1 Introduction

When are two spaces X,Y homeomorphic, that is, when is there a continuousmap f : X → Y with continuous inverse?

Let f, g : X → Y be continuous maps. f is homotopic to g if there exists ahomotopy, H : X × I → Y such that H(x, 0) = f(x) and H(x, 1) = g(x).

Put Ht : X → Y by Ht(x) = H(x, t), then H0 = f,H1 = g.Notation: f, g, ft with f0 = f, f1 = g from Hatcher. If f is homotopic to g,

then we write f ' g. Similarly for spaces.Let X,Y be two spaces. They have the same homotopy type if there are

continuous f : X → Y and g : Y → X such that g◦f : X → X and f ◦g : Y → Yare homotopic to the appropriate identity maps.

Examples: In Rn, we have Bn = Dn = {x ∈ Rn : ‖x‖ ≤ 1}.Bn has the homotopy type of a point, for example, {0}. Let i : {0} → Bn

be the inclusion, and r : Bn → {0} by r(x) = 0.Then r◦i = id{0} and i◦r : Bn → Bn is homotopic to idBn by H : Bn×I → I

by H : Bn × I → Bn, H(x, t) = (1− t)x.But for n > 1, Sn−1 = {x ∈ Rn : ‖x‖ = 1} are not homotopy equivalent to

a point.

2 Fundamental Group

Definition 2.1 (Path). Let X be a topological space. A path in X is f : I → Xa continuous map.

Two paths from x0 to x1 are homotopic if there exists H : I × I → S suchthat H(s, 0) = f(s), H(s, 1) = g(s), H(0, t) = x0 and H(1, t) = x1.

In Euclidean space (in fact, any convex subspace of Euclidean space) everypath is homotopic to a straight line path, by a straight line homotopy.

Let f be a path from x0 to x1 and g be a path from x1 to x2, then f · g is a

new path from x0 to x2 defined by (f · g)t) ={

f(2t) 0 ≤ t ≤ 1/2g(2t− 1) 1/2 ≤ t ≤ 1

If x0 ∈ X, then let cx0 : I → X be the constant path cx0(t) = x0.Easy: If f0 ' f1 from x0 to x1 and g0 ' g1 from x1 to x2, then f0 ·g0 ' f1 ·g1.

Definition 2.2 (Loop). Fix x0 ∈ X and call x0 the base point. A loop at x0 isa path f in X such that f(0) = f(1) = x0. Let [f ] be all loops at x0 which arepath homotopic to f . We denote this relation by f 'p g.

Definition 2.3. We define π1(X,x0) = {[f ] : f is a loop at x0}(We can also define π0(X,x0) = all path components of X with base point

being path component of x0.)

Example: O(2) has two path components, the base point is x0 = I.What is π1(O(2), I)? SO(2) is the set of all rotations of R2, and it is home-

omorphic to the circle.We claim that π1(X,x0) is a group, and call it the Fundamental Group of

X at x0.

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Theorem 2.1. π1(X,x0) with composition as above forms a group.

Proof. [f ][g] = [f · g] is the product. Take the identity to be [cx0 ]. For [f ]−1,we take [f ] where f = f(1 − t). We must check (f · g) · g 'p f · (g · h),f · cx0 'p f 'p cx0 · f and f · f 'p f · f 'p cx0 .

We can do these using reparameterizations. If f : I → X is any path, andγ : I → I is continuous with γ(0) = 0 and γ(1) = 1, then f(γ(t)) is a pathwhich is homotopic to f . F (s, t) = f((1− t)γ(s) + ts).

Now, see page 27.

Lemma 2.2 (Pasting Lemma). Let X be a topological space such that X =A1 ∪ . . . ∪ A`, each Ai is closed. Let f : X → Y be a function to anothertopological space. If ∀i, f |Ai : Ai → Y is continuous, then so is f .

Proof. Let C be a closed set in Y . Then f−1(C) = f |−1A1

(C) ∪ . . . ∪ f |−1A`

(C).Each term in this union is closed in the appropriate Ai, and as Ai is closed inX, each term is itself closed in X. As the finite union of closed sets is closed,f−1(C) is closed in X, and so f is continuous.

Let f, g, h be paths in X such that (f · g) · h is defined. We want to showthat this composition is associative. That is, that (f · g) · h 'p f · (g · h). Theright hand side is also equal to [(f · g) · h] ◦ γ, where

γ =

12 t 0 ≤ t ≤ 1

2t− 1

212 ≤ t ≤

34

2t− 1 34 ≤ t ≤ 1

And so, we define H(s, t) =

f( 4s

t+1 ) 0 ≤ s ≤ (t+ 1)/4g(4s− t− 1) (t+ 1)/4 ≤ s ≤ (t+ 2)/th( 4s−2−t

2−t ) (t+ 2)/4 ≤ s ≤ 1Change of Base PointLet X be a topological space, and x0, x1 ∈ X such that there is a path h

from x0 to x1. Then ∃βh : π1(X,x1) → π1(X,x0) given by βh([f ]) = [h · f · h]which is an isomorphism of groups.

And so, if X is path-connected, we can write π1(X).

Definition 2.4 (Simply-Connected). Call X simply connected if X is pathconnected and π1(X) is trivial.

Quotient TopologyI = [0, 1], and we want to identify 0 ∼ 1.So I/ ∼ is a space, and we believe it is homeomorphic (≈→) the circle S1 ⊂ C.So we define p : I → S1 such that p(t) = e2πit. p(0) = p(1) = 1 ∈ S1 ⊂ C.

Definition 2.5 (Quotient Topology). More generally, let X be a topologicalspace and ∼ be an equivalence relation. Let g : X → X/ ∼ take x to [x]. CallV ⊂ X/ ∼ open iff g−1(V ) is open in X. Then we get a topology on X/ ∼ andwe get that

2

Y

X X/ ∼.................................................................. .........

...∀f

....................................................................................................... ............g

...................................

............... ∃h

Proposition 2.3. π1(S1, 1) ' Z.

Proof. We want to define a homomorphism Φ : Z → π1(S1) be Φ(n) = [ωn]where ωn(s) = e2πins.

We need to show that this is a homomorphism. Let p : R → S1 by p(s) =e2πis. Note that ωn = p ◦ ωn, where ωn : I → R is given by ωn(s) = ns.

I S1

R

................................................................................................................. ............ωn

.................................................................................................................................................................................

ωn

.............................................................................................................................

p

Φ(n) = [p ◦ f ] whenever f is any path in R from 0 to n. Then ωn must behomotopic to f .

And so, Φ is a homomorphism.Let τm : R → R be a translation, τm : x 7→ x + m. Then ωm · τmωn is

a path in R from 0 (to m) to n + m. Thus Φ(n + m) = [p ◦ (ωm · τmωn] =[p ◦ ωm] · [p(τmωn)] = Φ(m) · Φ(n).

We need to show that Φ is surjective and that ker Φ = {0}.We will need two facts.

1. If f : I → S1 starting at x0 ∈ X and let x0 ∈ p−1(x0), then ∃ unique liftf : I → R starting at x0.

This implies that Φ is surjective, as we let [f ] ∈ π1(S1, 1) and takingx0 = 1, x0 = 0, we apply it and get f . Then p(f(1)) = f(1) = 1. Sof(1) = m ∈ Z, e2πif(1) = 1 so Φ(m) = [p ◦ f ] = [f ].

2. Let F : I×I → S1 be a homotopy of paths, all starting at x0 ∈ S1. Choosesome x0 ∈ p−1(x0). Then, ∃ a unique lifted homotopy F : I × I → R ofpaths starting at x0.

I × I (S1, x0)

(R, x0)

................................................................................ ............F......................................................................................................

F

........

........

........

........

........

........

........

........

........

........

........

........

.................

............

p

We note that p ◦ F = F , and so p ◦ ft = ft.The second fact implies that Φ is injective. Suppose that Φ(n) = e = [c1] ∈

(S1, 1).Assume that n 6= 0. Then [ωn] = e = [c1]. Then ωn is path homotopic

to the constant map, and so, by the second fact, we can lift this homotopy

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to ωn 'p ω0 = c0, but these have different endpoints, and so cannot be pathhomotopic, which is a contradiction.

All that remains is to prove these two facts.Observe the property of p : R → S1: there exists an open cover {Uα}

of S1 such that p−1(Uα) = ∪iVαi where the Vαi are disjoint open sets andp|Vαi : Vαi → Uα is a homeomorphism.

One such cover is the covering of S1 by two open sets with appropriateoverlap.

We will first prove fact 1:

(I, 0) (S1, x0)

(R, x0)

................................................................................ ............f...........

...........................................................................................

f

.............................................................................................................................

p

The Lebesgue Covering Lemma states that if X is a compact metric spaceand {Wα} is an open cover of X, then there exists λ > 0 such that every subsetof X with diameter less than λ lies in some Wα.

So, there exists a partition 0 = t0 < t1 < . . . , t` = 1 such that f([ti−1, ti]) ⊂some Uα. We construct a lifting f on [0, tk] by induction on k. We start witht0, [0, 0], f(0) = x0 Assume k < `, have f such that p ◦ f = f . We needto define f on [tk, tk+1]. We know that pf(tk) = f(tk), f(tk) ∈ Uα for someα, so f(tk) ∈ Vαi for some unique i. Put f(t) for tk ≤ t ≤ tk+1, equal top|Vαi f(t) ∈ Vα,i ⊂ R, so existence is done.

Now we prove uniqueness. If p : X → X is a covering projection, that is, pis onto and if given a continuous map

Y, y0 X,x0

X, x0

............................................................................................... ............f

.............................................................................................................................

p

......................................................................................................

h, g

and if g, h : (Y, y0)→ (X, x0) are two lifting maps of f , then g = h providedthat Y is connected, Y is locally connected. We prove this by putting A = {y ∈Y : g(y) = h(y)} so that y0 ∈ A. Put B = {y ∈ Y : g(y) 6= h(y)}. Both A andB are open.

We now must prove fact 2.

I × I, (0, 0) S1, x0

R, x0

.......................................................... ............F

.............................................................................................................................

p

......................................................................................................

F

Proof by picture:Divide the unit square into a smaller grid. F (lower left square) ⊂ Uα, and so

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F (segment I) ⊂ Vαi , and you can extend the map by continuing it from squareto adjacent square. Or see Hatcher.

Theorem 2.4 (Brouwer Fixed Point Theorem). Any continuous f : D2 → D2

has a fixed point.

Usually this is proved with the no retraction theorem, though can be provedthe other way. The no retraction theorem states that there is no continuousr : D2 → S1 such that r(x) = x for x ∈ S1.

Theorem 2.5 (Borsuk-Ulam Theorem). If f : Sn → Rn is continuous, thenthere exists x ∈ Sn such that f(x) = f(−x).

For n = 1, intermediate value theorem proves it. For n = 2, use the funda-mental group, see page 32. For n ≥ 2, need homology and transfer or cohomol-ogy.

Look at page 38, exercise 9.π1 as a functor.(X,x0) gets π1(X,x0) the fundamental group.

Also, (X,x0)φ→ (Y, y0) induces a homomorphism of groups π1(X,x0)

φ∗→π1(Y, y0) define by φ∗([f ]) = [φ ◦ f ]. Note φ = idX ⇒ φ∗ = idπ1(X,x0) and

(X,x0)φ→ (Y, y0)

ψ→ (Z, z0) gives ψ∗ ◦ φ∗ : π1(X,x0) → π1(Z, z0) equal to(ψ ◦ φ)∗.

Proposition 2.6. π1(X×Y ) ' π1(X)×π1(Y ), and the isomorphism is obtainedby the induced maps from the projection maps of X × Y to X,Y .

π1(S1 × . . .× S1︸ ︷︷ ︸n times

) ' Zn

S1 ∨ S1 is two circles intersecting at a point. Then π1(S1 ∨ S1) = F2. Wewill need the Seifert-Van Kampen theorem.

If we take C \ {−1, 1}, we see that it retracts onto S1 ∨ S1.

Proposition 2.7. π1Sn = 0 for n > 1.

Corollary 2.8. π1(Rn \ {0}) = 0 if n > 2.

Sn−1 ⊂ Rn \ {0} is a homotopy equivalence by r(x) = x‖x‖ , and rt(x) =

tx+ (1− t) x‖x‖ .

Definition 2.6 (Retraction). If A ⊆ X, then a retraction of X onto A isr : X → A such that r(a) = a for all a ∈ A. If i : A→ X be the inclusion, thenr ◦ i = idA.

Choose x0 ∈ A ⊆ X. Then π1(A, x0) i∗→ π1(X,x0) and r∗ in the otherdirection, we now have that r∗ ◦ i∗ = idπ1(A). And so, i∗ is injective.∃ no retraction r : D2 → S1. Why? If there were one, then there would be

a homomorphism π1(S1)→ π1(D2) which is injective, that is, an injective mapZ→ 0, which is impossible.

Notation: X ≈ Y is used to denote homeomorphic, X ' Y is homotopic.If f ' g : (X,x0)→ (Y, y0) then f∗ = g∗ : π1(X,x0)→ π1(Y, y0).

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Corollary 2.9. If f : (X,x0) → (Y, y0) is a homotopy equivalence of spaceswith base points, then f∗ : π1(X,x0)→ π1(Y, y0) is an isomorphism.

The reason is that ∃g : (Y, y0)→ (X,x0) such that g ◦ f ' idX , f ◦ g ' idYas based maps, and so π1(X,x0)→ π(Y, y0) has an inverse homomorphism, g∗,and so they are inverse isomorphisms.

Definition 2.7 (Deformation Retraction). Suppose A ⊆ X is a subspace. Adeformation retraction of X onto A is a homotopy H : X × I → X such thatfor all x ∈ X, H(x, 0) = x, H(x, 1) ∈ A and H(a, t) = a for all t if a ∈ A.

Then i : A → X is a homotopy equivalence, and r = H(x, 1) : X → A is ahomotopy inverse.

Note Sn ⊆ Rn+1 \ {0} shows, by x 7→ tx+ (1− t) x‖x‖ that Sn ' Rn+1 \ {0}

Proposition 2.10 (Prop 1.14). π1(Sn) = 0 for n ≥ 2.

Proof. Sn = U∪V where U, V are homeomorphic to Rn. U∩V is path connectedand homotopic to Sn−1 × (−ε, ε).

We now show that π1(Sn, x0) = {1}.Claim: If X = U ∪ V , U, V open, x0 ∈ U ∩ V , and U ∩ V path connected,

consider:

π1(U, x0) π1(V, x0)

π1(X,x0)

................................................................................................................................

i∗

................................................................................................................................

j∗

Then each element of π1(X,x0) is a finite product of elements in Im i∗ orIm j∗.

Proof of claim: Let f : I → X be a loop at x0, f : [0, 1]→ X. Use Lebesguecovering lemma to get partition 0 = t0 < t1 < . . . < t`−1 < t` = 1 such thatf([ti−1, ti]) is contained in U or V .

Assume that f([tt−1, ti]) ⊆ U , then f([ti, ti+1]) ⊆ V . We call the restrictionto each of these fi, and so f 'p f1 · . . . · f`, and it is homotopic to (f1 · g1) · . . . ·(g`−1 · f`).

f 'p f1 · f2 · f3 'p f1g1 · g1 · f2 · g2 · g2 · f3, that is, a loop in U , one in V ,and then one in U .

Theorem 2.11 (Easy Part of Van Kampen Theorem). Let X be ∪Aα with Aαpath connected and open in X, x0 ∈ Aα for all α. Assume that Aα ∩ Aβ isaways path connected. Then iα → X the inclusion induces iα∗ : π1(Aα, x0) →π1(X,x0) so that each Im(iα∗) is a subgroup of π1(X,x0).

Then, each element of π1(X,x0) is a product of elements in Im iα∗

Proposition 2.12. Let φ : X → Y be a homotopy equivalence, and x0 ∈ X.Then φ∗ : π1(X,x0)→ π1(Y, φ(x0)) is an isomorphism.

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Lemma 2.13. Let φt : X → Y be a homotopy, let x0 ∈ X, and let h(t) =φt(x0), a path in Y from φ0(x0) to φ1(x0). Then here’s a commutative triangle

π1(X,x0) π1(Y, φ0(x0))

π1(Y, φ1(x0))

............................ ............φ0∗

.........................................................................................................................................................................

βh

.............................................................................................................................

φ1∗

Proof. Put ht(s) = h(ts). Let f be a loop in X at x0. Inspect ht · (φtf) · ht, aloop at φ0(x0). If t = 0, get φ0∗([f ]). If t = 1, then get βh(φ1∗([f ])).

Now we prove the prop:

Proof. φ : X → Y , ψ : Y → X and ψ ◦ φ ' idX , φ ◦ ψ ' idY .The Lemma gives that ψ∗ ◦ φ∗ is an isomorphism, so φ∗ is injective.Also, it says that ψ∗ is an isomorphism, and so A α→ B

β→ C are groups, withα, β injective and β◦α isomorphism, then α is onto, and so an isomorphism.

We now go back to chapter zero to define CW-Complexes.

Definition 2.8 (CW-Complexes). Let X be a Hausdorff Space and assume thatX has an increasing filtration by closed subspaces, that is, there are X0 ⊂ X1 ⊂. . . ⊂ Xn ⊂ . . ., such that

1. X = ∪n≥0Xn We call Xn the n-skeleton

2. X0 is discrete called the vertices or 0-cells.

3. X1 the 1-skeleton adjoins the edges, the 1-cells, gives a graph.

4. Xn is obtained from Xn−1 by adjoining n-cells as follows: Xn = (Xn−1t∐αD

nα)/ ∼ if x ∈ ∂Dn

α = Sn−1α , then x ∼ φα(x) ∈ Xn−1 for φα : ∂Dn

α →Xn−1 continuous.

Further, a subset A of X is open (closed) in X iff A ∩Xn is open (closed)in Xn.

Note: When is A ⊂ Xn closed? True iff A ∩Xn−1 is closed and under thecharacteristic maps Φα : Dn

α ⊂ Xn−1 t∐β D

nβ → Xn,Φ−1

α (A) is closed in Dnα.

Also, enα = Φα(int of Dnα) ⊂ Xn is an open n-cell. Φα(Dn

α, Sn−1α ) →

(Xn, Xn−1), Φα|Sn−1α

= φα, Xn = Xn−1 ∪⋃α e

nα.

Third, enα = Φα(Dnα) may be identified with enα = enα \ enα = φα(Sn−1

α )Examples

1. Any graph, even infinite.

2.∧α S

1α for any index set.

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3. If X is a finite CW-complex, then χ(X) =∑i≥0(−1)iαi(X) where αi(X)

is the number of i-cells. For T 2, α0 = 1, α1 = 2, α2 = 1 so χ = 0.

4. Sn start with S1 the circle in C. χ(S1) = 0 because we get n edges and nvertices.

For S2, we see the half spheres are 2-cells, and then notice that we get 2i-cells for each, which gives χ(S2) = 2.

Sn is a CW complex with 2 i-cells for 0 ≤ i ≤ n. More economically,

Sn = e0 ∪ en, and so χ(Sn) = 1 + (−1)n ={

2 n ∈ 2Z0 n /∈ 2Z

5. RPn = Sn/ ∼ where x ∼ −x. It is also the set of lines through the originin Rn+1.

6. CPn = S2n+1 ⊂ Cn+1 with x ∼ y if x = λy for λ ∈ S1.

RPn = RPn−1 ∪ en? Well, RPn = Sn/ ∼ identifying antipodal points, whichis equal to Dn/ ∼, identification of antipodal points on the boundary. φ :Dn → Dn/ ∼ gives us the attaching map, because Sn−1 → Sn−1/ ∼= RPn−1

are contained in Dn.Example: {Sn}, Sn−1 ⊂ Sn. Then S∞ = ∪Sn is contractible.RP∞ = ∪RPn and CP∞ = ∪CPn.

Definition 2.9 (Subcomplex). Let X be a CW-complex. Call a subspace A ⊂ Xif A is closed, and if enα ∩A 6= ∅ then enα ⊆ A.

Put An = A ∩Xn for all n. Then A is a CW-complex. For example, eachXn is a subcomplex of X. We call (X,A) a CW-pair if X is a CW-complexand A is a subcomplex of X. Then X/A is what you get when you identify allpoints of A. X/A can be made naturally into a CW-complex.

FACT: If (X,A) is a CW-pair and A is contractible to a point, then g : X →X/A is a homotopy equivalence.

Homotopy Extension Property: Given A ⊂ X, A closed and f : A→ Y , canyou extend f to f : X → Y ?

Sometimes, given i : A → X and f : A → Y , there exists f ' g such thatg does have an extension g : X → Y . Want to be able to conclude that f hasextension f : X → Y .

Definition 2.10 (Homotopy Extension Property: Provisional Def). (X,A), Aclosed in X, has the HEP iff whenever f ' g : A → X and g has an extensionf : X → Y so does g.

Definition 2.11 (Homotopy Extension Property). (X,A) has HEP if any mapF : A× I ∪X × 0→ Y has extension F : X × I → Y .

When does (X,A) have HEP? Answer is iff A × I ∪ X × 0 is a retract ofX × I.

Example: (Dn, Sn−1) has HEP. Proof: Dn × I is a solid cylinder, Sn−1 ×I ∪Dn × 0 is the boundary, minus the top. It is a retraction because if we take

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a point higher up and hold it still, we draw a line through it to any point onthe sides or bottom, and everything on that line gets projected to that point.

This implies that any CW-pair (X,A) has HEP (prop 0.16). Prop 0.17 saysthat if (X,A) has HEP and A is contractible, then X → X/A is a homotopyequivalence.

Topological Properties of CW-complexes: X is path connected iff X1 is pathconnected. X is locally contractible. X is normal. X is compactly generated.If K is a compact subset of X, then ∃n such that K ⊆ Xn, even better, Kis contained in a finite union of closed cells, best, K is contained in a finitesubcomplex.

And now, we return to chapter 1.We need an algebraic aside so that we can do VK. Problem: Given groups

{Gα}α∈A and homomorphisms φα : Gα → H, H another group, want somegroup G together with homomorphisms iα : Gα → G such that ∃Φ : G → Hunique subject to

Gα G

G

................................................................................................................. ............iα

.............................................................. ............φα

..........................

.............

............... Φ

Then (G, iα) is the free product or coproduct of the Gα’s.Make G = ∗αGα as all words h1h2 . . . h` with hi ∈ Gαi \ {1}. We call it a

reduced word if it cannot be simplified further. G is the reduced words underconcatenation.

Theorem 2.14 (Seifert-van Kampen). Let X be a topological space, x0 ∈ Xsuch that X = ∪α∈IAα with Aα open, path-connected and x0 ∈ Aα for allα. If Aα ∩ Aβ ∩ Aγ is always path-connected, then we have a homomorphismΦ : ∗απ1(Aα) → π1(X) which is surjective, and N = ker Φ is the normalsubgroup of the free product generated by all elements iαβ∗(ω)i−1

βα∗(ω) whereiαβ : Aα ∩Aβ → Aα is the inclusion.

That is, π1(X) ' ∗α(Aα)/N .A universal property holds:

π1(Aα ∩Aβ)

π1(Aα) π1(Aβ)

π1(X)

G

......................

.........................................

...............................................................

......................................................................................................................................................................................................

φα

..............................................................................................................................................................................

φβ

...............................................................

......................

.........................................

........

.....

........

.....

........

.....

........

.....

............

............

∃!

Example: π1(∨αXα) with xα ∈ Xα and ∀α, xα ∈ Uα ⊂ Xα, Uα open and

Uα deformation retracts to xα. Then π1(X) ' ∗π1(Xα), where X = ∨αXα.Also, assume each Aα is path connected, then so is X.

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Proof. Take Aα = Xα∨(∨β Uβ) ' Xα. Then for α 6= β, Aα∩Aβ =

∨γ Uγ ' x0.

Now we apply VK Theorem trivially.

Using this, we find that π1(S1∨S1) = Z∗Z, the free group on two generators.π1(S1 ∨ S2) ' Z.In fact, if G is any connected graph, then a maximal tree can be contracted

to a point, and so has fundamental group free, and so is a bouquet of circles.π1(R2 \N pts) =free group on n generators. This space contracts to ∨Ni=1S

1.So π1(S2 \N) ' free group on n− 1 generators.

Suppose Rn \K is path connected, K is closed and bounded. π1(Rn \K)→π1(Sn \K) is an isomorphism when n > 2.

Proof. Take U = Rn \ K and V = Sn \ BR(0) where K ⊆ BR(0). Note thatU ∩ V = Rn \BR(0), which is path connected. (ADD DETAILS LATER)

Let A be a circle in the plane. Then R3 \A ' S1, so π1(R3 \A) = Z.

Proposition 2.15. Inspect Sp+q−1 ⊂ Pp+q. It has Sp−1, Sq−1 as subspheres.Then there exists a deformation retraction from Sp+q−1\Sq−1 onto Sp−1. Thus,π1(S3 \A) ' π1(S1).

Proof. Define ft : Sp+q−1 \ Sq−1 → Sp+q−1 \ Sq−1 by ft(x, y) = ((1− t)(x, y) +t( x‖x‖ ), 0)/(‖TOP‖).

R3 \ (A ∪B) where A,B are unknotted circles contracts to S1 ∨ S1.R3 \ (A∪B) where A,B are linked, then R3 \ (A∪B) ' S1×S1, and so the

fundamental group is Z× Z.What is π1(R3 \ (A∪B)), the complement of linked circles? It is isomorphic

to π1(S3 \ (A ∪ B)). Now, S3 = {(z, w) ∈ C2 : |z|2 + |w|2 = 1}. Put Tz ={(z, w) ∈ S3 : |w|2 ≤ 1/2} and Tw = {(z, w) ∈ S3 : |z|2 ≤ 1/2}. Note that{(z, 0) : |z| = 1} ⊂ Tz and {(0, w) : |w| = 1} ⊂ Tw. Now, S3 = Tz ∪ Tw andTz ∩ Tw = {(z, w) ∈ S3 : |z|2 = |w|2 + 1/2} ' S1 × S1. So Tz ' S1 ×D2 thesolid torus.

Tz ∩ Tw is a deformation retract of S3 \ (A ∪B).Tz = {(z, w) : |z|2 + |w|2 = 1, |w|2 ≤ 1/2}. z = |z|eiθ and |z|2 = 1− |w|2 so

(z, w) 7→ (eiθ = z/|z|, w) ∈ S1 × 1√2D2 ' S1 ×D2.

π1(Km,n) where (m,n) = 1 is the group 〈a, b|amb−n〉.Attaching 2-cells creates relations by how they are attached.Let X be a path connected space, x0 ∈ X Attach 2-cells via ϕα : (S1, s0)→

(X,x0) and put Y = (X t⊔αD

2α)/ ∼ where ϕα(x) ∼ x for all x ∈ S1

α.

Proposition 2.16. Then π1(Y, x0) ' π1(X,x0)/N , N is the smallest subgroupcontaining all [ϕα] ∈ π1(X,x0)

Proof. Case 1: a Single 2-cell. Φ : D2 → e2 ⊂ Y , y = Φ(0). Y = (Y \{y})∪e2 =e2 \ {y} ' S1. VK implies that π1(Y,Φ(d1)) ' π1(X \ {y},Φ(d1))/minimalsubgroup generated by [f ].

π1(Y,Φ(d1)) ' π1(Y, x0) ' π1(X,x0)/[ϕα].

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Case 2: Induction for the case of a finite number of 2-cells.Case 3: Any number at all:Let ϕα : S1 → X. Y = X ∪

⋃α∈A e

2α, yα = Φα(0).

Uα = X ∪⋃β(eβ \ yβ) ∪ Eα = Y = ∪β{yβ} ∪ eα.

Uα will be open and path connected subsets. Thus, α 6= β ⇒ Uα ∩ Uβ =Y − ∪β∈A{yβ} ' X.

Let F be all finite subsets of A. Put for F ∈ F , UF = ∪α∈FUα ' X ∪⋃αF

eα.We know π1(UF ) = π1(X)/normal subgroup generated by [ϕα] for α ∈ F .VK Theorem: Note that UF1∩UF2 = UF1∩F2 , and also for triple intersections.

Thus,

π1(UF1)

π1(UF1∩F2)

π1(VF2)

G

π1(Y )

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∃!

If F is the set of all finite subsets of A and we order F by inclusion, weobtain an directed ordered set. Given F1, F2 ∈ F , note that F1 ⊂ F1 ∪ F2 andF2 ⊂ F1 ∪ F2.

Thus, π1(Y ) is the colimit (or direct limit) of {π1(UF )}F∈F along with thehomomorphisms from inclusions in F .

An algebraic comment: If D is a directed set α ∈ D with Gα a group andd ≤ d′ implies there is a homomorphism id,d′ : Gd → Gd′ and d = d′ implies itis the identity, with d ≤ d′ ≤ d′′ givesGd Gd′′

Gd′

.......................................................... ............

.............................................................................................................. ............

......................................................................

Then lim−→ = colim(Gα, iα,α′) = ∗dGα/N where N is the normal subgroupgenerated by {g−1

α iα,α′(gα)} with gα ∈ Gα and α ≤ α′.NEXT: Covering Spaces, read the section

Definition 2.12 (Covering Space). A covering map p : X → X is a map suchthat X is covered by evenly covered open sets U , that is, p−1(U) = ∪αVα, Vα’sopen and disjoint in X with p : Vα → U a homeomorphism.

Remarks: p is a local homeomorphism, meaning for all x ∈ X, there is anopen nbhd V of x such that p|V : V → p(V ) is a homeomorphism p(V ) open inX.

p is an open mapping. p is a quotient mapping, provided p is surjective.

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Examples: ∅ and any X, id : X → X.The number of sheets is the index of the subgroup that is the fundamental

group of the covering space. It is normal if you get the same subgroup (ratherthan a conjugate one) from a change of basepoint.

Lifting Theorems

Theorem 2.17 (Homotopy Lifting Property). Given

Y × I

Y

X

X.............................................................................................................................

inc

.............................................................................................................................

p

.................................................................................................... ............{ft}

................................................................................................................. ............f0

Then there exist a unique {ft} : Y × I → X.

Theorem 2.18. If p : X → X is any covering map and p(x0) = x0 then

1. p∗ : π1(X, x0)→ π1(X,x0) is injective

2. Im p∗ = {[f ] : f based at x0 such that the lifting f such that f(0) = x0 isa loop}.

Proof. 1. how that ker p∗ = 1. Suppose that [f ] ∈ π1(X, x0) such that[p ◦ f ] = 1 ∈ π1(X,x0). So p ◦ f = f ' cx0 . The HLT says that we canlift to get f 'p cx0 , and so [f ] = 1.

2. ⊇ is clear. ⊆ is also clear.

Proposition 2.19. If X and X are path connected, then |p−1(x)|, x ∈ X isconstant and equal to [π1(X,x0) : p∗(π1(X, x0))].

Theorem 2.20 (Lifting Theorem). Assume that Y is path connected and locallypath connected, with f : (Y, y0) → (X,x0) and p : (X, x0) → (X,x0) is acovering map. A lift f : Y → X exists iff f∗(π1(Y, y0)) ⊆ p∗(π1(X, x0)).

Proof. It f exists, the inclusion is clear.Assume that the inclusion holds. Take y ∈ Y run a path γy = γ from y0 to

y. Then f ◦ γy is a path in X from x0 to f(y). Lift to ˜f ◦ γy which is a path inX from x0 to a point we shall name f(y). f(y) = ˜f ◦ γy(1) is set to f(y) by p.

Must check the independence of path from y0 to y, that f : Y → X iscontinuous and f(y0) = x0, though the last part is trivial.

So is ˜f ◦ γy(1) = ˜f ◦ δy(1) for δ, γ paths from y0 to y? (f ◦ γy) · f ◦ δy 'f ◦ (γy · δy) 'p p ◦ ε where ε is a loop at x0 in X. Thus f ◦ γy ' (p ◦ ε) · (f ◦ δy),and so ˜f ◦ γy 'p ε · ˜f ◦ δy. Thus, the endpoints are the same.

Why must f : Y → X be continuous? Let y ∈ Y , and check continuity at y.f(y) ∈ U evenly covered, f(g) ∈ Vα, p : Vα → U homeo. Show that there is a

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neighborhood W of y such that f(W ) ⊆ Vα. Since U is open and f(y) ∈ U , thereis an open nbhd W of y such that f(W ) ⊂ U hypotheses on Y let us assume thatW is path connected. Recall, f(y) = ˜f ◦ γy(1) ∈ Vα. Let w ∈W , run a path inW , ζw from y to w. Then f(w) = ˜f ◦ (γy · ζw)(1) = ˜(f ◦ γy) · (f ◦ ζw)(1). Thefirst part ends at f(y), and the second part happens in U .

And so, this equals ˜f ◦ γy ·p−1(f ·ζw)(1) ∈ Vα. And so, the map is continuous.

Next we construct the universal cover. If H ≤ π1(X,x0) then we constructXH → X such that pα(XH) = H. We will also attempt to classify all coveringspaces of some ”good” X up to isomorphism.

GraphsFor a graph, χ(X) = |V | − |E|, and if X is an n-sheeted cover of X, then

χ(X) = nχ(X).If we take X → X/T ≈

∨S1 is a homotopy equivalence, and so χ(X) =

χ(∨S1) = 1−k, where k is the number of circles. Thus, χ(X) = n(1−k) = 1−`

and ` = 1 + (n− 1)n.If X is some nice space, path conn and locally path conn, then an isomor-

phism of spaces over X is f : X1 → X2 such that p2 ◦ f = p1.We define Cov(X) =all isomorphism classes of coverings of X.

Y = p−1(X)

Y X

X

................................................................................................................. ............φ

.............................................................................................................................

p

.............................................................................................................................

p

then Y = {(y, x) ∈ Y × X : φ(y) = p(x)} maps down to Y by πY . Thus, weget a map φ∗ : Cov(X)→ Cov(Y )

Assume that X is as before and also that for each x ∈ X, there is an openU containing x such that every loop in U is homotopic to the constant loop inX. Then we call it SLSC (Semi-Locally Simply Connected).

[Losternik-Schrilmann Category of a Space]: RPn = Sn/(x ∼ −x). LetUi ⊂ RPn be given by xi 6= 0. So RPn is a union of n+ 1 contractible sets. Ingeneral, X = V0 ∪ . . . ∪ Vn is an open cover such that ιi : Vi → X is homotopicto a constant map. This is similar to SLSC.

Theorem 2.21. There exists a simply connected covering space if SLSC holdsfor X, and conversely.

Let pu : Xu → X be simply connected covering space of X and let X → X beany other covering space, X also path connected. Then the diagram commutes:

X,x0

Xu, xu

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pu

X, x0................................................................................................. p

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φ

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Note: §1.3 Ex 16 says that φ is also a covering map. Can a s.c. X havecovering space no isom to id : X → X? p : X → X given, then isomorphic as itmust be a one-sheeted covering space.

Easy: If 1-sheet, then p is a homeomorphism. Also if s is a section of thecovering map (p ◦ s = idX) and s(x) ∈ p−1(x) for all x ∈ X, then p is ahomeomorphism.

Proposition 2.22. If H is a subgroup of π1(X,x0) then there exists a coveringX → X such that the induced map p∗(π1(X, x0)) = H.

We define an automorphism of a covering space to be an isomorphism f :X → X. We also call these Deck Transformations. We name the group of theseas G(X, p) = G(X/X).

If for any two points x1, x2 such that p(x1) = p(x2) ∈ X, there existsφ ∈ G(X, p) such that φ(x1) = x2, call p normal.

Proposition 2.23. Let p : (X, x0) → (X,x0) be a covering with X,X pathconn and X locally path conn. Then

1. p is normal iff p∗π1(X, x0) is a normal subgroup of π1(X,x0).

2. Assuming p is normal, there exists an isomorphism G(X, p) ' π1(X,x0)/p∗π1(X, x0).

So for the universal covering, G(X, p) ' π1(X,x0)Assume X → X is a universal cover. So G(X, p) acts transitively on p−1(x0).

Define φ : π1(X,x0) → G(X, p) by if [γ] ∈ π1(X,x0) lift γ to γ a path in Xwith γ(0) = x0 and γ(1) ∈ p−1(x0). Then there exists g ∈ G(X, p) such thatg(x0) = γ(1). Put φ([γ])(x0) = γ(1), and φ([γ]) ∈ G(X, p).

This element is unique, as if g1, g2 ∈ G(X, p) and if x0 such that g1x0 = g2x0

then g1 = g2 by uniqueness of lifting.Then we can show easily that φ is a homomorphism of groups and that φ is

surjective. φ is also injective.Change point of view:If Y is a given space and G is a group, G × Y → Y an action such that

g : Y → Y is a homeomorphism of Y . Then y ∼ gy for g ∈ G gives anequivalence relation, and Y → Y/G is the space of orbits.

Favorite Example: Y = Sn, G = {±1} ' Z2 by (+1)y = y and (−1)y = −y,the antipodal map. Y/G = Sn/Z2 ' RPn. Also π1(RPn) ' G ' Z2.

Let G act on Y . Call the action even if ∀g ∈ Y , ∃ open neighborhood V ofy such that g 6= h⇒ gV ∩ hV = ∅.

Lemma 2.24. If G acts evenly on Y , then p : Y → Y/G is a covering map.

Proof. We know p is continuous and open since ifW ⊆ Y is open then p−1(p(W )) =∪g∈GgW . In fact, this is a disjoin union.

Claim: For V as in definition, then p(V ) is evenly covered. Note that p|gV :gV → V is bijective. [Clearly surjective, for injective, we say gv1, gv2 are in thesame orbit, so ∃h ∈ G such that gv1 = hgv2, disjointness implies that hg = g,so h = 1, hence v1 = v2.]

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We call such a covering space a G-covering.Isomorphism of G-coverings is φ : Y1 → Y2 over Y1/G ≈ X ≈ Y2/G such

that φ is a homeo and φ(gy) = gφ(y).Example, Y = S2n−1 ⊂ Cn, G = {z ∈ C : zk = 1}, then G×S2n−1 → S2n−1

by v 7→ zv. Then S2n−1 → S2n−1/G gives us a lens space.Trivial G-covering over a given X is X ×G→ X by projection where G has

the discrete topology on it. h(x, g) = (x, hg).

Lemma 2.25. Any G-covering Y → X is locally trivial as a G-covering. ie,X can be covered by open sets U such that gv ∈ p−1(U) → U is isomorphic toU ×G→ U by projection.

Proposition 2.26. Let p : Y → X be a G-covering. Then we have a homo-morphism G→ Aut(Y, p) which is injective. The the case where Y is connected,this is an isomorphism.

Proof. Pick y0 ∈ Y , then for ϕ ∈ Aut(Y, p) = G(Y, p), ϕ(Y0) satisfies p(ϕ(y0)) =py0 so ∃g ∈ G such that ϕ(y0) = gy0. So ϕ and y 7→ gy0 are 2 elements ofAut(Y, p) which agree on y0. Uniqueness props for lifting implies that φ sendsy → gy.

Proposition 2.27. Let p : Y → X be a covering, Y connected, then Aut(Y, p)acts evenly on Y . If Aut(Y, p) acts transitively on a fiber of p, then the coveringis a G-covering for G = Aut(Y, p).

Chapter 11 and 13 of Fulton.

Proposition 2.28. Let G act evenly on a simply connected and locally pathconnected space Y . Then for X = Y/G, π1X = G.

Proof. We’ve seen that π1(X) ' Aut(Y, p) for p : Y → X and prop aboveimplies that G ' Aut(Y, p), so π1(X) ' Aut(Y, p) ' G.

Chapter 14 - The Van Kampen Theoremx0 ∈ X = U ∪V , x0 ∈ U ∩V , U, V, U ∩V path connected, then we can define

the fundamental group and a map π1(X) → G which is unique to each G thatmakes a commutative diagram commute.

What is the meaning of a π1(X,x0)→ G?Answer: G-coverings of X up to isomorphism preserving basepoints.Suppose given a homomorphism ρ : π1(X,x0)→ G. Construct a G-covering

pρ : Yρ → X with yρ ∈ Yρ s.t. pρ(yρ) = x0.Give G the discrete topology. Consider X ×G and a left action of π1(X,x0)

on it by [σ] · (z, g) = ([σ] · z, g · ρ([σ])−1).Then, put Yρ = (X ×G)/π1(X).

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3 Homology

Let X be a topological space. We want to define Hi(X) the homology groupsi ≥ 0 with some good properties. We expect it to be functorial and to be useful.

∆-complexesWe look at T 2.

•v

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b

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c

And calling the inside of the triangles U and L. Let v0, . . . , vn ∈ Rm, assumethat they are affinely independent. That is, no three are collinear, no fourcoplanar, etc.

Equivalently, if c0v0 + . . . + cnvn = 0 and c0 + . . . + cn = 0 then ci = 0 forall i.

So, let v0, . . . , vn be affinely independent. Then define [v0, . . . , vn] = all∑ni=0 tivi where ti ≥ 0 and

∑ti = 1.

The Barycenter is 1n+1

∑vi.

The standard n-simplex is ∆n = [e0, . . . , en] ⊆ Rn+1.If v0, . . . , vn are affinely independent points, ∃ a homeomorphism ∆n →

[v0, . . . , vn], as any point in ∆n is of the form (t0, . . . , tn), and we can send it to∑tivi.We can orient the edges by the edge between vi and vj points to the greater

one.We define simplicial homology for a ∆-complex.We look at X = ∪enα open n-simplexes and this union is disjoint. We have

for each enα σα : ∆n → [v0, . . . , vn], which gives a map from int∆n → enα.We define ∆n(X) to be the free abelian group generated by all open n-

simplexes. We can identify enα ↔ σnα. So the elements look like finite sums ofsimplices or of these maps. We call each of these a simplicial chain.

We now attempt to define a boundary operator. ∂([v0, v1]) = v1 − v0,∂([v0, v1, v2]) = [v0, v1] + [v1, v2]− [v0, v2] = [v0, v1, v2]− [v0, v1, v2] + [v0, v1, v2].

In general ∂([v0, . . . , vn]) =∑

(−1)i[v0, . . . , vi, . . . , vn].And so, we have for ∆-complex X, ∆n(X) ∂→ ∆n−1(X).

Lemma 3.1. ∂∂ = 0.

We define Zn(C) = ker(∆n(X) ∂→ ∆n−1(X)) to be the cycles and Bn(X) =Im(∆n+1(X)→ ∆n(X)). Bn ⊆ Zn by the lemma.

We define H∆n (X) = Zn(X)/Bn(X).

For the torus above, ∆2(T 2) has basis U,L, ∆1(T 2) has basis a, b, c and∆0(T 2) has basis v.

∂(U) = ∂(V ) = a+ b− c, ∂(a) = ∂(b) = ∂(c) = 0.H0 = ∆0/0 ' Z. H1 = ∆1/(a+ b− c) ' Z⊕ Z. H2 = Z(U − L)/0 ' Z.

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Singular Homology [Eilenberg]LetX be any space. A singular n-simplex inX is a continuous map σ : ∆n →

X. Put Cn(X) = the free abelian group generated by all singular n-simplexes,so elements look like

∑nσσ where σ is a map ∆n → X where nσ ∈ Z and only

finitely many are nonzero.∂ : Cn(X) → Cn−1(X) is defined by, if σ : ∆n → X, ∂σ =

∑ni=0(−1)iσi

where σi : ∆n−1 → X is the inclusion of the ith face, φi followed by σ.φi(t0, . . . , tn−1) = (t0, . . . , ti−1, 0, ti, . . . , tn−1).

Lemma 3.2. ∂2 = 0.

If X ∼= Y , then Hn(X) ∼= Hn(Y ).

Proposition 3.3. If X = {pt}, then Hn(X) ={

Z n = 00 else

Proof. For n ≥ 0, σn : ∆n → {pt} is unique. Thus, ∂σn is the alternative sumof σn−1’s, which is 0 if n is odd, 1 if n is even.

Thus, we get the complex Z→ Z→ Z→ . . .→ Z. In each, either everythingis a cycle and everything is a boundary, or else none of either is. Except indimension zero, as everything is a cycle, but nothing is a boundary, so H1(X) =Z.

Proposition 3.4. If X has path components Xα, then Hn(X) ∼= ⊕αHn(Xα).

Proposition 3.5. If X is path connected and nonempty, then H0(X) ∼= Z.

Proof. Choose x0 ∈ X. Consider C0(X) =∑x∈X nxx.

C1(X) ∂→ C0(X)→ 0, but we can augment it, so instead we get C0(X) ε→ Z.ε(∑nxx) =

∑nx.

Note: ε∂ = 0 since ε∂σ1 = 1− 1 = 0. Claim: ker ε = Im ∂.We know that Im ∂ ⊂ ker ε. We need to show the converse.Let c = n1x1 + . . .+n`x` ∈ C0(X) with ε(c) = 0 so n1 + . . .+n` = 0. Choose

paths fi from x0 to xi (1 ≤ i ≤ `) so fi = σi : ∆1 → X with ∂σi = xi − x0 forall i.

Then σ(∑niσi) = σni(xi − x0) =

∑nixi −

∑nix0 = c

Definition 3.1 (Reduced Homology). Let X 6= ∅ be out space. We have . . .→

C2(X)→ C1(X)→ C0(X) ε→ Z. Define Hn(X) ={

Hn(X) n > 0ker ε/ Im ∂ n = 0

Fact: If X is path connected, x0 ∈ X, then there exists a homomorphismπ1(X,x0)→ H1(X) called the Hurewicz homomorphism which is surjective andhas kernel the commutator subgroup of π1.

Definition 3.2 (Induced Homomorphism). If f : X → Y is a continuous map,then f] : Cn(X)→ Cn(Y ) by composing f with σ : ∆n → X to get σ : ∆n → Y .

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For c ∈ Cn(X), ∂f](c) = f]∂(c).Note: f](Zn(X)) ⊆ Zn(Y ) and f](Bn(X)) ⊂ Bn(Y ) induces f∗ : Hn(X) →

Hn(Y ).0 → Bn(X) → Zn(X) → Hn(X) → 0, f] gives an homomorphism of short

exact sequences to 0→ Bn(Y )→ Zn(Y )→ Hn(Y )→ 0.

Theorem 3.6. f, g : X → Y and f ' g, then f∗ = g∗ : Hn(X)→ Hn(Y ).

Proof. Let F : X × I → Y continuous with F (x, 0) = f(x) and F (x, 1) = g(x)for all x ∈ X. We want to show that f∗ = g∗ : Hn(X)→ Hn(Y ).

Consider ∆n × I. ∆n × {0} = [v0, . . . , vn], ∆n × {1} = [w0, . . . , wn]. Wehave additional n-simplices [v0, . . . , vi, wi+1, . . . , wn].

Note: ∆n × I =⋃ni=0[v0, . . . , vi, wi, . . . , wn] is an n+ 1-simplex.

Define P : Cn(X)→ Cn+1(Y ) by σ 7→∑ni=0(−1)iF ◦(σ×id)|[v0,...,vi,wi,...,wn].

Check: ∂◦P = g]−f]−P ◦∂ as hom Cn(X)→ Cn(Y ). or ∂P+P∂ = g]−f].(P is a chain homotopy between f] and g])Let z ∈ Zn(X). Then z ∈ Hn(X). Then f∗z = f]z and g∗z = g]z, so to

show that f∗ = g∗, we must show that f]z − g]z is a boundary.f]z − g]z = −(∂Pz + P∂z) = ∂(−Pz).

Corollary 3.7. X ' Y implies Hn(X) ' Hn(Y )

What is Hn(Sk) when k, n > 0?Want relative homology long exact sequence, → Hn+1(X,A) ∆→ Hn(A) i∗→

Hn(X)j∗→ Hn(X,A)→

Given a space X, A ⊆ X, Z ⊂ A such that Z ⊂int(A).Have (X \ Z,A \ Z)→ (X,A).

Theorem 3.8. i∗ : Hn(X \ Z,A \ Z)→ Hn(X,A) is an isomorphism.

Use the above to determine the homology groups of Sk by induction on k;later (or read)

Definition 3.3 (Relative Homology). We define the relative chain complex of apair (X,A) where A ⊂ X as Cn(X,A) = Cn(X)/Cn(A). The relative homologyis the homology of this chain complex.

Proof of Theorem:

Proof. We have, easily, i∗, j∗, we need ∆ : Hn(X,A)→ Hn−1(A).Let z ∈ Zn(X,A) with z ∈ Hn(X,A). Pick x ∈ Cn(X) with x 7→ z. Then

there exists a unique a ∈ Cn−1(A) such that a 7→ ∂x. Note that ∂a = 0 soa ∈ Zn−1(A).

Put ∆(z) = a ∈ Hn−1(A). Must check that ∆ is well-defined.

Theorem 3.9. For (X,A), there exists a long exact sequence Hn+1X → Hn+1(X,A) ∆→Hn(A)→ Hn(X)→ Hn(X,A) δ→ Hn−1(A).

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Proof. Must verify exactness at Hn(A), Hn(X), Hn(X,A).At Hn(A): i∗∆z = 0 is clear from the definition of ∆z. Suppose a′ ∈ Hn(A).

a′ ∈ Zn(A) and i]a′ = 0 in Hn(X). Then i]a′ = ∂x′, x′ ∈ Cn+1(X). Exactness

follows by diagram chasing.At Hn(X): Clearly j∗i∗ = 0. Suppose given z ∈ Hn(X) such that z → 0 in

Hn(X,A), there exists a ∈ Cn(A) and n ∈ Cn+1X sch that z = ∂n+ i∗a checkthat a is a cycle, i∗n = z.

Proposition 3.10. Let f, g : (X,A) → (Y,B) be homotopic as maps of pairs.Then natural f∗ : Hn(X,A)→ Hn(Y,B) agrees with g∗.

Proof. ∃P : Cn(X) → Cn+1(Y ) for all n such that P∂ + ∂P = g] − f] a chainhomotopy. Observe that since f(A) ⊂ B, P : Cn(A) → Cn+1(B), we pass toquotients to get P : Cn(X,A)→ Cn+1(Y,B) still a chain homotopy between f]and g]. Thus, on relative homology, f∗, g∗ are the same.

Naturality of Long Exact Sequence: Suppose f : (X,A)→ (Y,B).

Hn+1(Y,B)Hn(B)Hn(Y )HnY,B

Hn+1(X,A)Hn(A)Hn(X)Hn(X,A)

..............................................................∆ ................................... ........................................

.................................................................∆ ..................................... ....................................................

........

........

........

..............

............f∗ ..................................................f∗ ........

........

........

..............

............f∗ ..................................................f∗

commutes.Reformulation of Excision TheoremX ⊃ A, Z ⊂ A put B = X \X, so A∩B = A\Z. Now, Z = X \ int(X \Z).

So Z ⊂ int(A) means that X \ int(B) ⊂ int(A), thus A =∫

(A) ∪ int(B).

Theorem 3.11. If U = {A,B} covers X so that X = int(A) ∪ int(B), thenHn(B,A ∩B)→ Hn(X,A) is an isomorphism.

Application: X is a space SX is a suspension of X,m that is, X× I/ ∼ with(x, 1) ∼ (x′, 1) and (x, 0) ∼ (x′, 0) for all x, x′ ∈ X.

Then there exists isomorphisms Hn+1(SX)→ Hn(X)

Corollary 3.12. Hk(Sn) is Z for k − n, 0 else.

Proof. We now prove the suspension theorem.A = SX \ {S}, B = SX \ {N}. A ∩ B ' X, so 0 → Hn+1(SX) →

Hn+1(SX,A)→ 0 exact (thus middle is an isomorphism), and alsoHn+1(SX,A)isomorphic to Hn+1(B,A∩B), and 0→ Hn+1(B,A∩B) ∆→ Hn(A∩B)→ 0.

Generalized Excision Theorem:X is given. Let U = {Uα} be a cover of X such that {int(Uα)} is a cover

of X. Let CUn (X) ⊂ Cn(X) be a subgroups generated by all U-small singularn-complexes σ : ∆n → X meaning σ(∆n) ⊂ Uα for some α.

Observe that σ U-small implies that ∂σ is U-small.Thus, ∂ maps CUn (X) → CUn−1(X). So get ι∗ : HUn (X) → Hn(X) is an

isomorphism.

19

More precisely, ι : CUn (X) → Cn(X) is an inclusion which has chain homo-topy inverse ρ : Cn(X)→ CUn (X), ι and ρ are both chain maps and ρ ◦ ι 'c idand ι ◦ ρ 'c id where 'c is chain homotopy.

Exercise: From this, you can get the excision theorem.Proof of the generalized excision theorem

Proof. We will prove that ι∗ : HUn (X) → Hn(X) is an isomorphism. We willprove it in four parts.

1. Subdivide simplexes, with iterated barycentric subdivision.

If [w0, . . . , wn] is an n-simplex in RN and b ∈ RN , then b · [w0, . . . , wn] =[b, w0, . . . , wn], an n+ 1 simplex.

Aim: Barycentric Subdivision of an n-simplex [v0, . . . , vn]. We define the−1 simplex to be [∅]. We will define this by induction on n.

For n = −1, nothing. For n = 0, also no change.

For n > 0, assume that each (n−1)-face of [v0, . . . , vn] has been subdividedinto n! (n− 1)-faces, which we will call [w0, . . . , wn−1]

Then we let b = 1n+1 (v0+. . .+vn) be the barycenter of [v0, . . . , vn] and con-

sider all n-simplexes [b, w0, . . . , wn−1] = b · [w0, . . . , wn−1] giving a grandtotal of n!(n+ 1) = (n+ 1)! n-simplexes.

Lemma 3.13. diam[b, w0, . . . , wn−1] ≤ nn+1diam[v0, . . . , vn].

2. Y ⊆ RN a convex set. Let LCn(Y ) ⊂ Cn(Y ) be the subgroup generatedby all linear maps σ : ∆n → Y .

Note: LCn(Y ) ∂→ LCn−1(Y ), so the linear chains give us a chain complex.If b ∈ Y , we get b : LCn → LCn+1 by b · [w0, . . . , wn] = [b, w0, . . . , wn].Check that ∂b(α) = α−b(∂α) for all α ∈ LCn(Y ). That is, ∂b+b∂ = id−0on LCn(Y ).

Take α = [w0, . . . , wn], then ∂b[w0, . . . , wn] = ∂[b, w0, . . . , wn] = [w0, . . . , wn]−[b, w1, . . . , wn] + . . ., so it works out.

Define S : LCn(Y ) → LCn(Y ) by induction on n, let λ : ∆n → Y , λ =[w0, . . . , wn] and let bλ = the barycenter of [w0, . . . , wn]. Put S([∅]) = [∅]and S([w0]) = [w0]. In general, S(λ) = bλ · (S∂λ) in LCn(Y ).

Check ∂S = S∂. Next we construct a chain homotopy T : LCn(Y ) →LCn+1(Y ) between S and id. We do this inductively by setting T = 0 onLC−1(Y ) and Tλ = bλ · (λ− T∂λ) for n ≥ 0.

3. We will now subdivide general chains. S : CnX → CnX by Sσ = σ]S ·[e0, . . . , en], where S[e0, . . . , en] ∈ LCn.

Check ∂Sσ = S∂σ and putting Tσ = σ]T∆n, then ∂T + T∂ = id−S.

20

4. Iterated subdivision. Recall that we have S, T such that ∂S = S∂ and∂T + T∂ = id−S.

This says that Si is a chain map, and that Si is even chain homotopic toid.

Dm =∑

0≤i≤m TSi = T (

∑Si), check that ∂Dm +Dm∂ = id−Sm.

For each σ : ∆n → X, there exists m such that Sm(σ) ∈ CUn (X). Put

m(σ) =least m that works. Define D : Cn(X) → Cn+1(X) by Dσ =Dm(σ)σ, then ∂Dσ +Dσ∂ = σ − ρ(σ). We discover that ρ(σ) ∈ CU

n (X).

Check ρ : Cn(X) → CUn (X) is a chain map, ∂ρ = ρ∂, and moreover

∂D +D∂ = id−ι ◦ ρ where ι is the inclusion of CU into C.

Note that ρι = id on CU , and so we are done.

Theorem 2.13

Definition 3.4 (Good Pair). Hn(X,A) → Hn(X/A) isomorphism if A 6= ∅, Aclosed, A ⊂ V open, A a deformation retract of V . Then we call (X,A) a goodpair.

Prop 2.21CU∗ (X) → C∗(X) the inclusion, p in the opposite direction, with pi 'c id

and ip 'c idRelative version, (X,U ), (A,U |A).

0 CU |A∗ (A) CU

∗ (X) CU∗ (X,A) 0............................................................ ............ ..................................................... ............................................................................................... ............ ........................................................................... ............

0 C∗(A) C∗(X) C∗(X,A) 0................................................................................ ............ ................................................................. .............................................................................................................. ............ ................................................................................... ............

.............................................................................................................................

.............................................................................................................................

.............................................................................................................................

implies that HU∗ (X,A)→ H∗(X,A) is an isomorphism.

Pf. of excision theorem in formX =∫

(A)∪∫

(B) implies thatH∗(B,A∩B) =H∗(X,A).

Proof. Let U = {A,B}0 C

U |A∗ CU

∗ (X) CU∗ (X,A) 0.............................................................................. ............ ..................................................... ................................................................................................................ ............ ........................................................................... ............

C∗(A) C∗(A)⊕ C∗(B)with vertical columns equivalences.Then CU

∗ (X,A) = C∗A+C∗BC∗A

' C∗BC∗(A)∩C∗(B) = C∗B

C∗(A∩B) = C∗(B,A∩B).

21

(X,A) a good pair implies that q : (X,A)→ (X/A,A/A) induces an isomor-phism H∗(X,A)→ H∗(X/A,A/A)→ H∗(X/A).

A closed subset of V open, A deformation retract of V implies that A/Aclosed subset of V/A, A/A a def retract of V/A.

(X,A,B) triple, X ⊃ A ⊃ B, then 0→ C∗(A,B)→ C∗(X,B)→ C∗(X,A)→0 is a short exact sequence of chain complexes, and so we get a long exact se-quence on relative homology groups.H∗(X,A) H∗(X,V )

H∗(X/A,A/A)H∗(X/A, V/A

................................................ ............'

.............................................'

.............................................................................................................................

q∗

.............................................................................................................................

In fact, we can excise A and get isomorphisms to H∗(X \ A, V \ A) and toH∗(X/A \A/A, V/A \A/A).

i : A → V is a homotopy equivalence, so Hn(A) ' Hn(V ) → Hn(V,A) =0→ Hn−1(A) ' Hn−1(V )

Excision holds for CW-pairs. X = A ∪B, X a CW complex and A,B CW-subcomplexes implies that H∗(B,A ∩ B) ' H∗(X,A). induces isomorphisms....?

If X =∨αXα where xα ∈ Xα for all α, (X, {xα}) is a good pair for all α,

then ια : Xα → X the inclusions induce ⊕αiα∗ : ⊕H∗(Xα)→ H∗(X).

Theorem 3.14. If U ⊆ Rm, V ⊆ Rn are open and V homeomorphic to U , thenm = n.

Proof. Let x ∈ U . Excision implies that Hm(U,U \ {x})→ Hm(Rm,Rm \ {x}),is an isomorphism, as are Hm(Sm−1) to Hm(Rm, Sm−1) to it, and so we getδ`mZ = Hm(Sm−1). Homeomorphisms induce isomorphisms on homology, som = n, as δ`mZ = δ`nZ.

Theorem 3.15. If X is a ∆-complex, then there exists a natural isomorphismH∆∗ ' H∗(X).

In fact, if A is a subcomplex of a ∆-complex X, then get isomorphismH∆∗ (X,A) = H∗(X,A).

Proof. Take dimX < ∞, A = ∅. Let Xk be the union of all simplexes in X ofdim less than or equal to k, call this the k-skeleton.

Induction on k, ∅ ⊂ X0 ⊂ . . . ⊂ Xn−1 ( Xn = X, dimX = n.See diagram on bottom of 128 for (Xk, Xk−1). By the 5-lemma, if the outside

maps are isomorphisms, then the inside one is.

Degree of a map f : Sn → Sn

Definition 3.5 (Degree). Hn(Sn) ' Z for all n ≥ 0. f induces f∗ : Hn(Sn)→Hn(Sn), so there exists d ∈ Z such that f∗(α) = dα for all α ∈ Hn(Sn). Setdeg f = d.

22

Here are some basic properties:

1. deg(id) = 1

2. f ' g ⇒ f∗ = g∗ ⇒ deg f = deg g.

3. f 'constant⇒ deg f = 0. So if f : Sn → Sn not surjective, then deg f =0.

4. deg fg = deg f deg g.

5. R : Sn → Sn a reflection implies degR = −1. (Why?)

6. degA = (−1)n+1, where Ax = −x for all x.

7. If f : Sn → Sn has no fixed points, then deg f = (−1)n−1

8. If A ∈ O(n+ 1), then degA = detA.

9. fn : z 7→ zn on S1 has deg fn = n.

10. f, g : Sn → Sn and deg f = deg g imply that f ' g.

11. f : Sn → Sn has suspension Sf : Sn+1 → Sn+1 indeed, SX for anyf : X → Y induces Sf : SX → SY . Then degSf = deg f . [Giverf : Sn+1 → Sn+1, it must be homotopic to a suspension [Sn+1, Sn+1] ' Z,there exists a map, the Hopf map, S3 h→ S2 such that h 6' constant map.]

12. f : Sn → Sn, y ∈ Sn, f−1(y) = {x1, . . . , xk}, degxi f = ±1, then deg f =∑degxi f .

Cellular HomologyX a CW-complex, ∅ ⊆ X0 ⊆ . . . ⊆ Xn ⊆ . . ., X =

⋃iX

i.Hk(X) ' HCW

k (X), the cellular homology given by

→ CCWn+1(X) d→ CCWn (X) d→ CCWn−1(X)→, d2 = 0.Put CCWn (X) = Hn(Xn, Xn−1) is free abelian, a generator for each n cell

of X.Why? (Xn, Xn−1) is a good pair, Xn \Xn−1 = ∪αenα where enα is homeo-

morphic to the open ball.Let xα correspond to the origin in the homeomorphism. Put V = Xn \{xα}

open and there exists a retraction of V onto Xn−1. Hence, Hn(Xn, Xn−1) 'Hn(Xn/Xn−1) = Hn(∨αBnα/sn−1

α ) ' ⊕Z with generators enα.We need d : Hn(Xn, Xn−1) → Hn−1(Xn−1, Xn−2), we get it by j∗ ◦ ∆

through Hn−1(Xn−1) where j : (Xn−1, ∅)→ (Xn−1, Xn−2) is the inclusion butas a map of pairs.

We must check that d2 = 0, d = ∆◦j∗ : Hn(Xn, Xn−1)→ Hn−1(Xn−1, Xn−2),which is true as j∗ ◦∆ = 0. Thus H∗(CCW∗ (X)) = HCW

∗ (X).

Lemma 3.16. 1. Hk(Xn, Xn−1) = 0 if k 6= n

23

2. Hk(Xn) = 0 if k > n and if dimX <∞, Hk(X) = 0 if k > dim(X)

3. i : Xn → X induces isomorphism i∗ : HkXn → HkX if k < n.

Proof. Proof of b: Use induction on n. For n = 0, clear. Suppose n > 0,result is known for smaller skeleton. Then Hk(Xn−1) = 0 → Hk(Xn) →Hk(Xn, Xn−1) = 0, so Hk(Xn) = 0.

Proof of c: First, suppose dimX < ∞, X = XN , N = dimX. Xn ⊆Xn+1 ⊂ . . . ⊂ XN = X. If n = N , good and simple. Suppose that n <N for k < n, we get 0 = Hk+1(Xn+1, Xn) → Hk(Xn) → Hk(Xn+1) →Hk(Xn+1, Xn) = 0 so we get isomorphism for all k.

Need the diagram on page 139. Proof by diagram chasing. We will needthe next result to do this. If k < n, Hk(Xn) → Hk(Xn+1) → . . . → Hk(XN )isomorphisms, but does not necessarily reach Hk(X).

Lemma 3.17. If K is a compact subset of a CW-complex X, then K ⊂ Xn forsome n.

Proof. Suppose not, then for each n, choose xn ∈ K with xn /∈ Xn. ThenK0 = {xn} ⊂ K is an infinite set. Note: K0 ∩Xp is finite for all p, so K0 ∩Xis closed. Thus, K0 is compact. Note that each subset of K0 is closed, and soK0 has the discrete topology, contradiction.

Easiest Applications1) Suppose there are no n-cells at all for some n. Then Hn(X) = 0.2) Suppose there are k n-cells. Then CCWn (X) ' Zk, so ZCWn (X) ' Z`,

` ≤ k, so Hn(X) has ≤ k generators. In particular, rank Hn(X) ≤ k.3) Suppose there are n-cells but no n + 1 or n − 1-cells. Then d = 0 in

CCW∗ (X), so CCW∗ (X) = Hn(X) ' free abelian group on the n-cells.4) H2k(CPn) ' Z for 0 ≤ k ≤ n, others are 0.Application of Degree

Theorem 3.18. Z2 is the only nontrivial group that can map freely on S2n.

Proof. Any homeomorphism h : S2n → S2n has degree ±1, so we get a ho-momorphism deg : G → {±1}. If g ∈ G, g 6= 1, then gx 6= x for all x. Sowe x 7→ gx is homotopic to A : x 7→ −x, so deg g = (−1)2n+1 = −1. Hence,ker deg = {1}, so degG→ {±1} is injective. Thus |G| ≤ 2.

Let f = fd : Sn → Sn be a degree d map. X = Sn ∪f en+1, find Hi(X).Have Φ : (Dn+1, Sn)→ (X,Sn), induces isometried on homology, so

Hn+1(Dn+1, Sn) Hn(Sn)

Hn+1(X,Sn) Hn(Sn)

.................. ............∆

...................................... ............∆

.............................................................................................................................

Φ∗

.............................................................................................................................

f∗

Hn+1(X) ............................ ............ Hn(X)...................................................................... ............

24

With the first map on the bottom injective and the last surjective. So weconclude that Hn(X) ' Z/dZ, Hn+1(X) = 0 and Hi(X) = 0 if i 6= n, n+ 1.

If f = f2 : S1 → S2 is a map of degree two, we can see that RPn hashomology Hn = Z/2 iff n = 1 and 0 otherwise.

IF X,Y are CW-Complexes and f : X → Y call f cellular if f(Xn) ⊂ Y n

for all n ≥ 0.FACT: Cellular Approximation Theorem: Any continuous f : X → Y be-

tween CW-complexes is homotopic to a cellular map.Ex: Any f : Sm → Sn with m < n is homotopic to a constant map.Then f : (Xn, Xn−1)→ (Y n, Y n−1) for all n, so get a chain mapHn(Xn, Xn−1)→

Hn(Y n, Y n−1), so we get HCWn (X) → HCW

n (Y ), thus we get a commutativesquare

HCWn (Y )

Hn(Y )

HCWn (X)

Hn(X)

........

........

........

........

........

........

........

........

........

........

........

........

.................

............

'

........

........

........

........

........

........

........

........

........

........

........

........

.................

............

'

............................................................

fCW∗

.......................................................................................f∗

Cellular Boundary FormulaPrelude: Choose gen of Hn(Sn) compatibly under isomorphisms Hn+1(Sn−1)→

Hn(Sn) starting with S0. Then also Hn(Dn, Sn−1) → Hn−1(Sn−1). Use {enα}as gens for CCWn (X). Then dn(enα) =

∑β dαβe

n−1β , dαβ ∈ Z the degree of

Sn−1α → Xn−1 → Xn−1

Xn−1\en−1β

' Sn−1β .

Homology with CoefficientsIf R is a commuative ring with identity (say a field) then we take Cn(X;R)

to be the free R-module generated by the singular simplexes, which is, in fact,R⊗ZCn(X), and every works out with Hn(X;R). So for example, Hn(X;R) 'R for X = Sn.

Let X be a finite complex, and αi the number of cells in each dimension.Then χ(X) =

∑dimXi=0 (−1)iαi.

Hi(X) = Zβi + Fi where Fi is finite. βi is called the ith Betti number.[Hi(X)⊗Z Q has dim βi].

Fact: For finitely generated abelian groups, if 0 → A → B → C → 0 isexact, then rankA− rankB + rankC = 0.

Theorem 3.19. χ(X) =∑dimXi=0 (−1)iβi.

Proof. 0 → BCWi (X) → ZCWi (X) → HCWi (X) → 0 is s.e.s. HCW

i (X) 'Hi(X), and 0→ ZCWi (X)→ CCWi (X)→ BCWi−1 (X)→ 0 s.e.s.

So∑

(−1)u rankHi(X) =∑

(−1)i rankHCWi (X) =

∑(−1)i[rankZCWi (X)−

rankBCWi (X)] =∑

(−1)i(rankCCWi (X) − rankBCWi−1 (X) − rankBCWi (X)) =∑(−1)iαi −

∑(−1)iBCWi (X) −

∑(−1)i rankBCWi−1 (X) =

∑(−1)iαi = χ(X).

Mayer-Vietoris Sequence

25

(Vietoris) Let A,B ⊂ X, int(A) ∪ int(B) = X. Our aim is a l.e.s. . . . →Hn(A ∩B) Φ→ Hn(A)⊕Hn(B) Ψ→ Hn(X) ∆→ Hn−1(A ∩B)→ . . .

Recall that Cn(A)+Cn(B)→ Cn(X) for all n gives C∗(A)+C∗(B)→ C∗(X)which induces an isomorphism on homology groups.

Next: 0→ C∗(A ∩B)→ C∗(A)⊕C∗(B)→ C∗(A) +C∗(B) ⊂ C∗(X) where+ is the join.

x 7→(x−x

),(xy

)7→ x+ y. This is a short exact sequence. This gives:

∆→ Hn(A ∩B)→ Hn(A)⊕Hn(B)→ Hn(C∗(A) + C∗(B)) ' Hn(X)→ . . .Snake Lemma and some applications

4 Cohomology

Why bother with Cohomology? Because it has a ring structure, given by a cupproduct Hp(X)⊗Hq(X)→ Hp+q(X). In fact, it is a graded ring. This is likedifferential forms (they are a primary example).

A cellular chain, if X is a CW-complex, looks like 0→ Cn → Cn−1 → . . .→C1 → C1 → 0.

Cochains: If A is an abelian group, so is hom(A,Z). Af→ B, then we get

hom(A,Z)f∗← hom(B,Z), so this is a contravariant functor. Similarly, if we have

Af→ B

g→ C, we get a similar sequence of homs.If A = Zk, B = Z`, then f : A → B a function. What is f∗ : hom(B,Z) →

hom(A,Z)? Well hom(Zk,Z) ' Zk, but this is not a natural isomorphism.So a cochain is 0 ← hom(Cn,Z) ← hom(Cn−1,Z) ← . . . ← hom(C1,Z) ←

hom(C0,Z) ← 0. Traditionally, we write the map as δ = ∂∗ and call it thecoboundary. We have δδ = 0.

And so, we define Hk(X) = ker δ/ Im δ. We often denote hom(Cn,Z) as Cn.Why did we do this? This comes from differential forms on manifolds. A

p-form is useful, because it can be integrated over a p-simplex to obtain a realnumber

∫∆p α. That is, α ∈ Cp(M,R). For differential forms, there is a map d

from p-forms to (p+ 1)-forms called the exterior differential. The big propertyis that

∫∆dα =

∫∂∆

α, called Stokes’ Theorem. If α ∈ Cp is not necessarily a

p-form, then what is δα? We have Cp+1∂→ Cp

α→ R. So δα = α ◦ ∂. That is,δα(∆) =

∑(−1)iα(∆i).

We hope that Hk(X) = hom(Hk(X),Z).Example: X = RP2. We have 0 → Z x2→ Z 0→ Z → 0, the chain complex

C∗(X).C∗(X) is 0 ← Z x2← Z 0← Z ← 0, so H0(X) = Z. H1(X) = 0 and H2(X) =

Z2. So our hopes fail...but we did get the same groups, just in the wrong places.For S2, we have C∗ is 0→ Z→ 0→ Z→ 0, so C∗ is 0← Z← 0← Z← 0,

so the cohomology groups are the same as the homology groups.In fact, this is a general phenomenon, for finitely generated abelian groups,

the free part stays when switching to cohomology, but the torsion part switches

26

dimensions.Let f∗ : A∗ → B∗ be a chain map between chain complexes. We want to

define the algebraic mapping cylinder.

Definition 4.1 (Mapping Cylinder). f : X → Y is a continuous map of topo-logical spaces. Take (X × I ∪ Y )/(x, 1) ∼ f(x). We call this M(f).

The mapping cylinder has nice properties, like M(f) is homotopy equivalentto Y an that X ↪→M(f), and the diagram

X

M(f)

Y.............................................................................................................................................................................

................................................................................................................. ............

.............................................................................................................................

'

It gives us a pair (M(f), X) for the map between X and Y , which we canuse to obtain long exact sequences on homology.

Definition 4.2 (Algebraic Mapping Cylinder). We define M(f)k = Ak⊕Bk⊕Ak−1. Then the map M(f)k → M(f)k−1 is given by ∂ on Ak, Bk and to ∂ onAk−1 to Ak−2, but also f∗ from Ak−1 to Bk−1 and the identity from Ak−1 toAk−1. That is, with rows summed across:Ak Ak−1 Bk

Ak−1 Ak−2 Bk−1

.............................................................................................................................

∂

.............................................................................................................................

-∂

.............................................................................................................................

∂

Error: Incorrect label specification Error: Incorrect label specification

.................................................a

Error: Incorrect label specification Error: Incorrect label specification

.................................................f

with the sign change made to make sure that this is a complex. We call thiscomplex the algebraic mapping cylinder, so ∂(a, a′, b) = (∂a+a′,−∂a′, ∂b+fa′).

Need to show that M(f)∗ is chain homotopy equivalent to B∗.We are aiming at the following:

Proposition 4.1. Suppose that C∗ and D∗ are chain complexes of free abeliangroups and f∗ : C∗ → D∗ is a chain map such that f∗ induces an isomorphismon homology. Then f∗ is a chain homotopy equivalence. That is, there existg∗ : D∗ → C∗, S : C∗ → C∗+1 and T : D∗ → D∗+1 so that ∂S + S∂ = 1− g∗f∗and ∂T + T∂ = 1− f∗g∗.

This implies that f∗ : H∗(D∗)→ H∗(D∗) is an isomorphism.

Proof. S∗δ + δS∗ = 1− f∗g∗ and T ∗δ + δT ∗ = 1− g∗f∗. Check this!

Proposition 4.2. Suppose that C∗ is a chain complex of free abelian groupsand Hk(C∗) is finitely gneerated for each k. Hk(C∗) ' Fk ⊕Tk where Fk is freeand Tk is torsion. Then Hk(C∗) ' Fk ⊕ Tk−1

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So we were at f∗ : A∗ → B∗, M(f)k = Ak ⊕ Ak−1 ⊕ Bk with ∂(a, a′, b) =(∂a+ a′,−∂a′, ∂b− fa′).

s : M(f)∗ →M(f)∗+1, s(a, a′, b) = (0, a, 0).∂S + S∂(a, a′, b) = ∂(0, a, 0) + s(∂a+ a′,−∂a′, ∂b− fa′) = (a,−∂a,−fa) +

(−, ∂a, a′, 0) = (a, a′,−fa) = 1− r. r(a, a′, b) = (0, 0, b+ fa).And so we get

0 A∗ M(f)∗ C∗(f) 0............................................................................................... ............ ................................................................................... ............................................................................................................................. ............ .................................................................................................... ............

B∗

........

........

........

........

........

........

........

........

........

........

........

........

.................

............

f∗

................................................................................................................................................................. ............

r∗

Get that f∗ induces isos on homology iff H∗(C(f)) = 0.C(f)∗ acyclic implies that f∗ is a chain equiv (later)

Definition 4.3 (Free Resolution). Let H be an abelian group. A resolution isa long exact sequence . . .→ F2 → F1 → F0 → H → 0.

Proof. Resolutions are unique up to chain homotopy.

G2 G1 G0 K 0

F2 F1 F0 H 0

................................................................................................................. ............ ................................................................................................................. ............ ................................................................................................................. ............ ................................................................................................................. ............ ................................................................................................................. ............

................................................................................................................. ............ ................................................................................................................. ............ ................................................................................................................. ............ ................................................................................................................. ............ ................................................................................................................. ............

.............................................................................................................................

φ

.............................................................................................................................

φ0

Extends to a chain map which is unique up to chain homotopy, as we cansuppose that there are φ’s such that the diagram also commutes with them.We want S : F∗ → G∗+1 with ∂S + S∂ = φ − φ. We want s : F0 → G1 with∂s = φ− φ. Want ∂sz = (φ− φ)z, this gets hit by ∂ iff ∂(φ− φ)z = 0, and thatis (φ∂ − φ∂)z = 0.

Want s : F1 → G2 such that ∂s + s∂ = φ − φ, z a generator in F1. Want∂sz = (φ−φ−s∂)z. There is an element sz with the right ∂ iff ∂(φ−φ−s∂)z = 0,and this is (∂φ− ∂φ− ∂s∂)z = (φ0∂ − φ0∂ − ∂s∂)z = (φ0 − φ0 − ∂s)∂z.

So we can take H and a free resolution of H, and take hom(−, G) of it, it

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may develop homology, and we thus obtain derived functors.MISSED A LECTURESuppose that A∗

f∗→ B∗ is such that f∗ induces an isomorphism on Homology.Then H∗(C(f)) = 0 because 0→ A∗ →M(f)∗ → C(f)∗ → 0 is exact.

H∗(C(f)) = 0 ⇒ C(f)∗ is chain contractible, as C(f)∗ is a resolution of 0,so chain equivalent to the zero complex by uniqueness of resolution.

C(f)∗ is chain contractible implies that f has a chain homotopy inverse,as S : C(f)∗ → C(F )∗+1 is a chain contraction, ∂ : C(f)∗ → C(f)∗01, then(−∂ 0−f ∂

): A∗−1 ⊕ B∗ → A∗−2 ⊕ B∗−1. S =

(k `m n

): A∗−1 ⊕ B∗ →

A∗ ⊕B∗+1. ∂S + S∂ = I, and can be worked out via matrix multiplication.There is a splitting Hn(X,G) ' hom(Hn(X), G)⊕Ext(Hn(X), G), but it is

not natural, so it must be used carefully, if at all.Cup Product(Cohomology is into a commutative ring Λ with identity)Let ϕ ∈ Ck(X,Λ) and ψ ∈ C`(X,Λ), then ϕ∪ψ ∈ Ck+`(X,Λ) by (ϕ∪ψ)σ =

ϕ(σ[v0, . . . , vk])ψ(σ[vk, . . . , vk+`])δ(ϕ ∪ ψ) = δϕ ∪ ψ + (−1)kϕ ∪ δψ.Why is this good?If ϕ,ψ are cocycles then ϕ ∪ ψ is a cocycle, ie, if δϕ = δψ = 0 then δ(ϕ ∪

ψ) = 0. Also (ϕ + δα) ∪ (ψ + δβ) = ϕ ∪ ψ + ϕ ∪ δβ + δα ∪ ψ + δα ∪ δβ =ϕ ∪ ψ + δ(α ∪ ψ) + δ(α ∪ δβ) + (−1)k(ϕ ∪ β), so it is ϕ ∪ ψ + δγ for some γ.

Thus, this defines Hk(X)×H`(X)→ Hk+`(X).We should check the formula for δ(ϕ ∪ ψ). We take σ = 〈v0, . . . , vk+`+1〉.

δ(ϕ ∪ ψ)(σ) = (ϕ ∪ ψ)δσ = (ϕ ∪ ψ)∑

(−1)i〈v0, . . . , vi, . . . , vk+`+1〉.δϕ∪ψ(σ) = δϕ[v0, . . . , vk+1]ψ[vk+1, . . . , vk+`+1] =

∑(−1)iϕ[v0, . . . , vi, . . . , vk+1]ψ[vk+1, . . . , vk+`+1]

ϕ ∪ δψ(σ) = ϕ[v0, . . . , vk]∑

(−1)iψ[vk, . . . , vi, . . . , vk+`+1]Sign counting to check that it’s right, see various sources for details.Poincare Duality: For a closed orientable manifold, there is an isomrophism

Hn−k(M)→ Hk(Mn) by a ”cap product”

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