to the subcategory of Fun(CopSet)consisting of the functors hA or that it is equivalent to the subcategory of Fun(CopSet) consistingof representable functors This functor C rarr Fun(CopSet) is usually called the Yoneda embeddingof C into Fun(CopSet)) Thus we can view Fun(CopSet) as an ldquoenlargmentrdquo of C and the ideawe are trying to make sense of is that other objects in this ldquolargerrdquo category should be interpretedas ldquonon-existent objectsrdquo of C

Just how far can we actually push this idea that arbitrary functors Cop rarr Set should be thoughtof as being ldquoobjectsrdquo of C Pretty far in fact For instance often times in geometry one comesacross a type of ldquosmoothrdquo space on which some group acts and one is interested in doing ldquocalculusrdquoon the resulting quotient space which is the space of orbits under the given group action Theproblem is that such quotient spaces are very often (usually in fact) not ldquosmoothrdquo themselves sothat the tools from calculus are not readily available The solution is to nonetheless treat thisquotient as if it were ldquosmoothrdquo and attempt to do calculus anyway To make this precise thisldquonon-smoothrdquo space actually turns out to have a natural interpretation as a ldquosmooth functorrdquo andit is through this functor which we can view this non-smooth space as actually being smooth afterall so we are forced to expand our notion of what we mean by ldquosmooth spacerdquo by allowing thingswhich are not spaces in the usual sense which certainly leads to more abstraction but the benefitis that it actually makes sense to do calculus on such ldquospacesrdquo interpreted as functors

The discussion above is pretty vague so here is one final more concrete example Recall thatequivalence

CHausop rarr ucClowast-Alg

between the category of compact Hausdorff spaces and the category of unital commutative Clowast-algebras mentioned at the start We now claim that a non-commutative (unital) Clowast-algebra shouldbe interpreted as the algebra of functions on some non-existent ldquononcommutative compact Haus-dorff spacerdquo and it is this idea which the subject of noncommutative geometry makes precise Solet B be a non-commutative Clowast-algebra Then the functor of points hB gives a functor

ucClowast-Alg rarr Set

which sends C(X) (where X is some compact Hausdorff space) to Hom(C(X) B) where Hom de-notes homomorphisms of Clowast-algebras whatever that means Composing with the previous functorwe get a functor wersquoll denote by B itself

B CHausop rarr Set

which on objects at least is given by X 983041rarr Hom(C(X) B) This functor is not representable sinceB is not commutative but the philosophy espoused above suggests we should think of it as giving a

25

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

with concatenation as the group operation (When we have to elements of G or two elements of Hnext to each other with use their respective group operations to turn these into single elements ofG or H) This was also the topic of a student presentation and here is the basic idea Given grouphomomorphisms g G rarr N and k H rarr N that G lowastH is the coproduct comes down to showingthat the map

h G lowastH rarr N defined by h(g1h1 gnhn) = f(g1)k(h1) middot middot middot f(gn)k(hn)

and similarly on other possible words in GlowastH is a group homomorphism which follows essentiallyfrom the definition of the group operation on G lowastH indeed this group operation arises preciselyfrom this requirement

Now if we instead tried to use GtimesH with its standard inclusions G rarr GtimesH and H rarr GtimesHas the coproduct the issue is that the analogous map

h GtimesH rarr N defined by h((g h)) = f(g)k(h)

would NOT be a homomorphism because h((g1 h1)(g2 h2)) is not equal to h(g1 h1)h(g2 h2) bythe way in which the group operation is defined on the direct product G times H However if westrict our category to only consist of abelian groups so that the N rsquos we consider in the defini-tion of coproduct are also abelian then G times H does serve as a coproduct we can always rewritef(g1)k(g1) middot middot middot f(gn)k(gn) as f(g1) middot middot middot f(gn)k(h1) middot middot middot k(hn) in N Irsquoll leave it to you to flesh out allof these details

Concrete categories The categories Set Top Vect and Grp all have the property thattheir objects are simply sets possibly equipped with extra structure and that their morphisms areordinary set-theoretic functions possibly satisfying some additional requirement Such categoriesare nice since we can often use intuition about sets and functions in their study A concrete categoryis intuitively one where we can think of objects as being ldquosetsrdquo and morphisms as being ldquofunctionsrdquoonly that we have to interpret this in the right way

Here is the definition a concrete category is a category C equipped with a faithful functorF C rarr Set which is part of the data The idea is that such a functor gives us way to turnan object A isin Ob(C) into a set F (A) and a morphism f isin MorC(AB) into a function Ff isinMorSet(F (A) F (B)) where the ldquofaithfulrdquo requirement says that we do not lose information aboutf when doing so so that we can (essentially) learn everything we need to know about f by studyingFf instead In the four examples above the required faithful functor is simply the forgetful functorand indeed in general we think of F as being a ldquoforgetful functorrdquo for the concrete category (C F )

Examples Here are two more examples of concrete categories Let G be a group and considerthe category BG with a single object and elements of G morphisms As written this does notappear to be concrete since morphisms are not honest functions lowast rarr lowast but the point is that thiscategory can indeed by ldquoconcretizedrdquo by specifying an appropriate ldquoforgetful functorrdquo A functorF BG rarr Set is the data of a (left) action of G on a set S = F (lowast) and saying this functor isfaithful is what it means to say that this action is faithful different g h isin G act on S in differentways or equivalently that there exists s isin S such that gs ∕= hs Any group has at least onefaithful left actionmdashnamely the action on itself given by left multiplicationmdashto BG can always beequipped with a faithful functor to Set This is all just a way of phrasing the fact that any groupis isomorphic to a subgroup of a permutation group

The category Rel of relations can also be ldquoconcretizedrdquo even though as written morphismsA rarr B are not honest functions Take F Rel rarr Set to be the power set functor which sends an

14

object in Rel to its power set and a relation f A rarr B to the induced function Ff P (A) rarr P (B)defined by

(Ff)(S) = b isin B | there exists a isin S such that (a b) isin f

In other words if you think of the relation f sube A times B as a ldquopossibly not everywhere-definedpossibly multivalued functionrdquo (Ff)(S) is analogous to taking the image of S That F is faithfulwill be left to the homework

In fact most categories yoursquoll see can be made concrete by specifying an appropriate forgetfulfunctor but not all categories can be made concrete in this way Here is the standard example of anon-concrete category which we mention only to say there is such a thing but which we wonrsquot doanything more with since understanding it better requires knowledge of algebraic topology Thehomotopy category of topological spaces is the category hTop whose objects are topological spacesand whose morphisms are given by equivalence classes of homotopic maps

MorhTop(XY ) = homotopy classes of continuous maps X rarr Y

(Saying that two maps are homotopic means that you can ldquodeformrdquo one into the other but wersquollleave the precise definition to a course in algebraic topology) It turns out that no functor fromhTop to Set can be faithful so that hTop cannot be made concrete but this is actually a deepand difficult thing to show so we make no attempt to do so here

Natural Isomorphisms Representability

Natural transformations The definition of a natural transformation η F rArr G between two(both covariant or both contravariant) functors FG C rarr D is given in Section 14 of the bookThe key idea is that a natural transformation gives a way to turn data about F into data aboutG in a way which is compatible with all possible morphisms as expressed by the commutativity ofthe diagrams

F (A) F (B)

G(A) G(B)

Ff

ηA ηB

Gf

arising from morphisms f A rarr B (or f B rarr A in the contravariant case) in C In the case of anatural isomoprhism η F rArr G so that each ηA F (A) rarr G(A) is actually an isomorphism in Dthe point is not only that F and G produce isomorphic objects but they do so in a way which iscompatible with all possible category-theoretic data

The first standard example of a natural isomorphism is the one between the identity functor onthe category of finite-dimensional and the functor which sends a finite-dimensional vector space Vto V lowastlowast the dual of its dual induced by the isomorphism V rarr V lowastlowast defined by

v 983041rarr (the linear map on V lowast which sends f to f(v))

This was the topic of a student presentation and can be found in Section 14 of the book Theessential reason why this isomorphism is ldquonaturalrdquo is that V rarr V lowastlowast can be defined without referenceto a basis Contrast this with the the functor which sends V to its (single) dual V lowast it is still truethat finite-dimensional vector space is isomorphic to its dual but specifying such an isomorphism

15

requires additional data which prevents the identity functor from being ldquonaturally isomorphicrdquo tothe single dual functor

Some history At this point it makes sense to say a bit about the interesting origins of categorytheory in the 1950rsquos Eilenberg and Mac Lane where studying cohomology (or possible homologyone of those two) in algebraic topology which associates algebraic data (a group ring or similar)to a topological space in a way which encodes some of the topology They had two cohomologicalconstructions which ended up proving isomorphic results but they noticed that not only werethe resulting objects the same but the actual constructions themselves were in some sense theldquosamerdquo They thus then needed a way to formalize this notion and invented the concept of anatural transformation to do so where they interpreted two constructions as being the same asbeing compatible with all morphisms

But this then leads to the question what types of objects should a natural transformationoccur between Eilenberg and Mac Lane then had to invent the notion of a ldquofunctorrdquo to describewhat a natural transformation went from and what it went to In their motivating setting theyinvented the notion of a functor to describe more formally properties which their cohomologicalconstructions were supposed to satisfy But this then leads to the question what types of objectsshould a functor map between In other words what type of data should a functor (cohomologicalconstruction in their example) take as input and should it output They finally had to thus inventthe notion of a ldquocategoryrdquo to describe the source and target of a functor so that their cohomologicalconstructions could be interpreted as functors from a category of topological spaces to a categoryof algebraic objects

Thus historically natural transformations came first then functors and then categories Aswith most things in mathematics categories thus arose from attempts to encode what was beingseen in some concrete examples in a more general and formal setting

Representable functors Representable functors are defined in Section 21 of the book but wersquollgive the required definitions here in a hopefully simpler to digest manner The point is that anyobject A in a (locally small) category C gives two ldquonaturalrdquo functors C rarr Set one covariant andthe other contravariant which as wersquoll see pretty much encode all data about A itself We definehA C rarr Set to be the covariant functor defined by

hA(B) = Mor(AB) on objects and

hA(Bfminusrarr C) = the map Mor(AB)

hAfminusminusrarr Mor(AC) defined by g 983041rarr f g

We define hA C rarr Set to the contravariant functor defined by

hA(B) = Mor(BA) on objects and

hA(Bfminusrarr C) = the map Mor(CA)

hAfminusminusrarr Mor(BA) defined by g 983041rarr g f

(The functor hA is called the functor of points of A and wersquoll usually think of it as being a covariantfunctor hA Cop rarr Set The functor hA does not have a standard name) So hA is defined bymorphisms from A and hA is defined by morphisms into A Wersquoll see later the sense in which suchfunctors encode all information about the object A

A functor F C rarr Set is representable if is naturally isomorphic to such a functor so F sim= hA

for some A in the covariant case and F sim= hA for some A in the contravariant case The idea wersquollbuild towards is that we can essentially treat such a functor as being A itself so that representable

16

functors can be thought as being actual objects in C For now here is a key example wersquove seenbefore which we can now formulate using the language of representability

Given AB isin Ob(C) let F C rarr Set be the contravariant functor defined by

F (C) = Mor(CA)timesMor(CB) and F (Cfminusrarr D) = [(g k) 983041rarr (g f k f)]

where (g k) 983041rarr (g f k f) gives a map Mor(DA) timesMor(DB) rarr Mor(CA) timesMor(CB) ieF (D) rarr F (C) In order for this functor to be representable requires the existence of an object inC satisfying

Mor(CA)timesMor(CB) sim= Mor(C representing object)

in a way which is compatible with all morphisms but we have actually already seen this notionelsewhere it is essentially the definition of a product AtimesB for A and B Indeed we claim that Fis naturally isomorphic to hAtimesB as we now show

First to define a natural isomorphism hAtimesBsim= F requires for each C isin Ob(C) a choice of a

bijection (ie isomorphism in Set)

ηC hAtimesB(C)sim=minusrarr F (C)

which in our case looks like

ηC Mor(CAtimesB))sim=minusrarr Mor(CA)timesMor(CB)

Take this to be the map

(Cfminusrarr AtimesB) 983041rarr (pA f pB f)

where pA A times B rarr A and pB A times B rarr B are the two projection morphisms in the definitionof a product The definition of product implies that this map is invertible since given k C rarr Aand ℓ C rarr B there exists a unique h C rarr A times B such that k = pA h and ℓ = pB h Forthese bijections to define a natural isomorphism hAtimesB

sim= F requires that they be compatible withall morphisms in C which means that given any g C rarr D in C the following diagram shouldcommute

hAtimesB(C) hAtimesB(D)

F (C) F (D)

hAtimesB(g)

ηC ηD

Fg

which in the case at hand looks like

Mor(CAtimesB) Mor(DAtimesB)

Mor(CA)timesMor(CB) Mor(DA)timesMor(DB)

minus g

ηC ηD

(minus gminus g)

Take an element k D rarr AtimesB in the upper right Applying the map on top gives the elementk g in the upper left and then applying ηC on the left side gives

(pA (k g) pB (k g))

17

in the lower left Instead starting with k D rarr AtimesB in the upper right applying ηD on the rightgives (pA k pB k) and applying the map on the bottom gives

((pA k) g (pB k) g)

in the lower left which is the same as the previous element in the lower left by the associativity ofcomposition Hence the isomorphisms ηC do define a natural isomorphism between hAtimesB and F Again the point is that you can then interpret the functor F as being the ldquosamerdquo as the productAtimesB since both encode the same data

More Representable Examples

Coproducts As in the case of products the definition of coproducts can also be phrased in termsof a representable functor Given objects A and B in C let F C rarr Set be the covariant functordefined by

F (C) = Mor(AC)timesMor(BC) and F (Cfminusrarr D) = [(g k) 983041rarr (f g f k)]

where the latter is a function Mor(AC)timesMor(BC) rarr Mor(AD)timesMor(BD) This functor isrepresentable if and only if a coproduct A ⊔ B for A and B exists and the representing object isthat coproduct the key is the requirement that we have bijections

Mor(AC)timesMor(BC) sim= Mor(A ⊔BC)

for all C which is essentially the defining property a coproduct for A and B should have

Power set functor contravariant version Let P Set rarr Set be the contravariant powerset functor defined on objects by sending a set to its power set and on morphisms by sendingf A rarr B to the preimage function Pf P (B) rarr P (A) given by S 983041rarr fminus1(S) In order for this tobe representable requires the existence of a set X with natural bijections

P (A) sim= Mor(AX)

for all sets A that is a set X so that mapping into X is the same data as specifying a subset ofthe domain We take X = 0 1 to be a two-element set and the required bijections

ηA P (A) rarr Mor(A 0 1) to be given by S 983041rarr χS

where XS A rarr 0 1 is the indicator or characteristic function S defined by sending elementsof S to 1 and elements of A minus S to 0 The inverse of ηA sends a function g A rarr 0 1 to thepreimage gminus1(1) sube A

To verify that the bijections ηA define the data of a natural isomorphism η P rArr h01 wecheck the commutativity of some diagrams Given a function f A rarr B consider

P (A) P (B)

h01(A) = Mor(A 0 1) h01(B) = Mor(B 0 1)

Pf = fminus1

ηA ηB

minus f

18

Take S isin P (B) in the upper right Applying the map on top gives fminus1(S) isin P (A) and thenapplying the map on the left gives χfminus1(S) isin Mor(A 0 1) On the other hand applying the mapon the right to S isin P (B) gives χS isin Mor(B 0 1) and then applying the map on the bottomgives χS f isin Mor(A 0 1) The two resulting functions χfminus1(S)χS f in the lower left are

χfminus1(S)(x) =

9830831 x isin fminus1(S)

0 otherwiseand χS(f(x)) =

9830831 f(x) isin S

0 otherwise

so these are the same since x isin fminus1(S) if and only if f(x) isin S Thus the diagram above commutesso P is naturally isomorphic to h01 and hence P is representable

Power set functor covariant version Now consider the covariant power set functor stilldenoted by P Set rarr Set which again sends a set to its power set but now sends a functionf A rarr B to the image function Pf P (A) rarr P (B) defined by S 983041rarr f(S) In order for this tobe representable there would have to exist a set X such that P sim= hX which translates into havingnatural bijections

P (S) sim= Mor(XS)

for all sets S However in this case there is no natural choice for X as there is no natural way todetect subsets of S by mapping into S as opposed to the case of P (S) sim= Mor(S 0 1)

That no such X exists can be made precise using a simple argument when S = empty P (S) hascardinality 1 soX would have to be empty as well since otherwise Mor(XS) would have cardinality0 but if X is empty then Mor(empty S) always cardinality 1 regardless of S which is clearly not trueof P (S) Thus the covariant power set functor is not representable

Forgetful on Top The forgetful functor F Top rarr Set is representable Indeed a representingobject would have to satisfy

X sim= Map(objectX)

as sets for any topological space X (where Map denotes the set of continuous maps) and it iseasy to see that a single point satisfies this property a continuous map f pt rarr X is completelydetermined by the element f(pt) of X and any such element gives a continuous map in this wayOf course checking that F sim= hpt requires checking that various diagrams commute but this issimple to work out in this case so we omit it here

Extracting a topology With an eye towards the idea that the functors hA hA should captureall information about an object A in a category note that the result above shows that in TophX first of all captures all information about X as a set since the set X can be extracted fromhX(pt) = Map(ptX)

We claim that hX actually captures all information about X as a topological space since thetopology on X can be extracted from hX in the following way As we saw in the power set examplewe have a bijection

P (X) sim= Mor(X 0 1)

Now equip 0 1 with the topology in which 1 is open but 0 is not Then a continuous mapf X rarr 0 1 is fully determined by the preimage fminus1(1) since to be continuous only requiresthat this single preimage be open in X Moreoever any open subset of X arises in this way bytaking f to be the indicator function of that open set Thus we get a bijection

Op(X) sim= Map(X 0 1)

19

where Op(X) denotes the set of open subsets of X ie the topology on X Hence the topology onX can be extracted from hX as claimed so hX and hX do encode all information about X

This fact can also be interpreted as saying that the contravariant functor which sends a topo-logical space X to its collection of open subsets (and which sends a continuous map to the mapinduced by taking preimages) is representable with representing object 0 1 equipped with thetopology given above This will be worked out in a homework problem

Forgetful on Grp The forgetful functor Grp rarr Set is also representable In this case arepresenting object should be a group giving set-theoretic bijections

G sim= Hom(objectG)

for all groups G where Hom denotes the set of group homomorphisms The group Z works

G sim= Hom(Z G)

since a homomorphism φ Z rarr G is completely determined by φ(1) since Z is generated by 1 andthere are no restrictions on what φ(1) isin G can actually be The inverse of the desired bijection isthus φ 983041rarr φ(1) and it is simple to check that morphisms behave appropriately so that the forgetfulfunctor in this setting is indeed isomoprhic to hZ

Forgetful on Ring The forgetful functor Ring rarr Set is also representable where by Ring wemean the category of rings with identity elements and where morphisms (ie ring homomorphisms)are required to preserve identities Here we need an object satisfying

R sim= Hom(object R)

for all R isin Ob(Ring) and we can see that Z[x] the ring of polynomials in the variable x withcoefficients in Z works Indeed given a ring homomorphism φ Z[x] rarr R the fact that φ isrequired to preserve identities fully determines φ(n) for any n isin Z (since φ(n) = nφ(1)) and hencesince φ preserves multiplication its behavior is fully determined by φ(x) which can be any elementof R The inverse of the bijection we want is thus φ 983041rarr φ(x) and it will follow that this forgetfulfunctor is isomorphic to hZ[x]

Equivalences between Categories

Definitions iso and equiv Now that we have a notion of a ldquomorphismrdquo between categorieswe can talk about the sense in which two categories are the ldquosamerdquo The first possible definitionis the ldquoobviousrdquo one you might expect given the definition of ldquoisomorphismrdquo wersquove seen in everyother context until now we say that C is isomorphic to D if there exist functors F C rarr D andG D rarr C such that G F = idC and F G = idD where idC and idD are identity functors

However this definition turns out to be too restrictive Indeed the requirement that the givencompositions equal identity functors says that given an object A in C the object G(F (A)) beliterally the same as A itself and similarly for objects in D But we know that two objects in acategory can be thought of as being the ldquosamerdquo without being literally the same as long as theyare isomorphic For instance productscoproducts are not unique in the literal sense but only inthe sense that they are isomorphic in a unique way To take another example the definition ofa functor being representable does not say that F (B) literally be the same as hB(A) only thatthey be isomorphic for sure the power set P (S) of a set S is not literally the same as the set offunctions S rarr 0 1mdashit is only the ldquosamerdquo in the sense that they encode the same data

20

Thus it makes sense to relax the requirement that A andG(F (A)) in the definition of isomorphiccategories be equal to the requirement that they be isomorphic This leads to the following betterdefinition of two categories being the ldquosamerdquo C is equivalent to D if there exist functors F C rarr D

and G D rarr C such that G F sim= idC and F G sim= idD So we require only that for instance thefunctor G F be naturally isomorphic to the identity functor on C which says that for any objectA G(F (A)) be isomorphic to A in a natural way Wersquoll see that this truly is the better notionof what it means for two categories to be the ldquosamerdquo and it reflects the idea that all categoricalproperties of one category can be transported to the other

Alternate characterization of equivalences A functor F C rarr D implementing an equiv-alence between categories is called unsurprisingly and equivalence and the ldquoinverserdquo functorG D rarr C is still referred to as being its inverse even though it is not technically an ldquoinverserdquoin the sense that compositions give literal identities But as in the case of Set where ldquoinvertiblerdquocan be characterized without reference to an inverse as ldquoinjectiverdquo and ldquosurjectiverdquo so too canequivalences between categories be characterized without explicit reference to an inverse

A functor F C rarr D is an equivalence if and only it is full faithful and essentiallysurjective

We have seen the notions of full and faithful before and essentially surjective means that any objectin D is isomorphic to something in the image of F for all D isin Ob(D) there exits C isin Ob(C) suchthat F (C) sim= D Justifying this result was the topic of a student presentation and can be foundin the book in Section 15 (Technically to construct an ldquoinverserdquo of F requires that we work withsmall categories but whatever) The bulk of the real work goes into showing that the ldquoinverserdquofunctor constructed is actually a functor which comes down to some abstract nonsense argumentvia commutative diagrams

Equivalences preserve everything We previously saw the notions of full faithful and es-sentially surjective back when thinking about what properties a functor could have which wouldguarantee it sent products to products so the result we proved back then was that equivalencespreserve products Similarly equivalences preserve coproducts and the proof is basically the sameafter reversing arrows A clean way of deriving this is as follows if F C rarr D is an equivalencethen so is F op Cop rarr Dop so since F op preserves products F preserves coproducts becausecoproducts in a category become products in the opposite category

In general an equivalence between categories will preserve whatever categorical notions youwant monomorphisms epimorphisms initial objects terminal objects etc so that equivalentcategories really do have essentially the ldquosamerdquo properties

Equivalence example Here is a first example (also found in the book) of equivalent categorieswhich are not isomorphic Let FVectR be the category whose objects are finite-dimensional realvector spaces and whose morphisms are linear transformations Let MatR be the category ofmatrices defined as follows

bull objects in MatR are nonnegative integers

bull a morphism n rarr m is an mtimes n matrix with real entries

bull composition in MatR is given by matrix multiplication given B n rarr m and A m rarr kmdashsoB is an mtimes n matrix and A is a k timesm matrixmdashthe composition A B n rarr k is the k times nmatrix AB

bull identity morphisms in MatR are given by identity matrices

21

We claim that FVectR and MatR are equivalent First note that they certainly cannot be isomo-prhic isomorphisms in MatR exist only between two objects which are literally the same (ie anmtimesn matrix can be invertible only when m = n) but isomorphisms in FVectn can exist betweenobjects which are not literally the same So there are in a sense not enough objects in MatR toallow it to be isomorphic to FVectR

To construct an equivalence choose one and for all an isomorphism TV V rarr RdimV forevery object in FVectR or equivalently choose once and for all a basis for every V DefineF FVectR rarr MatR by

V 983041rarr dimV on objects and

(T V rarr W ) 983041rarr (the standard matrix of SW T Sminus1V RdimV rarr RdimW ) on morphisms

Equivalently F sends T to the matrix of T relative to the chosen bases for V and W It isstraightforward to check that this is a functor The inverse functor G MatR rarr VectR is givenby

n 983041rarr Rn on objects and

(mtimes n matrix A) 983041rarr (the linear transformation A Rn rarr Rm induced by A) on morphisms

One can check that G is a functor and that F G and G F are naturally isomorphic to identitiesor instead one can argue that F alone is full faithful and essentially surjective The point is thatG(F (V )) gives back RdimV so not literally V itself but only something isomorphic to V which iswhy this gives an equivalence of categories and not an isomorphism

Yoneda Lemma Functors as Objects

Compact Hausdorff spaces and Clowast-algebras Last time we gave an example of equivalentcategories which were not isomorphic but in many ways that example is unsatisfactory when doinglinear algebra people for sure often work with matrix representations of linear transformations asthe example from last would suggest but no one in their right mind literally thinks about suchthings in terms of the equivalence FVectR rarr MatR itself In other words this equivalence isessentially ldquocooked uprdquo for the sake of giving an example but it does not really give any insightinto the actual mathematics of linear algebra

So with that in mind we will describe another equivalence which actually does have some realmeaning in the sense of producing new ways of thinking about some mathematics Given a compactHausdorff space X we let C(X) denote the space of continuous complex-valued functions on X

C(X) = f X rarr C | f is continuous

We can equip the set C(X) with various additional structures First it is a complex vector spaceunder ordinary addition and scalar multiplication of functions Second we can multiply two func-tions together

(fg)(p) = f(p)g(p)

which turns C(X) into whatrsquos called an algebra over C (We wonrsquot give the definition of anldquoalgebrardquo but the point is that this multiplication is in some sense ldquocompatiblerdquo with the vectorspace structure) Next any element f of C(X) has a conjugate element f obtained by takingcomplex conjugates of the values of f

f(p) = f(p)

22

And finally we can equip C(X) with a norm via

983042f983042 = suppisinX

|f(p)|

where |middot| denotes complex absolute value (This supremum exists sinceX is compact by the ExtremeValue Theorem) All of these structures (vector space algebra multiplication conjugation norm)turn C(X) into whatrsquos called a Clowast-algebra We wonrsquot define this precisely but the definition encodesvarious ways in which these structures are compatible Actually C(X) is a unital commutative Clowast-algebra where ldquounitalrdquo means that the multiplication has an identity element (namely the constantfunction 1) and commutative means that fg = gf There is a natural notion of homomorphismbetween Clowast-algebras which are maps between them which behave nicely with all the structuresinvolved In particular given a continuous map f X rarr Y between compact Hausdorff spaces weget a Clowast-algebra homomorphism flowast C(Y ) rarr C(X) given by pre-composition with f flowast sendsg isin C(Y ) to g f isin C(X)

We thus have a contravariant functor CHaus rarr ucClowast-Alg from the category of compactHausdorff spaces and the category of unital commutative Clowast-algebras It turns out that anyunital commutative Clowast-algebra arises in this way in the sense that given any unital commutativeClowast-algebra B there exists a compact Hausdorff space X such that C(X) sim= B as Clowast-algebrasand that morphisms between these algebra arises from continuous maps in the manner describeabove This says that the functor above is essentially surjective full and it turns out that it isalso faithful so that it is an equivalence Thus all information about compact Hausdorff spaces iscompletely encoded in information about unital commtuative Clowast-algebras and indeed one couldlearn everything there is to know about a compact Hausdorff space X from its Clowast-algebra C(X)of functions The fact that the categories CHaus and ucClowast-Alg are equivalent is the startingpoint in the subject known as noncommutative geometry which wersquoll say something about a bitlater on If we drop the requirement that our algebras be unital it turns out that the categoryof commutative Clowast-algebras in general is equivalent to the category of locally compact Hausdorffspaces via essentially the same functor where we take as morphisms in the category of locallycompact Hausdorff spaces to be continuous functions which ldquovanish at infinityrdquo

We are only introducing Clowast-algebras to give an example of an interesting and actually usefulequivalence between categories but just for the fun of it wersquoll note the following Examples ofnoncommutative Clowast-algebras come from what are known as ldquoalgebras of bounded operators onHilbert spacesrdquo which show up in quantum mechanics as the things which describe physicallyobservable quantities like position momentum and energy (In fact all Clowast-algebras arise fromHilbert spaces in this way) Thus Clowast-algebras show up in mathematical formulations of quantummechanics and the notion of noncommutative geometry mentioned above is at the core of someattempts to understand the elusive concept known as ldquoquantum geometryrdquo

Yoneda Lemma The Yoneda Lemma states that given a (covariant) functor F C rarr Set whereC is locally small for any A isin Ob(C) there exists a bijection

F (A) sim= Nat(hA F )

where the right side denotes the set of natural transformations from hA to F A similar result holdsin the contravariant case where we get bijections F (A) sim= Nat(hA F ) (There is also a sense inwhich these bijections are themselves natural but wersquoll ignore this bit) The proof of the YonedaLemma was the topic of a student presentation and can be found in Section 22 of the book

Here we will get a bit philosophical and talk about what the point of the Yoneda Lemmaactually is and later the sense in which it says that we can view arbitrary functors as ldquonon-existent

23

objectsrdquo in a category First we mention what we will call the ldquofunctor of pointsrdquo approach tomathematics a mathematical object can be fully characterized by all morphisms into it That isknowledge of Mor(SA) for all S is equivalent to knowledge of A itself Indeed this is an idea wementioned previously in relation to the functors hAmdashwhich we previously said is referred to as beingthe functor of points of Amdashand now we seek to push this idea further In some examples literalpoints of A can be extracted from hA as in the case of the bijection X sim= hX(pt) = Map(ptX) fora topological space X or G sim= hG(Z) = Mor(Z G) for a group G Moreover in these examples therest of the (topological) structure onX or (group) structure on G can also be extracted from hX andhG respectively In general we can view Mor(SA) as in a sense being ldquopoints of A parameterizedby Srdquo which comes from the idea that in say Set Mor(SA) = AS is ldquothe product of S-manycopies of Ardquo so literally points of A ldquoparameterizedrdquo by S In various areas of mathematics anelement of Mor(SA) is thought of as being a ldquogeneralizedrdquo point of A so that the functor of pointshA detects generalized points of A

So an object A is fully encoded by the functor hA Now we claim that morphisms betweenobjects are also encoded by such functors Indeed in the case where F = hB is itself a functor ofpoints the Yoneda result Nat(hA F ) sim= F (A) becomes

Nat(hA hB) sim= hB(A) = Mor(AB)

which says precisely that morphisms between the objects A and B can be obtained solely fromthe functors hA and hB (This fact is one reason why one should prefer the contravariant functorhA to the covariant functor hA in the covariant case Nat(hA F ) sim= F (A) for F = hB becomesNat(hA hB) sim= hB(A) = Mor(BA) so that morphisms between two objects get reversed whencharacterizing them in terms of representable functors)

Pushing our philosophy further we now know that not only objects but also morphisms betweenobjects are fully encoded by the functors hA The final step is to note that we should not only careabout objects and morphisms between them but that an important part of the data of an objectA is also how arbitrary contravariant functors F C rarr Set behave on A and that this too can beobtained from hA precisely via the Yoneda result that F (A) sim= Nat(hA F ) So objects morphismsbetween objects and how arbitrary functors behave on objects are all encoded in knowledge ofthe functors hA so that these functors truly do capture in the most general sense possible allinformation about A

Functors as nonexistent objects The discussion above is all well and good but it still doesnrsquotquite say why we should care about the fact that we can study on object A by studying hA insteadin that we still havenrsquot spoken about what type of insight about A can be gleamed from hA thatwasnrsquot readily apparent in A itself already Now we argue that the true point of the YonedaLemma is that it gives a way to think about arbitrary functors F Cop rarr Set as ldquoobjectsrdquo in C

itself or really as objects in an ldquoenlargementrdquo of C Certainly for those functors F sim= hA which arerepresentable we can indeed think about F as simply being the object A but now we claim thateven non-representable F can be thought of as being ldquononexistentrdquo objects in C The upshot is thatoften in mathematics we come across things which do not actually exist but the Yoneda Lemmagives a way to (indirectly) study such things anyway by interpreting them as functors instead

The starting point of course is that an actual object A is the same as hA as we have explainedabove Thus by analogy we can interpret a non-representable functor F as being an ldquoobjectrdquo in C

as long as we know what it means to ldquomaprdquo into F from other things which are honestly objects in CThat is if F were indeed representable then F (A) sim= Nat(hA F ) sim= Mor(AF ) would characterizethe functor of points of F (here we abuse notation by using the same letter F to denote both arepresentable functor and the object which represents it) so now we take Mor(AF ) = F (A) as

24

the definition of the ldquofunctor of pointsrdquo of the non-existent object F We are thus taking elementsof F (A) as being what it means to ldquomaprdquo from A into F even when F is not representable Thefact that the Yoneda Lemma gives a way to define maps from A into F thus says that F itself isequipped with the same data that a literal object of A would be equipped with so why shouldnrsquotwe interpret F as a being a type of ldquoobjectrdquo in C as well

To phrase this another way note that the Yoneda Lemma gives rise to a functor

C rarr Fun(CopSet)

where Fun(CopSet) denotes the category whose objects are functors from Cop to Set and morphismsare natural transformation defined on objects by A 983041rarr hA and on morphisms by the bijectionMor(AB) sim= Nat(hA hB) arising from the Yoneda Lemma That this is actually bijection saysthat this functor is full and faithful so that we can view C as being a subcategory of the functorcategory Fun(CopSet) (We cay say that C is isomorphic to the subcategory of Fun(CopSet)consisting of the functors hA or that it is equivalent to the subcategory of Fun(CopSet) consistingof representable functors This functor C rarr Fun(CopSet) is usually called the Yoneda embeddingof C into Fun(CopSet)) Thus we can view Fun(CopSet) as an ldquoenlargmentrdquo of C and the ideawe are trying to make sense of is that other objects in this ldquolargerrdquo category should be interpretedas ldquonon-existent objectsrdquo of C

Just how far can we actually push this idea that arbitrary functors Cop rarr Set should be thoughtof as being ldquoobjectsrdquo of C Pretty far in fact For instance often times in geometry one comesacross a type of ldquosmoothrdquo space on which some group acts and one is interested in doing ldquocalculusrdquoon the resulting quotient space which is the space of orbits under the given group action Theproblem is that such quotient spaces are very often (usually in fact) not ldquosmoothrdquo themselves sothat the tools from calculus are not readily available The solution is to nonetheless treat thisquotient as if it were ldquosmoothrdquo and attempt to do calculus anyway To make this precise thisldquonon-smoothrdquo space actually turns out to have a natural interpretation as a ldquosmooth functorrdquo andit is through this functor which we can view this non-smooth space as actually being smooth afterall so we are forced to expand our notion of what we mean by ldquosmooth spacerdquo by allowing thingswhich are not spaces in the usual sense which certainly leads to more abstraction but the benefitis that it actually makes sense to do calculus on such ldquospacesrdquo interpreted as functors

The discussion above is pretty vague so here is one final more concrete example Recall thatequivalence

CHausop rarr ucClowast-Alg

between the category of compact Hausdorff spaces and the category of unital commutative Clowast-algebras mentioned at the start We now claim that a non-commutative (unital) Clowast-algebra shouldbe interpreted as the algebra of functions on some non-existent ldquononcommutative compact Haus-dorff spacerdquo and it is this idea which the subject of noncommutative geometry makes precise Solet B be a non-commutative Clowast-algebra Then the functor of points hB gives a functor

ucClowast-Alg rarr Set

which sends C(X) (where X is some compact Hausdorff space) to Hom(C(X) B) where Hom de-notes homomorphisms of Clowast-algebras whatever that means Composing with the previous functorwe get a functor wersquoll denote by B itself

B CHausop rarr Set

which on objects at least is given by X 983041rarr Hom(C(X) B) This functor is not representable sinceB is not commutative but the philosophy espoused above suggests we should think of it as giving a

25

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

object in Rel to its power set and a relation f A rarr B to the induced function Ff P (A) rarr P (B)defined by

(Ff)(S) = b isin B | there exists a isin S such that (a b) isin f

In other words if you think of the relation f sube A times B as a ldquopossibly not everywhere-definedpossibly multivalued functionrdquo (Ff)(S) is analogous to taking the image of S That F is faithfulwill be left to the homework

In fact most categories yoursquoll see can be made concrete by specifying an appropriate forgetfulfunctor but not all categories can be made concrete in this way Here is the standard example of anon-concrete category which we mention only to say there is such a thing but which we wonrsquot doanything more with since understanding it better requires knowledge of algebraic topology Thehomotopy category of topological spaces is the category hTop whose objects are topological spacesand whose morphisms are given by equivalence classes of homotopic maps

MorhTop(XY ) = homotopy classes of continuous maps X rarr Y

(Saying that two maps are homotopic means that you can ldquodeformrdquo one into the other but wersquollleave the precise definition to a course in algebraic topology) It turns out that no functor fromhTop to Set can be faithful so that hTop cannot be made concrete but this is actually a deepand difficult thing to show so we make no attempt to do so here

Natural Isomorphisms Representability

Natural transformations The definition of a natural transformation η F rArr G between two(both covariant or both contravariant) functors FG C rarr D is given in Section 14 of the bookThe key idea is that a natural transformation gives a way to turn data about F into data aboutG in a way which is compatible with all possible morphisms as expressed by the commutativity ofthe diagrams

F (A) F (B)

G(A) G(B)

Ff

ηA ηB

Gf

arising from morphisms f A rarr B (or f B rarr A in the contravariant case) in C In the case of anatural isomoprhism η F rArr G so that each ηA F (A) rarr G(A) is actually an isomorphism in Dthe point is not only that F and G produce isomorphic objects but they do so in a way which iscompatible with all possible category-theoretic data

The first standard example of a natural isomorphism is the one between the identity functor onthe category of finite-dimensional and the functor which sends a finite-dimensional vector space Vto V lowastlowast the dual of its dual induced by the isomorphism V rarr V lowastlowast defined by

v 983041rarr (the linear map on V lowast which sends f to f(v))

This was the topic of a student presentation and can be found in Section 14 of the book Theessential reason why this isomorphism is ldquonaturalrdquo is that V rarr V lowastlowast can be defined without referenceto a basis Contrast this with the the functor which sends V to its (single) dual V lowast it is still truethat finite-dimensional vector space is isomorphic to its dual but specifying such an isomorphism

15

requires additional data which prevents the identity functor from being ldquonaturally isomorphicrdquo tothe single dual functor

Some history At this point it makes sense to say a bit about the interesting origins of categorytheory in the 1950rsquos Eilenberg and Mac Lane where studying cohomology (or possible homologyone of those two) in algebraic topology which associates algebraic data (a group ring or similar)to a topological space in a way which encodes some of the topology They had two cohomologicalconstructions which ended up proving isomorphic results but they noticed that not only werethe resulting objects the same but the actual constructions themselves were in some sense theldquosamerdquo They thus then needed a way to formalize this notion and invented the concept of anatural transformation to do so where they interpreted two constructions as being the same asbeing compatible with all morphisms

But this then leads to the question what types of objects should a natural transformationoccur between Eilenberg and Mac Lane then had to invent the notion of a ldquofunctorrdquo to describewhat a natural transformation went from and what it went to In their motivating setting theyinvented the notion of a functor to describe more formally properties which their cohomologicalconstructions were supposed to satisfy But this then leads to the question what types of objectsshould a functor map between In other words what type of data should a functor (cohomologicalconstruction in their example) take as input and should it output They finally had to thus inventthe notion of a ldquocategoryrdquo to describe the source and target of a functor so that their cohomologicalconstructions could be interpreted as functors from a category of topological spaces to a categoryof algebraic objects

Thus historically natural transformations came first then functors and then categories Aswith most things in mathematics categories thus arose from attempts to encode what was beingseen in some concrete examples in a more general and formal setting

Representable functors Representable functors are defined in Section 21 of the book but wersquollgive the required definitions here in a hopefully simpler to digest manner The point is that anyobject A in a (locally small) category C gives two ldquonaturalrdquo functors C rarr Set one covariant andthe other contravariant which as wersquoll see pretty much encode all data about A itself We definehA C rarr Set to be the covariant functor defined by

hA(B) = Mor(AB) on objects and

hA(Bfminusrarr C) = the map Mor(AB)

hAfminusminusrarr Mor(AC) defined by g 983041rarr f g

We define hA C rarr Set to the contravariant functor defined by

hA(B) = Mor(BA) on objects and

hA(Bfminusrarr C) = the map Mor(CA)

hAfminusminusrarr Mor(BA) defined by g 983041rarr g f

(The functor hA is called the functor of points of A and wersquoll usually think of it as being a covariantfunctor hA Cop rarr Set The functor hA does not have a standard name) So hA is defined bymorphisms from A and hA is defined by morphisms into A Wersquoll see later the sense in which suchfunctors encode all information about the object A

A functor F C rarr Set is representable if is naturally isomorphic to such a functor so F sim= hA

for some A in the covariant case and F sim= hA for some A in the contravariant case The idea wersquollbuild towards is that we can essentially treat such a functor as being A itself so that representable

16

functors can be thought as being actual objects in C For now here is a key example wersquove seenbefore which we can now formulate using the language of representability

Given AB isin Ob(C) let F C rarr Set be the contravariant functor defined by

F (C) = Mor(CA)timesMor(CB) and F (Cfminusrarr D) = [(g k) 983041rarr (g f k f)]

where (g k) 983041rarr (g f k f) gives a map Mor(DA) timesMor(DB) rarr Mor(CA) timesMor(CB) ieF (D) rarr F (C) In order for this functor to be representable requires the existence of an object inC satisfying

Mor(CA)timesMor(CB) sim= Mor(C representing object)

in a way which is compatible with all morphisms but we have actually already seen this notionelsewhere it is essentially the definition of a product AtimesB for A and B Indeed we claim that Fis naturally isomorphic to hAtimesB as we now show

First to define a natural isomorphism hAtimesBsim= F requires for each C isin Ob(C) a choice of a

bijection (ie isomorphism in Set)

ηC hAtimesB(C)sim=minusrarr F (C)

which in our case looks like

ηC Mor(CAtimesB))sim=minusrarr Mor(CA)timesMor(CB)

Take this to be the map

(Cfminusrarr AtimesB) 983041rarr (pA f pB f)

where pA A times B rarr A and pB A times B rarr B are the two projection morphisms in the definitionof a product The definition of product implies that this map is invertible since given k C rarr Aand ℓ C rarr B there exists a unique h C rarr A times B such that k = pA h and ℓ = pB h Forthese bijections to define a natural isomorphism hAtimesB

sim= F requires that they be compatible withall morphisms in C which means that given any g C rarr D in C the following diagram shouldcommute

hAtimesB(C) hAtimesB(D)

F (C) F (D)

hAtimesB(g)

ηC ηD

Fg

which in the case at hand looks like

Mor(CAtimesB) Mor(DAtimesB)

Mor(CA)timesMor(CB) Mor(DA)timesMor(DB)

minus g

ηC ηD

(minus gminus g)

Take an element k D rarr AtimesB in the upper right Applying the map on top gives the elementk g in the upper left and then applying ηC on the left side gives

(pA (k g) pB (k g))

17

in the lower left Instead starting with k D rarr AtimesB in the upper right applying ηD on the rightgives (pA k pB k) and applying the map on the bottom gives

((pA k) g (pB k) g)

in the lower left which is the same as the previous element in the lower left by the associativity ofcomposition Hence the isomorphisms ηC do define a natural isomorphism between hAtimesB and F Again the point is that you can then interpret the functor F as being the ldquosamerdquo as the productAtimesB since both encode the same data

More Representable Examples

Coproducts As in the case of products the definition of coproducts can also be phrased in termsof a representable functor Given objects A and B in C let F C rarr Set be the covariant functordefined by

F (C) = Mor(AC)timesMor(BC) and F (Cfminusrarr D) = [(g k) 983041rarr (f g f k)]

where the latter is a function Mor(AC)timesMor(BC) rarr Mor(AD)timesMor(BD) This functor isrepresentable if and only if a coproduct A ⊔ B for A and B exists and the representing object isthat coproduct the key is the requirement that we have bijections

Mor(AC)timesMor(BC) sim= Mor(A ⊔BC)

for all C which is essentially the defining property a coproduct for A and B should have

Power set functor contravariant version Let P Set rarr Set be the contravariant powerset functor defined on objects by sending a set to its power set and on morphisms by sendingf A rarr B to the preimage function Pf P (B) rarr P (A) given by S 983041rarr fminus1(S) In order for this tobe representable requires the existence of a set X with natural bijections

P (A) sim= Mor(AX)

for all sets A that is a set X so that mapping into X is the same data as specifying a subset ofthe domain We take X = 0 1 to be a two-element set and the required bijections

ηA P (A) rarr Mor(A 0 1) to be given by S 983041rarr χS

where XS A rarr 0 1 is the indicator or characteristic function S defined by sending elementsof S to 1 and elements of A minus S to 0 The inverse of ηA sends a function g A rarr 0 1 to thepreimage gminus1(1) sube A

To verify that the bijections ηA define the data of a natural isomorphism η P rArr h01 wecheck the commutativity of some diagrams Given a function f A rarr B consider

P (A) P (B)

h01(A) = Mor(A 0 1) h01(B) = Mor(B 0 1)

Pf = fminus1

ηA ηB

minus f

18

Take S isin P (B) in the upper right Applying the map on top gives fminus1(S) isin P (A) and thenapplying the map on the left gives χfminus1(S) isin Mor(A 0 1) On the other hand applying the mapon the right to S isin P (B) gives χS isin Mor(B 0 1) and then applying the map on the bottomgives χS f isin Mor(A 0 1) The two resulting functions χfminus1(S)χS f in the lower left are

χfminus1(S)(x) =

9830831 x isin fminus1(S)

0 otherwiseand χS(f(x)) =

9830831 f(x) isin S

0 otherwise

so these are the same since x isin fminus1(S) if and only if f(x) isin S Thus the diagram above commutesso P is naturally isomorphic to h01 and hence P is representable

Power set functor covariant version Now consider the covariant power set functor stilldenoted by P Set rarr Set which again sends a set to its power set but now sends a functionf A rarr B to the image function Pf P (A) rarr P (B) defined by S 983041rarr f(S) In order for this tobe representable there would have to exist a set X such that P sim= hX which translates into havingnatural bijections

P (S) sim= Mor(XS)

for all sets S However in this case there is no natural choice for X as there is no natural way todetect subsets of S by mapping into S as opposed to the case of P (S) sim= Mor(S 0 1)

That no such X exists can be made precise using a simple argument when S = empty P (S) hascardinality 1 soX would have to be empty as well since otherwise Mor(XS) would have cardinality0 but if X is empty then Mor(empty S) always cardinality 1 regardless of S which is clearly not trueof P (S) Thus the covariant power set functor is not representable

Forgetful on Top The forgetful functor F Top rarr Set is representable Indeed a representingobject would have to satisfy

X sim= Map(objectX)

as sets for any topological space X (where Map denotes the set of continuous maps) and it iseasy to see that a single point satisfies this property a continuous map f pt rarr X is completelydetermined by the element f(pt) of X and any such element gives a continuous map in this wayOf course checking that F sim= hpt requires checking that various diagrams commute but this issimple to work out in this case so we omit it here

Extracting a topology With an eye towards the idea that the functors hA hA should captureall information about an object A in a category note that the result above shows that in TophX first of all captures all information about X as a set since the set X can be extracted fromhX(pt) = Map(ptX)

We claim that hX actually captures all information about X as a topological space since thetopology on X can be extracted from hX in the following way As we saw in the power set examplewe have a bijection

P (X) sim= Mor(X 0 1)

Now equip 0 1 with the topology in which 1 is open but 0 is not Then a continuous mapf X rarr 0 1 is fully determined by the preimage fminus1(1) since to be continuous only requiresthat this single preimage be open in X Moreoever any open subset of X arises in this way bytaking f to be the indicator function of that open set Thus we get a bijection

Op(X) sim= Map(X 0 1)

19

where Op(X) denotes the set of open subsets of X ie the topology on X Hence the topology onX can be extracted from hX as claimed so hX and hX do encode all information about X

This fact can also be interpreted as saying that the contravariant functor which sends a topo-logical space X to its collection of open subsets (and which sends a continuous map to the mapinduced by taking preimages) is representable with representing object 0 1 equipped with thetopology given above This will be worked out in a homework problem

Forgetful on Grp The forgetful functor Grp rarr Set is also representable In this case arepresenting object should be a group giving set-theoretic bijections

G sim= Hom(objectG)

for all groups G where Hom denotes the set of group homomorphisms The group Z works

G sim= Hom(Z G)

since a homomorphism φ Z rarr G is completely determined by φ(1) since Z is generated by 1 andthere are no restrictions on what φ(1) isin G can actually be The inverse of the desired bijection isthus φ 983041rarr φ(1) and it is simple to check that morphisms behave appropriately so that the forgetfulfunctor in this setting is indeed isomoprhic to hZ

Forgetful on Ring The forgetful functor Ring rarr Set is also representable where by Ring wemean the category of rings with identity elements and where morphisms (ie ring homomorphisms)are required to preserve identities Here we need an object satisfying

R sim= Hom(object R)

for all R isin Ob(Ring) and we can see that Z[x] the ring of polynomials in the variable x withcoefficients in Z works Indeed given a ring homomorphism φ Z[x] rarr R the fact that φ isrequired to preserve identities fully determines φ(n) for any n isin Z (since φ(n) = nφ(1)) and hencesince φ preserves multiplication its behavior is fully determined by φ(x) which can be any elementof R The inverse of the bijection we want is thus φ 983041rarr φ(x) and it will follow that this forgetfulfunctor is isomorphic to hZ[x]

Equivalences between Categories

Definitions iso and equiv Now that we have a notion of a ldquomorphismrdquo between categorieswe can talk about the sense in which two categories are the ldquosamerdquo The first possible definitionis the ldquoobviousrdquo one you might expect given the definition of ldquoisomorphismrdquo wersquove seen in everyother context until now we say that C is isomorphic to D if there exist functors F C rarr D andG D rarr C such that G F = idC and F G = idD where idC and idD are identity functors

However this definition turns out to be too restrictive Indeed the requirement that the givencompositions equal identity functors says that given an object A in C the object G(F (A)) beliterally the same as A itself and similarly for objects in D But we know that two objects in acategory can be thought of as being the ldquosamerdquo without being literally the same as long as theyare isomorphic For instance productscoproducts are not unique in the literal sense but only inthe sense that they are isomorphic in a unique way To take another example the definition ofa functor being representable does not say that F (B) literally be the same as hB(A) only thatthey be isomorphic for sure the power set P (S) of a set S is not literally the same as the set offunctions S rarr 0 1mdashit is only the ldquosamerdquo in the sense that they encode the same data

20

Thus it makes sense to relax the requirement that A andG(F (A)) in the definition of isomorphiccategories be equal to the requirement that they be isomorphic This leads to the following betterdefinition of two categories being the ldquosamerdquo C is equivalent to D if there exist functors F C rarr D

and G D rarr C such that G F sim= idC and F G sim= idD So we require only that for instance thefunctor G F be naturally isomorphic to the identity functor on C which says that for any objectA G(F (A)) be isomorphic to A in a natural way Wersquoll see that this truly is the better notionof what it means for two categories to be the ldquosamerdquo and it reflects the idea that all categoricalproperties of one category can be transported to the other

Alternate characterization of equivalences A functor F C rarr D implementing an equiv-alence between categories is called unsurprisingly and equivalence and the ldquoinverserdquo functorG D rarr C is still referred to as being its inverse even though it is not technically an ldquoinverserdquoin the sense that compositions give literal identities But as in the case of Set where ldquoinvertiblerdquocan be characterized without reference to an inverse as ldquoinjectiverdquo and ldquosurjectiverdquo so too canequivalences between categories be characterized without explicit reference to an inverse

A functor F C rarr D is an equivalence if and only it is full faithful and essentiallysurjective

We have seen the notions of full and faithful before and essentially surjective means that any objectin D is isomorphic to something in the image of F for all D isin Ob(D) there exits C isin Ob(C) suchthat F (C) sim= D Justifying this result was the topic of a student presentation and can be foundin the book in Section 15 (Technically to construct an ldquoinverserdquo of F requires that we work withsmall categories but whatever) The bulk of the real work goes into showing that the ldquoinverserdquofunctor constructed is actually a functor which comes down to some abstract nonsense argumentvia commutative diagrams

Equivalences preserve everything We previously saw the notions of full faithful and es-sentially surjective back when thinking about what properties a functor could have which wouldguarantee it sent products to products so the result we proved back then was that equivalencespreserve products Similarly equivalences preserve coproducts and the proof is basically the sameafter reversing arrows A clean way of deriving this is as follows if F C rarr D is an equivalencethen so is F op Cop rarr Dop so since F op preserves products F preserves coproducts becausecoproducts in a category become products in the opposite category

In general an equivalence between categories will preserve whatever categorical notions youwant monomorphisms epimorphisms initial objects terminal objects etc so that equivalentcategories really do have essentially the ldquosamerdquo properties

Equivalence example Here is a first example (also found in the book) of equivalent categorieswhich are not isomorphic Let FVectR be the category whose objects are finite-dimensional realvector spaces and whose morphisms are linear transformations Let MatR be the category ofmatrices defined as follows

bull objects in MatR are nonnegative integers

bull a morphism n rarr m is an mtimes n matrix with real entries

bull composition in MatR is given by matrix multiplication given B n rarr m and A m rarr kmdashsoB is an mtimes n matrix and A is a k timesm matrixmdashthe composition A B n rarr k is the k times nmatrix AB

bull identity morphisms in MatR are given by identity matrices

21

We claim that FVectR and MatR are equivalent First note that they certainly cannot be isomo-prhic isomorphisms in MatR exist only between two objects which are literally the same (ie anmtimesn matrix can be invertible only when m = n) but isomorphisms in FVectn can exist betweenobjects which are not literally the same So there are in a sense not enough objects in MatR toallow it to be isomorphic to FVectR

To construct an equivalence choose one and for all an isomorphism TV V rarr RdimV forevery object in FVectR or equivalently choose once and for all a basis for every V DefineF FVectR rarr MatR by

V 983041rarr dimV on objects and

(T V rarr W ) 983041rarr (the standard matrix of SW T Sminus1V RdimV rarr RdimW ) on morphisms

Equivalently F sends T to the matrix of T relative to the chosen bases for V and W It isstraightforward to check that this is a functor The inverse functor G MatR rarr VectR is givenby

n 983041rarr Rn on objects and

(mtimes n matrix A) 983041rarr (the linear transformation A Rn rarr Rm induced by A) on morphisms

One can check that G is a functor and that F G and G F are naturally isomorphic to identitiesor instead one can argue that F alone is full faithful and essentially surjective The point is thatG(F (V )) gives back RdimV so not literally V itself but only something isomorphic to V which iswhy this gives an equivalence of categories and not an isomorphism

Yoneda Lemma Functors as Objects

Compact Hausdorff spaces and Clowast-algebras Last time we gave an example of equivalentcategories which were not isomorphic but in many ways that example is unsatisfactory when doinglinear algebra people for sure often work with matrix representations of linear transformations asthe example from last would suggest but no one in their right mind literally thinks about suchthings in terms of the equivalence FVectR rarr MatR itself In other words this equivalence isessentially ldquocooked uprdquo for the sake of giving an example but it does not really give any insightinto the actual mathematics of linear algebra

So with that in mind we will describe another equivalence which actually does have some realmeaning in the sense of producing new ways of thinking about some mathematics Given a compactHausdorff space X we let C(X) denote the space of continuous complex-valued functions on X

C(X) = f X rarr C | f is continuous

We can equip the set C(X) with various additional structures First it is a complex vector spaceunder ordinary addition and scalar multiplication of functions Second we can multiply two func-tions together

(fg)(p) = f(p)g(p)

which turns C(X) into whatrsquos called an algebra over C (We wonrsquot give the definition of anldquoalgebrardquo but the point is that this multiplication is in some sense ldquocompatiblerdquo with the vectorspace structure) Next any element f of C(X) has a conjugate element f obtained by takingcomplex conjugates of the values of f

f(p) = f(p)

22

And finally we can equip C(X) with a norm via

983042f983042 = suppisinX

|f(p)|

where |middot| denotes complex absolute value (This supremum exists sinceX is compact by the ExtremeValue Theorem) All of these structures (vector space algebra multiplication conjugation norm)turn C(X) into whatrsquos called a Clowast-algebra We wonrsquot define this precisely but the definition encodesvarious ways in which these structures are compatible Actually C(X) is a unital commutative Clowast-algebra where ldquounitalrdquo means that the multiplication has an identity element (namely the constantfunction 1) and commutative means that fg = gf There is a natural notion of homomorphismbetween Clowast-algebras which are maps between them which behave nicely with all the structuresinvolved In particular given a continuous map f X rarr Y between compact Hausdorff spaces weget a Clowast-algebra homomorphism flowast C(Y ) rarr C(X) given by pre-composition with f flowast sendsg isin C(Y ) to g f isin C(X)

We thus have a contravariant functor CHaus rarr ucClowast-Alg from the category of compactHausdorff spaces and the category of unital commutative Clowast-algebras It turns out that anyunital commutative Clowast-algebra arises in this way in the sense that given any unital commutativeClowast-algebra B there exists a compact Hausdorff space X such that C(X) sim= B as Clowast-algebrasand that morphisms between these algebra arises from continuous maps in the manner describeabove This says that the functor above is essentially surjective full and it turns out that it isalso faithful so that it is an equivalence Thus all information about compact Hausdorff spaces iscompletely encoded in information about unital commtuative Clowast-algebras and indeed one couldlearn everything there is to know about a compact Hausdorff space X from its Clowast-algebra C(X)of functions The fact that the categories CHaus and ucClowast-Alg are equivalent is the startingpoint in the subject known as noncommutative geometry which wersquoll say something about a bitlater on If we drop the requirement that our algebras be unital it turns out that the categoryof commutative Clowast-algebras in general is equivalent to the category of locally compact Hausdorffspaces via essentially the same functor where we take as morphisms in the category of locallycompact Hausdorff spaces to be continuous functions which ldquovanish at infinityrdquo

We are only introducing Clowast-algebras to give an example of an interesting and actually usefulequivalence between categories but just for the fun of it wersquoll note the following Examples ofnoncommutative Clowast-algebras come from what are known as ldquoalgebras of bounded operators onHilbert spacesrdquo which show up in quantum mechanics as the things which describe physicallyobservable quantities like position momentum and energy (In fact all Clowast-algebras arise fromHilbert spaces in this way) Thus Clowast-algebras show up in mathematical formulations of quantummechanics and the notion of noncommutative geometry mentioned above is at the core of someattempts to understand the elusive concept known as ldquoquantum geometryrdquo

Yoneda Lemma The Yoneda Lemma states that given a (covariant) functor F C rarr Set whereC is locally small for any A isin Ob(C) there exists a bijection

F (A) sim= Nat(hA F )

where the right side denotes the set of natural transformations from hA to F A similar result holdsin the contravariant case where we get bijections F (A) sim= Nat(hA F ) (There is also a sense inwhich these bijections are themselves natural but wersquoll ignore this bit) The proof of the YonedaLemma was the topic of a student presentation and can be found in Section 22 of the book

Here we will get a bit philosophical and talk about what the point of the Yoneda Lemmaactually is and later the sense in which it says that we can view arbitrary functors as ldquonon-existent

23

objectsrdquo in a category First we mention what we will call the ldquofunctor of pointsrdquo approach tomathematics a mathematical object can be fully characterized by all morphisms into it That isknowledge of Mor(SA) for all S is equivalent to knowledge of A itself Indeed this is an idea wementioned previously in relation to the functors hAmdashwhich we previously said is referred to as beingthe functor of points of Amdashand now we seek to push this idea further In some examples literalpoints of A can be extracted from hA as in the case of the bijection X sim= hX(pt) = Map(ptX) fora topological space X or G sim= hG(Z) = Mor(Z G) for a group G Moreover in these examples therest of the (topological) structure onX or (group) structure on G can also be extracted from hX andhG respectively In general we can view Mor(SA) as in a sense being ldquopoints of A parameterizedby Srdquo which comes from the idea that in say Set Mor(SA) = AS is ldquothe product of S-manycopies of Ardquo so literally points of A ldquoparameterizedrdquo by S In various areas of mathematics anelement of Mor(SA) is thought of as being a ldquogeneralizedrdquo point of A so that the functor of pointshA detects generalized points of A

So an object A is fully encoded by the functor hA Now we claim that morphisms betweenobjects are also encoded by such functors Indeed in the case where F = hB is itself a functor ofpoints the Yoneda result Nat(hA F ) sim= F (A) becomes

Nat(hA hB) sim= hB(A) = Mor(AB)

which says precisely that morphisms between the objects A and B can be obtained solely fromthe functors hA and hB (This fact is one reason why one should prefer the contravariant functorhA to the covariant functor hA in the covariant case Nat(hA F ) sim= F (A) for F = hB becomesNat(hA hB) sim= hB(A) = Mor(BA) so that morphisms between two objects get reversed whencharacterizing them in terms of representable functors)

Pushing our philosophy further we now know that not only objects but also morphisms betweenobjects are fully encoded by the functors hA The final step is to note that we should not only careabout objects and morphisms between them but that an important part of the data of an objectA is also how arbitrary contravariant functors F C rarr Set behave on A and that this too can beobtained from hA precisely via the Yoneda result that F (A) sim= Nat(hA F ) So objects morphismsbetween objects and how arbitrary functors behave on objects are all encoded in knowledge ofthe functors hA so that these functors truly do capture in the most general sense possible allinformation about A

Functors as nonexistent objects The discussion above is all well and good but it still doesnrsquotquite say why we should care about the fact that we can study on object A by studying hA insteadin that we still havenrsquot spoken about what type of insight about A can be gleamed from hA thatwasnrsquot readily apparent in A itself already Now we argue that the true point of the YonedaLemma is that it gives a way to think about arbitrary functors F Cop rarr Set as ldquoobjectsrdquo in C

itself or really as objects in an ldquoenlargementrdquo of C Certainly for those functors F sim= hA which arerepresentable we can indeed think about F as simply being the object A but now we claim thateven non-representable F can be thought of as being ldquononexistentrdquo objects in C The upshot is thatoften in mathematics we come across things which do not actually exist but the Yoneda Lemmagives a way to (indirectly) study such things anyway by interpreting them as functors instead

The starting point of course is that an actual object A is the same as hA as we have explainedabove Thus by analogy we can interpret a non-representable functor F as being an ldquoobjectrdquo in C

as long as we know what it means to ldquomaprdquo into F from other things which are honestly objects in CThat is if F were indeed representable then F (A) sim= Nat(hA F ) sim= Mor(AF ) would characterizethe functor of points of F (here we abuse notation by using the same letter F to denote both arepresentable functor and the object which represents it) so now we take Mor(AF ) = F (A) as

24

the definition of the ldquofunctor of pointsrdquo of the non-existent object F We are thus taking elementsof F (A) as being what it means to ldquomaprdquo from A into F even when F is not representable Thefact that the Yoneda Lemma gives a way to define maps from A into F thus says that F itself isequipped with the same data that a literal object of A would be equipped with so why shouldnrsquotwe interpret F as a being a type of ldquoobjectrdquo in C as well

To phrase this another way note that the Yoneda Lemma gives rise to a functor

C rarr Fun(CopSet)

where Fun(CopSet) denotes the category whose objects are functors from Cop to Set and morphismsare natural transformation defined on objects by A 983041rarr hA and on morphisms by the bijectionMor(AB) sim= Nat(hA hB) arising from the Yoneda Lemma That this is actually bijection saysthat this functor is full and faithful so that we can view C as being a subcategory of the functorcategory Fun(CopSet) (We cay say that C is isomorphic to the subcategory of Fun(CopSet)consisting of the functors hA or that it is equivalent to the subcategory of Fun(CopSet) consistingof representable functors This functor C rarr Fun(CopSet) is usually called the Yoneda embeddingof C into Fun(CopSet)) Thus we can view Fun(CopSet) as an ldquoenlargmentrdquo of C and the ideawe are trying to make sense of is that other objects in this ldquolargerrdquo category should be interpretedas ldquonon-existent objectsrdquo of C

Just how far can we actually push this idea that arbitrary functors Cop rarr Set should be thoughtof as being ldquoobjectsrdquo of C Pretty far in fact For instance often times in geometry one comesacross a type of ldquosmoothrdquo space on which some group acts and one is interested in doing ldquocalculusrdquoon the resulting quotient space which is the space of orbits under the given group action Theproblem is that such quotient spaces are very often (usually in fact) not ldquosmoothrdquo themselves sothat the tools from calculus are not readily available The solution is to nonetheless treat thisquotient as if it were ldquosmoothrdquo and attempt to do calculus anyway To make this precise thisldquonon-smoothrdquo space actually turns out to have a natural interpretation as a ldquosmooth functorrdquo andit is through this functor which we can view this non-smooth space as actually being smooth afterall so we are forced to expand our notion of what we mean by ldquosmooth spacerdquo by allowing thingswhich are not spaces in the usual sense which certainly leads to more abstraction but the benefitis that it actually makes sense to do calculus on such ldquospacesrdquo interpreted as functors

The discussion above is pretty vague so here is one final more concrete example Recall thatequivalence

CHausop rarr ucClowast-Alg

between the category of compact Hausdorff spaces and the category of unital commutative Clowast-algebras mentioned at the start We now claim that a non-commutative (unital) Clowast-algebra shouldbe interpreted as the algebra of functions on some non-existent ldquononcommutative compact Haus-dorff spacerdquo and it is this idea which the subject of noncommutative geometry makes precise Solet B be a non-commutative Clowast-algebra Then the functor of points hB gives a functor

ucClowast-Alg rarr Set

which sends C(X) (where X is some compact Hausdorff space) to Hom(C(X) B) where Hom de-notes homomorphisms of Clowast-algebras whatever that means Composing with the previous functorwe get a functor wersquoll denote by B itself

B CHausop rarr Set

which on objects at least is given by X 983041rarr Hom(C(X) B) This functor is not representable sinceB is not commutative but the philosophy espoused above suggests we should think of it as giving a

25

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

requires additional data which prevents the identity functor from being ldquonaturally isomorphicrdquo tothe single dual functor

Some history At this point it makes sense to say a bit about the interesting origins of categorytheory in the 1950rsquos Eilenberg and Mac Lane where studying cohomology (or possible homologyone of those two) in algebraic topology which associates algebraic data (a group ring or similar)to a topological space in a way which encodes some of the topology They had two cohomologicalconstructions which ended up proving isomorphic results but they noticed that not only werethe resulting objects the same but the actual constructions themselves were in some sense theldquosamerdquo They thus then needed a way to formalize this notion and invented the concept of anatural transformation to do so where they interpreted two constructions as being the same asbeing compatible with all morphisms

But this then leads to the question what types of objects should a natural transformationoccur between Eilenberg and Mac Lane then had to invent the notion of a ldquofunctorrdquo to describewhat a natural transformation went from and what it went to In their motivating setting theyinvented the notion of a functor to describe more formally properties which their cohomologicalconstructions were supposed to satisfy But this then leads to the question what types of objectsshould a functor map between In other words what type of data should a functor (cohomologicalconstruction in their example) take as input and should it output They finally had to thus inventthe notion of a ldquocategoryrdquo to describe the source and target of a functor so that their cohomologicalconstructions could be interpreted as functors from a category of topological spaces to a categoryof algebraic objects

Thus historically natural transformations came first then functors and then categories Aswith most things in mathematics categories thus arose from attempts to encode what was beingseen in some concrete examples in a more general and formal setting

Representable functors Representable functors are defined in Section 21 of the book but wersquollgive the required definitions here in a hopefully simpler to digest manner The point is that anyobject A in a (locally small) category C gives two ldquonaturalrdquo functors C rarr Set one covariant andthe other contravariant which as wersquoll see pretty much encode all data about A itself We definehA C rarr Set to be the covariant functor defined by

hA(B) = Mor(AB) on objects and

hA(Bfminusrarr C) = the map Mor(AB)

hAfminusminusrarr Mor(AC) defined by g 983041rarr f g

We define hA C rarr Set to the contravariant functor defined by

hA(B) = Mor(BA) on objects and

hA(Bfminusrarr C) = the map Mor(CA)

hAfminusminusrarr Mor(BA) defined by g 983041rarr g f

(The functor hA is called the functor of points of A and wersquoll usually think of it as being a covariantfunctor hA Cop rarr Set The functor hA does not have a standard name) So hA is defined bymorphisms from A and hA is defined by morphisms into A Wersquoll see later the sense in which suchfunctors encode all information about the object A

A functor F C rarr Set is representable if is naturally isomorphic to such a functor so F sim= hA

for some A in the covariant case and F sim= hA for some A in the contravariant case The idea wersquollbuild towards is that we can essentially treat such a functor as being A itself so that representable

16

functors can be thought as being actual objects in C For now here is a key example wersquove seenbefore which we can now formulate using the language of representability

Given AB isin Ob(C) let F C rarr Set be the contravariant functor defined by

F (C) = Mor(CA)timesMor(CB) and F (Cfminusrarr D) = [(g k) 983041rarr (g f k f)]

where (g k) 983041rarr (g f k f) gives a map Mor(DA) timesMor(DB) rarr Mor(CA) timesMor(CB) ieF (D) rarr F (C) In order for this functor to be representable requires the existence of an object inC satisfying

Mor(CA)timesMor(CB) sim= Mor(C representing object)

in a way which is compatible with all morphisms but we have actually already seen this notionelsewhere it is essentially the definition of a product AtimesB for A and B Indeed we claim that Fis naturally isomorphic to hAtimesB as we now show

First to define a natural isomorphism hAtimesBsim= F requires for each C isin Ob(C) a choice of a

bijection (ie isomorphism in Set)

ηC hAtimesB(C)sim=minusrarr F (C)

which in our case looks like

ηC Mor(CAtimesB))sim=minusrarr Mor(CA)timesMor(CB)

Take this to be the map

(Cfminusrarr AtimesB) 983041rarr (pA f pB f)

where pA A times B rarr A and pB A times B rarr B are the two projection morphisms in the definitionof a product The definition of product implies that this map is invertible since given k C rarr Aand ℓ C rarr B there exists a unique h C rarr A times B such that k = pA h and ℓ = pB h Forthese bijections to define a natural isomorphism hAtimesB

sim= F requires that they be compatible withall morphisms in C which means that given any g C rarr D in C the following diagram shouldcommute

hAtimesB(C) hAtimesB(D)

F (C) F (D)

hAtimesB(g)

ηC ηD

Fg

which in the case at hand looks like

Mor(CAtimesB) Mor(DAtimesB)

Mor(CA)timesMor(CB) Mor(DA)timesMor(DB)

minus g

ηC ηD

(minus gminus g)

Take an element k D rarr AtimesB in the upper right Applying the map on top gives the elementk g in the upper left and then applying ηC on the left side gives

(pA (k g) pB (k g))

17

in the lower left Instead starting with k D rarr AtimesB in the upper right applying ηD on the rightgives (pA k pB k) and applying the map on the bottom gives

((pA k) g (pB k) g)

in the lower left which is the same as the previous element in the lower left by the associativity ofcomposition Hence the isomorphisms ηC do define a natural isomorphism between hAtimesB and F Again the point is that you can then interpret the functor F as being the ldquosamerdquo as the productAtimesB since both encode the same data

More Representable Examples

Coproducts As in the case of products the definition of coproducts can also be phrased in termsof a representable functor Given objects A and B in C let F C rarr Set be the covariant functordefined by

F (C) = Mor(AC)timesMor(BC) and F (Cfminusrarr D) = [(g k) 983041rarr (f g f k)]

where the latter is a function Mor(AC)timesMor(BC) rarr Mor(AD)timesMor(BD) This functor isrepresentable if and only if a coproduct A ⊔ B for A and B exists and the representing object isthat coproduct the key is the requirement that we have bijections

Mor(AC)timesMor(BC) sim= Mor(A ⊔BC)

for all C which is essentially the defining property a coproduct for A and B should have

Power set functor contravariant version Let P Set rarr Set be the contravariant powerset functor defined on objects by sending a set to its power set and on morphisms by sendingf A rarr B to the preimage function Pf P (B) rarr P (A) given by S 983041rarr fminus1(S) In order for this tobe representable requires the existence of a set X with natural bijections

P (A) sim= Mor(AX)

for all sets A that is a set X so that mapping into X is the same data as specifying a subset ofthe domain We take X = 0 1 to be a two-element set and the required bijections

ηA P (A) rarr Mor(A 0 1) to be given by S 983041rarr χS

where XS A rarr 0 1 is the indicator or characteristic function S defined by sending elementsof S to 1 and elements of A minus S to 0 The inverse of ηA sends a function g A rarr 0 1 to thepreimage gminus1(1) sube A

To verify that the bijections ηA define the data of a natural isomorphism η P rArr h01 wecheck the commutativity of some diagrams Given a function f A rarr B consider

P (A) P (B)

h01(A) = Mor(A 0 1) h01(B) = Mor(B 0 1)

Pf = fminus1

ηA ηB

minus f

18

Take S isin P (B) in the upper right Applying the map on top gives fminus1(S) isin P (A) and thenapplying the map on the left gives χfminus1(S) isin Mor(A 0 1) On the other hand applying the mapon the right to S isin P (B) gives χS isin Mor(B 0 1) and then applying the map on the bottomgives χS f isin Mor(A 0 1) The two resulting functions χfminus1(S)χS f in the lower left are

χfminus1(S)(x) =

9830831 x isin fminus1(S)

0 otherwiseand χS(f(x)) =

9830831 f(x) isin S

0 otherwise

so these are the same since x isin fminus1(S) if and only if f(x) isin S Thus the diagram above commutesso P is naturally isomorphic to h01 and hence P is representable

Power set functor covariant version Now consider the covariant power set functor stilldenoted by P Set rarr Set which again sends a set to its power set but now sends a functionf A rarr B to the image function Pf P (A) rarr P (B) defined by S 983041rarr f(S) In order for this tobe representable there would have to exist a set X such that P sim= hX which translates into havingnatural bijections

P (S) sim= Mor(XS)

for all sets S However in this case there is no natural choice for X as there is no natural way todetect subsets of S by mapping into S as opposed to the case of P (S) sim= Mor(S 0 1)

That no such X exists can be made precise using a simple argument when S = empty P (S) hascardinality 1 soX would have to be empty as well since otherwise Mor(XS) would have cardinality0 but if X is empty then Mor(empty S) always cardinality 1 regardless of S which is clearly not trueof P (S) Thus the covariant power set functor is not representable

Forgetful on Top The forgetful functor F Top rarr Set is representable Indeed a representingobject would have to satisfy

X sim= Map(objectX)

as sets for any topological space X (where Map denotes the set of continuous maps) and it iseasy to see that a single point satisfies this property a continuous map f pt rarr X is completelydetermined by the element f(pt) of X and any such element gives a continuous map in this wayOf course checking that F sim= hpt requires checking that various diagrams commute but this issimple to work out in this case so we omit it here

Extracting a topology With an eye towards the idea that the functors hA hA should captureall information about an object A in a category note that the result above shows that in TophX first of all captures all information about X as a set since the set X can be extracted fromhX(pt) = Map(ptX)

We claim that hX actually captures all information about X as a topological space since thetopology on X can be extracted from hX in the following way As we saw in the power set examplewe have a bijection

P (X) sim= Mor(X 0 1)

Now equip 0 1 with the topology in which 1 is open but 0 is not Then a continuous mapf X rarr 0 1 is fully determined by the preimage fminus1(1) since to be continuous only requiresthat this single preimage be open in X Moreoever any open subset of X arises in this way bytaking f to be the indicator function of that open set Thus we get a bijection

Op(X) sim= Map(X 0 1)

19

where Op(X) denotes the set of open subsets of X ie the topology on X Hence the topology onX can be extracted from hX as claimed so hX and hX do encode all information about X

This fact can also be interpreted as saying that the contravariant functor which sends a topo-logical space X to its collection of open subsets (and which sends a continuous map to the mapinduced by taking preimages) is representable with representing object 0 1 equipped with thetopology given above This will be worked out in a homework problem

Forgetful on Grp The forgetful functor Grp rarr Set is also representable In this case arepresenting object should be a group giving set-theoretic bijections

G sim= Hom(objectG)

for all groups G where Hom denotes the set of group homomorphisms The group Z works

G sim= Hom(Z G)

since a homomorphism φ Z rarr G is completely determined by φ(1) since Z is generated by 1 andthere are no restrictions on what φ(1) isin G can actually be The inverse of the desired bijection isthus φ 983041rarr φ(1) and it is simple to check that morphisms behave appropriately so that the forgetfulfunctor in this setting is indeed isomoprhic to hZ

Forgetful on Ring The forgetful functor Ring rarr Set is also representable where by Ring wemean the category of rings with identity elements and where morphisms (ie ring homomorphisms)are required to preserve identities Here we need an object satisfying

R sim= Hom(object R)

for all R isin Ob(Ring) and we can see that Z[x] the ring of polynomials in the variable x withcoefficients in Z works Indeed given a ring homomorphism φ Z[x] rarr R the fact that φ isrequired to preserve identities fully determines φ(n) for any n isin Z (since φ(n) = nφ(1)) and hencesince φ preserves multiplication its behavior is fully determined by φ(x) which can be any elementof R The inverse of the bijection we want is thus φ 983041rarr φ(x) and it will follow that this forgetfulfunctor is isomorphic to hZ[x]

Equivalences between Categories

Definitions iso and equiv Now that we have a notion of a ldquomorphismrdquo between categorieswe can talk about the sense in which two categories are the ldquosamerdquo The first possible definitionis the ldquoobviousrdquo one you might expect given the definition of ldquoisomorphismrdquo wersquove seen in everyother context until now we say that C is isomorphic to D if there exist functors F C rarr D andG D rarr C such that G F = idC and F G = idD where idC and idD are identity functors

However this definition turns out to be too restrictive Indeed the requirement that the givencompositions equal identity functors says that given an object A in C the object G(F (A)) beliterally the same as A itself and similarly for objects in D But we know that two objects in acategory can be thought of as being the ldquosamerdquo without being literally the same as long as theyare isomorphic For instance productscoproducts are not unique in the literal sense but only inthe sense that they are isomorphic in a unique way To take another example the definition ofa functor being representable does not say that F (B) literally be the same as hB(A) only thatthey be isomorphic for sure the power set P (S) of a set S is not literally the same as the set offunctions S rarr 0 1mdashit is only the ldquosamerdquo in the sense that they encode the same data

20

Thus it makes sense to relax the requirement that A andG(F (A)) in the definition of isomorphiccategories be equal to the requirement that they be isomorphic This leads to the following betterdefinition of two categories being the ldquosamerdquo C is equivalent to D if there exist functors F C rarr D

and G D rarr C such that G F sim= idC and F G sim= idD So we require only that for instance thefunctor G F be naturally isomorphic to the identity functor on C which says that for any objectA G(F (A)) be isomorphic to A in a natural way Wersquoll see that this truly is the better notionof what it means for two categories to be the ldquosamerdquo and it reflects the idea that all categoricalproperties of one category can be transported to the other

Alternate characterization of equivalences A functor F C rarr D implementing an equiv-alence between categories is called unsurprisingly and equivalence and the ldquoinverserdquo functorG D rarr C is still referred to as being its inverse even though it is not technically an ldquoinverserdquoin the sense that compositions give literal identities But as in the case of Set where ldquoinvertiblerdquocan be characterized without reference to an inverse as ldquoinjectiverdquo and ldquosurjectiverdquo so too canequivalences between categories be characterized without explicit reference to an inverse

A functor F C rarr D is an equivalence if and only it is full faithful and essentiallysurjective

We have seen the notions of full and faithful before and essentially surjective means that any objectin D is isomorphic to something in the image of F for all D isin Ob(D) there exits C isin Ob(C) suchthat F (C) sim= D Justifying this result was the topic of a student presentation and can be foundin the book in Section 15 (Technically to construct an ldquoinverserdquo of F requires that we work withsmall categories but whatever) The bulk of the real work goes into showing that the ldquoinverserdquofunctor constructed is actually a functor which comes down to some abstract nonsense argumentvia commutative diagrams

Equivalences preserve everything We previously saw the notions of full faithful and es-sentially surjective back when thinking about what properties a functor could have which wouldguarantee it sent products to products so the result we proved back then was that equivalencespreserve products Similarly equivalences preserve coproducts and the proof is basically the sameafter reversing arrows A clean way of deriving this is as follows if F C rarr D is an equivalencethen so is F op Cop rarr Dop so since F op preserves products F preserves coproducts becausecoproducts in a category become products in the opposite category

In general an equivalence between categories will preserve whatever categorical notions youwant monomorphisms epimorphisms initial objects terminal objects etc so that equivalentcategories really do have essentially the ldquosamerdquo properties

Equivalence example Here is a first example (also found in the book) of equivalent categorieswhich are not isomorphic Let FVectR be the category whose objects are finite-dimensional realvector spaces and whose morphisms are linear transformations Let MatR be the category ofmatrices defined as follows

bull objects in MatR are nonnegative integers

bull a morphism n rarr m is an mtimes n matrix with real entries

bull composition in MatR is given by matrix multiplication given B n rarr m and A m rarr kmdashsoB is an mtimes n matrix and A is a k timesm matrixmdashthe composition A B n rarr k is the k times nmatrix AB

bull identity morphisms in MatR are given by identity matrices

21

We claim that FVectR and MatR are equivalent First note that they certainly cannot be isomo-prhic isomorphisms in MatR exist only between two objects which are literally the same (ie anmtimesn matrix can be invertible only when m = n) but isomorphisms in FVectn can exist betweenobjects which are not literally the same So there are in a sense not enough objects in MatR toallow it to be isomorphic to FVectR

To construct an equivalence choose one and for all an isomorphism TV V rarr RdimV forevery object in FVectR or equivalently choose once and for all a basis for every V DefineF FVectR rarr MatR by

V 983041rarr dimV on objects and

(T V rarr W ) 983041rarr (the standard matrix of SW T Sminus1V RdimV rarr RdimW ) on morphisms

Equivalently F sends T to the matrix of T relative to the chosen bases for V and W It isstraightforward to check that this is a functor The inverse functor G MatR rarr VectR is givenby

n 983041rarr Rn on objects and

(mtimes n matrix A) 983041rarr (the linear transformation A Rn rarr Rm induced by A) on morphisms

One can check that G is a functor and that F G and G F are naturally isomorphic to identitiesor instead one can argue that F alone is full faithful and essentially surjective The point is thatG(F (V )) gives back RdimV so not literally V itself but only something isomorphic to V which iswhy this gives an equivalence of categories and not an isomorphism

Yoneda Lemma Functors as Objects

Compact Hausdorff spaces and Clowast-algebras Last time we gave an example of equivalentcategories which were not isomorphic but in many ways that example is unsatisfactory when doinglinear algebra people for sure often work with matrix representations of linear transformations asthe example from last would suggest but no one in their right mind literally thinks about suchthings in terms of the equivalence FVectR rarr MatR itself In other words this equivalence isessentially ldquocooked uprdquo for the sake of giving an example but it does not really give any insightinto the actual mathematics of linear algebra

So with that in mind we will describe another equivalence which actually does have some realmeaning in the sense of producing new ways of thinking about some mathematics Given a compactHausdorff space X we let C(X) denote the space of continuous complex-valued functions on X

C(X) = f X rarr C | f is continuous

We can equip the set C(X) with various additional structures First it is a complex vector spaceunder ordinary addition and scalar multiplication of functions Second we can multiply two func-tions together

(fg)(p) = f(p)g(p)

which turns C(X) into whatrsquos called an algebra over C (We wonrsquot give the definition of anldquoalgebrardquo but the point is that this multiplication is in some sense ldquocompatiblerdquo with the vectorspace structure) Next any element f of C(X) has a conjugate element f obtained by takingcomplex conjugates of the values of f

f(p) = f(p)

22

And finally we can equip C(X) with a norm via

983042f983042 = suppisinX

|f(p)|

where |middot| denotes complex absolute value (This supremum exists sinceX is compact by the ExtremeValue Theorem) All of these structures (vector space algebra multiplication conjugation norm)turn C(X) into whatrsquos called a Clowast-algebra We wonrsquot define this precisely but the definition encodesvarious ways in which these structures are compatible Actually C(X) is a unital commutative Clowast-algebra where ldquounitalrdquo means that the multiplication has an identity element (namely the constantfunction 1) and commutative means that fg = gf There is a natural notion of homomorphismbetween Clowast-algebras which are maps between them which behave nicely with all the structuresinvolved In particular given a continuous map f X rarr Y between compact Hausdorff spaces weget a Clowast-algebra homomorphism flowast C(Y ) rarr C(X) given by pre-composition with f flowast sendsg isin C(Y ) to g f isin C(X)

We thus have a contravariant functor CHaus rarr ucClowast-Alg from the category of compactHausdorff spaces and the category of unital commutative Clowast-algebras It turns out that anyunital commutative Clowast-algebra arises in this way in the sense that given any unital commutativeClowast-algebra B there exists a compact Hausdorff space X such that C(X) sim= B as Clowast-algebrasand that morphisms between these algebra arises from continuous maps in the manner describeabove This says that the functor above is essentially surjective full and it turns out that it isalso faithful so that it is an equivalence Thus all information about compact Hausdorff spaces iscompletely encoded in information about unital commtuative Clowast-algebras and indeed one couldlearn everything there is to know about a compact Hausdorff space X from its Clowast-algebra C(X)of functions The fact that the categories CHaus and ucClowast-Alg are equivalent is the startingpoint in the subject known as noncommutative geometry which wersquoll say something about a bitlater on If we drop the requirement that our algebras be unital it turns out that the categoryof commutative Clowast-algebras in general is equivalent to the category of locally compact Hausdorffspaces via essentially the same functor where we take as morphisms in the category of locallycompact Hausdorff spaces to be continuous functions which ldquovanish at infinityrdquo

We are only introducing Clowast-algebras to give an example of an interesting and actually usefulequivalence between categories but just for the fun of it wersquoll note the following Examples ofnoncommutative Clowast-algebras come from what are known as ldquoalgebras of bounded operators onHilbert spacesrdquo which show up in quantum mechanics as the things which describe physicallyobservable quantities like position momentum and energy (In fact all Clowast-algebras arise fromHilbert spaces in this way) Thus Clowast-algebras show up in mathematical formulations of quantummechanics and the notion of noncommutative geometry mentioned above is at the core of someattempts to understand the elusive concept known as ldquoquantum geometryrdquo

Yoneda Lemma The Yoneda Lemma states that given a (covariant) functor F C rarr Set whereC is locally small for any A isin Ob(C) there exists a bijection

F (A) sim= Nat(hA F )

where the right side denotes the set of natural transformations from hA to F A similar result holdsin the contravariant case where we get bijections F (A) sim= Nat(hA F ) (There is also a sense inwhich these bijections are themselves natural but wersquoll ignore this bit) The proof of the YonedaLemma was the topic of a student presentation and can be found in Section 22 of the book

Here we will get a bit philosophical and talk about what the point of the Yoneda Lemmaactually is and later the sense in which it says that we can view arbitrary functors as ldquonon-existent

23

objectsrdquo in a category First we mention what we will call the ldquofunctor of pointsrdquo approach tomathematics a mathematical object can be fully characterized by all morphisms into it That isknowledge of Mor(SA) for all S is equivalent to knowledge of A itself Indeed this is an idea wementioned previously in relation to the functors hAmdashwhich we previously said is referred to as beingthe functor of points of Amdashand now we seek to push this idea further In some examples literalpoints of A can be extracted from hA as in the case of the bijection X sim= hX(pt) = Map(ptX) fora topological space X or G sim= hG(Z) = Mor(Z G) for a group G Moreover in these examples therest of the (topological) structure onX or (group) structure on G can also be extracted from hX andhG respectively In general we can view Mor(SA) as in a sense being ldquopoints of A parameterizedby Srdquo which comes from the idea that in say Set Mor(SA) = AS is ldquothe product of S-manycopies of Ardquo so literally points of A ldquoparameterizedrdquo by S In various areas of mathematics anelement of Mor(SA) is thought of as being a ldquogeneralizedrdquo point of A so that the functor of pointshA detects generalized points of A

So an object A is fully encoded by the functor hA Now we claim that morphisms betweenobjects are also encoded by such functors Indeed in the case where F = hB is itself a functor ofpoints the Yoneda result Nat(hA F ) sim= F (A) becomes

Nat(hA hB) sim= hB(A) = Mor(AB)

which says precisely that morphisms between the objects A and B can be obtained solely fromthe functors hA and hB (This fact is one reason why one should prefer the contravariant functorhA to the covariant functor hA in the covariant case Nat(hA F ) sim= F (A) for F = hB becomesNat(hA hB) sim= hB(A) = Mor(BA) so that morphisms between two objects get reversed whencharacterizing them in terms of representable functors)

Pushing our philosophy further we now know that not only objects but also morphisms betweenobjects are fully encoded by the functors hA The final step is to note that we should not only careabout objects and morphisms between them but that an important part of the data of an objectA is also how arbitrary contravariant functors F C rarr Set behave on A and that this too can beobtained from hA precisely via the Yoneda result that F (A) sim= Nat(hA F ) So objects morphismsbetween objects and how arbitrary functors behave on objects are all encoded in knowledge ofthe functors hA so that these functors truly do capture in the most general sense possible allinformation about A

Functors as nonexistent objects The discussion above is all well and good but it still doesnrsquotquite say why we should care about the fact that we can study on object A by studying hA insteadin that we still havenrsquot spoken about what type of insight about A can be gleamed from hA thatwasnrsquot readily apparent in A itself already Now we argue that the true point of the YonedaLemma is that it gives a way to think about arbitrary functors F Cop rarr Set as ldquoobjectsrdquo in C

itself or really as objects in an ldquoenlargementrdquo of C Certainly for those functors F sim= hA which arerepresentable we can indeed think about F as simply being the object A but now we claim thateven non-representable F can be thought of as being ldquononexistentrdquo objects in C The upshot is thatoften in mathematics we come across things which do not actually exist but the Yoneda Lemmagives a way to (indirectly) study such things anyway by interpreting them as functors instead

The starting point of course is that an actual object A is the same as hA as we have explainedabove Thus by analogy we can interpret a non-representable functor F as being an ldquoobjectrdquo in C

as long as we know what it means to ldquomaprdquo into F from other things which are honestly objects in CThat is if F were indeed representable then F (A) sim= Nat(hA F ) sim= Mor(AF ) would characterizethe functor of points of F (here we abuse notation by using the same letter F to denote both arepresentable functor and the object which represents it) so now we take Mor(AF ) = F (A) as

24

the definition of the ldquofunctor of pointsrdquo of the non-existent object F We are thus taking elementsof F (A) as being what it means to ldquomaprdquo from A into F even when F is not representable Thefact that the Yoneda Lemma gives a way to define maps from A into F thus says that F itself isequipped with the same data that a literal object of A would be equipped with so why shouldnrsquotwe interpret F as a being a type of ldquoobjectrdquo in C as well

To phrase this another way note that the Yoneda Lemma gives rise to a functor

C rarr Fun(CopSet)

where Fun(CopSet) denotes the category whose objects are functors from Cop to Set and morphismsare natural transformation defined on objects by A 983041rarr hA and on morphisms by the bijectionMor(AB) sim= Nat(hA hB) arising from the Yoneda Lemma That this is actually bijection saysthat this functor is full and faithful so that we can view C as being a subcategory of the functorcategory Fun(CopSet) (We cay say that C is isomorphic to the subcategory of Fun(CopSet)consisting of the functors hA or that it is equivalent to the subcategory of Fun(CopSet) consistingof representable functors This functor C rarr Fun(CopSet) is usually called the Yoneda embeddingof C into Fun(CopSet)) Thus we can view Fun(CopSet) as an ldquoenlargmentrdquo of C and the ideawe are trying to make sense of is that other objects in this ldquolargerrdquo category should be interpretedas ldquonon-existent objectsrdquo of C

Just how far can we actually push this idea that arbitrary functors Cop rarr Set should be thoughtof as being ldquoobjectsrdquo of C Pretty far in fact For instance often times in geometry one comesacross a type of ldquosmoothrdquo space on which some group acts and one is interested in doing ldquocalculusrdquoon the resulting quotient space which is the space of orbits under the given group action Theproblem is that such quotient spaces are very often (usually in fact) not ldquosmoothrdquo themselves sothat the tools from calculus are not readily available The solution is to nonetheless treat thisquotient as if it were ldquosmoothrdquo and attempt to do calculus anyway To make this precise thisldquonon-smoothrdquo space actually turns out to have a natural interpretation as a ldquosmooth functorrdquo andit is through this functor which we can view this non-smooth space as actually being smooth afterall so we are forced to expand our notion of what we mean by ldquosmooth spacerdquo by allowing thingswhich are not spaces in the usual sense which certainly leads to more abstraction but the benefitis that it actually makes sense to do calculus on such ldquospacesrdquo interpreted as functors

The discussion above is pretty vague so here is one final more concrete example Recall thatequivalence

CHausop rarr ucClowast-Alg

between the category of compact Hausdorff spaces and the category of unital commutative Clowast-algebras mentioned at the start We now claim that a non-commutative (unital) Clowast-algebra shouldbe interpreted as the algebra of functions on some non-existent ldquononcommutative compact Haus-dorff spacerdquo and it is this idea which the subject of noncommutative geometry makes precise Solet B be a non-commutative Clowast-algebra Then the functor of points hB gives a functor

ucClowast-Alg rarr Set

which sends C(X) (where X is some compact Hausdorff space) to Hom(C(X) B) where Hom de-notes homomorphisms of Clowast-algebras whatever that means Composing with the previous functorwe get a functor wersquoll denote by B itself

B CHausop rarr Set

which on objects at least is given by X 983041rarr Hom(C(X) B) This functor is not representable sinceB is not commutative but the philosophy espoused above suggests we should think of it as giving a

25

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

functors can be thought as being actual objects in C For now here is a key example wersquove seenbefore which we can now formulate using the language of representability

Given AB isin Ob(C) let F C rarr Set be the contravariant functor defined by

F (C) = Mor(CA)timesMor(CB) and F (Cfminusrarr D) = [(g k) 983041rarr (g f k f)]

where (g k) 983041rarr (g f k f) gives a map Mor(DA) timesMor(DB) rarr Mor(CA) timesMor(CB) ieF (D) rarr F (C) In order for this functor to be representable requires the existence of an object inC satisfying

Mor(CA)timesMor(CB) sim= Mor(C representing object)

in a way which is compatible with all morphisms but we have actually already seen this notionelsewhere it is essentially the definition of a product AtimesB for A and B Indeed we claim that Fis naturally isomorphic to hAtimesB as we now show

First to define a natural isomorphism hAtimesBsim= F requires for each C isin Ob(C) a choice of a

bijection (ie isomorphism in Set)

ηC hAtimesB(C)sim=minusrarr F (C)

which in our case looks like

ηC Mor(CAtimesB))sim=minusrarr Mor(CA)timesMor(CB)

Take this to be the map

(Cfminusrarr AtimesB) 983041rarr (pA f pB f)

where pA A times B rarr A and pB A times B rarr B are the two projection morphisms in the definitionof a product The definition of product implies that this map is invertible since given k C rarr Aand ℓ C rarr B there exists a unique h C rarr A times B such that k = pA h and ℓ = pB h Forthese bijections to define a natural isomorphism hAtimesB

sim= F requires that they be compatible withall morphisms in C which means that given any g C rarr D in C the following diagram shouldcommute

hAtimesB(C) hAtimesB(D)

F (C) F (D)

hAtimesB(g)

ηC ηD

Fg

which in the case at hand looks like

Mor(CAtimesB) Mor(DAtimesB)

Mor(CA)timesMor(CB) Mor(DA)timesMor(DB)

minus g

ηC ηD

(minus gminus g)

Take an element k D rarr AtimesB in the upper right Applying the map on top gives the elementk g in the upper left and then applying ηC on the left side gives

(pA (k g) pB (k g))

17

in the lower left Instead starting with k D rarr AtimesB in the upper right applying ηD on the rightgives (pA k pB k) and applying the map on the bottom gives

((pA k) g (pB k) g)

in the lower left which is the same as the previous element in the lower left by the associativity ofcomposition Hence the isomorphisms ηC do define a natural isomorphism between hAtimesB and F Again the point is that you can then interpret the functor F as being the ldquosamerdquo as the productAtimesB since both encode the same data

More Representable Examples

Coproducts As in the case of products the definition of coproducts can also be phrased in termsof a representable functor Given objects A and B in C let F C rarr Set be the covariant functordefined by

F (C) = Mor(AC)timesMor(BC) and F (Cfminusrarr D) = [(g k) 983041rarr (f g f k)]

where the latter is a function Mor(AC)timesMor(BC) rarr Mor(AD)timesMor(BD) This functor isrepresentable if and only if a coproduct A ⊔ B for A and B exists and the representing object isthat coproduct the key is the requirement that we have bijections

Mor(AC)timesMor(BC) sim= Mor(A ⊔BC)

for all C which is essentially the defining property a coproduct for A and B should have

Power set functor contravariant version Let P Set rarr Set be the contravariant powerset functor defined on objects by sending a set to its power set and on morphisms by sendingf A rarr B to the preimage function Pf P (B) rarr P (A) given by S 983041rarr fminus1(S) In order for this tobe representable requires the existence of a set X with natural bijections

P (A) sim= Mor(AX)

for all sets A that is a set X so that mapping into X is the same data as specifying a subset ofthe domain We take X = 0 1 to be a two-element set and the required bijections

ηA P (A) rarr Mor(A 0 1) to be given by S 983041rarr χS

where XS A rarr 0 1 is the indicator or characteristic function S defined by sending elementsof S to 1 and elements of A minus S to 0 The inverse of ηA sends a function g A rarr 0 1 to thepreimage gminus1(1) sube A

To verify that the bijections ηA define the data of a natural isomorphism η P rArr h01 wecheck the commutativity of some diagrams Given a function f A rarr B consider

P (A) P (B)

h01(A) = Mor(A 0 1) h01(B) = Mor(B 0 1)

Pf = fminus1

ηA ηB

minus f

18

Take S isin P (B) in the upper right Applying the map on top gives fminus1(S) isin P (A) and thenapplying the map on the left gives χfminus1(S) isin Mor(A 0 1) On the other hand applying the mapon the right to S isin P (B) gives χS isin Mor(B 0 1) and then applying the map on the bottomgives χS f isin Mor(A 0 1) The two resulting functions χfminus1(S)χS f in the lower left are

χfminus1(S)(x) =

9830831 x isin fminus1(S)

0 otherwiseand χS(f(x)) =

9830831 f(x) isin S

0 otherwise

so these are the same since x isin fminus1(S) if and only if f(x) isin S Thus the diagram above commutesso P is naturally isomorphic to h01 and hence P is representable

Power set functor covariant version Now consider the covariant power set functor stilldenoted by P Set rarr Set which again sends a set to its power set but now sends a functionf A rarr B to the image function Pf P (A) rarr P (B) defined by S 983041rarr f(S) In order for this tobe representable there would have to exist a set X such that P sim= hX which translates into havingnatural bijections

P (S) sim= Mor(XS)

for all sets S However in this case there is no natural choice for X as there is no natural way todetect subsets of S by mapping into S as opposed to the case of P (S) sim= Mor(S 0 1)

That no such X exists can be made precise using a simple argument when S = empty P (S) hascardinality 1 soX would have to be empty as well since otherwise Mor(XS) would have cardinality0 but if X is empty then Mor(empty S) always cardinality 1 regardless of S which is clearly not trueof P (S) Thus the covariant power set functor is not representable

Forgetful on Top The forgetful functor F Top rarr Set is representable Indeed a representingobject would have to satisfy

X sim= Map(objectX)

as sets for any topological space X (where Map denotes the set of continuous maps) and it iseasy to see that a single point satisfies this property a continuous map f pt rarr X is completelydetermined by the element f(pt) of X and any such element gives a continuous map in this wayOf course checking that F sim= hpt requires checking that various diagrams commute but this issimple to work out in this case so we omit it here

Extracting a topology With an eye towards the idea that the functors hA hA should captureall information about an object A in a category note that the result above shows that in TophX first of all captures all information about X as a set since the set X can be extracted fromhX(pt) = Map(ptX)

We claim that hX actually captures all information about X as a topological space since thetopology on X can be extracted from hX in the following way As we saw in the power set examplewe have a bijection

P (X) sim= Mor(X 0 1)

Now equip 0 1 with the topology in which 1 is open but 0 is not Then a continuous mapf X rarr 0 1 is fully determined by the preimage fminus1(1) since to be continuous only requiresthat this single preimage be open in X Moreoever any open subset of X arises in this way bytaking f to be the indicator function of that open set Thus we get a bijection

Op(X) sim= Map(X 0 1)

19

where Op(X) denotes the set of open subsets of X ie the topology on X Hence the topology onX can be extracted from hX as claimed so hX and hX do encode all information about X

This fact can also be interpreted as saying that the contravariant functor which sends a topo-logical space X to its collection of open subsets (and which sends a continuous map to the mapinduced by taking preimages) is representable with representing object 0 1 equipped with thetopology given above This will be worked out in a homework problem

Forgetful on Grp The forgetful functor Grp rarr Set is also representable In this case arepresenting object should be a group giving set-theoretic bijections

G sim= Hom(objectG)

for all groups G where Hom denotes the set of group homomorphisms The group Z works

G sim= Hom(Z G)

since a homomorphism φ Z rarr G is completely determined by φ(1) since Z is generated by 1 andthere are no restrictions on what φ(1) isin G can actually be The inverse of the desired bijection isthus φ 983041rarr φ(1) and it is simple to check that morphisms behave appropriately so that the forgetfulfunctor in this setting is indeed isomoprhic to hZ

Forgetful on Ring The forgetful functor Ring rarr Set is also representable where by Ring wemean the category of rings with identity elements and where morphisms (ie ring homomorphisms)are required to preserve identities Here we need an object satisfying

R sim= Hom(object R)

for all R isin Ob(Ring) and we can see that Z[x] the ring of polynomials in the variable x withcoefficients in Z works Indeed given a ring homomorphism φ Z[x] rarr R the fact that φ isrequired to preserve identities fully determines φ(n) for any n isin Z (since φ(n) = nφ(1)) and hencesince φ preserves multiplication its behavior is fully determined by φ(x) which can be any elementof R The inverse of the bijection we want is thus φ 983041rarr φ(x) and it will follow that this forgetfulfunctor is isomorphic to hZ[x]

Equivalences between Categories

Definitions iso and equiv Now that we have a notion of a ldquomorphismrdquo between categorieswe can talk about the sense in which two categories are the ldquosamerdquo The first possible definitionis the ldquoobviousrdquo one you might expect given the definition of ldquoisomorphismrdquo wersquove seen in everyother context until now we say that C is isomorphic to D if there exist functors F C rarr D andG D rarr C such that G F = idC and F G = idD where idC and idD are identity functors

However this definition turns out to be too restrictive Indeed the requirement that the givencompositions equal identity functors says that given an object A in C the object G(F (A)) beliterally the same as A itself and similarly for objects in D But we know that two objects in acategory can be thought of as being the ldquosamerdquo without being literally the same as long as theyare isomorphic For instance productscoproducts are not unique in the literal sense but only inthe sense that they are isomorphic in a unique way To take another example the definition ofa functor being representable does not say that F (B) literally be the same as hB(A) only thatthey be isomorphic for sure the power set P (S) of a set S is not literally the same as the set offunctions S rarr 0 1mdashit is only the ldquosamerdquo in the sense that they encode the same data

20

Thus it makes sense to relax the requirement that A andG(F (A)) in the definition of isomorphiccategories be equal to the requirement that they be isomorphic This leads to the following betterdefinition of two categories being the ldquosamerdquo C is equivalent to D if there exist functors F C rarr D

and G D rarr C such that G F sim= idC and F G sim= idD So we require only that for instance thefunctor G F be naturally isomorphic to the identity functor on C which says that for any objectA G(F (A)) be isomorphic to A in a natural way Wersquoll see that this truly is the better notionof what it means for two categories to be the ldquosamerdquo and it reflects the idea that all categoricalproperties of one category can be transported to the other

Alternate characterization of equivalences A functor F C rarr D implementing an equiv-alence between categories is called unsurprisingly and equivalence and the ldquoinverserdquo functorG D rarr C is still referred to as being its inverse even though it is not technically an ldquoinverserdquoin the sense that compositions give literal identities But as in the case of Set where ldquoinvertiblerdquocan be characterized without reference to an inverse as ldquoinjectiverdquo and ldquosurjectiverdquo so too canequivalences between categories be characterized without explicit reference to an inverse

A functor F C rarr D is an equivalence if and only it is full faithful and essentiallysurjective

We have seen the notions of full and faithful before and essentially surjective means that any objectin D is isomorphic to something in the image of F for all D isin Ob(D) there exits C isin Ob(C) suchthat F (C) sim= D Justifying this result was the topic of a student presentation and can be foundin the book in Section 15 (Technically to construct an ldquoinverserdquo of F requires that we work withsmall categories but whatever) The bulk of the real work goes into showing that the ldquoinverserdquofunctor constructed is actually a functor which comes down to some abstract nonsense argumentvia commutative diagrams

Equivalences preserve everything We previously saw the notions of full faithful and es-sentially surjective back when thinking about what properties a functor could have which wouldguarantee it sent products to products so the result we proved back then was that equivalencespreserve products Similarly equivalences preserve coproducts and the proof is basically the sameafter reversing arrows A clean way of deriving this is as follows if F C rarr D is an equivalencethen so is F op Cop rarr Dop so since F op preserves products F preserves coproducts becausecoproducts in a category become products in the opposite category

In general an equivalence between categories will preserve whatever categorical notions youwant monomorphisms epimorphisms initial objects terminal objects etc so that equivalentcategories really do have essentially the ldquosamerdquo properties

Equivalence example Here is a first example (also found in the book) of equivalent categorieswhich are not isomorphic Let FVectR be the category whose objects are finite-dimensional realvector spaces and whose morphisms are linear transformations Let MatR be the category ofmatrices defined as follows

bull objects in MatR are nonnegative integers

bull a morphism n rarr m is an mtimes n matrix with real entries

bull composition in MatR is given by matrix multiplication given B n rarr m and A m rarr kmdashsoB is an mtimes n matrix and A is a k timesm matrixmdashthe composition A B n rarr k is the k times nmatrix AB

bull identity morphisms in MatR are given by identity matrices

21

We claim that FVectR and MatR are equivalent First note that they certainly cannot be isomo-prhic isomorphisms in MatR exist only between two objects which are literally the same (ie anmtimesn matrix can be invertible only when m = n) but isomorphisms in FVectn can exist betweenobjects which are not literally the same So there are in a sense not enough objects in MatR toallow it to be isomorphic to FVectR

To construct an equivalence choose one and for all an isomorphism TV V rarr RdimV forevery object in FVectR or equivalently choose once and for all a basis for every V DefineF FVectR rarr MatR by

V 983041rarr dimV on objects and

(T V rarr W ) 983041rarr (the standard matrix of SW T Sminus1V RdimV rarr RdimW ) on morphisms

Equivalently F sends T to the matrix of T relative to the chosen bases for V and W It isstraightforward to check that this is a functor The inverse functor G MatR rarr VectR is givenby

n 983041rarr Rn on objects and

(mtimes n matrix A) 983041rarr (the linear transformation A Rn rarr Rm induced by A) on morphisms

One can check that G is a functor and that F G and G F are naturally isomorphic to identitiesor instead one can argue that F alone is full faithful and essentially surjective The point is thatG(F (V )) gives back RdimV so not literally V itself but only something isomorphic to V which iswhy this gives an equivalence of categories and not an isomorphism

Yoneda Lemma Functors as Objects

Compact Hausdorff spaces and Clowast-algebras Last time we gave an example of equivalentcategories which were not isomorphic but in many ways that example is unsatisfactory when doinglinear algebra people for sure often work with matrix representations of linear transformations asthe example from last would suggest but no one in their right mind literally thinks about suchthings in terms of the equivalence FVectR rarr MatR itself In other words this equivalence isessentially ldquocooked uprdquo for the sake of giving an example but it does not really give any insightinto the actual mathematics of linear algebra

So with that in mind we will describe another equivalence which actually does have some realmeaning in the sense of producing new ways of thinking about some mathematics Given a compactHausdorff space X we let C(X) denote the space of continuous complex-valued functions on X

C(X) = f X rarr C | f is continuous

We can equip the set C(X) with various additional structures First it is a complex vector spaceunder ordinary addition and scalar multiplication of functions Second we can multiply two func-tions together

(fg)(p) = f(p)g(p)

which turns C(X) into whatrsquos called an algebra over C (We wonrsquot give the definition of anldquoalgebrardquo but the point is that this multiplication is in some sense ldquocompatiblerdquo with the vectorspace structure) Next any element f of C(X) has a conjugate element f obtained by takingcomplex conjugates of the values of f

f(p) = f(p)

22

And finally we can equip C(X) with a norm via

983042f983042 = suppisinX

|f(p)|

where |middot| denotes complex absolute value (This supremum exists sinceX is compact by the ExtremeValue Theorem) All of these structures (vector space algebra multiplication conjugation norm)turn C(X) into whatrsquos called a Clowast-algebra We wonrsquot define this precisely but the definition encodesvarious ways in which these structures are compatible Actually C(X) is a unital commutative Clowast-algebra where ldquounitalrdquo means that the multiplication has an identity element (namely the constantfunction 1) and commutative means that fg = gf There is a natural notion of homomorphismbetween Clowast-algebras which are maps between them which behave nicely with all the structuresinvolved In particular given a continuous map f X rarr Y between compact Hausdorff spaces weget a Clowast-algebra homomorphism flowast C(Y ) rarr C(X) given by pre-composition with f flowast sendsg isin C(Y ) to g f isin C(X)

We thus have a contravariant functor CHaus rarr ucClowast-Alg from the category of compactHausdorff spaces and the category of unital commutative Clowast-algebras It turns out that anyunital commutative Clowast-algebra arises in this way in the sense that given any unital commutativeClowast-algebra B there exists a compact Hausdorff space X such that C(X) sim= B as Clowast-algebrasand that morphisms between these algebra arises from continuous maps in the manner describeabove This says that the functor above is essentially surjective full and it turns out that it isalso faithful so that it is an equivalence Thus all information about compact Hausdorff spaces iscompletely encoded in information about unital commtuative Clowast-algebras and indeed one couldlearn everything there is to know about a compact Hausdorff space X from its Clowast-algebra C(X)of functions The fact that the categories CHaus and ucClowast-Alg are equivalent is the startingpoint in the subject known as noncommutative geometry which wersquoll say something about a bitlater on If we drop the requirement that our algebras be unital it turns out that the categoryof commutative Clowast-algebras in general is equivalent to the category of locally compact Hausdorffspaces via essentially the same functor where we take as morphisms in the category of locallycompact Hausdorff spaces to be continuous functions which ldquovanish at infinityrdquo

We are only introducing Clowast-algebras to give an example of an interesting and actually usefulequivalence between categories but just for the fun of it wersquoll note the following Examples ofnoncommutative Clowast-algebras come from what are known as ldquoalgebras of bounded operators onHilbert spacesrdquo which show up in quantum mechanics as the things which describe physicallyobservable quantities like position momentum and energy (In fact all Clowast-algebras arise fromHilbert spaces in this way) Thus Clowast-algebras show up in mathematical formulations of quantummechanics and the notion of noncommutative geometry mentioned above is at the core of someattempts to understand the elusive concept known as ldquoquantum geometryrdquo

Yoneda Lemma The Yoneda Lemma states that given a (covariant) functor F C rarr Set whereC is locally small for any A isin Ob(C) there exists a bijection

F (A) sim= Nat(hA F )

where the right side denotes the set of natural transformations from hA to F A similar result holdsin the contravariant case where we get bijections F (A) sim= Nat(hA F ) (There is also a sense inwhich these bijections are themselves natural but wersquoll ignore this bit) The proof of the YonedaLemma was the topic of a student presentation and can be found in Section 22 of the book

Here we will get a bit philosophical and talk about what the point of the Yoneda Lemmaactually is and later the sense in which it says that we can view arbitrary functors as ldquonon-existent

23

objectsrdquo in a category First we mention what we will call the ldquofunctor of pointsrdquo approach tomathematics a mathematical object can be fully characterized by all morphisms into it That isknowledge of Mor(SA) for all S is equivalent to knowledge of A itself Indeed this is an idea wementioned previously in relation to the functors hAmdashwhich we previously said is referred to as beingthe functor of points of Amdashand now we seek to push this idea further In some examples literalpoints of A can be extracted from hA as in the case of the bijection X sim= hX(pt) = Map(ptX) fora topological space X or G sim= hG(Z) = Mor(Z G) for a group G Moreover in these examples therest of the (topological) structure onX or (group) structure on G can also be extracted from hX andhG respectively In general we can view Mor(SA) as in a sense being ldquopoints of A parameterizedby Srdquo which comes from the idea that in say Set Mor(SA) = AS is ldquothe product of S-manycopies of Ardquo so literally points of A ldquoparameterizedrdquo by S In various areas of mathematics anelement of Mor(SA) is thought of as being a ldquogeneralizedrdquo point of A so that the functor of pointshA detects generalized points of A

So an object A is fully encoded by the functor hA Now we claim that morphisms betweenobjects are also encoded by such functors Indeed in the case where F = hB is itself a functor ofpoints the Yoneda result Nat(hA F ) sim= F (A) becomes

Nat(hA hB) sim= hB(A) = Mor(AB)

which says precisely that morphisms between the objects A and B can be obtained solely fromthe functors hA and hB (This fact is one reason why one should prefer the contravariant functorhA to the covariant functor hA in the covariant case Nat(hA F ) sim= F (A) for F = hB becomesNat(hA hB) sim= hB(A) = Mor(BA) so that morphisms between two objects get reversed whencharacterizing them in terms of representable functors)

Pushing our philosophy further we now know that not only objects but also morphisms betweenobjects are fully encoded by the functors hA The final step is to note that we should not only careabout objects and morphisms between them but that an important part of the data of an objectA is also how arbitrary contravariant functors F C rarr Set behave on A and that this too can beobtained from hA precisely via the Yoneda result that F (A) sim= Nat(hA F ) So objects morphismsbetween objects and how arbitrary functors behave on objects are all encoded in knowledge ofthe functors hA so that these functors truly do capture in the most general sense possible allinformation about A

Functors as nonexistent objects The discussion above is all well and good but it still doesnrsquotquite say why we should care about the fact that we can study on object A by studying hA insteadin that we still havenrsquot spoken about what type of insight about A can be gleamed from hA thatwasnrsquot readily apparent in A itself already Now we argue that the true point of the YonedaLemma is that it gives a way to think about arbitrary functors F Cop rarr Set as ldquoobjectsrdquo in C

itself or really as objects in an ldquoenlargementrdquo of C Certainly for those functors F sim= hA which arerepresentable we can indeed think about F as simply being the object A but now we claim thateven non-representable F can be thought of as being ldquononexistentrdquo objects in C The upshot is thatoften in mathematics we come across things which do not actually exist but the Yoneda Lemmagives a way to (indirectly) study such things anyway by interpreting them as functors instead

The starting point of course is that an actual object A is the same as hA as we have explainedabove Thus by analogy we can interpret a non-representable functor F as being an ldquoobjectrdquo in C

as long as we know what it means to ldquomaprdquo into F from other things which are honestly objects in CThat is if F were indeed representable then F (A) sim= Nat(hA F ) sim= Mor(AF ) would characterizethe functor of points of F (here we abuse notation by using the same letter F to denote both arepresentable functor and the object which represents it) so now we take Mor(AF ) = F (A) as

24

the definition of the ldquofunctor of pointsrdquo of the non-existent object F We are thus taking elementsof F (A) as being what it means to ldquomaprdquo from A into F even when F is not representable Thefact that the Yoneda Lemma gives a way to define maps from A into F thus says that F itself isequipped with the same data that a literal object of A would be equipped with so why shouldnrsquotwe interpret F as a being a type of ldquoobjectrdquo in C as well

To phrase this another way note that the Yoneda Lemma gives rise to a functor

C rarr Fun(CopSet)

where Fun(CopSet) denotes the category whose objects are functors from Cop to Set and morphismsare natural transformation defined on objects by A 983041rarr hA and on morphisms by the bijectionMor(AB) sim= Nat(hA hB) arising from the Yoneda Lemma That this is actually bijection saysthat this functor is full and faithful so that we can view C as being a subcategory of the functorcategory Fun(CopSet) (We cay say that C is isomorphic to the subcategory of Fun(CopSet)consisting of the functors hA or that it is equivalent to the subcategory of Fun(CopSet) consistingof representable functors This functor C rarr Fun(CopSet) is usually called the Yoneda embeddingof C into Fun(CopSet)) Thus we can view Fun(CopSet) as an ldquoenlargmentrdquo of C and the ideawe are trying to make sense of is that other objects in this ldquolargerrdquo category should be interpretedas ldquonon-existent objectsrdquo of C

Just how far can we actually push this idea that arbitrary functors Cop rarr Set should be thoughtof as being ldquoobjectsrdquo of C Pretty far in fact For instance often times in geometry one comesacross a type of ldquosmoothrdquo space on which some group acts and one is interested in doing ldquocalculusrdquoon the resulting quotient space which is the space of orbits under the given group action Theproblem is that such quotient spaces are very often (usually in fact) not ldquosmoothrdquo themselves sothat the tools from calculus are not readily available The solution is to nonetheless treat thisquotient as if it were ldquosmoothrdquo and attempt to do calculus anyway To make this precise thisldquonon-smoothrdquo space actually turns out to have a natural interpretation as a ldquosmooth functorrdquo andit is through this functor which we can view this non-smooth space as actually being smooth afterall so we are forced to expand our notion of what we mean by ldquosmooth spacerdquo by allowing thingswhich are not spaces in the usual sense which certainly leads to more abstraction but the benefitis that it actually makes sense to do calculus on such ldquospacesrdquo interpreted as functors

The discussion above is pretty vague so here is one final more concrete example Recall thatequivalence

CHausop rarr ucClowast-Alg

between the category of compact Hausdorff spaces and the category of unital commutative Clowast-algebras mentioned at the start We now claim that a non-commutative (unital) Clowast-algebra shouldbe interpreted as the algebra of functions on some non-existent ldquononcommutative compact Haus-dorff spacerdquo and it is this idea which the subject of noncommutative geometry makes precise Solet B be a non-commutative Clowast-algebra Then the functor of points hB gives a functor

ucClowast-Alg rarr Set

which sends C(X) (where X is some compact Hausdorff space) to Hom(C(X) B) where Hom de-notes homomorphisms of Clowast-algebras whatever that means Composing with the previous functorwe get a functor wersquoll denote by B itself

B CHausop rarr Set

which on objects at least is given by X 983041rarr Hom(C(X) B) This functor is not representable sinceB is not commutative but the philosophy espoused above suggests we should think of it as giving a

25

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

in the lower left Instead starting with k D rarr AtimesB in the upper right applying ηD on the rightgives (pA k pB k) and applying the map on the bottom gives

((pA k) g (pB k) g)

in the lower left which is the same as the previous element in the lower left by the associativity ofcomposition Hence the isomorphisms ηC do define a natural isomorphism between hAtimesB and F Again the point is that you can then interpret the functor F as being the ldquosamerdquo as the productAtimesB since both encode the same data

More Representable Examples

Coproducts As in the case of products the definition of coproducts can also be phrased in termsof a representable functor Given objects A and B in C let F C rarr Set be the covariant functordefined by

F (C) = Mor(AC)timesMor(BC) and F (Cfminusrarr D) = [(g k) 983041rarr (f g f k)]

where the latter is a function Mor(AC)timesMor(BC) rarr Mor(AD)timesMor(BD) This functor isrepresentable if and only if a coproduct A ⊔ B for A and B exists and the representing object isthat coproduct the key is the requirement that we have bijections

Mor(AC)timesMor(BC) sim= Mor(A ⊔BC)

for all C which is essentially the defining property a coproduct for A and B should have

Power set functor contravariant version Let P Set rarr Set be the contravariant powerset functor defined on objects by sending a set to its power set and on morphisms by sendingf A rarr B to the preimage function Pf P (B) rarr P (A) given by S 983041rarr fminus1(S) In order for this tobe representable requires the existence of a set X with natural bijections

P (A) sim= Mor(AX)

for all sets A that is a set X so that mapping into X is the same data as specifying a subset ofthe domain We take X = 0 1 to be a two-element set and the required bijections

ηA P (A) rarr Mor(A 0 1) to be given by S 983041rarr χS

where XS A rarr 0 1 is the indicator or characteristic function S defined by sending elementsof S to 1 and elements of A minus S to 0 The inverse of ηA sends a function g A rarr 0 1 to thepreimage gminus1(1) sube A

To verify that the bijections ηA define the data of a natural isomorphism η P rArr h01 wecheck the commutativity of some diagrams Given a function f A rarr B consider

P (A) P (B)

h01(A) = Mor(A 0 1) h01(B) = Mor(B 0 1)

Pf = fminus1

ηA ηB

minus f

18

Take S isin P (B) in the upper right Applying the map on top gives fminus1(S) isin P (A) and thenapplying the map on the left gives χfminus1(S) isin Mor(A 0 1) On the other hand applying the mapon the right to S isin P (B) gives χS isin Mor(B 0 1) and then applying the map on the bottomgives χS f isin Mor(A 0 1) The two resulting functions χfminus1(S)χS f in the lower left are

χfminus1(S)(x) =

9830831 x isin fminus1(S)

0 otherwiseand χS(f(x)) =

9830831 f(x) isin S

0 otherwise

so these are the same since x isin fminus1(S) if and only if f(x) isin S Thus the diagram above commutesso P is naturally isomorphic to h01 and hence P is representable

Power set functor covariant version Now consider the covariant power set functor stilldenoted by P Set rarr Set which again sends a set to its power set but now sends a functionf A rarr B to the image function Pf P (A) rarr P (B) defined by S 983041rarr f(S) In order for this tobe representable there would have to exist a set X such that P sim= hX which translates into havingnatural bijections

P (S) sim= Mor(XS)

for all sets S However in this case there is no natural choice for X as there is no natural way todetect subsets of S by mapping into S as opposed to the case of P (S) sim= Mor(S 0 1)

That no such X exists can be made precise using a simple argument when S = empty P (S) hascardinality 1 soX would have to be empty as well since otherwise Mor(XS) would have cardinality0 but if X is empty then Mor(empty S) always cardinality 1 regardless of S which is clearly not trueof P (S) Thus the covariant power set functor is not representable

Forgetful on Top The forgetful functor F Top rarr Set is representable Indeed a representingobject would have to satisfy

X sim= Map(objectX)

as sets for any topological space X (where Map denotes the set of continuous maps) and it iseasy to see that a single point satisfies this property a continuous map f pt rarr X is completelydetermined by the element f(pt) of X and any such element gives a continuous map in this wayOf course checking that F sim= hpt requires checking that various diagrams commute but this issimple to work out in this case so we omit it here

Extracting a topology With an eye towards the idea that the functors hA hA should captureall information about an object A in a category note that the result above shows that in TophX first of all captures all information about X as a set since the set X can be extracted fromhX(pt) = Map(ptX)

We claim that hX actually captures all information about X as a topological space since thetopology on X can be extracted from hX in the following way As we saw in the power set examplewe have a bijection

P (X) sim= Mor(X 0 1)

Now equip 0 1 with the topology in which 1 is open but 0 is not Then a continuous mapf X rarr 0 1 is fully determined by the preimage fminus1(1) since to be continuous only requiresthat this single preimage be open in X Moreoever any open subset of X arises in this way bytaking f to be the indicator function of that open set Thus we get a bijection

Op(X) sim= Map(X 0 1)

19

where Op(X) denotes the set of open subsets of X ie the topology on X Hence the topology onX can be extracted from hX as claimed so hX and hX do encode all information about X

This fact can also be interpreted as saying that the contravariant functor which sends a topo-logical space X to its collection of open subsets (and which sends a continuous map to the mapinduced by taking preimages) is representable with representing object 0 1 equipped with thetopology given above This will be worked out in a homework problem

Forgetful on Grp The forgetful functor Grp rarr Set is also representable In this case arepresenting object should be a group giving set-theoretic bijections

G sim= Hom(objectG)

for all groups G where Hom denotes the set of group homomorphisms The group Z works

G sim= Hom(Z G)

since a homomorphism φ Z rarr G is completely determined by φ(1) since Z is generated by 1 andthere are no restrictions on what φ(1) isin G can actually be The inverse of the desired bijection isthus φ 983041rarr φ(1) and it is simple to check that morphisms behave appropriately so that the forgetfulfunctor in this setting is indeed isomoprhic to hZ

Forgetful on Ring The forgetful functor Ring rarr Set is also representable where by Ring wemean the category of rings with identity elements and where morphisms (ie ring homomorphisms)are required to preserve identities Here we need an object satisfying

R sim= Hom(object R)

for all R isin Ob(Ring) and we can see that Z[x] the ring of polynomials in the variable x withcoefficients in Z works Indeed given a ring homomorphism φ Z[x] rarr R the fact that φ isrequired to preserve identities fully determines φ(n) for any n isin Z (since φ(n) = nφ(1)) and hencesince φ preserves multiplication its behavior is fully determined by φ(x) which can be any elementof R The inverse of the bijection we want is thus φ 983041rarr φ(x) and it will follow that this forgetfulfunctor is isomorphic to hZ[x]

Equivalences between Categories

Definitions iso and equiv Now that we have a notion of a ldquomorphismrdquo between categorieswe can talk about the sense in which two categories are the ldquosamerdquo The first possible definitionis the ldquoobviousrdquo one you might expect given the definition of ldquoisomorphismrdquo wersquove seen in everyother context until now we say that C is isomorphic to D if there exist functors F C rarr D andG D rarr C such that G F = idC and F G = idD where idC and idD are identity functors

However this definition turns out to be too restrictive Indeed the requirement that the givencompositions equal identity functors says that given an object A in C the object G(F (A)) beliterally the same as A itself and similarly for objects in D But we know that two objects in acategory can be thought of as being the ldquosamerdquo without being literally the same as long as theyare isomorphic For instance productscoproducts are not unique in the literal sense but only inthe sense that they are isomorphic in a unique way To take another example the definition ofa functor being representable does not say that F (B) literally be the same as hB(A) only thatthey be isomorphic for sure the power set P (S) of a set S is not literally the same as the set offunctions S rarr 0 1mdashit is only the ldquosamerdquo in the sense that they encode the same data

20

Thus it makes sense to relax the requirement that A andG(F (A)) in the definition of isomorphiccategories be equal to the requirement that they be isomorphic This leads to the following betterdefinition of two categories being the ldquosamerdquo C is equivalent to D if there exist functors F C rarr D

and G D rarr C such that G F sim= idC and F G sim= idD So we require only that for instance thefunctor G F be naturally isomorphic to the identity functor on C which says that for any objectA G(F (A)) be isomorphic to A in a natural way Wersquoll see that this truly is the better notionof what it means for two categories to be the ldquosamerdquo and it reflects the idea that all categoricalproperties of one category can be transported to the other

Alternate characterization of equivalences A functor F C rarr D implementing an equiv-alence between categories is called unsurprisingly and equivalence and the ldquoinverserdquo functorG D rarr C is still referred to as being its inverse even though it is not technically an ldquoinverserdquoin the sense that compositions give literal identities But as in the case of Set where ldquoinvertiblerdquocan be characterized without reference to an inverse as ldquoinjectiverdquo and ldquosurjectiverdquo so too canequivalences between categories be characterized without explicit reference to an inverse

A functor F C rarr D is an equivalence if and only it is full faithful and essentiallysurjective

We have seen the notions of full and faithful before and essentially surjective means that any objectin D is isomorphic to something in the image of F for all D isin Ob(D) there exits C isin Ob(C) suchthat F (C) sim= D Justifying this result was the topic of a student presentation and can be foundin the book in Section 15 (Technically to construct an ldquoinverserdquo of F requires that we work withsmall categories but whatever) The bulk of the real work goes into showing that the ldquoinverserdquofunctor constructed is actually a functor which comes down to some abstract nonsense argumentvia commutative diagrams

Equivalences preserve everything We previously saw the notions of full faithful and es-sentially surjective back when thinking about what properties a functor could have which wouldguarantee it sent products to products so the result we proved back then was that equivalencespreserve products Similarly equivalences preserve coproducts and the proof is basically the sameafter reversing arrows A clean way of deriving this is as follows if F C rarr D is an equivalencethen so is F op Cop rarr Dop so since F op preserves products F preserves coproducts becausecoproducts in a category become products in the opposite category

In general an equivalence between categories will preserve whatever categorical notions youwant monomorphisms epimorphisms initial objects terminal objects etc so that equivalentcategories really do have essentially the ldquosamerdquo properties

Equivalence example Here is a first example (also found in the book) of equivalent categorieswhich are not isomorphic Let FVectR be the category whose objects are finite-dimensional realvector spaces and whose morphisms are linear transformations Let MatR be the category ofmatrices defined as follows

bull objects in MatR are nonnegative integers

bull a morphism n rarr m is an mtimes n matrix with real entries

bull composition in MatR is given by matrix multiplication given B n rarr m and A m rarr kmdashsoB is an mtimes n matrix and A is a k timesm matrixmdashthe composition A B n rarr k is the k times nmatrix AB

bull identity morphisms in MatR are given by identity matrices

21

We claim that FVectR and MatR are equivalent First note that they certainly cannot be isomo-prhic isomorphisms in MatR exist only between two objects which are literally the same (ie anmtimesn matrix can be invertible only when m = n) but isomorphisms in FVectn can exist betweenobjects which are not literally the same So there are in a sense not enough objects in MatR toallow it to be isomorphic to FVectR

To construct an equivalence choose one and for all an isomorphism TV V rarr RdimV forevery object in FVectR or equivalently choose once and for all a basis for every V DefineF FVectR rarr MatR by

V 983041rarr dimV on objects and

(T V rarr W ) 983041rarr (the standard matrix of SW T Sminus1V RdimV rarr RdimW ) on morphisms

Equivalently F sends T to the matrix of T relative to the chosen bases for V and W It isstraightforward to check that this is a functor The inverse functor G MatR rarr VectR is givenby

n 983041rarr Rn on objects and

(mtimes n matrix A) 983041rarr (the linear transformation A Rn rarr Rm induced by A) on morphisms

One can check that G is a functor and that F G and G F are naturally isomorphic to identitiesor instead one can argue that F alone is full faithful and essentially surjective The point is thatG(F (V )) gives back RdimV so not literally V itself but only something isomorphic to V which iswhy this gives an equivalence of categories and not an isomorphism

Yoneda Lemma Functors as Objects

Compact Hausdorff spaces and Clowast-algebras Last time we gave an example of equivalentcategories which were not isomorphic but in many ways that example is unsatisfactory when doinglinear algebra people for sure often work with matrix representations of linear transformations asthe example from last would suggest but no one in their right mind literally thinks about suchthings in terms of the equivalence FVectR rarr MatR itself In other words this equivalence isessentially ldquocooked uprdquo for the sake of giving an example but it does not really give any insightinto the actual mathematics of linear algebra

So with that in mind we will describe another equivalence which actually does have some realmeaning in the sense of producing new ways of thinking about some mathematics Given a compactHausdorff space X we let C(X) denote the space of continuous complex-valued functions on X

C(X) = f X rarr C | f is continuous

We can equip the set C(X) with various additional structures First it is a complex vector spaceunder ordinary addition and scalar multiplication of functions Second we can multiply two func-tions together

(fg)(p) = f(p)g(p)

which turns C(X) into whatrsquos called an algebra over C (We wonrsquot give the definition of anldquoalgebrardquo but the point is that this multiplication is in some sense ldquocompatiblerdquo with the vectorspace structure) Next any element f of C(X) has a conjugate element f obtained by takingcomplex conjugates of the values of f

f(p) = f(p)

22

And finally we can equip C(X) with a norm via

983042f983042 = suppisinX

|f(p)|

where |middot| denotes complex absolute value (This supremum exists sinceX is compact by the ExtremeValue Theorem) All of these structures (vector space algebra multiplication conjugation norm)turn C(X) into whatrsquos called a Clowast-algebra We wonrsquot define this precisely but the definition encodesvarious ways in which these structures are compatible Actually C(X) is a unital commutative Clowast-algebra where ldquounitalrdquo means that the multiplication has an identity element (namely the constantfunction 1) and commutative means that fg = gf There is a natural notion of homomorphismbetween Clowast-algebras which are maps between them which behave nicely with all the structuresinvolved In particular given a continuous map f X rarr Y between compact Hausdorff spaces weget a Clowast-algebra homomorphism flowast C(Y ) rarr C(X) given by pre-composition with f flowast sendsg isin C(Y ) to g f isin C(X)

We thus have a contravariant functor CHaus rarr ucClowast-Alg from the category of compactHausdorff spaces and the category of unital commutative Clowast-algebras It turns out that anyunital commutative Clowast-algebra arises in this way in the sense that given any unital commutativeClowast-algebra B there exists a compact Hausdorff space X such that C(X) sim= B as Clowast-algebrasand that morphisms between these algebra arises from continuous maps in the manner describeabove This says that the functor above is essentially surjective full and it turns out that it isalso faithful so that it is an equivalence Thus all information about compact Hausdorff spaces iscompletely encoded in information about unital commtuative Clowast-algebras and indeed one couldlearn everything there is to know about a compact Hausdorff space X from its Clowast-algebra C(X)of functions The fact that the categories CHaus and ucClowast-Alg are equivalent is the startingpoint in the subject known as noncommutative geometry which wersquoll say something about a bitlater on If we drop the requirement that our algebras be unital it turns out that the categoryof commutative Clowast-algebras in general is equivalent to the category of locally compact Hausdorffspaces via essentially the same functor where we take as morphisms in the category of locallycompact Hausdorff spaces to be continuous functions which ldquovanish at infinityrdquo

We are only introducing Clowast-algebras to give an example of an interesting and actually usefulequivalence between categories but just for the fun of it wersquoll note the following Examples ofnoncommutative Clowast-algebras come from what are known as ldquoalgebras of bounded operators onHilbert spacesrdquo which show up in quantum mechanics as the things which describe physicallyobservable quantities like position momentum and energy (In fact all Clowast-algebras arise fromHilbert spaces in this way) Thus Clowast-algebras show up in mathematical formulations of quantummechanics and the notion of noncommutative geometry mentioned above is at the core of someattempts to understand the elusive concept known as ldquoquantum geometryrdquo

Yoneda Lemma The Yoneda Lemma states that given a (covariant) functor F C rarr Set whereC is locally small for any A isin Ob(C) there exists a bijection

F (A) sim= Nat(hA F )

where the right side denotes the set of natural transformations from hA to F A similar result holdsin the contravariant case where we get bijections F (A) sim= Nat(hA F ) (There is also a sense inwhich these bijections are themselves natural but wersquoll ignore this bit) The proof of the YonedaLemma was the topic of a student presentation and can be found in Section 22 of the book

Here we will get a bit philosophical and talk about what the point of the Yoneda Lemmaactually is and later the sense in which it says that we can view arbitrary functors as ldquonon-existent

23

objectsrdquo in a category First we mention what we will call the ldquofunctor of pointsrdquo approach tomathematics a mathematical object can be fully characterized by all morphisms into it That isknowledge of Mor(SA) for all S is equivalent to knowledge of A itself Indeed this is an idea wementioned previously in relation to the functors hAmdashwhich we previously said is referred to as beingthe functor of points of Amdashand now we seek to push this idea further In some examples literalpoints of A can be extracted from hA as in the case of the bijection X sim= hX(pt) = Map(ptX) fora topological space X or G sim= hG(Z) = Mor(Z G) for a group G Moreover in these examples therest of the (topological) structure onX or (group) structure on G can also be extracted from hX andhG respectively In general we can view Mor(SA) as in a sense being ldquopoints of A parameterizedby Srdquo which comes from the idea that in say Set Mor(SA) = AS is ldquothe product of S-manycopies of Ardquo so literally points of A ldquoparameterizedrdquo by S In various areas of mathematics anelement of Mor(SA) is thought of as being a ldquogeneralizedrdquo point of A so that the functor of pointshA detects generalized points of A

So an object A is fully encoded by the functor hA Now we claim that morphisms betweenobjects are also encoded by such functors Indeed in the case where F = hB is itself a functor ofpoints the Yoneda result Nat(hA F ) sim= F (A) becomes

Nat(hA hB) sim= hB(A) = Mor(AB)

which says precisely that morphisms between the objects A and B can be obtained solely fromthe functors hA and hB (This fact is one reason why one should prefer the contravariant functorhA to the covariant functor hA in the covariant case Nat(hA F ) sim= F (A) for F = hB becomesNat(hA hB) sim= hB(A) = Mor(BA) so that morphisms between two objects get reversed whencharacterizing them in terms of representable functors)

Pushing our philosophy further we now know that not only objects but also morphisms betweenobjects are fully encoded by the functors hA The final step is to note that we should not only careabout objects and morphisms between them but that an important part of the data of an objectA is also how arbitrary contravariant functors F C rarr Set behave on A and that this too can beobtained from hA precisely via the Yoneda result that F (A) sim= Nat(hA F ) So objects morphismsbetween objects and how arbitrary functors behave on objects are all encoded in knowledge ofthe functors hA so that these functors truly do capture in the most general sense possible allinformation about A

Functors as nonexistent objects The discussion above is all well and good but it still doesnrsquotquite say why we should care about the fact that we can study on object A by studying hA insteadin that we still havenrsquot spoken about what type of insight about A can be gleamed from hA thatwasnrsquot readily apparent in A itself already Now we argue that the true point of the YonedaLemma is that it gives a way to think about arbitrary functors F Cop rarr Set as ldquoobjectsrdquo in C

itself or really as objects in an ldquoenlargementrdquo of C Certainly for those functors F sim= hA which arerepresentable we can indeed think about F as simply being the object A but now we claim thateven non-representable F can be thought of as being ldquononexistentrdquo objects in C The upshot is thatoften in mathematics we come across things which do not actually exist but the Yoneda Lemmagives a way to (indirectly) study such things anyway by interpreting them as functors instead

The starting point of course is that an actual object A is the same as hA as we have explainedabove Thus by analogy we can interpret a non-representable functor F as being an ldquoobjectrdquo in C

as long as we know what it means to ldquomaprdquo into F from other things which are honestly objects in CThat is if F were indeed representable then F (A) sim= Nat(hA F ) sim= Mor(AF ) would characterizethe functor of points of F (here we abuse notation by using the same letter F to denote both arepresentable functor and the object which represents it) so now we take Mor(AF ) = F (A) as

24

the definition of the ldquofunctor of pointsrdquo of the non-existent object F We are thus taking elementsof F (A) as being what it means to ldquomaprdquo from A into F even when F is not representable Thefact that the Yoneda Lemma gives a way to define maps from A into F thus says that F itself isequipped with the same data that a literal object of A would be equipped with so why shouldnrsquotwe interpret F as a being a type of ldquoobjectrdquo in C as well

To phrase this another way note that the Yoneda Lemma gives rise to a functor

C rarr Fun(CopSet)

where Fun(CopSet) denotes the category whose objects are functors from Cop to Set and morphismsare natural transformation defined on objects by A 983041rarr hA and on morphisms by the bijectionMor(AB) sim= Nat(hA hB) arising from the Yoneda Lemma That this is actually bijection saysthat this functor is full and faithful so that we can view C as being a subcategory of the functorcategory Fun(CopSet) (We cay say that C is isomorphic to the subcategory of Fun(CopSet)consisting of the functors hA or that it is equivalent to the subcategory of Fun(CopSet) consistingof representable functors This functor C rarr Fun(CopSet) is usually called the Yoneda embeddingof C into Fun(CopSet)) Thus we can view Fun(CopSet) as an ldquoenlargmentrdquo of C and the ideawe are trying to make sense of is that other objects in this ldquolargerrdquo category should be interpretedas ldquonon-existent objectsrdquo of C

Just how far can we actually push this idea that arbitrary functors Cop rarr Set should be thoughtof as being ldquoobjectsrdquo of C Pretty far in fact For instance often times in geometry one comesacross a type of ldquosmoothrdquo space on which some group acts and one is interested in doing ldquocalculusrdquoon the resulting quotient space which is the space of orbits under the given group action Theproblem is that such quotient spaces are very often (usually in fact) not ldquosmoothrdquo themselves sothat the tools from calculus are not readily available The solution is to nonetheless treat thisquotient as if it were ldquosmoothrdquo and attempt to do calculus anyway To make this precise thisldquonon-smoothrdquo space actually turns out to have a natural interpretation as a ldquosmooth functorrdquo andit is through this functor which we can view this non-smooth space as actually being smooth afterall so we are forced to expand our notion of what we mean by ldquosmooth spacerdquo by allowing thingswhich are not spaces in the usual sense which certainly leads to more abstraction but the benefitis that it actually makes sense to do calculus on such ldquospacesrdquo interpreted as functors

The discussion above is pretty vague so here is one final more concrete example Recall thatequivalence

CHausop rarr ucClowast-Alg

between the category of compact Hausdorff spaces and the category of unital commutative Clowast-algebras mentioned at the start We now claim that a non-commutative (unital) Clowast-algebra shouldbe interpreted as the algebra of functions on some non-existent ldquononcommutative compact Haus-dorff spacerdquo and it is this idea which the subject of noncommutative geometry makes precise Solet B be a non-commutative Clowast-algebra Then the functor of points hB gives a functor

ucClowast-Alg rarr Set

which sends C(X) (where X is some compact Hausdorff space) to Hom(C(X) B) where Hom de-notes homomorphisms of Clowast-algebras whatever that means Composing with the previous functorwe get a functor wersquoll denote by B itself

B CHausop rarr Set

which on objects at least is given by X 983041rarr Hom(C(X) B) This functor is not representable sinceB is not commutative but the philosophy espoused above suggests we should think of it as giving a

25

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

Take S isin P (B) in the upper right Applying the map on top gives fminus1(S) isin P (A) and thenapplying the map on the left gives χfminus1(S) isin Mor(A 0 1) On the other hand applying the mapon the right to S isin P (B) gives χS isin Mor(B 0 1) and then applying the map on the bottomgives χS f isin Mor(A 0 1) The two resulting functions χfminus1(S)χS f in the lower left are

χfminus1(S)(x) =

9830831 x isin fminus1(S)

0 otherwiseand χS(f(x)) =

9830831 f(x) isin S

0 otherwise

so these are the same since x isin fminus1(S) if and only if f(x) isin S Thus the diagram above commutesso P is naturally isomorphic to h01 and hence P is representable

Power set functor covariant version Now consider the covariant power set functor stilldenoted by P Set rarr Set which again sends a set to its power set but now sends a functionf A rarr B to the image function Pf P (A) rarr P (B) defined by S 983041rarr f(S) In order for this tobe representable there would have to exist a set X such that P sim= hX which translates into havingnatural bijections

P (S) sim= Mor(XS)

for all sets S However in this case there is no natural choice for X as there is no natural way todetect subsets of S by mapping into S as opposed to the case of P (S) sim= Mor(S 0 1)

That no such X exists can be made precise using a simple argument when S = empty P (S) hascardinality 1 soX would have to be empty as well since otherwise Mor(XS) would have cardinality0 but if X is empty then Mor(empty S) always cardinality 1 regardless of S which is clearly not trueof P (S) Thus the covariant power set functor is not representable

Forgetful on Top The forgetful functor F Top rarr Set is representable Indeed a representingobject would have to satisfy

X sim= Map(objectX)

as sets for any topological space X (where Map denotes the set of continuous maps) and it iseasy to see that a single point satisfies this property a continuous map f pt rarr X is completelydetermined by the element f(pt) of X and any such element gives a continuous map in this wayOf course checking that F sim= hpt requires checking that various diagrams commute but this issimple to work out in this case so we omit it here

Extracting a topology With an eye towards the idea that the functors hA hA should captureall information about an object A in a category note that the result above shows that in TophX first of all captures all information about X as a set since the set X can be extracted fromhX(pt) = Map(ptX)

We claim that hX actually captures all information about X as a topological space since thetopology on X can be extracted from hX in the following way As we saw in the power set examplewe have a bijection

P (X) sim= Mor(X 0 1)

Now equip 0 1 with the topology in which 1 is open but 0 is not Then a continuous mapf X rarr 0 1 is fully determined by the preimage fminus1(1) since to be continuous only requiresthat this single preimage be open in X Moreoever any open subset of X arises in this way bytaking f to be the indicator function of that open set Thus we get a bijection

Op(X) sim= Map(X 0 1)

19

where Op(X) denotes the set of open subsets of X ie the topology on X Hence the topology onX can be extracted from hX as claimed so hX and hX do encode all information about X

This fact can also be interpreted as saying that the contravariant functor which sends a topo-logical space X to its collection of open subsets (and which sends a continuous map to the mapinduced by taking preimages) is representable with representing object 0 1 equipped with thetopology given above This will be worked out in a homework problem

Forgetful on Grp The forgetful functor Grp rarr Set is also representable In this case arepresenting object should be a group giving set-theoretic bijections

G sim= Hom(objectG)

for all groups G where Hom denotes the set of group homomorphisms The group Z works

G sim= Hom(Z G)

since a homomorphism φ Z rarr G is completely determined by φ(1) since Z is generated by 1 andthere are no restrictions on what φ(1) isin G can actually be The inverse of the desired bijection isthus φ 983041rarr φ(1) and it is simple to check that morphisms behave appropriately so that the forgetfulfunctor in this setting is indeed isomoprhic to hZ

Forgetful on Ring The forgetful functor Ring rarr Set is also representable where by Ring wemean the category of rings with identity elements and where morphisms (ie ring homomorphisms)are required to preserve identities Here we need an object satisfying

R sim= Hom(object R)

for all R isin Ob(Ring) and we can see that Z[x] the ring of polynomials in the variable x withcoefficients in Z works Indeed given a ring homomorphism φ Z[x] rarr R the fact that φ isrequired to preserve identities fully determines φ(n) for any n isin Z (since φ(n) = nφ(1)) and hencesince φ preserves multiplication its behavior is fully determined by φ(x) which can be any elementof R The inverse of the bijection we want is thus φ 983041rarr φ(x) and it will follow that this forgetfulfunctor is isomorphic to hZ[x]

Equivalences between Categories

Definitions iso and equiv Now that we have a notion of a ldquomorphismrdquo between categorieswe can talk about the sense in which two categories are the ldquosamerdquo The first possible definitionis the ldquoobviousrdquo one you might expect given the definition of ldquoisomorphismrdquo wersquove seen in everyother context until now we say that C is isomorphic to D if there exist functors F C rarr D andG D rarr C such that G F = idC and F G = idD where idC and idD are identity functors

However this definition turns out to be too restrictive Indeed the requirement that the givencompositions equal identity functors says that given an object A in C the object G(F (A)) beliterally the same as A itself and similarly for objects in D But we know that two objects in acategory can be thought of as being the ldquosamerdquo without being literally the same as long as theyare isomorphic For instance productscoproducts are not unique in the literal sense but only inthe sense that they are isomorphic in a unique way To take another example the definition ofa functor being representable does not say that F (B) literally be the same as hB(A) only thatthey be isomorphic for sure the power set P (S) of a set S is not literally the same as the set offunctions S rarr 0 1mdashit is only the ldquosamerdquo in the sense that they encode the same data

20

Thus it makes sense to relax the requirement that A andG(F (A)) in the definition of isomorphiccategories be equal to the requirement that they be isomorphic This leads to the following betterdefinition of two categories being the ldquosamerdquo C is equivalent to D if there exist functors F C rarr D

and G D rarr C such that G F sim= idC and F G sim= idD So we require only that for instance thefunctor G F be naturally isomorphic to the identity functor on C which says that for any objectA G(F (A)) be isomorphic to A in a natural way Wersquoll see that this truly is the better notionof what it means for two categories to be the ldquosamerdquo and it reflects the idea that all categoricalproperties of one category can be transported to the other

Alternate characterization of equivalences A functor F C rarr D implementing an equiv-alence between categories is called unsurprisingly and equivalence and the ldquoinverserdquo functorG D rarr C is still referred to as being its inverse even though it is not technically an ldquoinverserdquoin the sense that compositions give literal identities But as in the case of Set where ldquoinvertiblerdquocan be characterized without reference to an inverse as ldquoinjectiverdquo and ldquosurjectiverdquo so too canequivalences between categories be characterized without explicit reference to an inverse

A functor F C rarr D is an equivalence if and only it is full faithful and essentiallysurjective

We have seen the notions of full and faithful before and essentially surjective means that any objectin D is isomorphic to something in the image of F for all D isin Ob(D) there exits C isin Ob(C) suchthat F (C) sim= D Justifying this result was the topic of a student presentation and can be foundin the book in Section 15 (Technically to construct an ldquoinverserdquo of F requires that we work withsmall categories but whatever) The bulk of the real work goes into showing that the ldquoinverserdquofunctor constructed is actually a functor which comes down to some abstract nonsense argumentvia commutative diagrams

Equivalences preserve everything We previously saw the notions of full faithful and es-sentially surjective back when thinking about what properties a functor could have which wouldguarantee it sent products to products so the result we proved back then was that equivalencespreserve products Similarly equivalences preserve coproducts and the proof is basically the sameafter reversing arrows A clean way of deriving this is as follows if F C rarr D is an equivalencethen so is F op Cop rarr Dop so since F op preserves products F preserves coproducts becausecoproducts in a category become products in the opposite category

In general an equivalence between categories will preserve whatever categorical notions youwant monomorphisms epimorphisms initial objects terminal objects etc so that equivalentcategories really do have essentially the ldquosamerdquo properties

Equivalence example Here is a first example (also found in the book) of equivalent categorieswhich are not isomorphic Let FVectR be the category whose objects are finite-dimensional realvector spaces and whose morphisms are linear transformations Let MatR be the category ofmatrices defined as follows

bull objects in MatR are nonnegative integers

bull a morphism n rarr m is an mtimes n matrix with real entries

bull composition in MatR is given by matrix multiplication given B n rarr m and A m rarr kmdashsoB is an mtimes n matrix and A is a k timesm matrixmdashthe composition A B n rarr k is the k times nmatrix AB

bull identity morphisms in MatR are given by identity matrices

21

We claim that FVectR and MatR are equivalent First note that they certainly cannot be isomo-prhic isomorphisms in MatR exist only between two objects which are literally the same (ie anmtimesn matrix can be invertible only when m = n) but isomorphisms in FVectn can exist betweenobjects which are not literally the same So there are in a sense not enough objects in MatR toallow it to be isomorphic to FVectR

To construct an equivalence choose one and for all an isomorphism TV V rarr RdimV forevery object in FVectR or equivalently choose once and for all a basis for every V DefineF FVectR rarr MatR by

V 983041rarr dimV on objects and

(T V rarr W ) 983041rarr (the standard matrix of SW T Sminus1V RdimV rarr RdimW ) on morphisms

Equivalently F sends T to the matrix of T relative to the chosen bases for V and W It isstraightforward to check that this is a functor The inverse functor G MatR rarr VectR is givenby

n 983041rarr Rn on objects and

(mtimes n matrix A) 983041rarr (the linear transformation A Rn rarr Rm induced by A) on morphisms

One can check that G is a functor and that F G and G F are naturally isomorphic to identitiesor instead one can argue that F alone is full faithful and essentially surjective The point is thatG(F (V )) gives back RdimV so not literally V itself but only something isomorphic to V which iswhy this gives an equivalence of categories and not an isomorphism

Yoneda Lemma Functors as Objects

Compact Hausdorff spaces and Clowast-algebras Last time we gave an example of equivalentcategories which were not isomorphic but in many ways that example is unsatisfactory when doinglinear algebra people for sure often work with matrix representations of linear transformations asthe example from last would suggest but no one in their right mind literally thinks about suchthings in terms of the equivalence FVectR rarr MatR itself In other words this equivalence isessentially ldquocooked uprdquo for the sake of giving an example but it does not really give any insightinto the actual mathematics of linear algebra

So with that in mind we will describe another equivalence which actually does have some realmeaning in the sense of producing new ways of thinking about some mathematics Given a compactHausdorff space X we let C(X) denote the space of continuous complex-valued functions on X

C(X) = f X rarr C | f is continuous

We can equip the set C(X) with various additional structures First it is a complex vector spaceunder ordinary addition and scalar multiplication of functions Second we can multiply two func-tions together

(fg)(p) = f(p)g(p)

which turns C(X) into whatrsquos called an algebra over C (We wonrsquot give the definition of anldquoalgebrardquo but the point is that this multiplication is in some sense ldquocompatiblerdquo with the vectorspace structure) Next any element f of C(X) has a conjugate element f obtained by takingcomplex conjugates of the values of f

f(p) = f(p)

22

And finally we can equip C(X) with a norm via

983042f983042 = suppisinX

|f(p)|

where |middot| denotes complex absolute value (This supremum exists sinceX is compact by the ExtremeValue Theorem) All of these structures (vector space algebra multiplication conjugation norm)turn C(X) into whatrsquos called a Clowast-algebra We wonrsquot define this precisely but the definition encodesvarious ways in which these structures are compatible Actually C(X) is a unital commutative Clowast-algebra where ldquounitalrdquo means that the multiplication has an identity element (namely the constantfunction 1) and commutative means that fg = gf There is a natural notion of homomorphismbetween Clowast-algebras which are maps between them which behave nicely with all the structuresinvolved In particular given a continuous map f X rarr Y between compact Hausdorff spaces weget a Clowast-algebra homomorphism flowast C(Y ) rarr C(X) given by pre-composition with f flowast sendsg isin C(Y ) to g f isin C(X)

We thus have a contravariant functor CHaus rarr ucClowast-Alg from the category of compactHausdorff spaces and the category of unital commutative Clowast-algebras It turns out that anyunital commutative Clowast-algebra arises in this way in the sense that given any unital commutativeClowast-algebra B there exists a compact Hausdorff space X such that C(X) sim= B as Clowast-algebrasand that morphisms between these algebra arises from continuous maps in the manner describeabove This says that the functor above is essentially surjective full and it turns out that it isalso faithful so that it is an equivalence Thus all information about compact Hausdorff spaces iscompletely encoded in information about unital commtuative Clowast-algebras and indeed one couldlearn everything there is to know about a compact Hausdorff space X from its Clowast-algebra C(X)of functions The fact that the categories CHaus and ucClowast-Alg are equivalent is the startingpoint in the subject known as noncommutative geometry which wersquoll say something about a bitlater on If we drop the requirement that our algebras be unital it turns out that the categoryof commutative Clowast-algebras in general is equivalent to the category of locally compact Hausdorffspaces via essentially the same functor where we take as morphisms in the category of locallycompact Hausdorff spaces to be continuous functions which ldquovanish at infinityrdquo

We are only introducing Clowast-algebras to give an example of an interesting and actually usefulequivalence between categories but just for the fun of it wersquoll note the following Examples ofnoncommutative Clowast-algebras come from what are known as ldquoalgebras of bounded operators onHilbert spacesrdquo which show up in quantum mechanics as the things which describe physicallyobservable quantities like position momentum and energy (In fact all Clowast-algebras arise fromHilbert spaces in this way) Thus Clowast-algebras show up in mathematical formulations of quantummechanics and the notion of noncommutative geometry mentioned above is at the core of someattempts to understand the elusive concept known as ldquoquantum geometryrdquo

Yoneda Lemma The Yoneda Lemma states that given a (covariant) functor F C rarr Set whereC is locally small for any A isin Ob(C) there exists a bijection

F (A) sim= Nat(hA F )

where the right side denotes the set of natural transformations from hA to F A similar result holdsin the contravariant case where we get bijections F (A) sim= Nat(hA F ) (There is also a sense inwhich these bijections are themselves natural but wersquoll ignore this bit) The proof of the YonedaLemma was the topic of a student presentation and can be found in Section 22 of the book

Here we will get a bit philosophical and talk about what the point of the Yoneda Lemmaactually is and later the sense in which it says that we can view arbitrary functors as ldquonon-existent

23

objectsrdquo in a category First we mention what we will call the ldquofunctor of pointsrdquo approach tomathematics a mathematical object can be fully characterized by all morphisms into it That isknowledge of Mor(SA) for all S is equivalent to knowledge of A itself Indeed this is an idea wementioned previously in relation to the functors hAmdashwhich we previously said is referred to as beingthe functor of points of Amdashand now we seek to push this idea further In some examples literalpoints of A can be extracted from hA as in the case of the bijection X sim= hX(pt) = Map(ptX) fora topological space X or G sim= hG(Z) = Mor(Z G) for a group G Moreover in these examples therest of the (topological) structure onX or (group) structure on G can also be extracted from hX andhG respectively In general we can view Mor(SA) as in a sense being ldquopoints of A parameterizedby Srdquo which comes from the idea that in say Set Mor(SA) = AS is ldquothe product of S-manycopies of Ardquo so literally points of A ldquoparameterizedrdquo by S In various areas of mathematics anelement of Mor(SA) is thought of as being a ldquogeneralizedrdquo point of A so that the functor of pointshA detects generalized points of A

So an object A is fully encoded by the functor hA Now we claim that morphisms betweenobjects are also encoded by such functors Indeed in the case where F = hB is itself a functor ofpoints the Yoneda result Nat(hA F ) sim= F (A) becomes

Nat(hA hB) sim= hB(A) = Mor(AB)

which says precisely that morphisms between the objects A and B can be obtained solely fromthe functors hA and hB (This fact is one reason why one should prefer the contravariant functorhA to the covariant functor hA in the covariant case Nat(hA F ) sim= F (A) for F = hB becomesNat(hA hB) sim= hB(A) = Mor(BA) so that morphisms between two objects get reversed whencharacterizing them in terms of representable functors)

Pushing our philosophy further we now know that not only objects but also morphisms betweenobjects are fully encoded by the functors hA The final step is to note that we should not only careabout objects and morphisms between them but that an important part of the data of an objectA is also how arbitrary contravariant functors F C rarr Set behave on A and that this too can beobtained from hA precisely via the Yoneda result that F (A) sim= Nat(hA F ) So objects morphismsbetween objects and how arbitrary functors behave on objects are all encoded in knowledge ofthe functors hA so that these functors truly do capture in the most general sense possible allinformation about A

Functors as nonexistent objects The discussion above is all well and good but it still doesnrsquotquite say why we should care about the fact that we can study on object A by studying hA insteadin that we still havenrsquot spoken about what type of insight about A can be gleamed from hA thatwasnrsquot readily apparent in A itself already Now we argue that the true point of the YonedaLemma is that it gives a way to think about arbitrary functors F Cop rarr Set as ldquoobjectsrdquo in C

itself or really as objects in an ldquoenlargementrdquo of C Certainly for those functors F sim= hA which arerepresentable we can indeed think about F as simply being the object A but now we claim thateven non-representable F can be thought of as being ldquononexistentrdquo objects in C The upshot is thatoften in mathematics we come across things which do not actually exist but the Yoneda Lemmagives a way to (indirectly) study such things anyway by interpreting them as functors instead

The starting point of course is that an actual object A is the same as hA as we have explainedabove Thus by analogy we can interpret a non-representable functor F as being an ldquoobjectrdquo in C

as long as we know what it means to ldquomaprdquo into F from other things which are honestly objects in CThat is if F were indeed representable then F (A) sim= Nat(hA F ) sim= Mor(AF ) would characterizethe functor of points of F (here we abuse notation by using the same letter F to denote both arepresentable functor and the object which represents it) so now we take Mor(AF ) = F (A) as

24

the definition of the ldquofunctor of pointsrdquo of the non-existent object F We are thus taking elementsof F (A) as being what it means to ldquomaprdquo from A into F even when F is not representable Thefact that the Yoneda Lemma gives a way to define maps from A into F thus says that F itself isequipped with the same data that a literal object of A would be equipped with so why shouldnrsquotwe interpret F as a being a type of ldquoobjectrdquo in C as well

To phrase this another way note that the Yoneda Lemma gives rise to a functor

C rarr Fun(CopSet)

where Fun(CopSet) denotes the category whose objects are functors from Cop to Set and morphismsare natural transformation defined on objects by A 983041rarr hA and on morphisms by the bijectionMor(AB) sim= Nat(hA hB) arising from the Yoneda Lemma That this is actually bijection saysthat this functor is full and faithful so that we can view C as being a subcategory of the functorcategory Fun(CopSet) (We cay say that C is isomorphic to the subcategory of Fun(CopSet)consisting of the functors hA or that it is equivalent to the subcategory of Fun(CopSet) consistingof representable functors This functor C rarr Fun(CopSet) is usually called the Yoneda embeddingof C into Fun(CopSet)) Thus we can view Fun(CopSet) as an ldquoenlargmentrdquo of C and the ideawe are trying to make sense of is that other objects in this ldquolargerrdquo category should be interpretedas ldquonon-existent objectsrdquo of C

Just how far can we actually push this idea that arbitrary functors Cop rarr Set should be thoughtof as being ldquoobjectsrdquo of C Pretty far in fact For instance often times in geometry one comesacross a type of ldquosmoothrdquo space on which some group acts and one is interested in doing ldquocalculusrdquoon the resulting quotient space which is the space of orbits under the given group action Theproblem is that such quotient spaces are very often (usually in fact) not ldquosmoothrdquo themselves sothat the tools from calculus are not readily available The solution is to nonetheless treat thisquotient as if it were ldquosmoothrdquo and attempt to do calculus anyway To make this precise thisldquonon-smoothrdquo space actually turns out to have a natural interpretation as a ldquosmooth functorrdquo andit is through this functor which we can view this non-smooth space as actually being smooth afterall so we are forced to expand our notion of what we mean by ldquosmooth spacerdquo by allowing thingswhich are not spaces in the usual sense which certainly leads to more abstraction but the benefitis that it actually makes sense to do calculus on such ldquospacesrdquo interpreted as functors

The discussion above is pretty vague so here is one final more concrete example Recall thatequivalence

CHausop rarr ucClowast-Alg

between the category of compact Hausdorff spaces and the category of unital commutative Clowast-algebras mentioned at the start We now claim that a non-commutative (unital) Clowast-algebra shouldbe interpreted as the algebra of functions on some non-existent ldquononcommutative compact Haus-dorff spacerdquo and it is this idea which the subject of noncommutative geometry makes precise Solet B be a non-commutative Clowast-algebra Then the functor of points hB gives a functor

ucClowast-Alg rarr Set

which sends C(X) (where X is some compact Hausdorff space) to Hom(C(X) B) where Hom de-notes homomorphisms of Clowast-algebras whatever that means Composing with the previous functorwe get a functor wersquoll denote by B itself

B CHausop rarr Set

which on objects at least is given by X 983041rarr Hom(C(X) B) This functor is not representable sinceB is not commutative but the philosophy espoused above suggests we should think of it as giving a

25

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

where Op(X) denotes the set of open subsets of X ie the topology on X Hence the topology onX can be extracted from hX as claimed so hX and hX do encode all information about X

This fact can also be interpreted as saying that the contravariant functor which sends a topo-logical space X to its collection of open subsets (and which sends a continuous map to the mapinduced by taking preimages) is representable with representing object 0 1 equipped with thetopology given above This will be worked out in a homework problem

Forgetful on Grp The forgetful functor Grp rarr Set is also representable In this case arepresenting object should be a group giving set-theoretic bijections

G sim= Hom(objectG)

for all groups G where Hom denotes the set of group homomorphisms The group Z works

G sim= Hom(Z G)

since a homomorphism φ Z rarr G is completely determined by φ(1) since Z is generated by 1 andthere are no restrictions on what φ(1) isin G can actually be The inverse of the desired bijection isthus φ 983041rarr φ(1) and it is simple to check that morphisms behave appropriately so that the forgetfulfunctor in this setting is indeed isomoprhic to hZ

Forgetful on Ring The forgetful functor Ring rarr Set is also representable where by Ring wemean the category of rings with identity elements and where morphisms (ie ring homomorphisms)are required to preserve identities Here we need an object satisfying

R sim= Hom(object R)

for all R isin Ob(Ring) and we can see that Z[x] the ring of polynomials in the variable x withcoefficients in Z works Indeed given a ring homomorphism φ Z[x] rarr R the fact that φ isrequired to preserve identities fully determines φ(n) for any n isin Z (since φ(n) = nφ(1)) and hencesince φ preserves multiplication its behavior is fully determined by φ(x) which can be any elementof R The inverse of the bijection we want is thus φ 983041rarr φ(x) and it will follow that this forgetfulfunctor is isomorphic to hZ[x]

Equivalences between Categories

Definitions iso and equiv Now that we have a notion of a ldquomorphismrdquo between categorieswe can talk about the sense in which two categories are the ldquosamerdquo The first possible definitionis the ldquoobviousrdquo one you might expect given the definition of ldquoisomorphismrdquo wersquove seen in everyother context until now we say that C is isomorphic to D if there exist functors F C rarr D andG D rarr C such that G F = idC and F G = idD where idC and idD are identity functors

However this definition turns out to be too restrictive Indeed the requirement that the givencompositions equal identity functors says that given an object A in C the object G(F (A)) beliterally the same as A itself and similarly for objects in D But we know that two objects in acategory can be thought of as being the ldquosamerdquo without being literally the same as long as theyare isomorphic For instance productscoproducts are not unique in the literal sense but only inthe sense that they are isomorphic in a unique way To take another example the definition ofa functor being representable does not say that F (B) literally be the same as hB(A) only thatthey be isomorphic for sure the power set P (S) of a set S is not literally the same as the set offunctions S rarr 0 1mdashit is only the ldquosamerdquo in the sense that they encode the same data

20

Thus it makes sense to relax the requirement that A andG(F (A)) in the definition of isomorphiccategories be equal to the requirement that they be isomorphic This leads to the following betterdefinition of two categories being the ldquosamerdquo C is equivalent to D if there exist functors F C rarr D

and G D rarr C such that G F sim= idC and F G sim= idD So we require only that for instance thefunctor G F be naturally isomorphic to the identity functor on C which says that for any objectA G(F (A)) be isomorphic to A in a natural way Wersquoll see that this truly is the better notionof what it means for two categories to be the ldquosamerdquo and it reflects the idea that all categoricalproperties of one category can be transported to the other

Alternate characterization of equivalences A functor F C rarr D implementing an equiv-alence between categories is called unsurprisingly and equivalence and the ldquoinverserdquo functorG D rarr C is still referred to as being its inverse even though it is not technically an ldquoinverserdquoin the sense that compositions give literal identities But as in the case of Set where ldquoinvertiblerdquocan be characterized without reference to an inverse as ldquoinjectiverdquo and ldquosurjectiverdquo so too canequivalences between categories be characterized without explicit reference to an inverse

A functor F C rarr D is an equivalence if and only it is full faithful and essentiallysurjective

We have seen the notions of full and faithful before and essentially surjective means that any objectin D is isomorphic to something in the image of F for all D isin Ob(D) there exits C isin Ob(C) suchthat F (C) sim= D Justifying this result was the topic of a student presentation and can be foundin the book in Section 15 (Technically to construct an ldquoinverserdquo of F requires that we work withsmall categories but whatever) The bulk of the real work goes into showing that the ldquoinverserdquofunctor constructed is actually a functor which comes down to some abstract nonsense argumentvia commutative diagrams

Equivalences preserve everything We previously saw the notions of full faithful and es-sentially surjective back when thinking about what properties a functor could have which wouldguarantee it sent products to products so the result we proved back then was that equivalencespreserve products Similarly equivalences preserve coproducts and the proof is basically the sameafter reversing arrows A clean way of deriving this is as follows if F C rarr D is an equivalencethen so is F op Cop rarr Dop so since F op preserves products F preserves coproducts becausecoproducts in a category become products in the opposite category

In general an equivalence between categories will preserve whatever categorical notions youwant monomorphisms epimorphisms initial objects terminal objects etc so that equivalentcategories really do have essentially the ldquosamerdquo properties

Equivalence example Here is a first example (also found in the book) of equivalent categorieswhich are not isomorphic Let FVectR be the category whose objects are finite-dimensional realvector spaces and whose morphisms are linear transformations Let MatR be the category ofmatrices defined as follows

bull objects in MatR are nonnegative integers

bull a morphism n rarr m is an mtimes n matrix with real entries

bull composition in MatR is given by matrix multiplication given B n rarr m and A m rarr kmdashsoB is an mtimes n matrix and A is a k timesm matrixmdashthe composition A B n rarr k is the k times nmatrix AB

bull identity morphisms in MatR are given by identity matrices

21

We claim that FVectR and MatR are equivalent First note that they certainly cannot be isomo-prhic isomorphisms in MatR exist only between two objects which are literally the same (ie anmtimesn matrix can be invertible only when m = n) but isomorphisms in FVectn can exist betweenobjects which are not literally the same So there are in a sense not enough objects in MatR toallow it to be isomorphic to FVectR

To construct an equivalence choose one and for all an isomorphism TV V rarr RdimV forevery object in FVectR or equivalently choose once and for all a basis for every V DefineF FVectR rarr MatR by

V 983041rarr dimV on objects and

(T V rarr W ) 983041rarr (the standard matrix of SW T Sminus1V RdimV rarr RdimW ) on morphisms

Equivalently F sends T to the matrix of T relative to the chosen bases for V and W It isstraightforward to check that this is a functor The inverse functor G MatR rarr VectR is givenby

n 983041rarr Rn on objects and

(mtimes n matrix A) 983041rarr (the linear transformation A Rn rarr Rm induced by A) on morphisms

One can check that G is a functor and that F G and G F are naturally isomorphic to identitiesor instead one can argue that F alone is full faithful and essentially surjective The point is thatG(F (V )) gives back RdimV so not literally V itself but only something isomorphic to V which iswhy this gives an equivalence of categories and not an isomorphism

Yoneda Lemma Functors as Objects

Compact Hausdorff spaces and Clowast-algebras Last time we gave an example of equivalentcategories which were not isomorphic but in many ways that example is unsatisfactory when doinglinear algebra people for sure often work with matrix representations of linear transformations asthe example from last would suggest but no one in their right mind literally thinks about suchthings in terms of the equivalence FVectR rarr MatR itself In other words this equivalence isessentially ldquocooked uprdquo for the sake of giving an example but it does not really give any insightinto the actual mathematics of linear algebra

So with that in mind we will describe another equivalence which actually does have some realmeaning in the sense of producing new ways of thinking about some mathematics Given a compactHausdorff space X we let C(X) denote the space of continuous complex-valued functions on X

C(X) = f X rarr C | f is continuous

We can equip the set C(X) with various additional structures First it is a complex vector spaceunder ordinary addition and scalar multiplication of functions Second we can multiply two func-tions together

(fg)(p) = f(p)g(p)

which turns C(X) into whatrsquos called an algebra over C (We wonrsquot give the definition of anldquoalgebrardquo but the point is that this multiplication is in some sense ldquocompatiblerdquo with the vectorspace structure) Next any element f of C(X) has a conjugate element f obtained by takingcomplex conjugates of the values of f

f(p) = f(p)

22

And finally we can equip C(X) with a norm via

983042f983042 = suppisinX

|f(p)|

where |middot| denotes complex absolute value (This supremum exists sinceX is compact by the ExtremeValue Theorem) All of these structures (vector space algebra multiplication conjugation norm)turn C(X) into whatrsquos called a Clowast-algebra We wonrsquot define this precisely but the definition encodesvarious ways in which these structures are compatible Actually C(X) is a unital commutative Clowast-algebra where ldquounitalrdquo means that the multiplication has an identity element (namely the constantfunction 1) and commutative means that fg = gf There is a natural notion of homomorphismbetween Clowast-algebras which are maps between them which behave nicely with all the structuresinvolved In particular given a continuous map f X rarr Y between compact Hausdorff spaces weget a Clowast-algebra homomorphism flowast C(Y ) rarr C(X) given by pre-composition with f flowast sendsg isin C(Y ) to g f isin C(X)

We thus have a contravariant functor CHaus rarr ucClowast-Alg from the category of compactHausdorff spaces and the category of unital commutative Clowast-algebras It turns out that anyunital commutative Clowast-algebra arises in this way in the sense that given any unital commutativeClowast-algebra B there exists a compact Hausdorff space X such that C(X) sim= B as Clowast-algebrasand that morphisms between these algebra arises from continuous maps in the manner describeabove This says that the functor above is essentially surjective full and it turns out that it isalso faithful so that it is an equivalence Thus all information about compact Hausdorff spaces iscompletely encoded in information about unital commtuative Clowast-algebras and indeed one couldlearn everything there is to know about a compact Hausdorff space X from its Clowast-algebra C(X)of functions The fact that the categories CHaus and ucClowast-Alg are equivalent is the startingpoint in the subject known as noncommutative geometry which wersquoll say something about a bitlater on If we drop the requirement that our algebras be unital it turns out that the categoryof commutative Clowast-algebras in general is equivalent to the category of locally compact Hausdorffspaces via essentially the same functor where we take as morphisms in the category of locallycompact Hausdorff spaces to be continuous functions which ldquovanish at infinityrdquo

We are only introducing Clowast-algebras to give an example of an interesting and actually usefulequivalence between categories but just for the fun of it wersquoll note the following Examples ofnoncommutative Clowast-algebras come from what are known as ldquoalgebras of bounded operators onHilbert spacesrdquo which show up in quantum mechanics as the things which describe physicallyobservable quantities like position momentum and energy (In fact all Clowast-algebras arise fromHilbert spaces in this way) Thus Clowast-algebras show up in mathematical formulations of quantummechanics and the notion of noncommutative geometry mentioned above is at the core of someattempts to understand the elusive concept known as ldquoquantum geometryrdquo

Yoneda Lemma The Yoneda Lemma states that given a (covariant) functor F C rarr Set whereC is locally small for any A isin Ob(C) there exists a bijection

F (A) sim= Nat(hA F )

where the right side denotes the set of natural transformations from hA to F A similar result holdsin the contravariant case where we get bijections F (A) sim= Nat(hA F ) (There is also a sense inwhich these bijections are themselves natural but wersquoll ignore this bit) The proof of the YonedaLemma was the topic of a student presentation and can be found in Section 22 of the book

Here we will get a bit philosophical and talk about what the point of the Yoneda Lemmaactually is and later the sense in which it says that we can view arbitrary functors as ldquonon-existent

23

objectsrdquo in a category First we mention what we will call the ldquofunctor of pointsrdquo approach tomathematics a mathematical object can be fully characterized by all morphisms into it That isknowledge of Mor(SA) for all S is equivalent to knowledge of A itself Indeed this is an idea wementioned previously in relation to the functors hAmdashwhich we previously said is referred to as beingthe functor of points of Amdashand now we seek to push this idea further In some examples literalpoints of A can be extracted from hA as in the case of the bijection X sim= hX(pt) = Map(ptX) fora topological space X or G sim= hG(Z) = Mor(Z G) for a group G Moreover in these examples therest of the (topological) structure onX or (group) structure on G can also be extracted from hX andhG respectively In general we can view Mor(SA) as in a sense being ldquopoints of A parameterizedby Srdquo which comes from the idea that in say Set Mor(SA) = AS is ldquothe product of S-manycopies of Ardquo so literally points of A ldquoparameterizedrdquo by S In various areas of mathematics anelement of Mor(SA) is thought of as being a ldquogeneralizedrdquo point of A so that the functor of pointshA detects generalized points of A

So an object A is fully encoded by the functor hA Now we claim that morphisms betweenobjects are also encoded by such functors Indeed in the case where F = hB is itself a functor ofpoints the Yoneda result Nat(hA F ) sim= F (A) becomes

Nat(hA hB) sim= hB(A) = Mor(AB)

which says precisely that morphisms between the objects A and B can be obtained solely fromthe functors hA and hB (This fact is one reason why one should prefer the contravariant functorhA to the covariant functor hA in the covariant case Nat(hA F ) sim= F (A) for F = hB becomesNat(hA hB) sim= hB(A) = Mor(BA) so that morphisms between two objects get reversed whencharacterizing them in terms of representable functors)

Pushing our philosophy further we now know that not only objects but also morphisms betweenobjects are fully encoded by the functors hA The final step is to note that we should not only careabout objects and morphisms between them but that an important part of the data of an objectA is also how arbitrary contravariant functors F C rarr Set behave on A and that this too can beobtained from hA precisely via the Yoneda result that F (A) sim= Nat(hA F ) So objects morphismsbetween objects and how arbitrary functors behave on objects are all encoded in knowledge ofthe functors hA so that these functors truly do capture in the most general sense possible allinformation about A

Functors as nonexistent objects The discussion above is all well and good but it still doesnrsquotquite say why we should care about the fact that we can study on object A by studying hA insteadin that we still havenrsquot spoken about what type of insight about A can be gleamed from hA thatwasnrsquot readily apparent in A itself already Now we argue that the true point of the YonedaLemma is that it gives a way to think about arbitrary functors F Cop rarr Set as ldquoobjectsrdquo in C

itself or really as objects in an ldquoenlargementrdquo of C Certainly for those functors F sim= hA which arerepresentable we can indeed think about F as simply being the object A but now we claim thateven non-representable F can be thought of as being ldquononexistentrdquo objects in C The upshot is thatoften in mathematics we come across things which do not actually exist but the Yoneda Lemmagives a way to (indirectly) study such things anyway by interpreting them as functors instead

The starting point of course is that an actual object A is the same as hA as we have explainedabove Thus by analogy we can interpret a non-representable functor F as being an ldquoobjectrdquo in C

as long as we know what it means to ldquomaprdquo into F from other things which are honestly objects in CThat is if F were indeed representable then F (A) sim= Nat(hA F ) sim= Mor(AF ) would characterizethe functor of points of F (here we abuse notation by using the same letter F to denote both arepresentable functor and the object which represents it) so now we take Mor(AF ) = F (A) as

24

the definition of the ldquofunctor of pointsrdquo of the non-existent object F We are thus taking elementsof F (A) as being what it means to ldquomaprdquo from A into F even when F is not representable Thefact that the Yoneda Lemma gives a way to define maps from A into F thus says that F itself isequipped with the same data that a literal object of A would be equipped with so why shouldnrsquotwe interpret F as a being a type of ldquoobjectrdquo in C as well

To phrase this another way note that the Yoneda Lemma gives rise to a functor

C rarr Fun(CopSet)

where Fun(CopSet) denotes the category whose objects are functors from Cop to Set and morphismsare natural transformation defined on objects by A 983041rarr hA and on morphisms by the bijectionMor(AB) sim= Nat(hA hB) arising from the Yoneda Lemma That this is actually bijection saysthat this functor is full and faithful so that we can view C as being a subcategory of the functorcategory Fun(CopSet) (We cay say that C is isomorphic to the subcategory of Fun(CopSet)consisting of the functors hA or that it is equivalent to the subcategory of Fun(CopSet) consistingof representable functors This functor C rarr Fun(CopSet) is usually called the Yoneda embeddingof C into Fun(CopSet)) Thus we can view Fun(CopSet) as an ldquoenlargmentrdquo of C and the ideawe are trying to make sense of is that other objects in this ldquolargerrdquo category should be interpretedas ldquonon-existent objectsrdquo of C

Just how far can we actually push this idea that arbitrary functors Cop rarr Set should be thoughtof as being ldquoobjectsrdquo of C Pretty far in fact For instance often times in geometry one comesacross a type of ldquosmoothrdquo space on which some group acts and one is interested in doing ldquocalculusrdquoon the resulting quotient space which is the space of orbits under the given group action Theproblem is that such quotient spaces are very often (usually in fact) not ldquosmoothrdquo themselves sothat the tools from calculus are not readily available The solution is to nonetheless treat thisquotient as if it were ldquosmoothrdquo and attempt to do calculus anyway To make this precise thisldquonon-smoothrdquo space actually turns out to have a natural interpretation as a ldquosmooth functorrdquo andit is through this functor which we can view this non-smooth space as actually being smooth afterall so we are forced to expand our notion of what we mean by ldquosmooth spacerdquo by allowing thingswhich are not spaces in the usual sense which certainly leads to more abstraction but the benefitis that it actually makes sense to do calculus on such ldquospacesrdquo interpreted as functors

The discussion above is pretty vague so here is one final more concrete example Recall thatequivalence

CHausop rarr ucClowast-Alg

between the category of compact Hausdorff spaces and the category of unital commutative Clowast-algebras mentioned at the start We now claim that a non-commutative (unital) Clowast-algebra shouldbe interpreted as the algebra of functions on some non-existent ldquononcommutative compact Haus-dorff spacerdquo and it is this idea which the subject of noncommutative geometry makes precise Solet B be a non-commutative Clowast-algebra Then the functor of points hB gives a functor

ucClowast-Alg rarr Set

which sends C(X) (where X is some compact Hausdorff space) to Hom(C(X) B) where Hom de-notes homomorphisms of Clowast-algebras whatever that means Composing with the previous functorwe get a functor wersquoll denote by B itself

B CHausop rarr Set

which on objects at least is given by X 983041rarr Hom(C(X) B) This functor is not representable sinceB is not commutative but the philosophy espoused above suggests we should think of it as giving a

25

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

Thus it makes sense to relax the requirement that A andG(F (A)) in the definition of isomorphiccategories be equal to the requirement that they be isomorphic This leads to the following betterdefinition of two categories being the ldquosamerdquo C is equivalent to D if there exist functors F C rarr D

and G D rarr C such that G F sim= idC and F G sim= idD So we require only that for instance thefunctor G F be naturally isomorphic to the identity functor on C which says that for any objectA G(F (A)) be isomorphic to A in a natural way Wersquoll see that this truly is the better notionof what it means for two categories to be the ldquosamerdquo and it reflects the idea that all categoricalproperties of one category can be transported to the other

Alternate characterization of equivalences A functor F C rarr D implementing an equiv-alence between categories is called unsurprisingly and equivalence and the ldquoinverserdquo functorG D rarr C is still referred to as being its inverse even though it is not technically an ldquoinverserdquoin the sense that compositions give literal identities But as in the case of Set where ldquoinvertiblerdquocan be characterized without reference to an inverse as ldquoinjectiverdquo and ldquosurjectiverdquo so too canequivalences between categories be characterized without explicit reference to an inverse

A functor F C rarr D is an equivalence if and only it is full faithful and essentiallysurjective

We have seen the notions of full and faithful before and essentially surjective means that any objectin D is isomorphic to something in the image of F for all D isin Ob(D) there exits C isin Ob(C) suchthat F (C) sim= D Justifying this result was the topic of a student presentation and can be foundin the book in Section 15 (Technically to construct an ldquoinverserdquo of F requires that we work withsmall categories but whatever) The bulk of the real work goes into showing that the ldquoinverserdquofunctor constructed is actually a functor which comes down to some abstract nonsense argumentvia commutative diagrams

Equivalences preserve everything We previously saw the notions of full faithful and es-sentially surjective back when thinking about what properties a functor could have which wouldguarantee it sent products to products so the result we proved back then was that equivalencespreserve products Similarly equivalences preserve coproducts and the proof is basically the sameafter reversing arrows A clean way of deriving this is as follows if F C rarr D is an equivalencethen so is F op Cop rarr Dop so since F op preserves products F preserves coproducts becausecoproducts in a category become products in the opposite category

In general an equivalence between categories will preserve whatever categorical notions youwant monomorphisms epimorphisms initial objects terminal objects etc so that equivalentcategories really do have essentially the ldquosamerdquo properties

Equivalence example Here is a first example (also found in the book) of equivalent categorieswhich are not isomorphic Let FVectR be the category whose objects are finite-dimensional realvector spaces and whose morphisms are linear transformations Let MatR be the category ofmatrices defined as follows

bull objects in MatR are nonnegative integers

bull a morphism n rarr m is an mtimes n matrix with real entries

bull composition in MatR is given by matrix multiplication given B n rarr m and A m rarr kmdashsoB is an mtimes n matrix and A is a k timesm matrixmdashthe composition A B n rarr k is the k times nmatrix AB

bull identity morphisms in MatR are given by identity matrices

21

We claim that FVectR and MatR are equivalent First note that they certainly cannot be isomo-prhic isomorphisms in MatR exist only between two objects which are literally the same (ie anmtimesn matrix can be invertible only when m = n) but isomorphisms in FVectn can exist betweenobjects which are not literally the same So there are in a sense not enough objects in MatR toallow it to be isomorphic to FVectR

To construct an equivalence choose one and for all an isomorphism TV V rarr RdimV forevery object in FVectR or equivalently choose once and for all a basis for every V DefineF FVectR rarr MatR by

V 983041rarr dimV on objects and

(T V rarr W ) 983041rarr (the standard matrix of SW T Sminus1V RdimV rarr RdimW ) on morphisms

Equivalently F sends T to the matrix of T relative to the chosen bases for V and W It isstraightforward to check that this is a functor The inverse functor G MatR rarr VectR is givenby

n 983041rarr Rn on objects and

(mtimes n matrix A) 983041rarr (the linear transformation A Rn rarr Rm induced by A) on morphisms

One can check that G is a functor and that F G and G F are naturally isomorphic to identitiesor instead one can argue that F alone is full faithful and essentially surjective The point is thatG(F (V )) gives back RdimV so not literally V itself but only something isomorphic to V which iswhy this gives an equivalence of categories and not an isomorphism

Yoneda Lemma Functors as Objects

Compact Hausdorff spaces and Clowast-algebras Last time we gave an example of equivalentcategories which were not isomorphic but in many ways that example is unsatisfactory when doinglinear algebra people for sure often work with matrix representations of linear transformations asthe example from last would suggest but no one in their right mind literally thinks about suchthings in terms of the equivalence FVectR rarr MatR itself In other words this equivalence isessentially ldquocooked uprdquo for the sake of giving an example but it does not really give any insightinto the actual mathematics of linear algebra

So with that in mind we will describe another equivalence which actually does have some realmeaning in the sense of producing new ways of thinking about some mathematics Given a compactHausdorff space X we let C(X) denote the space of continuous complex-valued functions on X

C(X) = f X rarr C | f is continuous

We can equip the set C(X) with various additional structures First it is a complex vector spaceunder ordinary addition and scalar multiplication of functions Second we can multiply two func-tions together

(fg)(p) = f(p)g(p)

which turns C(X) into whatrsquos called an algebra over C (We wonrsquot give the definition of anldquoalgebrardquo but the point is that this multiplication is in some sense ldquocompatiblerdquo with the vectorspace structure) Next any element f of C(X) has a conjugate element f obtained by takingcomplex conjugates of the values of f

f(p) = f(p)

22

And finally we can equip C(X) with a norm via

983042f983042 = suppisinX

|f(p)|

where |middot| denotes complex absolute value (This supremum exists sinceX is compact by the ExtremeValue Theorem) All of these structures (vector space algebra multiplication conjugation norm)turn C(X) into whatrsquos called a Clowast-algebra We wonrsquot define this precisely but the definition encodesvarious ways in which these structures are compatible Actually C(X) is a unital commutative Clowast-algebra where ldquounitalrdquo means that the multiplication has an identity element (namely the constantfunction 1) and commutative means that fg = gf There is a natural notion of homomorphismbetween Clowast-algebras which are maps between them which behave nicely with all the structuresinvolved In particular given a continuous map f X rarr Y between compact Hausdorff spaces weget a Clowast-algebra homomorphism flowast C(Y ) rarr C(X) given by pre-composition with f flowast sendsg isin C(Y ) to g f isin C(X)

We thus have a contravariant functor CHaus rarr ucClowast-Alg from the category of compactHausdorff spaces and the category of unital commutative Clowast-algebras It turns out that anyunital commutative Clowast-algebra arises in this way in the sense that given any unital commutativeClowast-algebra B there exists a compact Hausdorff space X such that C(X) sim= B as Clowast-algebrasand that morphisms between these algebra arises from continuous maps in the manner describeabove This says that the functor above is essentially surjective full and it turns out that it isalso faithful so that it is an equivalence Thus all information about compact Hausdorff spaces iscompletely encoded in information about unital commtuative Clowast-algebras and indeed one couldlearn everything there is to know about a compact Hausdorff space X from its Clowast-algebra C(X)of functions The fact that the categories CHaus and ucClowast-Alg are equivalent is the startingpoint in the subject known as noncommutative geometry which wersquoll say something about a bitlater on If we drop the requirement that our algebras be unital it turns out that the categoryof commutative Clowast-algebras in general is equivalent to the category of locally compact Hausdorffspaces via essentially the same functor where we take as morphisms in the category of locallycompact Hausdorff spaces to be continuous functions which ldquovanish at infinityrdquo

We are only introducing Clowast-algebras to give an example of an interesting and actually usefulequivalence between categories but just for the fun of it wersquoll note the following Examples ofnoncommutative Clowast-algebras come from what are known as ldquoalgebras of bounded operators onHilbert spacesrdquo which show up in quantum mechanics as the things which describe physicallyobservable quantities like position momentum and energy (In fact all Clowast-algebras arise fromHilbert spaces in this way) Thus Clowast-algebras show up in mathematical formulations of quantummechanics and the notion of noncommutative geometry mentioned above is at the core of someattempts to understand the elusive concept known as ldquoquantum geometryrdquo

Yoneda Lemma The Yoneda Lemma states that given a (covariant) functor F C rarr Set whereC is locally small for any A isin Ob(C) there exists a bijection

F (A) sim= Nat(hA F )

where the right side denotes the set of natural transformations from hA to F A similar result holdsin the contravariant case where we get bijections F (A) sim= Nat(hA F ) (There is also a sense inwhich these bijections are themselves natural but wersquoll ignore this bit) The proof of the YonedaLemma was the topic of a student presentation and can be found in Section 22 of the book

Here we will get a bit philosophical and talk about what the point of the Yoneda Lemmaactually is and later the sense in which it says that we can view arbitrary functors as ldquonon-existent

23

objectsrdquo in a category First we mention what we will call the ldquofunctor of pointsrdquo approach tomathematics a mathematical object can be fully characterized by all morphisms into it That isknowledge of Mor(SA) for all S is equivalent to knowledge of A itself Indeed this is an idea wementioned previously in relation to the functors hAmdashwhich we previously said is referred to as beingthe functor of points of Amdashand now we seek to push this idea further In some examples literalpoints of A can be extracted from hA as in the case of the bijection X sim= hX(pt) = Map(ptX) fora topological space X or G sim= hG(Z) = Mor(Z G) for a group G Moreover in these examples therest of the (topological) structure onX or (group) structure on G can also be extracted from hX andhG respectively In general we can view Mor(SA) as in a sense being ldquopoints of A parameterizedby Srdquo which comes from the idea that in say Set Mor(SA) = AS is ldquothe product of S-manycopies of Ardquo so literally points of A ldquoparameterizedrdquo by S In various areas of mathematics anelement of Mor(SA) is thought of as being a ldquogeneralizedrdquo point of A so that the functor of pointshA detects generalized points of A

So an object A is fully encoded by the functor hA Now we claim that morphisms betweenobjects are also encoded by such functors Indeed in the case where F = hB is itself a functor ofpoints the Yoneda result Nat(hA F ) sim= F (A) becomes

Nat(hA hB) sim= hB(A) = Mor(AB)

which says precisely that morphisms between the objects A and B can be obtained solely fromthe functors hA and hB (This fact is one reason why one should prefer the contravariant functorhA to the covariant functor hA in the covariant case Nat(hA F ) sim= F (A) for F = hB becomesNat(hA hB) sim= hB(A) = Mor(BA) so that morphisms between two objects get reversed whencharacterizing them in terms of representable functors)

Pushing our philosophy further we now know that not only objects but also morphisms betweenobjects are fully encoded by the functors hA The final step is to note that we should not only careabout objects and morphisms between them but that an important part of the data of an objectA is also how arbitrary contravariant functors F C rarr Set behave on A and that this too can beobtained from hA precisely via the Yoneda result that F (A) sim= Nat(hA F ) So objects morphismsbetween objects and how arbitrary functors behave on objects are all encoded in knowledge ofthe functors hA so that these functors truly do capture in the most general sense possible allinformation about A

Functors as nonexistent objects The discussion above is all well and good but it still doesnrsquotquite say why we should care about the fact that we can study on object A by studying hA insteadin that we still havenrsquot spoken about what type of insight about A can be gleamed from hA thatwasnrsquot readily apparent in A itself already Now we argue that the true point of the YonedaLemma is that it gives a way to think about arbitrary functors F Cop rarr Set as ldquoobjectsrdquo in C

itself or really as objects in an ldquoenlargementrdquo of C Certainly for those functors F sim= hA which arerepresentable we can indeed think about F as simply being the object A but now we claim thateven non-representable F can be thought of as being ldquononexistentrdquo objects in C The upshot is thatoften in mathematics we come across things which do not actually exist but the Yoneda Lemmagives a way to (indirectly) study such things anyway by interpreting them as functors instead

The starting point of course is that an actual object A is the same as hA as we have explainedabove Thus by analogy we can interpret a non-representable functor F as being an ldquoobjectrdquo in C

as long as we know what it means to ldquomaprdquo into F from other things which are honestly objects in CThat is if F were indeed representable then F (A) sim= Nat(hA F ) sim= Mor(AF ) would characterizethe functor of points of F (here we abuse notation by using the same letter F to denote both arepresentable functor and the object which represents it) so now we take Mor(AF ) = F (A) as

24

the definition of the ldquofunctor of pointsrdquo of the non-existent object F We are thus taking elementsof F (A) as being what it means to ldquomaprdquo from A into F even when F is not representable Thefact that the Yoneda Lemma gives a way to define maps from A into F thus says that F itself isequipped with the same data that a literal object of A would be equipped with so why shouldnrsquotwe interpret F as a being a type of ldquoobjectrdquo in C as well

To phrase this another way note that the Yoneda Lemma gives rise to a functor

C rarr Fun(CopSet)

where Fun(CopSet) denotes the category whose objects are functors from Cop to Set and morphismsare natural transformation defined on objects by A 983041rarr hA and on morphisms by the bijectionMor(AB) sim= Nat(hA hB) arising from the Yoneda Lemma That this is actually bijection saysthat this functor is full and faithful so that we can view C as being a subcategory of the functorcategory Fun(CopSet) (We cay say that C is isomorphic to the subcategory of Fun(CopSet)consisting of the functors hA or that it is equivalent to the subcategory of Fun(CopSet) consistingof representable functors This functor C rarr Fun(CopSet) is usually called the Yoneda embeddingof C into Fun(CopSet)) Thus we can view Fun(CopSet) as an ldquoenlargmentrdquo of C and the ideawe are trying to make sense of is that other objects in this ldquolargerrdquo category should be interpretedas ldquonon-existent objectsrdquo of C

Just how far can we actually push this idea that arbitrary functors Cop rarr Set should be thoughtof as being ldquoobjectsrdquo of C Pretty far in fact For instance often times in geometry one comesacross a type of ldquosmoothrdquo space on which some group acts and one is interested in doing ldquocalculusrdquoon the resulting quotient space which is the space of orbits under the given group action Theproblem is that such quotient spaces are very often (usually in fact) not ldquosmoothrdquo themselves sothat the tools from calculus are not readily available The solution is to nonetheless treat thisquotient as if it were ldquosmoothrdquo and attempt to do calculus anyway To make this precise thisldquonon-smoothrdquo space actually turns out to have a natural interpretation as a ldquosmooth functorrdquo andit is through this functor which we can view this non-smooth space as actually being smooth afterall so we are forced to expand our notion of what we mean by ldquosmooth spacerdquo by allowing thingswhich are not spaces in the usual sense which certainly leads to more abstraction but the benefitis that it actually makes sense to do calculus on such ldquospacesrdquo interpreted as functors

The discussion above is pretty vague so here is one final more concrete example Recall thatequivalence

CHausop rarr ucClowast-Alg

between the category of compact Hausdorff spaces and the category of unital commutative Clowast-algebras mentioned at the start We now claim that a non-commutative (unital) Clowast-algebra shouldbe interpreted as the algebra of functions on some non-existent ldquononcommutative compact Haus-dorff spacerdquo and it is this idea which the subject of noncommutative geometry makes precise Solet B be a non-commutative Clowast-algebra Then the functor of points hB gives a functor

ucClowast-Alg rarr Set

which sends C(X) (where X is some compact Hausdorff space) to Hom(C(X) B) where Hom de-notes homomorphisms of Clowast-algebras whatever that means Composing with the previous functorwe get a functor wersquoll denote by B itself

B CHausop rarr Set

which on objects at least is given by X 983041rarr Hom(C(X) B) This functor is not representable sinceB is not commutative but the philosophy espoused above suggests we should think of it as giving a

25

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

We claim that FVectR and MatR are equivalent First note that they certainly cannot be isomo-prhic isomorphisms in MatR exist only between two objects which are literally the same (ie anmtimesn matrix can be invertible only when m = n) but isomorphisms in FVectn can exist betweenobjects which are not literally the same So there are in a sense not enough objects in MatR toallow it to be isomorphic to FVectR

To construct an equivalence choose one and for all an isomorphism TV V rarr RdimV forevery object in FVectR or equivalently choose once and for all a basis for every V DefineF FVectR rarr MatR by

V 983041rarr dimV on objects and

(T V rarr W ) 983041rarr (the standard matrix of SW T Sminus1V RdimV rarr RdimW ) on morphisms

Equivalently F sends T to the matrix of T relative to the chosen bases for V and W It isstraightforward to check that this is a functor The inverse functor G MatR rarr VectR is givenby

n 983041rarr Rn on objects and

(mtimes n matrix A) 983041rarr (the linear transformation A Rn rarr Rm induced by A) on morphisms

One can check that G is a functor and that F G and G F are naturally isomorphic to identitiesor instead one can argue that F alone is full faithful and essentially surjective The point is thatG(F (V )) gives back RdimV so not literally V itself but only something isomorphic to V which iswhy this gives an equivalence of categories and not an isomorphism

Yoneda Lemma Functors as Objects

Compact Hausdorff spaces and Clowast-algebras Last time we gave an example of equivalentcategories which were not isomorphic but in many ways that example is unsatisfactory when doinglinear algebra people for sure often work with matrix representations of linear transformations asthe example from last would suggest but no one in their right mind literally thinks about suchthings in terms of the equivalence FVectR rarr MatR itself In other words this equivalence isessentially ldquocooked uprdquo for the sake of giving an example but it does not really give any insightinto the actual mathematics of linear algebra

So with that in mind we will describe another equivalence which actually does have some realmeaning in the sense of producing new ways of thinking about some mathematics Given a compactHausdorff space X we let C(X) denote the space of continuous complex-valued functions on X

C(X) = f X rarr C | f is continuous

We can equip the set C(X) with various additional structures First it is a complex vector spaceunder ordinary addition and scalar multiplication of functions Second we can multiply two func-tions together

(fg)(p) = f(p)g(p)

which turns C(X) into whatrsquos called an algebra over C (We wonrsquot give the definition of anldquoalgebrardquo but the point is that this multiplication is in some sense ldquocompatiblerdquo with the vectorspace structure) Next any element f of C(X) has a conjugate element f obtained by takingcomplex conjugates of the values of f

f(p) = f(p)

22

And finally we can equip C(X) with a norm via

983042f983042 = suppisinX

|f(p)|

where |middot| denotes complex absolute value (This supremum exists sinceX is compact by the ExtremeValue Theorem) All of these structures (vector space algebra multiplication conjugation norm)turn C(X) into whatrsquos called a Clowast-algebra We wonrsquot define this precisely but the definition encodesvarious ways in which these structures are compatible Actually C(X) is a unital commutative Clowast-algebra where ldquounitalrdquo means that the multiplication has an identity element (namely the constantfunction 1) and commutative means that fg = gf There is a natural notion of homomorphismbetween Clowast-algebras which are maps between them which behave nicely with all the structuresinvolved In particular given a continuous map f X rarr Y between compact Hausdorff spaces weget a Clowast-algebra homomorphism flowast C(Y ) rarr C(X) given by pre-composition with f flowast sendsg isin C(Y ) to g f isin C(X)

We thus have a contravariant functor CHaus rarr ucClowast-Alg from the category of compactHausdorff spaces and the category of unital commutative Clowast-algebras It turns out that anyunital commutative Clowast-algebra arises in this way in the sense that given any unital commutativeClowast-algebra B there exists a compact Hausdorff space X such that C(X) sim= B as Clowast-algebrasand that morphisms between these algebra arises from continuous maps in the manner describeabove This says that the functor above is essentially surjective full and it turns out that it isalso faithful so that it is an equivalence Thus all information about compact Hausdorff spaces iscompletely encoded in information about unital commtuative Clowast-algebras and indeed one couldlearn everything there is to know about a compact Hausdorff space X from its Clowast-algebra C(X)of functions The fact that the categories CHaus and ucClowast-Alg are equivalent is the startingpoint in the subject known as noncommutative geometry which wersquoll say something about a bitlater on If we drop the requirement that our algebras be unital it turns out that the categoryof commutative Clowast-algebras in general is equivalent to the category of locally compact Hausdorffspaces via essentially the same functor where we take as morphisms in the category of locallycompact Hausdorff spaces to be continuous functions which ldquovanish at infinityrdquo

We are only introducing Clowast-algebras to give an example of an interesting and actually usefulequivalence between categories but just for the fun of it wersquoll note the following Examples ofnoncommutative Clowast-algebras come from what are known as ldquoalgebras of bounded operators onHilbert spacesrdquo which show up in quantum mechanics as the things which describe physicallyobservable quantities like position momentum and energy (In fact all Clowast-algebras arise fromHilbert spaces in this way) Thus Clowast-algebras show up in mathematical formulations of quantummechanics and the notion of noncommutative geometry mentioned above is at the core of someattempts to understand the elusive concept known as ldquoquantum geometryrdquo

Yoneda Lemma The Yoneda Lemma states that given a (covariant) functor F C rarr Set whereC is locally small for any A isin Ob(C) there exists a bijection

F (A) sim= Nat(hA F )

where the right side denotes the set of natural transformations from hA to F A similar result holdsin the contravariant case where we get bijections F (A) sim= Nat(hA F ) (There is also a sense inwhich these bijections are themselves natural but wersquoll ignore this bit) The proof of the YonedaLemma was the topic of a student presentation and can be found in Section 22 of the book

Here we will get a bit philosophical and talk about what the point of the Yoneda Lemmaactually is and later the sense in which it says that we can view arbitrary functors as ldquonon-existent

23

objectsrdquo in a category First we mention what we will call the ldquofunctor of pointsrdquo approach tomathematics a mathematical object can be fully characterized by all morphisms into it That isknowledge of Mor(SA) for all S is equivalent to knowledge of A itself Indeed this is an idea wementioned previously in relation to the functors hAmdashwhich we previously said is referred to as beingthe functor of points of Amdashand now we seek to push this idea further In some examples literalpoints of A can be extracted from hA as in the case of the bijection X sim= hX(pt) = Map(ptX) fora topological space X or G sim= hG(Z) = Mor(Z G) for a group G Moreover in these examples therest of the (topological) structure onX or (group) structure on G can also be extracted from hX andhG respectively In general we can view Mor(SA) as in a sense being ldquopoints of A parameterizedby Srdquo which comes from the idea that in say Set Mor(SA) = AS is ldquothe product of S-manycopies of Ardquo so literally points of A ldquoparameterizedrdquo by S In various areas of mathematics anelement of Mor(SA) is thought of as being a ldquogeneralizedrdquo point of A so that the functor of pointshA detects generalized points of A

So an object A is fully encoded by the functor hA Now we claim that morphisms betweenobjects are also encoded by such functors Indeed in the case where F = hB is itself a functor ofpoints the Yoneda result Nat(hA F ) sim= F (A) becomes

Nat(hA hB) sim= hB(A) = Mor(AB)

which says precisely that morphisms between the objects A and B can be obtained solely fromthe functors hA and hB (This fact is one reason why one should prefer the contravariant functorhA to the covariant functor hA in the covariant case Nat(hA F ) sim= F (A) for F = hB becomesNat(hA hB) sim= hB(A) = Mor(BA) so that morphisms between two objects get reversed whencharacterizing them in terms of representable functors)

Pushing our philosophy further we now know that not only objects but also morphisms betweenobjects are fully encoded by the functors hA The final step is to note that we should not only careabout objects and morphisms between them but that an important part of the data of an objectA is also how arbitrary contravariant functors F C rarr Set behave on A and that this too can beobtained from hA precisely via the Yoneda result that F (A) sim= Nat(hA F ) So objects morphismsbetween objects and how arbitrary functors behave on objects are all encoded in knowledge ofthe functors hA so that these functors truly do capture in the most general sense possible allinformation about A

Functors as nonexistent objects The discussion above is all well and good but it still doesnrsquotquite say why we should care about the fact that we can study on object A by studying hA insteadin that we still havenrsquot spoken about what type of insight about A can be gleamed from hA thatwasnrsquot readily apparent in A itself already Now we argue that the true point of the YonedaLemma is that it gives a way to think about arbitrary functors F Cop rarr Set as ldquoobjectsrdquo in C

itself or really as objects in an ldquoenlargementrdquo of C Certainly for those functors F sim= hA which arerepresentable we can indeed think about F as simply being the object A but now we claim thateven non-representable F can be thought of as being ldquononexistentrdquo objects in C The upshot is thatoften in mathematics we come across things which do not actually exist but the Yoneda Lemmagives a way to (indirectly) study such things anyway by interpreting them as functors instead

The starting point of course is that an actual object A is the same as hA as we have explainedabove Thus by analogy we can interpret a non-representable functor F as being an ldquoobjectrdquo in C

as long as we know what it means to ldquomaprdquo into F from other things which are honestly objects in CThat is if F were indeed representable then F (A) sim= Nat(hA F ) sim= Mor(AF ) would characterizethe functor of points of F (here we abuse notation by using the same letter F to denote both arepresentable functor and the object which represents it) so now we take Mor(AF ) = F (A) as

24

the definition of the ldquofunctor of pointsrdquo of the non-existent object F We are thus taking elementsof F (A) as being what it means to ldquomaprdquo from A into F even when F is not representable Thefact that the Yoneda Lemma gives a way to define maps from A into F thus says that F itself isequipped with the same data that a literal object of A would be equipped with so why shouldnrsquotwe interpret F as a being a type of ldquoobjectrdquo in C as well

To phrase this another way note that the Yoneda Lemma gives rise to a functor

C rarr Fun(CopSet)

where Fun(CopSet) denotes the category whose objects are functors from Cop to Set and morphismsare natural transformation defined on objects by A 983041rarr hA and on morphisms by the bijectionMor(AB) sim= Nat(hA hB) arising from the Yoneda Lemma That this is actually bijection saysthat this functor is full and faithful so that we can view C as being a subcategory of the functorcategory Fun(CopSet) (We cay say that C is isomorphic to the subcategory of Fun(CopSet)consisting of the functors hA or that it is equivalent to the subcategory of Fun(CopSet) consistingof representable functors This functor C rarr Fun(CopSet) is usually called the Yoneda embeddingof C into Fun(CopSet)) Thus we can view Fun(CopSet) as an ldquoenlargmentrdquo of C and the ideawe are trying to make sense of is that other objects in this ldquolargerrdquo category should be interpretedas ldquonon-existent objectsrdquo of C

Just how far can we actually push this idea that arbitrary functors Cop rarr Set should be thoughtof as being ldquoobjectsrdquo of C Pretty far in fact For instance often times in geometry one comesacross a type of ldquosmoothrdquo space on which some group acts and one is interested in doing ldquocalculusrdquoon the resulting quotient space which is the space of orbits under the given group action Theproblem is that such quotient spaces are very often (usually in fact) not ldquosmoothrdquo themselves sothat the tools from calculus are not readily available The solution is to nonetheless treat thisquotient as if it were ldquosmoothrdquo and attempt to do calculus anyway To make this precise thisldquonon-smoothrdquo space actually turns out to have a natural interpretation as a ldquosmooth functorrdquo andit is through this functor which we can view this non-smooth space as actually being smooth afterall so we are forced to expand our notion of what we mean by ldquosmooth spacerdquo by allowing thingswhich are not spaces in the usual sense which certainly leads to more abstraction but the benefitis that it actually makes sense to do calculus on such ldquospacesrdquo interpreted as functors

The discussion above is pretty vague so here is one final more concrete example Recall thatequivalence

CHausop rarr ucClowast-Alg

between the category of compact Hausdorff spaces and the category of unital commutative Clowast-algebras mentioned at the start We now claim that a non-commutative (unital) Clowast-algebra shouldbe interpreted as the algebra of functions on some non-existent ldquononcommutative compact Haus-dorff spacerdquo and it is this idea which the subject of noncommutative geometry makes precise Solet B be a non-commutative Clowast-algebra Then the functor of points hB gives a functor

ucClowast-Alg rarr Set

which sends C(X) (where X is some compact Hausdorff space) to Hom(C(X) B) where Hom de-notes homomorphisms of Clowast-algebras whatever that means Composing with the previous functorwe get a functor wersquoll denote by B itself

B CHausop rarr Set

which on objects at least is given by X 983041rarr Hom(C(X) B) This functor is not representable sinceB is not commutative but the philosophy espoused above suggests we should think of it as giving a

25

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

And finally we can equip C(X) with a norm via

983042f983042 = suppisinX

|f(p)|

where |middot| denotes complex absolute value (This supremum exists sinceX is compact by the ExtremeValue Theorem) All of these structures (vector space algebra multiplication conjugation norm)turn C(X) into whatrsquos called a Clowast-algebra We wonrsquot define this precisely but the definition encodesvarious ways in which these structures are compatible Actually C(X) is a unital commutative Clowast-algebra where ldquounitalrdquo means that the multiplication has an identity element (namely the constantfunction 1) and commutative means that fg = gf There is a natural notion of homomorphismbetween Clowast-algebras which are maps between them which behave nicely with all the structuresinvolved In particular given a continuous map f X rarr Y between compact Hausdorff spaces weget a Clowast-algebra homomorphism flowast C(Y ) rarr C(X) given by pre-composition with f flowast sendsg isin C(Y ) to g f isin C(X)

We thus have a contravariant functor CHaus rarr ucClowast-Alg from the category of compactHausdorff spaces and the category of unital commutative Clowast-algebras It turns out that anyunital commutative Clowast-algebra arises in this way in the sense that given any unital commutativeClowast-algebra B there exists a compact Hausdorff space X such that C(X) sim= B as Clowast-algebrasand that morphisms between these algebra arises from continuous maps in the manner describeabove This says that the functor above is essentially surjective full and it turns out that it isalso faithful so that it is an equivalence Thus all information about compact Hausdorff spaces iscompletely encoded in information about unital commtuative Clowast-algebras and indeed one couldlearn everything there is to know about a compact Hausdorff space X from its Clowast-algebra C(X)of functions The fact that the categories CHaus and ucClowast-Alg are equivalent is the startingpoint in the subject known as noncommutative geometry which wersquoll say something about a bitlater on If we drop the requirement that our algebras be unital it turns out that the categoryof commutative Clowast-algebras in general is equivalent to the category of locally compact Hausdorffspaces via essentially the same functor where we take as morphisms in the category of locallycompact Hausdorff spaces to be continuous functions which ldquovanish at infinityrdquo

We are only introducing Clowast-algebras to give an example of an interesting and actually usefulequivalence between categories but just for the fun of it wersquoll note the following Examples ofnoncommutative Clowast-algebras come from what are known as ldquoalgebras of bounded operators onHilbert spacesrdquo which show up in quantum mechanics as the things which describe physicallyobservable quantities like position momentum and energy (In fact all Clowast-algebras arise fromHilbert spaces in this way) Thus Clowast-algebras show up in mathematical formulations of quantummechanics and the notion of noncommutative geometry mentioned above is at the core of someattempts to understand the elusive concept known as ldquoquantum geometryrdquo

Yoneda Lemma The Yoneda Lemma states that given a (covariant) functor F C rarr Set whereC is locally small for any A isin Ob(C) there exists a bijection

F (A) sim= Nat(hA F )

where the right side denotes the set of natural transformations from hA to F A similar result holdsin the contravariant case where we get bijections F (A) sim= Nat(hA F ) (There is also a sense inwhich these bijections are themselves natural but wersquoll ignore this bit) The proof of the YonedaLemma was the topic of a student presentation and can be found in Section 22 of the book

Here we will get a bit philosophical and talk about what the point of the Yoneda Lemmaactually is and later the sense in which it says that we can view arbitrary functors as ldquonon-existent

23

objectsrdquo in a category First we mention what we will call the ldquofunctor of pointsrdquo approach tomathematics a mathematical object can be fully characterized by all morphisms into it That isknowledge of Mor(SA) for all S is equivalent to knowledge of A itself Indeed this is an idea wementioned previously in relation to the functors hAmdashwhich we previously said is referred to as beingthe functor of points of Amdashand now we seek to push this idea further In some examples literalpoints of A can be extracted from hA as in the case of the bijection X sim= hX(pt) = Map(ptX) fora topological space X or G sim= hG(Z) = Mor(Z G) for a group G Moreover in these examples therest of the (topological) structure onX or (group) structure on G can also be extracted from hX andhG respectively In general we can view Mor(SA) as in a sense being ldquopoints of A parameterizedby Srdquo which comes from the idea that in say Set Mor(SA) = AS is ldquothe product of S-manycopies of Ardquo so literally points of A ldquoparameterizedrdquo by S In various areas of mathematics anelement of Mor(SA) is thought of as being a ldquogeneralizedrdquo point of A so that the functor of pointshA detects generalized points of A

So an object A is fully encoded by the functor hA Now we claim that morphisms betweenobjects are also encoded by such functors Indeed in the case where F = hB is itself a functor ofpoints the Yoneda result Nat(hA F ) sim= F (A) becomes

Nat(hA hB) sim= hB(A) = Mor(AB)

which says precisely that morphisms between the objects A and B can be obtained solely fromthe functors hA and hB (This fact is one reason why one should prefer the contravariant functorhA to the covariant functor hA in the covariant case Nat(hA F ) sim= F (A) for F = hB becomesNat(hA hB) sim= hB(A) = Mor(BA) so that morphisms between two objects get reversed whencharacterizing them in terms of representable functors)

Pushing our philosophy further we now know that not only objects but also morphisms betweenobjects are fully encoded by the functors hA The final step is to note that we should not only careabout objects and morphisms between them but that an important part of the data of an objectA is also how arbitrary contravariant functors F C rarr Set behave on A and that this too can beobtained from hA precisely via the Yoneda result that F (A) sim= Nat(hA F ) So objects morphismsbetween objects and how arbitrary functors behave on objects are all encoded in knowledge ofthe functors hA so that these functors truly do capture in the most general sense possible allinformation about A

Functors as nonexistent objects The discussion above is all well and good but it still doesnrsquotquite say why we should care about the fact that we can study on object A by studying hA insteadin that we still havenrsquot spoken about what type of insight about A can be gleamed from hA thatwasnrsquot readily apparent in A itself already Now we argue that the true point of the YonedaLemma is that it gives a way to think about arbitrary functors F Cop rarr Set as ldquoobjectsrdquo in C

itself or really as objects in an ldquoenlargementrdquo of C Certainly for those functors F sim= hA which arerepresentable we can indeed think about F as simply being the object A but now we claim thateven non-representable F can be thought of as being ldquononexistentrdquo objects in C The upshot is thatoften in mathematics we come across things which do not actually exist but the Yoneda Lemmagives a way to (indirectly) study such things anyway by interpreting them as functors instead

The starting point of course is that an actual object A is the same as hA as we have explainedabove Thus by analogy we can interpret a non-representable functor F as being an ldquoobjectrdquo in C

as long as we know what it means to ldquomaprdquo into F from other things which are honestly objects in CThat is if F were indeed representable then F (A) sim= Nat(hA F ) sim= Mor(AF ) would characterizethe functor of points of F (here we abuse notation by using the same letter F to denote both arepresentable functor and the object which represents it) so now we take Mor(AF ) = F (A) as

24

the definition of the ldquofunctor of pointsrdquo of the non-existent object F We are thus taking elementsof F (A) as being what it means to ldquomaprdquo from A into F even when F is not representable Thefact that the Yoneda Lemma gives a way to define maps from A into F thus says that F itself isequipped with the same data that a literal object of A would be equipped with so why shouldnrsquotwe interpret F as a being a type of ldquoobjectrdquo in C as well

To phrase this another way note that the Yoneda Lemma gives rise to a functor

C rarr Fun(CopSet)

where Fun(CopSet) denotes the category whose objects are functors from Cop to Set and morphismsare natural transformation defined on objects by A 983041rarr hA and on morphisms by the bijectionMor(AB) sim= Nat(hA hB) arising from the Yoneda Lemma That this is actually bijection saysthat this functor is full and faithful so that we can view C as being a subcategory of the functorcategory Fun(CopSet) (We cay say that C is isomorphic to the subcategory of Fun(CopSet)consisting of the functors hA or that it is equivalent to the subcategory of Fun(CopSet) consistingof representable functors This functor C rarr Fun(CopSet) is usually called the Yoneda embeddingof C into Fun(CopSet)) Thus we can view Fun(CopSet) as an ldquoenlargmentrdquo of C and the ideawe are trying to make sense of is that other objects in this ldquolargerrdquo category should be interpretedas ldquonon-existent objectsrdquo of C

Just how far can we actually push this idea that arbitrary functors Cop rarr Set should be thoughtof as being ldquoobjectsrdquo of C Pretty far in fact For instance often times in geometry one comesacross a type of ldquosmoothrdquo space on which some group acts and one is interested in doing ldquocalculusrdquoon the resulting quotient space which is the space of orbits under the given group action Theproblem is that such quotient spaces are very often (usually in fact) not ldquosmoothrdquo themselves sothat the tools from calculus are not readily available The solution is to nonetheless treat thisquotient as if it were ldquosmoothrdquo and attempt to do calculus anyway To make this precise thisldquonon-smoothrdquo space actually turns out to have a natural interpretation as a ldquosmooth functorrdquo andit is through this functor which we can view this non-smooth space as actually being smooth afterall so we are forced to expand our notion of what we mean by ldquosmooth spacerdquo by allowing thingswhich are not spaces in the usual sense which certainly leads to more abstraction but the benefitis that it actually makes sense to do calculus on such ldquospacesrdquo interpreted as functors

The discussion above is pretty vague so here is one final more concrete example Recall thatequivalence

CHausop rarr ucClowast-Alg

between the category of compact Hausdorff spaces and the category of unital commutative Clowast-algebras mentioned at the start We now claim that a non-commutative (unital) Clowast-algebra shouldbe interpreted as the algebra of functions on some non-existent ldquononcommutative compact Haus-dorff spacerdquo and it is this idea which the subject of noncommutative geometry makes precise Solet B be a non-commutative Clowast-algebra Then the functor of points hB gives a functor

ucClowast-Alg rarr Set

which sends C(X) (where X is some compact Hausdorff space) to Hom(C(X) B) where Hom de-notes homomorphisms of Clowast-algebras whatever that means Composing with the previous functorwe get a functor wersquoll denote by B itself

B CHausop rarr Set

which on objects at least is given by X 983041rarr Hom(C(X) B) This functor is not representable sinceB is not commutative but the philosophy espoused above suggests we should think of it as giving a

25

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

objectsrdquo in a category First we mention what we will call the ldquofunctor of pointsrdquo approach tomathematics a mathematical object can be fully characterized by all morphisms into it That isknowledge of Mor(SA) for all S is equivalent to knowledge of A itself Indeed this is an idea wementioned previously in relation to the functors hAmdashwhich we previously said is referred to as beingthe functor of points of Amdashand now we seek to push this idea further In some examples literalpoints of A can be extracted from hA as in the case of the bijection X sim= hX(pt) = Map(ptX) fora topological space X or G sim= hG(Z) = Mor(Z G) for a group G Moreover in these examples therest of the (topological) structure onX or (group) structure on G can also be extracted from hX andhG respectively In general we can view Mor(SA) as in a sense being ldquopoints of A parameterizedby Srdquo which comes from the idea that in say Set Mor(SA) = AS is ldquothe product of S-manycopies of Ardquo so literally points of A ldquoparameterizedrdquo by S In various areas of mathematics anelement of Mor(SA) is thought of as being a ldquogeneralizedrdquo point of A so that the functor of pointshA detects generalized points of A

So an object A is fully encoded by the functor hA Now we claim that morphisms betweenobjects are also encoded by such functors Indeed in the case where F = hB is itself a functor ofpoints the Yoneda result Nat(hA F ) sim= F (A) becomes

Nat(hA hB) sim= hB(A) = Mor(AB)

which says precisely that morphisms between the objects A and B can be obtained solely fromthe functors hA and hB (This fact is one reason why one should prefer the contravariant functorhA to the covariant functor hA in the covariant case Nat(hA F ) sim= F (A) for F = hB becomesNat(hA hB) sim= hB(A) = Mor(BA) so that morphisms between two objects get reversed whencharacterizing them in terms of representable functors)

Pushing our philosophy further we now know that not only objects but also morphisms betweenobjects are fully encoded by the functors hA The final step is to note that we should not only careabout objects and morphisms between them but that an important part of the data of an objectA is also how arbitrary contravariant functors F C rarr Set behave on A and that this too can beobtained from hA precisely via the Yoneda result that F (A) sim= Nat(hA F ) So objects morphismsbetween objects and how arbitrary functors behave on objects are all encoded in knowledge ofthe functors hA so that these functors truly do capture in the most general sense possible allinformation about A

Functors as nonexistent objects The discussion above is all well and good but it still doesnrsquotquite say why we should care about the fact that we can study on object A by studying hA insteadin that we still havenrsquot spoken about what type of insight about A can be gleamed from hA thatwasnrsquot readily apparent in A itself already Now we argue that the true point of the YonedaLemma is that it gives a way to think about arbitrary functors F Cop rarr Set as ldquoobjectsrdquo in C

itself or really as objects in an ldquoenlargementrdquo of C Certainly for those functors F sim= hA which arerepresentable we can indeed think about F as simply being the object A but now we claim thateven non-representable F can be thought of as being ldquononexistentrdquo objects in C The upshot is thatoften in mathematics we come across things which do not actually exist but the Yoneda Lemmagives a way to (indirectly) study such things anyway by interpreting them as functors instead

The starting point of course is that an actual object A is the same as hA as we have explainedabove Thus by analogy we can interpret a non-representable functor F as being an ldquoobjectrdquo in C

as long as we know what it means to ldquomaprdquo into F from other things which are honestly objects in CThat is if F were indeed representable then F (A) sim= Nat(hA F ) sim= Mor(AF ) would characterizethe functor of points of F (here we abuse notation by using the same letter F to denote both arepresentable functor and the object which represents it) so now we take Mor(AF ) = F (A) as

24

the definition of the ldquofunctor of pointsrdquo of the non-existent object F We are thus taking elementsof F (A) as being what it means to ldquomaprdquo from A into F even when F is not representable Thefact that the Yoneda Lemma gives a way to define maps from A into F thus says that F itself isequipped with the same data that a literal object of A would be equipped with so why shouldnrsquotwe interpret F as a being a type of ldquoobjectrdquo in C as well

To phrase this another way note that the Yoneda Lemma gives rise to a functor

C rarr Fun(CopSet)

where Fun(CopSet) denotes the category whose objects are functors from Cop to Set and morphismsare natural transformation defined on objects by A 983041rarr hA and on morphisms by the bijectionMor(AB) sim= Nat(hA hB) arising from the Yoneda Lemma That this is actually bijection saysthat this functor is full and faithful so that we can view C as being a subcategory of the functorcategory Fun(CopSet) (We cay say that C is isomorphic to the subcategory of Fun(CopSet)consisting of the functors hA or that it is equivalent to the subcategory of Fun(CopSet) consistingof representable functors This functor C rarr Fun(CopSet) is usually called the Yoneda embeddingof C into Fun(CopSet)) Thus we can view Fun(CopSet) as an ldquoenlargmentrdquo of C and the ideawe are trying to make sense of is that other objects in this ldquolargerrdquo category should be interpretedas ldquonon-existent objectsrdquo of C

Just how far can we actually push this idea that arbitrary functors Cop rarr Set should be thoughtof as being ldquoobjectsrdquo of C Pretty far in fact For instance often times in geometry one comesacross a type of ldquosmoothrdquo space on which some group acts and one is interested in doing ldquocalculusrdquoon the resulting quotient space which is the space of orbits under the given group action Theproblem is that such quotient spaces are very often (usually in fact) not ldquosmoothrdquo themselves sothat the tools from calculus are not readily available The solution is to nonetheless treat thisquotient as if it were ldquosmoothrdquo and attempt to do calculus anyway To make this precise thisldquonon-smoothrdquo space actually turns out to have a natural interpretation as a ldquosmooth functorrdquo andit is through this functor which we can view this non-smooth space as actually being smooth afterall so we are forced to expand our notion of what we mean by ldquosmooth spacerdquo by allowing thingswhich are not spaces in the usual sense which certainly leads to more abstraction but the benefitis that it actually makes sense to do calculus on such ldquospacesrdquo interpreted as functors

The discussion above is pretty vague so here is one final more concrete example Recall thatequivalence

CHausop rarr ucClowast-Alg

between the category of compact Hausdorff spaces and the category of unital commutative Clowast-algebras mentioned at the start We now claim that a non-commutative (unital) Clowast-algebra shouldbe interpreted as the algebra of functions on some non-existent ldquononcommutative compact Haus-dorff spacerdquo and it is this idea which the subject of noncommutative geometry makes precise Solet B be a non-commutative Clowast-algebra Then the functor of points hB gives a functor

ucClowast-Alg rarr Set

which sends C(X) (where X is some compact Hausdorff space) to Hom(C(X) B) where Hom de-notes homomorphisms of Clowast-algebras whatever that means Composing with the previous functorwe get a functor wersquoll denote by B itself

B CHausop rarr Set

which on objects at least is given by X 983041rarr Hom(C(X) B) This functor is not representable sinceB is not commutative but the philosophy espoused above suggests we should think of it as giving a

25

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

the definition of the ldquofunctor of pointsrdquo of the non-existent object F We are thus taking elementsof F (A) as being what it means to ldquomaprdquo from A into F even when F is not representable Thefact that the Yoneda Lemma gives a way to define maps from A into F thus says that F itself isequipped with the same data that a literal object of A would be equipped with so why shouldnrsquotwe interpret F as a being a type of ldquoobjectrdquo in C as well

To phrase this another way note that the Yoneda Lemma gives rise to a functor

C rarr Fun(CopSet)

where Fun(CopSet) denotes the category whose objects are functors from Cop to Set and morphismsare natural transformation defined on objects by A 983041rarr hA and on morphisms by the bijectionMor(AB) sim= Nat(hA hB) arising from the Yoneda Lemma That this is actually bijection saysthat this functor is full and faithful so that we can view C as being a subcategory of the functorcategory Fun(CopSet) (We cay say that C is isomorphic to the subcategory of Fun(CopSet)consisting of the functors hA or that it is equivalent to the subcategory of Fun(CopSet) consistingof representable functors This functor C rarr Fun(CopSet) is usually called the Yoneda embeddingof C into Fun(CopSet)) Thus we can view Fun(CopSet) as an ldquoenlargmentrdquo of C and the ideawe are trying to make sense of is that other objects in this ldquolargerrdquo category should be interpretedas ldquonon-existent objectsrdquo of C

Just how far can we actually push this idea that arbitrary functors Cop rarr Set should be thoughtof as being ldquoobjectsrdquo of C Pretty far in fact For instance often times in geometry one comesacross a type of ldquosmoothrdquo space on which some group acts and one is interested in doing ldquocalculusrdquoon the resulting quotient space which is the space of orbits under the given group action Theproblem is that such quotient spaces are very often (usually in fact) not ldquosmoothrdquo themselves sothat the tools from calculus are not readily available The solution is to nonetheless treat thisquotient as if it were ldquosmoothrdquo and attempt to do calculus anyway To make this precise thisldquonon-smoothrdquo space actually turns out to have a natural interpretation as a ldquosmooth functorrdquo andit is through this functor which we can view this non-smooth space as actually being smooth afterall so we are forced to expand our notion of what we mean by ldquosmooth spacerdquo by allowing thingswhich are not spaces in the usual sense which certainly leads to more abstraction but the benefitis that it actually makes sense to do calculus on such ldquospacesrdquo interpreted as functors

The discussion above is pretty vague so here is one final more concrete example Recall thatequivalence

CHausop rarr ucClowast-Alg

between the category of compact Hausdorff spaces and the category of unital commutative Clowast-algebras mentioned at the start We now claim that a non-commutative (unital) Clowast-algebra shouldbe interpreted as the algebra of functions on some non-existent ldquononcommutative compact Haus-dorff spacerdquo and it is this idea which the subject of noncommutative geometry makes precise Solet B be a non-commutative Clowast-algebra Then the functor of points hB gives a functor

ucClowast-Alg rarr Set

which sends C(X) (where X is some compact Hausdorff space) to Hom(C(X) B) where Hom de-notes homomorphisms of Clowast-algebras whatever that means Composing with the previous functorwe get a functor wersquoll denote by B itself

B CHausop rarr Set

which on objects at least is given by X 983041rarr Hom(C(X) B) This functor is not representable sinceB is not commutative but the philosophy espoused above suggests we should think of it as giving a

25

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

way to think of B as an ldquoobjectrdquo in CHaus anyway ie as a ldquononcommutative compact Hausdorffspacerdquo We define a ldquocontinuous maprdquo X rarr B from a true compact Hausdorff space X intothis ldquonon-existentrdquo one B to be precisely the data of a Clowast-algebra homomorphism B rarr C(X)Again if B were actually commutative this would literally be true by the equivalence of categoriesgiven at the start It turns out that many constructions in CHaus can then be carried over tosuch a functor B giving a way to do ldquogeometryrdquo on this non-existent space solely in terms of thenon-commutative Clowast-algebra B itself Good stuff

Equalizers and Coequalizers

More on Yoneda and functors Let us mention two more things based on the Yoneda LemmaFirst consider what the Yoneda Lemma says in the case of the category BG for a group G In thiscase there is only one representable functor hlowast to consider where lowast is the unique object of BGThe functor hlowast BG rarr Set behaves on objects via hlowast(lowast) = Mor(lowast lowast) = G and on morphisms by

g 983041rarr (h 983041rarr hg)

where the latter is the map Mor(lowast lowast) rarr Mor(lowast lowast) given by ldquopre-compositionrdquo with g Thus thefunctor hlowast BG rarr Set is precisely the one which encodes the right action of G on itself bymultiplication The Yoneda Embedding

BG rarr Fun(BGopSet)

thus ldquoembedsrdquo this action into a larger category of ldquoactionsrdquo The image of this embedding actuallylands in the subcategory of functors F such that F (lowast) = G is the underlying set of the given groupso we get an embedding of BG into the functors which encode permutations of the set G Hencethe Yoneda Embedding in this case amounts to an injective group homomorphism G rarr Perm(G)and the fact that such a group homomorphism exists is precisely the statement of Cayleyrsquos Theoremfrom group theory In this way then the Yoneda Embedding can be viewed as a vast generalizationof Cayleyrsquos Theorem which says that an arbitrary category can be viewed as subcategory of aldquocategory of symmetriesrdquo of itself

Second if G is still a group recall that the functor hG is meant to encode the same data asG But if H is another group the set of morphisms hG(H) = Hom(HG) can itself be given agroup structure defined by pointwise multiplication the product of φψ isin hG(H) is defined by(φψ)(h) = φ(h)ψ(h) The idea is that these group structures all together uniquely characterize thegroup structure on G itself Now suppose we had a non-representable functor F Grpop rarr Setwhich we wanted to view as a ldquonon-existentrdquo object in Grp so we want to view F as a type ofldquogrouprdquo How can we make this precise

The idea is as in the case where F actually is representable a ldquogrouprdquo structure on F shouldbe characterized by group structures on each F (A) = Mor(AF ) which are all compatible in somesense But a group structure on each set F (A) amounts to functions

F (A)times F (A) rarr F (A) for each A

which satisfy the group axioms The required ldquocompatibilityrdquo of these is expressed by saying thatthe data of all such group operations forms a natural transformation F times F rArr F where F times Fdenotes the functor which acts on objects as A 983041rarr F (A) times F (A) Thus in order to think of F asa type of ldquogrouprdquo what we actually need is a ldquomultiplicationrdquo F times F rArr F expressed as a naturaltransformation which obeys something similar to the group axioms Wersquoll talk about such things

26

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

later on where the point is that F should be whatrsquos called a group object in a category of functorsWersquoll see that the definition of an ordinary group can be expressed solely in terms of arrows anddiagrams so that we can take such diagrams as the definition of ldquogrouprdquo in other categories Inthis way the idea that F can be thought of as a ldquonon-existent grouprdquo will be made precise

Universal properties Now we return to considering various constructions where the commonthread underlying all is that of defining an object via a universal property We have already seentwo examples of such properties the one defining products and the one defining coproducts In thecase of a product A times B for AB isin Ob(C) the point is that we have morphisms A times B rarr A andAtimesB rarr B and the definition of product says precisely that AtimesB equipped with these morphismsis ldquouniversalrdquo among all morphisms from an arbitrary object D into A and B in the sense that themorphisms D rarr A and D rarr B can be obtained in a unique way from the morphisms AtimesB rarr Aand A times B rarr B defining the product Indeed this is the point behind the ldquothere exists a uniquemorphism h D rarr A times Brdquo part of the definition of A times B which does describe how to obtainD rarr A and D rarr B from the unique h D rarr AtimesB

Said another way the point is that we have a diagram

A B

AtimesBpA pB

which is universal among all such diagrams in the sense that any other diagram

A B

Df g

can be obtained from the first in a unique way via a map D rarr AtimesB

A B

D

A B

AtimesB

f = pA h g = pB h pA pB

existh

Similarly the definition of a coproduct A⊔B for A and B says that we have a diagram of the form

A B

A ⊔B

iA iB

which is universal among all such diagrams there is a unique way to obtain any

A B

D

f g

27

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

from the first

A B

A ⊔B

A B

D

iA iBf = h iA g = h iB

existh

By abstract nonsense whenever an object satisfies a universal property it will be unique up tounique isomorphism (The term universal property of universal element can be defined more pre-cisely using the Yoneda Lemma and the language of representable functors but for us the informalidea expressed above will be enough You can check the book for a more precise formulation ifinterested The point is that a universal property will be characterized by some diagram and thefunctor which assigns to an object all possible such diagrams will be representable the the objectsatisfying that universal property if it exists)

Equalizers and coequalizers The first universal properties we will consider beyond that ofproducts or coproducts are the ones defining the notion of an equalizer and a coequalizer Given apair of morphisms

A Bf

g

an equalizer for the pair is an object E and morphism

E A Be f

g

satisfying f e = g e which is universal among all such things a coequalizer for the pair is anobject Q and morphism

QA Bqf

g

satisfying q f = q g which is universal among all such things (We say that e ldquoequalizesrdquo f andg and q ldquocoequalizesrdquo f and g) To be clear saying that the equalizer diagram is universal amongall such diagrams means that given

Eprime A Beprime f

g

where f eprime = g eprime there exists a unique morphism Eprime rarr E such that the diagram for Eprime can beobtained from the one for E ie so that the following commutes

Eprime E A Bexist e f

g

eprime

The same is true for the definition of a coequalizer only with all arrows reversedThe main examples are in Set where the equalizer is given by

E = a isin A | f(a) = g(a)

28

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

and the coequalizer is the quotient (ie set of equivalence classes)

Q = B sim

where sim is the equivalence generated by the requirement that f(a) sim g(a) for all a isin A

Regular monomorphisms and epimorphisms It turns out that the morphism E rarr A definingan equalizer is always a monomorphism and the morphism B rarr Q defining a coequalizer is alwaysan epimorphism as you will show on an upcoming homework set In the case of Set the equalizerE is a subset of A and the coequalizer Q is a quotient of B We say that a monomorphism E rarr Ais a regular monomorphism if it is the equalizer of some pair of morphisms from A into B and anepimorphism B rarr Q is a regular epimorphism if it is the coequalizer of some pair of morphismsA into B We will see that regular epimorphisms can be thought of a generalizations of variousldquoquotientrdquo constructions and regular monomorphisms are generalizations of ldquosubrdquo constructionsFor instance a regular monomorphism Y rarr X in Top is precisely (up to homeomorphism) theinclusion of a subspace Y sube X and in Grp regular monomorphisms give subgroups

Some Functor Properties An Equivalence Example

Some Functor Properties Problem 5 from Homework 2 was the topic of a student presentationConcretely this problem covered Exercises 15iv 16iii and 16iv in the book which deal withfunctors which preservereflectcreate various types of morphisms The point was to understandhow to use faithfulnessfullness conditions

An Equivalence Example Problem 4 from Homework 2 was the topic of another studentpresentation which is Exercise 15ii in the book This asked to show that the category of finitepointed sets Finlowast was equivalent to the opposite of a certain category Γ defined by Segal Γ hasfinite sets as objects and a morphism S rarr T is by definition a function

f S rarr P (T )

from S into the power set of T such that if s1 ∕= s2 isin S then f(s1) cap f(s2) = empty Here we describethe desired equivalence but will leave checking all details to the reader

For n ge 0 let [n] = 0 1 2 n For simplicity we will take our finite sets to be of the formSn = [n]minus0 which is enough since any finite set is isomorphic to one of these We will then takeour finite pointed sets to be [n] for n ge 1 where 0 isin [n] will always denote the specified basepointThe functor F Γ rarr Finlowast is defined as

Sn 983041rarr [n] on objects and

(f Sn rarr P (Sm)) 983041rarr (Ff [m] rarr [n]) on morphisms

where Ff [m] rarr [n] is given by

(Ff)(i) =

983083j if i isin f(j) for some j isin Sn

0 if no such j exists

Note that Ff is well-defined since if there is a j isin Sn such that i isin f(j) there is only one such jby the requirement that f(a1) cap f(a2) = empty when a1 ∕= a2 The point is that we look at the subsets

29

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30

f(j) sube Sm for j isin Sn to see if any contain i isin [m] if so we send i to that j if not we send i to 0In particular (Ff)(0) = 0 since 0 appears in none of the Snrsquos so Ff does preserve the basepoint

Define the functor G Finlowast rarr Γ by

[n] 983041rarr Sn on objects and

(g [n] rarr [m]) 983041rarr (k 983041rarr gminus1(k)minus 0) on morphisms

where gminus1(k) minus 0 sube P (Sn) simply denotes the preimage of k isin Sm under g with zero excludedwhich defines a function Sm rarr P (Sn) as required of a morphism in Γ The check that F and Gare functors and that they are ldquoinverserdquo to one another is as stated before left to the reader

Segalrsquos Category Coequalizer Examples

Segalrsquos Category

ldquoSpacesrdquo via functors

Quotient groups as coequalizers

Quotient spaces as coequalizers

Limits and Colimits

Diagrams

Cones and cocones

Limits and colimits

Examples

Pullbacks and pushouts

More on LimitsColimits

Constructing limits in Set

Back to pullbacks

Inverse limits

Gluing constructions

More LimitColimit Examples

Coproducts in Vect

Supremums and infimums

Rinfin and Sinfin

Solenoids

30