Surface Topology - University of Illinois at Urbana–Champaignclein/BrendlePCMILecturespost.pdf ·...

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Surface Topology Tara E. Brendle July 22, 2011 Contents 1 Lecture 1: What is a surface? 3 1.1 Definition of a surface ................................. 3 1.2 Surfaces with boundary ................................ 4 1.3 Closed sets ....................................... 4 1.4 Compactness ...................................... 4 1.5 Connectedness, path-connectedness .......................... 6 1.6 Where we’re heading .................................. 7 2 Lecture 2: Quotients 8 2.1 Quotients as sets without metrics ........................... 8 2.2 What happens to the metric? ............................. 8 2.3 Pathification ...................................... 8 2.4 Examples ........................................ 9 3 Group theory 11 3.1 Groups and homomorphisms ............................. 11 3.2 Group actions ...................................... 12 3.3 Forming quotients of spaces via group actions .................... 13 4 Lecture 4: Euler characteristic 14 4.1 Euler characteristic: definition/examples ....................... 14 4.2 Euler characteristic and genus ............................. 15 5 Lecture 5: Fundamental Groups 18 5.1 Homotopy ........................................ 18 5.2 Fundamental Groups .................................. 18 5.3 Basepoints and functoriality .............................. 19 6 Lecture 6: Functoriality, Covering Spaces and Deck Transformations 21 6.1 Functoriality ...................................... 21 6.2 First examples: Convex sets in R n ; the circle S 1 .................. 21 6.3 Covering spaces ..................................... 22 6.4 Deck transformations .................................. 22 6.5 Lifting paths and homotopies ............................. 23 6.6 Finishing the proof of the Isomorphism Theorem .................. 24 1

Transcript of Surface Topology - University of Illinois at Urbana–Champaignclein/BrendlePCMILecturespost.pdf ·...

Page 1: Surface Topology - University of Illinois at Urbana–Champaignclein/BrendlePCMILecturespost.pdf · topology: Armstrong’s Basic Topology and Goodman’s Beginning Topology. 1Lecture1:Whatisasurface?

Surface Topology

Tara E. Brendle

July 22, 2011

Contents

1 Lecture 1: What is a surface? 31.1 Definition of a surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Surfaces with boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Closed sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4 Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.5 Connectedness, path-connectedness . . . . . . . . . . . . . . . . . . . . . . . . . . 61.6 Where we’re heading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Lecture 2: Quotients 82.1 Quotients as sets without metrics . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.2 What happens to the metric? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.3 Pathification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3 Group theory 113.1 Groups and homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.2 Group actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.3 Forming quotients of spaces via group actions . . . . . . . . . . . . . . . . . . . . 13

4 Lecture 4: Euler characteristic 144.1 Euler characteristic: definition/examples . . . . . . . . . . . . . . . . . . . . . . . 144.2 Euler characteristic and genus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

5 Lecture 5: Fundamental Groups 185.1 Homotopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185.2 Fundamental Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185.3 Basepoints and functoriality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

6 Lecture 6: Functoriality, Covering Spaces and Deck Transformations 216.1 Functoriality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216.2 First examples: Convex sets in Rn; the circle S1 . . . . . . . . . . . . . . . . . . 216.3 Covering spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226.4 Deck transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226.5 Lifting paths and homotopies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236.6 Finishing the proof of the Isomorphism Theorem . . . . . . . . . . . . . . . . . . 24

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6.7 The Torus, Again . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

7 Lecture 7: More on Fundamental Groups and Covering Spaces 267.1 Homotopy Equivalence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267.2 Action of the deck group on the cover . . . . . . . . . . . . . . . . . . . . . . . . 277.3 Fundamental groups of surfaces given as plane models . . . . . . . . . . . . . . . 28

8 Lecture 8: Jordan Curve Theorem 298.1 Graph Theory Cheat Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298.2 Polygonal Jordan Curve Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 308.3 General Jordan Curve Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

9 Lecture 9: Triangulability of Surfaces 339.1 Finishing up Jordan Curve Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 339.2 Corollaries of Jordan-Schoenflies . . . . . . . . . . . . . . . . . . . . . . . . . . . 349.3 Proof of Triangulability, assuming Jordan-Schoenflies. . . . . . . . . . . . . . . . 359.4 Well-definedness of connected sum . . . . . . . . . . . . . . . . . . . . . . . . . . 369.5 Finishing Classification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

10 Lecture 10: Proof of Triangulability of Surfaces 3710.1 Well-definedness of connected sum . . . . . . . . . . . . . . . . . . . . . . . . . . 3710.2 Finishing Classification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3710.3 Classification of simple closed curves on surfaces . . . . . . . . . . . . . . . . . . 38

11 Lecture 11: Basics of curves on surfaces 4011.1 Essential Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4011.2 Algebraic intersection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4011.3 Geodesic representatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4011.4 Torus example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4111.5 Bigon criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

12 Lecture 12: Mapping Class Groups I 4312.1 Punctures versus Boundary Components . . . . . . . . . . . . . . . . . . . . . . . 4312.2 Finite order examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4312.3 Some warmup examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

13 Lecture 13: Mapping Class Groups II 4513.1 Dehn twists . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4513.2 Torus, again . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4613.3 Exact sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

14 Lecture 14: Mapping Class Groups III 4714.1 Generating sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4714.2 The curve complex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4714.3 Variations on the curve complex . . . . . . . . . . . . . . . . . . . . . . . . . . . 4714.4 Generation of Mod(S) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

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Acknowledgement: This lecture follows two excellent treatments of the basics of surfacetopology: Armstrong’s Basic Topology and Goodman’s Beginning Topology.

1 Lecture 1: What is a surface?

Note that all “spaces” are metric spaces. Those with a background in topology can think ofthese as topological spaces, with the topology generated by the open sets induced by the metric.

1.1 Definition of a surface

The OED’s first definition of a surface is: “The outermost boundary (or one of the boundaries)of any material body, immediately adjacent to the air or empty space, or to another body.”Compare with the mathematical definition:

A surface S is a metric space which is locally homeomorphic to a disk. In other words,for every point p ∈ S, there exists an open set U ⊆ S such that p ∈ U and such that U ishomeomorphic to the open disk {(x, y) | x2 + y2 < 1}.

A topological space X is called Hausdorff, if for every a, b ∈ X with a #= b, there existdisjoint open sets U(a), U(b) containing a, b, respectively. In a topology course, a surface isusually defined with the word “Hausdorff” replacing “metric”, which gives a slightly moregeneral notion of a surface. Note that every metric space is automatically Hausdorff, since ifd(a, b) = ε, then we can take U(a), U(b) each to be balls of radius less than ε

2 centered at a, b,respectively.

Note to those with topology background: a compact Hausdorff space admits a metric if andonly if it’s second countable. Also, a locally metrizable space is metrizable iff it is Hausdorffand paracompact.

Examples.

• The plane R2 with the standard metric.

• The sphere S2: : by symmetry it’s enough to find a neighborhood of (0, 0, 1) homeomor-phic to a disk. Then take, say, all points with z > 0 and show that projection to xy-planeis a homeomorphism.

• The torus T 2, version one: surface of revolution in R2

• The torus T 2, version two: S1 × S1 in R4

Nonexample.

• To see why being locally homeomorphic to a disk is not a sufficient condition: Start withthe plane. Remove (0, 0) and replace it with two “copies”.

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1.2 Surfaces with boundary

A surface with boundary S is a metric space such that for every point p ∈ S, there exists anopen set U ⊆ S such that p ∈ U and such that U is homeomorphic to an open disk or to thehalf disk {(x, y) | x2 + y2 < 1 and x ≥ 0}. If p has a neighborhood of the first type it is aninterior point, otherwise p is a boundary point.

Examples of surfaces with boundary.

• The disk D2 = {(x, y) | x2 + y2 ≤ 1}

• The annulus A = {(x, y) | 1 ≤ x2 + y2 ≤ 2}

Exercise 1.1. Which of the following are surfaces? Surfaces with boundary?

1. {(x, y, z) ∈ R3 | x2 + y2 + z2 = 1 and z ≥ 0}

2. {(x, y, z) ∈ R3 | 4x2 + y2 + 9z2 = 1}

3. {(x, y, z) ∈ R3 | x = 0 or z = 0}

1.3 Closed sets

A subset of a space is closed if its complement is open.

Exercise 1.2. Think of an example of a subset of R2 with the usual metric which is both openand closed, and one which is neither open nor closed.

Let S be a subset of a space X . Then p ∈ X is a limit point of S (sometimes known asa point of accumulation) if every open set which contains p also contains at least one point ofS − {p}.

Exercise 1.3. Determine the limit points of [0, 1) in R.

Exercise 1.4. Determine the limit points of the set of points of the form 1nin R, for n ∈ N.

The union of a set S and its limit points is called the closure of S.

Exercise 1.5. A set is closed if and only if it contains all its limit points. Therefore a set isclosed if and only if it is equal to its closure.

1.4 Compactness

A space is compact if every open cover has a finite subcover. That is, if X = ∪α∈AVα, whereVα is open for all α in some (possibly infinite) index set A, then X = ∪n

i=1Vαifor some finite

subset of the Vα’s.

Note about terminology. Just to ensure constant confusion, mathematicians often use theterm “surface” to refer to both surfaces and surfaces with boundary as defined above, and usethe term “closed surface” to mean “compact surface without boundary”, i.e., as opposed to aset that’s closed in the sense that its complement is open.

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Theorem 1.6 (Heine-Borel for R). A closed interval of the real line is compact.

Proof. Let C be an open cover of [a, b]. Consider the set X of all x ∈ [a, b] such that [a, x] iscontained in a finite subcover of C. First note that X has a least upper bound s since it isnonempty and bounded above. We will show that s ∈ X and that s = b. Let U ∈ C be an openset containing s. If s < b we can choose ε such that (s − ε, s+ ε) ⊆ U . (If s = b, we can stillassume (s− ε, s] ⊆ U .)

Now, note that if x ∈ X and if a ≤ y ≤ x, then y ∈ X . Then, by definition of s, we musthave that s− ε

2 ∈ X , i.e., there is a finite subcover C′ of C for the interval [a, s− ε2 ]. But then

C′ ∪ U is a finite cover of [a, s], and hence s ∈ X .Finally, to see that s = b, we observe that if s < b, then C′ ∪ U actually covers [a, s + ε

2 ],which contradicts the fact that s is a least upper bound for X .

Note that this proof relied heavily on the completeness property of the real numbers, specif-ically that a nonempty set of real numbers which has an upper bound has a least upperbound.

The next two results aren’t hard to prove, but they are important enough to call themtheorems.

Theorem 1.7. A closed subset C of a compact space X is compact.

Proof. Let F be an open cover of C. Then F ∪ (X −C) is an open cover of X , and hence has afinite subcover F′. Then F′ (or possibly F′− (X−C)) is the finite subcover of C we are lookingfor.

Theorem 1.8. Prove that if f : X → Y is continuous, and then f(A) is compact if A iscompact.

Proof. Without loss of generality, we may assume that X is compact and that f is surjective.Let F be an open cover of Y . Then since f is continuous, {f−1(U) | U ∈ F} is an opencover of X and has a finite subcover of the form f−1(U1), . . . , f−1(Un). Since f is surjective,f(f−1(U)) = U for all U , hence {U1, . . . , Un} is the finite subcover of Y we require.

Theorem 1.9 (Heine-Borel for Rn). A subset of Rn is compact if and only if it is closed andbounded.

We’ll leave the “forward” direction as a series of exercises. To prove the converse, we’ll needthe following proposition:

Proposition 1.10. A finite product of closed intervals in Rn is compact.

Proof of Proposition. We’ll show that S = [a, b] × [c, d] is compact and leave the general caseas an exercise (it’s virtually identical). Let F be an arbitrary open cover of S. Without lossof generality, we may assume F is consists of sets of the form U × V , where U and V are bothopen intervals. (Exercise: explain why this is true.)

Now, let (x, y) be a point in S. Since F also covers {x} × [c, d], and since {x} × [c, d] ishomeomorphic to [c, d], an interval, there must be a finite subcover of {x}× [c, d] consisting ofelements of the form Ux

i × V xi , for i = 1, . . . ,m(x). Let Ux denote the intersection of all the

Uxi .

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But then the union of all the Ux, over all x ∈ [a, b], is an open cover of [a, b], which musthave a finite subcover Uxj , for j = 1, . . . , n. Now we have

[a, b]× [c, d] ⊂ (Ux1 × [c, d]) ∪ · · · ∪ (Uxn × [c, d])

⊂ ((Ux1

1 × V x1

1 ) ∪ · · · ∪ (Ux1

m(x1)× V x1

m(x1))) ∪ · · · ∪ ((Uxn

1 × V xn

1 ) ∪ · · · ∪ (Uxn

m(xn)× V xn

m(xn)))

and we’re done.

Closed and bounded ⇒ compact. Now, suppose we have a closed and bounded subset S of Rn.Since S is bounded it must be contained in the n-fold product [−a, a]× [−a, a]× · · ·× [−a, a]for some a ∈ R. (We can take this as the definition of a bounded set in Rn, if we like.) By thepreceding proposition, this set is compact. Hence S is a closed subset of a compact space, andby the theorem above, we’re done.

Exercise 1.11. Finishing the proof of Heine-Borel for Rn:

1. Recalling that metric spaces satisfy the Hausdorff property, prove that a compact subset Kof a metric space X is closed. (HINT: Show that if x ∈ X −K, then there exist disjointopen sets containing x and K, respectively. )

2. Complete the proof of the general Heine-Borel Theorem by considering the open cover ofRn by open balls of radius n ∈ Z, centered at the origin.

1.5 Connectedness, path-connectedness

A space is disconnected if it is the disjoint union of (at least) two nonempty open sets. Otherwise,it is connected. Though the definition is easy to state, there are two caveats: first, it can bevery difficult to prove that a space is connected, and second, our intuition about connectednessdoesn’t always serve as good guide (see the example of the topologist’s sine curve).

Exercise 1.12. Prove each of the following statements.

1. The continuous image of a connected space is connected.

2. (Finite) products of connected spaces are connected.

Some examples without proof.

• Rn is connected for all n ≥ 1.

• Intervals of all kinds (open, closed, half-open) in R are connected.

• Sn for all n ≥ 1 is connected (S0 is disconnected).

• T 2 is connected.

• Topologist’s sine curve {(x, sin( 1x)) | 0 < x ≤ 1} ∪ (0× [−1, 1]) is connected.

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The last example above (the topologist’s sine curve) seems to defy our sense of what “con-nected” should mean, perhaps, and if you have never done so, it is well worth carefully workingthrough the proof that it satisfies the above definition of connectedness in order to fully appre-ciate what this definition is trying to capture. For our purposes, however, we will mostly usewhat is a more natural notion of connectivity (at least to Prof. Brendle’s way of thinking):

[Recall: Prof. Leininger defined a path at the beginning of the previous lecture.]A space is path-connected if for any two points x, y, there is a path from x to y. Note that

the topologist’s sine curve fails to be path-connected. Prof. Leininger will say more about thisin his next lecture (I think!), and we’ll revisit this notion probably in Week 2.

1.6 Where we’re heading

A simple closed curve in a spaceX is a map f : S1 → X such that the restriction f : S1 → f(S1)is a homeomorphism.

1. Jordan Curve Theorem: The complement of any simple closed curve in R2 has twoconnected components.

2. Classification of Surfaces: A complete list of all closed orientable surfaces.

3. Mapping class groups of surfaces: The appropriate notion of a group of “symmetries”or “automorphisms” of a surface.

Note that there other versions of the Jordan Curve Theorem appearing in the literature.We will end up using a stronger statement, known as the Jordan-Schoenflies Theorem.

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2 Lecture 2: Quotients

Acknowledgement: Lecture 2 will follow the treatment of quotient spaces given in Schwartz’swonderful new book “Mostly Surfaces”.

2.1 Quotients as sets without metrics

An equivalence relation on a set X is binary relation ∼ which is reflexive, symmetric, andtransitive.

If ∼ is an equivalence relation on a set X , and if x ∈ X , then the equivalence class [x] of xunder ∼ is defined to be the set of all y ∈ X such that y ∼ x. The quotient of X by ∼, denotedX/ ∼, is the set of all equivalence classes (which form a partition of X). A priori, X/ ∼ is justa set, but in all the cases we care about, X/ ∼ will also inherit a metric from X .

Exercise 2.1. Explain what is wrong with the following “proof” that the reflexive property isredundant in the above definition: By symmetry, we have x ∼ y ⇒ y ∼ x. Thus, by transitivity,x ∼ x. Hence any relation that is symmetric and transitive is also reflexive.

Exercise 2.2. For every possible choice of two out of the above three properties, give an exampleof a set, and a relation on that set, which satisifies those two properties but not the third.

2.2 What happens to the metric?

Let X be a space with metric d′. One natural way to try to put a metric on the quotientspace X/ ∼ is as follows. Let S1, S2 be two equivalence classes in X/ ∼. Define δ(S1, S2) tobe the infimum of the set {d′(s1, s2) | s1 ∈ S1 and s2 ∈ S2}. The intuition here is that thedistance between the sets S1 and S2 is taken to be the ”shortest” distance between individualelements in the two sets. Since the construction involves infimums of a metric, we can see thatδ satisfies the axioms of nonnegativity and of symmetry. But it’s no longer necessarily the casethat δ(S1, S2) = 0 if and only if S1 = S2, nor that the triangle inequality is satisfied.

There’s a general construction for taking such a function and creating a new one whichretains most of the information encoded in and also satisfies the triangle inequality, which wediscuss next.

2.3 Pathification

The idea here is to start with a function that’s a little bit like a metric, and build slowly towardssomething that is a metric. Pathification is one step in this process.

Let Y be a set (think Y = X/ ∼)and let δ : Y × Y → R be a function satisfying symmetryand nonnegativity, but not necessarily the triangle inequality, and also not necessarily requiringthat δ(x, y) = 0 if and only if x = y. Now, we’re going to add in the triangle inequality (so westill won’t quite have a metric).

If we have two points x, y ∈ Y , then a chain from x to y is a finite sequence of points in Y :x = x0, x1, . . . , xn = y. If Cx,y is such a chain, then define δ(Cx,y) = δ(x0, x1) + δ(x1, x2) +· · ·+ δ(xn−1, xn). Define

d(x, y) = infCx,yδ(Cx,y)

Here we’re taking the infimum over all possible chains from x to y. The function d is sometimescalled the pathification of δ. It is clear from the definition of d that it satisfies nonnegativityand symmetry again.

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Exercise 2.3. Show that d satisfies the triangle inequality.

Metrics for Quotients. Back to quotient spaces: we now take the pathification d of thefunction δ we defined on the quotient X/ ∼ using the original metric d′ on X . The only way dcan fail to be a metric now is that we might have d(x, y) = 0 for x #= y. We will see an exampleof this soon.

If, however, d satisfies d(x, y) = 0 if and only if x = y, then d is a metric, which we will referto as the quotient metric for X/ ∼ derived from the original metric d′ on X , or we say thatX/ ∼ “inherits” the metric d from the metric d′. In this case, we will also say X/ ∼ is a goodquotient. Otherwise, X/ ∼ is a bad quotient and inherits a metric from X in no “natural” way.

2.4 Examples

Examples of good quotients.

1. The standard example of a torus. This generalizes to the standard 4g-gon “plane model”of a surface of genus g (see definition of genus below).

2. Mobius band

3. Any “plane model” of a surface or surface with boundary.

4. Cell complexes, simplicial complexes, δ-complexes.

5. Connected sum (we will prove that this is independent of choices involved later).

Exercise 2.4. Prove that the standard representation of a torus by identifying opposite sides ofthe unit square is a good quotient. Calculate the distance from (0, 0) to (34 ,

34 ) in this quotient.

Exercise 2.5. Prove that the Mobius band is a good quotient.

Exercise 2.6. Prove that the hexagon with sides abca−1b−1c−1 is homeomorphic to a torus.(A cutting and pasting argument is sufficient at this point.) Generalize this example to find a4g + 2-gon plane model of a connected sum of g copies of the torus.

Exercise 2.7. Forming S2 as a quotient of two disks: Let D1 and D2 be two copies of thestandard unit disk, each with the usual metric inherited from R2. Let X = D1 ∪ D2, withd(x, y) = 2 if x, y are in different disks, or d(x, y) is the standard distance in the disk if x andy live in the same copy. Now, define x ∼ y if and only if x = y or if x, y are the “same” pointon the boundaries of D1 and D2. Prove that X/ ∼ is a good quotient which is homeomorphicto S2. In this exercise, why is it important that the distance between points in the two differentdisks in X is 2? What goes wrong if it’s, say, 1 instead?

Exercise 2.8. The projective plane is the quotient S2/ ∼, where x ∼ y if x, y are antipodalpoints (i.e., x = −y. Prove that the projective plane is a surface.

Orientability and genus. A surface is non-orientable if it contains a copy of the Mobius band.Otherwise it is orientable. If a closed orientable surface S is homeomorphic to a connected sumof g tori, we say that it has genus g. The classification of surfaces tells us the genus of anorientable surface is a well-defined invariant. (We can also talk about genus of a non-orientablesurface, but won’t do that here.)

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Exercise 2.9. A Bad Quotient.Consider R with the usual metric, with x ∼ y if and only if x− y ∈ Q. Prove that not only doesthis quotient fail to be good, but it fails in a fairly spectacular way, in that the pathification ofthe inherited ”pre-distance” function on R/ ∼ is actually the zero map.

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3 Group theory

There are deep connections between understanding the structure of surfaces and group theory,so we’ll start with a review of some basics.

3.1 Groups and homomorphisms

A group is a pair (G, *), where G is a set and ∗ : G×G → G is a binary operation satisfying:

• (G1: Associativity) The operation ∗ is associative.

• (G2: Identity) There exists an element eG ∈ G such that e ∗ g = g ∗ e = g for all g ∈ G.

• (G3: Inverses) For every g ∈ G, there exists an element h ∈ G such that g ∗h = h ∗ g =eG.

An additional axiom (G0: Closure) is often included in the definition. “Closure” meansthat g ∗h ∈ G for all g, h ∈ G. I haven’t included it above since this is covered by our definitionof ∗ as a binary operation from G×G into G, but I’m mentioning it anyway because studentsoften forget to check it.

Notation: The element h in (G3) is the inverse of g in G and is usually denoted g−1 or g.

Exercise 3.1. Prove that the identity element eG is unique. Prove that inverses are unique.

Exercise 3.2. Let g ∈ G be an arbitrary but fixed element. Prove that the function µg : G → Gdefined by µg(x) = g ∗ x is a bijection.

Examples:

• Z, under addition

• R∗ = R− {0}, operation is the usual multiplication of real numbers.

• Sn, the group of bijections of a set with n elements (operation is function composition)

• GLn(R) (operation is matrix multiplication)

A function f : G → G′ satisfying f(gh) = f(g)f(h) for all g, h ∈ G is called a homomor-phism. A bijective homomorphism is an isomorphism. An isomorphism from a group G to itselfis an automorphism.

Examples:

• The determinant homomorphism from GLn(R) to R∗ is surjective but not injective.

• If G is any group, then for a fixed g ∈ G, the map ig : G → G given by ig(x) = gxg−1 isan automorphism of G.

• The set of all nonzero matrices of the form

(

a b−b a

)

where a and b are real numbers

forms a group under matrix multiplication. The map from C∗ to this matrix group given

by a+ bi .→

(

a b−b a

)

is an isomorphism.

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Exercise 3.3. Prove that the set of all automorphisms of G is itself a group under functioncomposition.

The kernel of a homomorphism f : G → H is the set of all g ∈ G such that f(g) = eH .

Exercise 3.4. Prove that the kernel is a subgroup of G, i.e., a group in its own right underthe restriction of the multiplication in G.

Exercise 3.5. Prove that a group homomorphism is injective if and only if its kernel is trivial.

3.2 Group actions

We mainly care about groups because they act on sets.Let X be a set and G a group. A (left) action of G on X is a function · : G×X → X such

that for all x, y ∈ X :

1. e · x = x, and

2. g · (h · x) = (gh) · x.

We sometimes say in this case that X is a G-set.Examples:

• Sn acts on the set {1, 2, 3, ..., n}.

• GLn(Z) acts on the set Zn and on the vector space Rn (note that for the purposes of thisexample, we only need that Rn is a set and can forget that it also has the extra structureof a vector space).

• The multiplicative group R∗ also acts on Rn. In fact, if F is the scalar field for a vectorspace V , then F ∗ = F − {0} acts on V . (This is built into the axioms of a vector space,in fact.)

• Every group acts on itself by left multiplication, (also by right multiplication)

Let x ∈ X . We define Gx = {g ∈ G | g · x = x}. This is known as the stabilizer of x in G,or the isotropy subgroup of x.

Exercise 3.6. Prove that for any x ∈ X, with X a G-set, then Gx is indeed a subgroup of G.

Let x ∼ y if there exists a g ∈ G such that g · x = y.

Exercise 3.7. Prove that ∼ is an equivalence relation on G and hence partitions X intoequivalence classes. Each equivalence class is known as an orbit in X under the action of G.If X has just one orbit under the action of G, then we say that G acts transitively on X.

Exercise 3.8. Prove that Sn acts transitively on {1, 2, . . . , n}

Exercise 3.9. Prove that Z acts on R by the operation of addition. Describe the orbits in R

under this action.

Exercise 3.10. Prove that Z2 acts on R2 by the operation of addition. Describe the orbits inR under this action.

We say that G acts faithfully on X if g · x = x for all x ∈ X implies that g = eG.

Exercise 3.11. Which of the examples of group actions we’ve described in this lecture arefaithful, if any?

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3.3 Forming quotients of spaces via group actions

A group action on a space gives us a natural way to form a quotient of that space in themanner described in the previous lecture, and we can talk about “good” and “bad” quotientsif the metric on the original space descends to a metric on the quotient, as before.

Examples:

• Z on R. What is the quotient space? Is it a good quotient? What is the metric?

• Q on R. What is the quotient space? Is it a good quotient? What is the metric?

• Z2 on R2. What is the quotient space? Is it a good quotient? What is the metric?

• Now consider the action of 2Z × 2Z on R2 by addition. What is the quotient space? Isit a good quotient? What is the metric? Note that 2Z× 2Z is isomorphic to Z × Z as agroup. Describe this example instead as the group Z× Z acting on R2. What is the keydifference between this example and the previous example?

Exercise 3.12. Generalize the previous result to describe all possible different metrics that onecould obtain on T 2 in this way.

NOTE: In fact all higher genus orientable surfaces arise this way. We’ll see more about thislater.

Some more terminology: if a group G acts on a space X , we usually insist that G acts byhomeomorphisms (in other words for every fixed g, the map on X defined by x .→ g · x is ahomeomorphism). If the result is a compact quotient space, we say that G acts cocompactly onX . So we see from the above examples that Z2 acts cocompactly (in infinitely many ways) onR2.

Another adjective that is frequently attached to group actions is “properly discontinuous”.The foundation of everything we’ve done so far is the notion of an open set, where open sets

are defined with reference to some metric. However, we could cheat and take any given set Sand just say that every subset of S is open (and hence every subset is closed). In this case weare “giving S the discrete topology”.

We often want to do this in the case that S is a group, since many groups do not comeequipped with any obvious metric of their own. But in some cases, for example the group R,we already have a more natural metric at hand for defining our open sets, which does not resultin every subset being both open and closed. However, it may be the case for some subset ofthis space that the “usual” metric actually induces the discrete topology on the subset. Forexample, the subgroup Z in R inherits the discrete topology, but Q does not. (Prove!)

If a group is being considered with the discrete topology, whether it inherits this in somenatural way or not, we call it a discrete group. We say that a discrete group G acting on a(locally compact Hausdorff) space X acts properly discontinuously if for all compact K ⊂ X ,we have that (g · K) ∩ K #= ∅ for at most finitely many g ∈ G. This condition might seemsomewhat strange at first, but the take-home point is that it guarantees nice properties in thequotient space, which you’ll hear more about from Prof. Leininger.

Back to examples, note that any surface is a locally compact Hausdorff space. Again, wesee that Z2 acts properly discontinuously on R2.

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4 Lecture 4: Euler characteristic

We begin by giving a complete statement of the classification of surfaces.

Theorem 4.1 (Classification of Surfaces). Every closed, connected, orientable surface ishomeomorphic to the connected sum of g tori for precisely one g ∈ Z≥0.

Recall from Lecture 2 notes that a surface S has genus g if S is homeomorphic to the con-nected sum of g tori. A priori, a surface could have more than one genus, but the classificationtheorem tells us that genus is a complete invariant of closed, connected, orientable surfaces.Also, at this point we haven’t proved well-definedness of connected sum. So right now, “genus”is not necessarily a meaningful concept.

It is important to realize that the classification of surfaces is really a multi-part theorem,so to make this point crystal clear, and to also keep track of where we use each part, we willrestate the theorem as follows:

Theorem 4.2 (Classification of Surfaces). Let S be a closed, connected orientable surface.Then:

1. The surface S is homeomorphic to a surface of genus g.

2. Any surface of genus g is homeomorphic to any other surface of genus g. In other words,the operation of connected sum is well defined.

3. If T is a surface of genus g′ #= g, then T is not homeomorphic to S.

In this lecture we will define an incredibly useful tool for studying surfaces known as theEuler characteristic and prove its basic properties.

Let S be a surface. A cell decomposition of S is a CW-complex C together with a homeomor-phism h : C → S. (Think: plane model for a surface.) Note that S admits a cell decomposition ifand only if S is triangulable (which means replacing “cell complex” with “simplicial 2-complex”in the above definition).

Theorem 4.3 (Kerekjarto, Rado). Every compact surface is triangulable.

The triangulability of surfaces is a deep result which we will use heavily in what follows.We will be proving this carefully next week. This is also a key ingredient in proving Part I ofthe Classification Theorem.

4.1 Euler characteristic: definition/examples

Let S be a surface equipped with a cell decomposition C with v vertices, e edges, and f faces.We define the Euler characteristic of S with respect to C by

χC(S) = v − e + f.

In fact, we will see that in fact the Euler characteristic does not depend on the particularcell decomposition of S, so we will usually just write χ.

Examples.

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• The disk D2, with cell decomposition given by a single 2-simplex, has Euler characteristicequal to 3− 3 + 1 = 1.

• The sphere S2, with cell decomposition given by two 2-simplices glued along their commonboundaries, has Euler characteristic equal to 3− 3 + 2 = 2.

• The torus T 2, with cell decomposition given by the usual identification of the sides of theunit square, has Euler characteristic equal to 1− 2 + 1 = 0.

Exercise 4.4. Find a different cell decomposition for each of the above and check explicitlythat you get the same result.

Exercise 4.5. Show that each of the following 3 “moves” (and their inverses) on a given celldecomposition leaves the Euler characteristic invariant:

1. Subdividing an edge by adding a vertex.

2. Subdividing a face by connecting two vertices with a new edge.

3. Introducing a new vertex in the interior of a face and a new edge connecting that vertexto an existing vertex adjacent to that face.

Exercise 4.6. Nontrivial Fact: Any two cell decompositions on the same surfaceoverlap in just a finite number of points, possibly after perturbing them slightly.

We will see a technique for proving this later on. For now, assume the fact, and use it toprove that Euler characteristic is independent of the choice of triangulation, by showing that afinite number of the 3 types of moves above are sufficient to “transform” any cell decompositioninto any other.

4.2 Euler characteristic and genus

The next formula is one of the most important in all of topology. Remember that at this stage,“having genus g” means “homeomorphic to some connected sum of g tori”, a constructionwhich a priori depends on choices of disks and gluing maps, although we will later see thatthese choices don’t end up mattering.

Theorem 4.7 (Euler’s Formula). If S is a closed surface of genus g with respect to a celldecomposition C, then χC(S) = 2− 2g.

Proof. WLOG we can consider the case where S is given as a plane model. (Here we areusing triangulability in a big way.) This is because the process of “forgetting” an edge in acell decomposition doesn’t change the Euler characteristic. (See exercise above, Type 2 move:You’re exchanging 2 faces for 1 face, while eliminating one edge, so the net change in f − e+ vis 0. ) Recall that a plane model has exactly 1 face, but the edges are paired up in some waywhich makes it slightly unclear how many distinct vertices there are, and we also want to clarifythe relationship between the number of edges and the genus of the surface.

Begin by (cyclically) reading the boundary word for the plane model for S. If it contains asubword of the form . . . a . . . a . . . then it contains a Mobius band and is not orientable. So ifwe see a in the boundary word it can only be followed by an a−1. If it (cyclically) contains asubword of the form aa−1, then we can glue these together and delete the corresponding edgeand vertex, leaving Euler characteristic unchanged. Repeat this until no such subwords exist.

Now, if the new boundary word contains a pattern of the form . . . a . . . b . . . a−1 . . . b−1 . . .we say that the boundary word contains a cross.

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Lemma 4.8. If the boundary word contains a cross, then S is not a sphere.

Proof. Think of the unit square with the usual toroidal identification of sides, with small bitof each corner deleted. So we see that our surface S has a decomposition as the connectedsum of a torus and some other surface. But then S is not homeomorphic to S2 since thereexists a simple closed curve C in the connected sum of a torus and another surface such thatS−C is still connected and path-connected. However, by the Jordan Curve Theorem (which wewill prove independently next week), any simple closed curve separates S2 into two connectedcomponents.

Exercise 4.9. Prove the converse: if the boundary word does not contain a cross, then S is asphere.

Lemma 4.10. The Euler characteristic of S2 is 2.

Proof. The proof is by induction on the number of edges in our plane model. Our base case isa plane model with boundary word aa−1. (Exercise: check that this really is a “good” quotienthomeomorphic to S2.) Now, by the previous lemma (or really, its converse, which was givenas an exercise immediately following the lemma), we know that our boundary word does notcontain a cross. So if we see something of the form a . . . b . . . a−1, it must actually be of theform a . . . b . . . b−1 . . . a−1. Now the “distance” from b to b−1 is shorter than from a to a−1, sowe keep going in this way until we must eventually arrive at a subword of the form cc−1. Wemake the usual identification here, which reduces the number of edges in our plane model by 2without changing the Euler characteristic. By induction, we are done now.

Now, to finish the proof of the theorem, we induct on the number of edges in the planemodel. If S is not a sphere, then by a previous lemma, its plane model has a cross. (You canprove this similarly to the above.) Let’s reproduce the construction we did in the proof of thatlemma, so that we can reinterpret our plane model. Let’s reintroduce enough edges in our celldecomposition so that we can “cut out” the torus and replace it with a disk. The resultingplane model (after forgetting some interior edges again) has 4 fewer edges, identified in pairs,which means our total edge count for Euler characteristic has gone down by 2. But the planemodel has the same set of vertices, glued together in the same way.

Scholium 4.11 (Classification Part I). Every surface is homeomorphic to some connected sumof some number of tori (possibly zero).

Unfortunately, the construction in the previous proof depended heavily on an initial, specificchoice of triangulation, not just on the fact that the surface was triangulable. If we knew thatχ is an invariant of the underlying space (see exercise above which proves this, assuming aNontrivial Fact), then we would also have Classification Part II. Similarly, we could prove thatχ is independent of the choice of triangulation by using Classification Part II. The exerciseabove gives you an idea of one approach to the first direction, and we will later encounter somemachinery that would equip us for doing this. We will also learn some different machinery thatwill enable us to prove Classification Part II without knowing that χ is an invariant of theunderlying space. So basically, we’ll be covered.

Exercise 4.12. Assuming the full classification of surfaces, identify the surface given by aplane model with boundary word dbec−1e−1cb−1a−1d−1a.

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Exercise 4.13. Prove that a connected compact surface of genus g with b boundary componentshas Euler characteristic equal to 2− 2g− b. (Feel free to assume that Euler characteristic is aninvariant of the underlying space here – just be aware that you’re using it!)

Exercise 4.14. It turns out that surfaces in each of the three cases χ > 0,χ = 0,χ < 0 sharecertain fundamental properties. Assuming the full classification theorem of surfaces, write downa complete list of compact connected surfaces with each of these properties.

Also, once you know that Euler characteristic is independent of the choice of cell decom-position of a surface, it’s easy to see how Euler characteristic behaves under connected sum.Suppose we have two surfaces S1, S2, with

Proposition 4.15. χ(S1%S2) = χ(S1) + χ(S2)− 2

Proof. Suppose D1 and D2 are the two disks you’re going to use for the connect sum. Choosecell compositions for S1, S2, for which D1, D2 are 2-cells, with their respective boundarieseach decomposed into n edges and n vertices. Then the connected sum has two fewer faces(corresponding to D1 and D2) than S1.

Exercise 4.16. Let P denote the nonorientable surface given by the plane model with boundaryword aa. This surface is called the projective plane. (We have seen a different descriptionpreviously.) Prove that P %P %P ∼= P %T 2.

Exercise 4.17. Redo everything in this lecture for nonorientable surfaces. In other words,formulate a reasonable version of the Classification Theorem for nonorientable surfaces. Includea reasonable notion of “genus” and a version of Euler’s formula.

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5 Lecture 5: Fundamental Groups

Throughout this lecture, “map” means “continuous map”.

5.1 Homotopy

Let X,Y be spaces, and let I = [0, 1]. We say that two (continuous) maps f0, f1 : X → Y arehomotopic if there exists a third (continuous) map F : X × I → Y such that:

F (x, 0) = f0(x) for all x ∈ X

F (x, 1) = f1(x) for all x ∈ X

We often denote F (x, t) by ft(x). The map F is called a homotopy from f0 to f1, and ifsuch an F exists, we write f0 ∼ f1.

Proposition 5.1. Homotopy gives an equivalence relation on the set of all continuous functionsfrom X to Y (for a given X,Y ).

Proof. The map F (x, t) = f(x) is a homotopy from f to itself, so ∼ is reflexive. Also, if F is ahomotopy from f0 to f1, then G(x, t) = F (x, 1− t) is a homotopy from f1 to f0.

Exercise 5.2. Finish the proof by showing that ∼ is transitive.

Exercise 5.3. Let X be any space, and let f : X → Rn be a map. Prove that f is homotopicto the zero map defined by ζ(x) = 0 for all x ∈ X. Conclude that all maps from X into R arehomotopic.

For our purposes, we will nearly always use I = [0, 1] as the domain X .

5.2 Fundamental Groups

A loop based at x0 in a space S is a path f : [0, 1] → S such that f(0) = f(1) = x0. The pointx0 is called the basepoint. Two loops f0, f1 are loop homotopic or homotopic rel x0 if thereexists a homotopy F from f0 to f1 such that ft is a loop based at x0 for all 0 ≤ t ≤ 1. Notethat this is definitely different from saying that f0 and f1 are homotopic – the difference is thata loop homotopy doesn’t allow the basepoint to move at all during the homotopy. We denotethe loop homotopy class of a loop f by [f ].

Having a basepoint that doesn’t move can be an annoying restriction sometimes, but itallows us to put a group structure on loop homotopy classes of loops as follows. Let [f ], [g] betwo homotopy classes of loops based at a point x0 in a space X . Note that this notation meansthat f, g are actual loops based at x0 representing their respective classes. We define [f ] ∗ [g]roughly by the rule “first do f , then do g, only do them each twice as fast so that we still finishin a unit’s worth of time”. More precisely, we define a loop h by

h(t) =

f(2t) for 0 ≤ t ≤ 12

g(2t− 1) for 12 ≤ t ≤ 1

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The problem with this definition is that it depends on a choice of representative of the loophomotopy classes [f ], [g]. We need to prove that if f ′ ∼ f and if g′ ∼ g, then f ′ ∗ g′ ∼ f ∗ g.Let F be the homotopy between f ′ and f and let G be the homotopy between g′ and g. Thenwe can define a homotopy J as follows:

J(s, t) =

F (2s, t) for 0 ≤ s ≤ 12

G(2s− 1, t) for 12 ≤ s ≤ 1

Basically we’re just doing each individual homotopy twice as fast.So we have a well defined operation on our set of based loops. The fundamental group of a

space X based at x0 is the set of all loop homotopy classes of loops based at x0 together withthe operation of loop multiplication, and is denoted π1(X,x0).

We need to prove that this is actually a group, but this is perhaps more difficult than itfirst appears. For example, it’s fairly clear that the constant loop defined by e(s) = x0 for alls ∈ [0, 1] should represent the identity element, but the composed map f ◦ e (meaning first dof , then do e) is not the same loop as the map f , so again, we need to show the existence of ahomotopy between them.

F (s, t) =

f(

2st+1

)

for 0 ≤ s ≤ t+12

x0(= e(s)) for t+12 ≤ s ≤ 1

Think of this as f being parametrized by s, but multiplied by 2t+1 , so that at t = 0, we are

doing f twice as fast, and then sitting at x0 for the remaining half-unit of time, but at t = 1,we are just doing f at the usual speed.

Exercise 5.4. Write an explicit formula for a homotopy between e ◦ f and f .

Similarly, it’s intuitively clear that the inverse of a loop should be “do the loop backwards”.So let f be a loop based at x0. Define f−1(s) = f(1− s). We need to show that [f ] ∗ [f−1] =[f−1] ∗ [f ] = [e]. To show that [f ] ∗ [f−1] = [e], we have the following homotopy:

F (s, t) =

f(2s) for 0 ≤ s ≤ 1−t2

f(1− t) for 1−t2 ≤ s ≤ 1+t

2

f(2− 2s) for 1+t2 ≤ s ≤ 1

Exercise 5.5. Write an explicit formula for a homotopy between f−1 ◦ f and e.

Exercise 5.6. Complete the proof that the fundamental group of a surface actually is a groupby proving that multiplication of loops is associative by writing down an explicit homotopy.

5.3 Basepoints and functoriality

In defining “the” fundamental group of a space Y , we needed to make a choice of basepointsy0. But how much does the fundamental group actually depend on this choice? Obviously theelements themselves will be different – a loop based at y0 is not the same as a loop based aty1 #= y0. But it turns out that in all of the cases we care about, the resulting group will be the

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same no matter what choice of basepoint we make, at least, up to isomorphism. In fact, oneshould really think of changing basepoints as being analogous to a change of basis for a vectorspace.

Theorem 5.7. Let Y be a space with y0, y1 ∈ Y such that there exists a path α from y0 to y1.Then π1(Y, y0) ∼= π1(Y, y1).

Corollary 5.8. If Y is path-connected, then we can safely talk about “the” fundamental groupof Y and we often denote it just by π1(Y ).

Proof. Let α be a path from y0 to y1. Then we define the following map:

φ : π1(Y, y1) → π1(Y, y0)

[f ] .→ [αfα−1]

Exercise 5.9. Prove that φ is a well defined isomorphism of groups. (The techniques are verysimilar to those we used to prove that the fundamental group is a group, and a complete proofcan be found in any good textbook, e.g., Munkres, Armstrong, Goodman, or Hatcher. To provebijectivity, it is usually easier to write down an inverse map rather than use methods such asproving the kernel is trivial, etc.)

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6 Lecture 6: Functoriality, Covering Spaces and DeckTransformations

Some terminology: A path-connected space X is simply connected if π1(X) is the trivial group.

6.1 Functoriality

When dealing with basepoints, it is standard to write the space and the basepoint as a pair(Y, y0) and refer to the pair as a pointed space.

Proposition 6.1. Let (Y, y0), (Z, z0) be two pointed spaces. Let φ : Y → Z be a (continu-ous) map such that φ(y0) = z0 (this is called a basepoint-preserving map). Then there is ahomomorphism φ∗ : π1(Y, y0) → π1(Z, z0).

Proof. Let [f ] ∈ π1(Y, y0). Then φ(f) is a loop based at z0 in Z, so we define φ∗([f ]) = [φ(f)].If [f ] = [g], where F is a homotopy from f to g, then φ ◦ F is a homotopy from φ(f) to φ(g),so this is a well defined function.

Exercise 6.2. Prove that φ∗ is a homomorphism. (This is actually even easier than well-definedness.)

The next theorem tells us that we have constructed a useful algebraic invariant for distin-guishing between topological spaces.

Theorem 6.3. If Y, Z are path-connected homeomorphic spaces, then π1(Y ) ∼= π1(Z).

The theorem is proved by the following two (very short) exercises.

Exercise 6.4. Let φ : (X,x0) → (Y, y0) and ψ : (Y, y0) → (Z, z0) be two basepoint-preservingmaps. Prove that (ψ ◦ φ)∗ = ψ∗ ◦ φ∗.

Exercise 6.5. Prove that if the map φ in the above proposition is actually a homeomorphism,then φ∗ is an isomorphism of groups.

When we pass from the continuous map φ of spaces to the induced homomorphism of groupsφ∗, we see that we are somehow respecting the natural structures and transformations that existbetween spaces and groups. Going from φ to φ∗ is an example of a functor from the categoryof spaces to the category of groups. We also use the term naturality to describe this kind ofprocess.

6.2 First examples: Convex sets in Rn; the circle S1

Actually computing fundamental groups can be difficult without some more machinery that wehaven’t developed yet.

We’ve already proved that EVERY map of EVERY space into R2 is homotopic to the zeromap ζ. If we take a convex subset A of Rn, we can use the so-called straight-line homotopy toprove that π1(A) ∼= 1. So if f : [0, 1] → A is a loop based at a0 ∈ A, then the map

F (s, t) = t ∗ a0 + (1− t)f(s)

gives a homotopy to the constant loop at a0. This works by sliding each point f(s) to a0, whichwe can do by convexity. In particular, we have shown:

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Proposition 6.6. The fundamental group π1(Rn) is trivial for all n.

The standard first “interesting” example is actually the circle, and indeed, it’s often easierto find non-trivial loops than it is to prove that there aren’t any.

Winding number. Define p : R → S1 by p(x) = (cos(2πx), sin(2πx)). If we think of S1 inthe complex plane C, we can write this as p(x) = e2πix. If we let x0 = (1, 0) ∈ S2, then we seethat p−1(x0) = Z.

Proposition 6.7. Let f be a loop in S1. Then there is a unique map f : [0, 1] → R such thatf(0) = 0 and p ◦ f = f on [0, 1]. If f0, f1 are homotopic loops in S1, then f0(1) = f1(1)

This proposition will follow directly from more general results which will be proved shortly.In light of the proposition, we define the winding number of the homotopy class [f ] to be

the integer f(1) which is uniquely determined by f . We denote this by w(f).Now consider the map

Φ : π1(S1, x0) → Z

given by Φ([f ]) = w(f).

Exercise 6.8. Prove that Φ is an isomorphism of groups. (HINT: Use what you already knowabout π1(R).)

Exercise 6.9. Let X,Y be path-connected spaces. Prove that π1(X × Y ) ∼= π1(X)× π1(Y ).

Exercise 6.10. Calculate π1(T 2).

6.3 Covering spaces

Let X,X be path connected spaces. Let p : X → X be a continuous map. Then we say thatp is a covering map if every point in X has a path-connected neighborhood U such that thepreimage p−1(U) consists of countably many disjoint sets U1, U2, . . . such that the restrictionof p to each component Ui is a homeomorphism. In this case, we say that X is a covering spaceor a cover of X , and we say that the neighborhood U is evenly covered by the Ui.

In the previous lecture we defined a map p : R → S1 by p(x) = (cos(2πx), sin(2πx)). Thismap is a covering map, and hence R covers S1.

6.4 Deck transformations

Let p : X → X be a covering map. A deck transformation (or covering automorphism) is ahomeomorphism φ : X → X such that p ◦ φ = p.

Proposition 6.11. The set of all deck transformations of a given covering map p : X → Xforms a group.

Exercise 6.12. Prove the proposition.

We call this group the deck group or the automorphism group of the covering map p, andwe denote it Aut(X, p) or just Aut(X).

Let’s revisit the covering map p(x) = (cos(2πx), sin(2πx)). Since cos(2πx), sin(2πx) are bothperiodic with period 1, we have that any function φn : R → R where φn(x) = x + n for some

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n ∈ Z is a deck transformation for this covering space. Conversely, any deck transformationclearly must be of this form. So the deck transformation group for p(x) = (cos(2πx), sin(2πx))is just Z.

We’ve seen the group Z appearing in association with S1 before, namely, as its fundamentalgroup. This is not a coincidence, as the next theorem tells us.

Theorem 6.13 (The Isomorphism Theorem). Suppose that p : X → X is a covering mapwhere X,X are path-connected and X is simply connected. Then Aut(X, p) ∼= π1(X).

Proof that there is a homomorphism. First we choose a basepoint x for X . Since X is path-connected, our choice doesn’t really matter.

Choose any point x ∈ p−1(x). Let ψ ∈ Aut(X, p). Then y = ψ(x) is another point inp−1(x), and since X is also path-connected there exists some path f from x to y. Then p ◦ f isa loop in X based at x. So we define

Φ : Aut(X, p) → π(X,x)

ψ .→ [p ◦ f ]

In order to prove that Φ is well-defined, we need to know that the particular choice of path fdoesn’t affect the final outcome. This follows easily from the next exercise.

Exercise 6.14. Let Y be a simply connected space, and let x, y ∈ Y . Then any two paths fromx to y are homotopic. (HINT: If f, g are the two paths, then fg−1 is a loop based at x.)

The next step is to show that Φ is a group homomorphism. Let ψ1,ψ2 be two deck trans-formations. Choose paths f1, f2 from x to ψ1(x),ψ2(x), respectively, and let f1 = p ◦ f1 andsimilarly f2 = p ◦ f2. Choose a third path g from x to (ψ1 ◦ ψ2)(x), and let g = p ◦ g. We needto show that f1f2 is loop homotopic to g.

To see this, note that ψ1 ◦ f2 is a path from ψ1(x) to ψ1 ◦ ψ2(x). Therefore the productf1 ∗ (ψ1 ◦ f2) is a path from x to (ψ1 ◦ψ2)(x). But since we proved above that Φ is independentof choice of path, we have that

Φ(ψ1 ◦ ψ2) = [p ◦ (f1 ∗ (ψ1 ◦ f2))]

= [(p ◦ f1) ∗ (p ◦ ψ1 ◦ f2)]

= [f1 ∗ (p ◦ f2)]

= [f1 ∗ f2]

= [f1] ∗ [f2]

= Φ(ψ1)Φ(ψ2)

In order to complete the proof of the Isomorphism Theorem, we need a bit more machinery.

6.5 Lifting paths and homotopies

Suppose we have a covering map p : X → X . Suppose we have a map h from a space Y intothe base space X . Then we can ask the general question of whether there exists a lifting of hto X, that is, a map h : Y → X such that p ◦ h = h. We will be particular interested in thecase where Y is either an interval I or a square I × I (the latter because we will be interested

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in lifting homotopies of paths). So we will use Q to denote either I or I × I, but actually thisworks for any cube, i.e., finite product of I’s. We will say that v is a vertex of Q if v is anendpoint of I or if v ∈ {(0, 0), (0, 1), (1, 0), (1, 1)}.

The theorem that we want to prove is the following.

Theorem 6.15. Let p : X → X be a covering map. Let f : Q → X be a continuous map. Letv be a vertex of Q, let x = f(v), and let x ∈ p−1(x). Then there is a unique lift f : Q → Xsuch that f(v) = x.

In other words, paths and homotopies can always be lifted, and this lifting is unique as soonas we specify where a vertex goes.

In order to prove this theorem, we will need two lemmas.

Lemma 6.16. Consider the case where Q ∼= I × I. There exists a real number N such that ifQ′ is a subcube of Q with sidelength less than 1

N, then f(Q′) is contained in an evenly covered

neighborhood of X.

Exercise 6.17. Prove this lemma. (HINT: Consider a sequence {Qn} of subcubes with diametergoing to 0 with Qj not contained in an evenly covered neighborhood. Use Bolzano-Weierstrass.)

Lemma 6.18. Suppose that f(Q) is contained in an evenly covered neighborhood. Then thereis a unique lift f : Q → X such that f(v) = x.

Proof. Let U be the evenly covered neighborhood containing Q. Let Uj be the component“upstairs” containing x. Let pj denote the restriction of p to Uj. Then pj is a homeomorphismand hence has an inverse p−1

j . Then f = p−1j ◦ f is the unique lift of f we are looking for.

Now we’re ready to complete the proof of the Lifting Property for Paths and Homotopies(and of maps of cubes in general).

Proof. First we partition Q into subcubes which are small enough to invoke the first lemmaabove which lets us find evenly covered neighborhoods for each of them. Let’s label these smallsubcubes Q1, . . . , Qm. WLOG we can order them so that Qk shares at least one vertex vk withsome Qj for j < k, and also so that v1 = v, the specified vertex. We define f one piece at atime, using the second lemma. Applying it to Q1, for example, we learn what the value of fhas to be on v2, which then determines what f has to look like on the next subcube, and so on.The pieces are guaranteed to fit together thanks to the uniqueness guaranteed by the lemma,resulting in one unique lift f of the entire map f .

6.6 Finishing the proof of the Isomorphism Theorem

We now have the necessary technology to prove injectivity and surjectivity of the homomorphismΦ defined above.

Injectivity. Let Φ(ψ) = 1 in π(X,x). Consider a path f joining x to ψ(x), and let f = p ◦ f .Then by definition Φ(ψ) = [f ]. But since we assumed ψ is in the kernel of Φ, there is a homotopyF from f to the constant loop. In this case, the homotopy F : Q → X is a continuous map ofthe square Q = I × I into X , so we can use the lifting theorem to find a unique lift F : Q → Xsuch that F (0, 0) = x.

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Now, one “side” of F is the path f , so by uniqueness of lifts, that “side” of F must be f ,which joins x to ψ(x). Similarly, the other “side” of F must be the constant path at x. Further,F (0, t) = F (1, t) = x, and hence F (0, t) and F (1, t) are constant maps. But this must meanthat in fact ψ(x) = x.

So, we have shown that any deck transformation which is in the kernel of Φ must preservethe point x. We now wish to show that in fact any such deck transformation must be theidentity element. Let y be any other point of X, and choose any path f from x to y. As usual,let x = p(x), y = p(y), and f = p ◦ f , so that f is a path in X from x to y.

Now, f is a lift of f with initial point x. But similarly, ψ ◦ f is also a lift of f and also hasinitial point x. By uniqueness of path lifting, we must have that these two paths are the same;in particular, they have the same terminal point. This means that y = ψ ◦ f(1) = ψ(y).

Surjectivity. Let [f ] ∈ π1(X,x). We need to find an element ψ ∈ Aut(X, p) which maps to[f ]. So let y ∈ X, and let α be a path from x to y. Then α = p ◦ α is a path from x to y = p(y)in X . That means that f ◦ α is a path from x to y, which we can lift uniquely to a path ˜f ◦ αfrom x to some other point z. So we will define ψ(y) = z. We need to check that the map ψwe have defined is actually a deck transformation and also that it maps to [f ] under Φ.

Recall that to compute Φ(ψ), we need to compute ψ(x). In this case, we can just take α tobe a constant path, so we are just lifting f to the unique lift f with initial point x. But thenΦ(ψ) = [p ◦ f ] = [f ].

Exercise 6.19. It remains to show that the map ψ we just defined is actually a deck transfor-mation. Prove first that p ◦ ψ = p, and then finish the proof of the Isomorphism Theorem byshowing that ψ is a homeomorphism of X.

Exercise 6.20. Prove that any deck transformation is completely determined by its action onjust one point; in particular, the only deck transformation that fixed any point of X is theidentity.

6.7 The Torus, Again

Compute the deck group of the covering of the torus T 2 by the plane R2 directly, and also byusing the Isomorphism Theorem.

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7 Lecture 7: More on Fundamental Groups and CoveringSpaces

The first part of this lecture follows Hatcher’s Algebraic Topology.

7.1 Homotopy Equivalence

Recall that: Let X,Y be spaces, and let I = [0, 1]. We say that two (continuous) mapsf0, f1 : X → Y are homotopic if there exists a third (continuous) map F : X × I → Y suchthat:

F (x, 0) = f0(x) for all x ∈ X

F (x, 1) = f1(x) for all x ∈ X

A map f : X → Y is a homotopy equivalence if there is another map g : Y → X such thatthe composition f ◦ g ∼ IdY and g ◦ f ∼ IdX . If a homotopy equivalence exists from X to Y ,then we write XY , and say that X and Y are homotopy equivalent and that they have the samehomotopy type.

Observation. A homeomorphism from X to Y is a homotopy equivalence.

Exercise 7.1. Prove that homotopy equivalence is an equivalence relation on spaces.

Deformation retracts. A special case of homotopy equivalence.Let X be a space with subspace A. Then a deformation retract of X to A is a (continuous)

map F : X × I → X such that

• F (x, 0) is the identity map IdX ,

• F (x, 1) maps X onto A

• For all t ∈ I, the map F (x, t) is equal to IdA when restricted to A.

Example: the symbols + and - are homotopy equivalent but not homeomorphic.

Exercise 7.2. Prove that a deformation retract gives a homotopy equivalence between spaces.

We’ve already seen that two homeomorphic spaces have isomorphic fundamental groups. Itturns out that we can extend this to a more general result.

Theorem 7.3. If f : X → Y is a homotopy equivalence, then the induced map f∗ : π1(X,x) →π1(Y, f(x)) is an isomorphism.

This theorem is fairly straightforward to prove, with the following lemma/exercise:

Exercise 7.4. Suppose that F : X × I → Y is a homotopy from f = F (x, 0) to g = F (x, 1).Let x0 ∈ X, and let α denote the path F (x0, t). Then we get induced maps f∗ : π1(X,x0) →π1(Y, f(x0)) and g∗ : π1(X,x0) → π1(Y, g(x0)), and we have an isomorphism of fundamentalgroups induced by the path α: φα : π1(Y, f(x0)) → π1(Y, g(x0)). Prove that φα ◦ g∗ = f∗.

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Proof of Theorem. Let g : Y → X be a homotopy inverse for the homotopy equivalence f :X → Y . Consider the composition f∗ ◦ g∗ ◦ f∗ : π1(X,x0) → π1(X, f ◦ g ◦ f(x0)). The previousexercise implies that the composition of the first two maps: g ◦ f : π1(X,x0) → π1(X, g ◦ f(x0))is φα for some α. In particular, this implies that f∗ is one-to-one. Similarly, we can see that g∗is also one-to-one, which in turn implies that f∗ is also surjective.

7.2 Action of the deck group on the cover

Notation: if a group G acts on a space X , we often denote the quotient by X/G.Let p : X → X be a covering map where as usual X,X are path-connected and X is simply

connected. The deck group Aut(X) clearly acts on the space X, and it acts by homeomorphisms(see end of Lecture 2 notes). In particular, for G = Aut(X), the following condition holds:

(*) Each x ∈ X has a neighborhood U such that g · U ∩ U = ∅ for all nontrivial g ∈ G.

Exercise 7.5. Prove this.

Recall that any deck transformation is completely determined by its action on just one point;in particular, the only deck transformation that fixed any point of X is the identity.

We have seen that Aut(X) ∼= π1(X). So this means that π1(X) also acts on X by homeo-morphisms. There is a natural way to define this action, as follows. Let x ∈ X, let x = p(x),and let f ∈ π1(X,x). Then there is a unique lift of f to a path f with initial point x. Let y beits terminal point, and define f · x = y.

Exercise 7.6. Prove that this action is compatible with the action of Aut(X) under the Iso-morphism Theorem.

Some terminology: a covering p : X → X is normal if for every x ∈ X and every pair ofpoints x, x′ ∈ p−1(x), there exists a deck transformation ψ such that ψ(x) = x′.

Theorem 7.7. If an action of a group G on a path-connected space X satisfies the conditionabove, then the following hold:

1. The map X → X/G defined by x .→ [x] is a normal covering map. (This is true even ifX is not path connected.)

2. The group G is the deck group for the covering X → X/G.

3. If X is also simply connected, then G ∼= π1(X/G).

Proof. To prove the first statement, let [x] ∈ X/G. Then take the neighborhood U of x in Xguaranteed by the condition (*) above. Then p(U) is a neighborhood of [x] which is evenlycovered. Clearly G is at least a subgroup of the deck group Aut(X), and the covering is normalbecause each pair of points in the preimage of a single point “downstairs” lives in a g1(U), g2(U),respectively, so g2g

−11 takes one to the other.

Now, to prove that G is the full set of a deck transformations, suppose that ψ is an arbitraryelement of Aut(X). Let [x] ∈ X/G, with x a point in the preimage p−1([x]). Then by definition,ψ(x) is in the same orbit as x, so there exists an element g ∈ G < Aut(X) such that ψ(x) = g ·x.But since an automorphism is completely determined by where it sends one point, we must havethat ψ = g ∈ G.

The third part follows from the first two by the Isomorphism Theorem.

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7.3 Fundamental groups of surfaces given as plane models

Recall that a subgroup N is normal in a group G if gng−1 ∈ N for all g ∈ G. This is preciselythe condition required to make the multiplication of cosets aN · bN = abN be well defined,hence giving a group structure on G/N , the set of all (left) cosets of N in G.

If R1, . . . , Rm are elements in a group H , then the normal closure of R1, . . . , Rm ∈ H is thesmallest normal subgroup of H containing the elements R1, . . . , Rm.

We say that a group G has presentation 〈g1, . . . , gn | R1, . . . , Rm〉 if the group G is generatedby the elements g1, . . . , gn and if G ∼= F/N , where F is the free group on {g1, . . . , gn} and N isthe normal closure of the elements R1, . . . , Rm. (Typically Ri is given as a word in the gj .)

Theorem 7.8 (Seifert-van Kampen Theorem). Let X = A ∪ B, where A,B,A ∩ B are allnonempty, open, and path-connected. Let jA, jB denote the respective inclusion maps of A ∩Binto A,B. Choose a basepoint x0 ∈ A ∩B. Now, suppose we know the following:

π1(A, x0) = 〈a1, . . . , an | R1, . . . Rp〉

π1(B, x0) = 〈b1, . . . , bm | S1, . . . Sq〉

π1(A ∩B, x0) = 〈c1, . . . , ck | T1, . . . Tr〉

Then we can write down the following presentation for the fundamental group of X:

π1(X,x0) ∼= 〈a1, . . . , an, b1, . . . , bm | R1, . . . Rp, S1, . . . Sq, (jA)∗(c1) = (jB)∗(c1), . . . , (jA)∗(ck) = (jB)∗(ck)〉

Proving this is nontrivial. Goodman contains a good sketch. Armstrong proves a versionusing simplicial approximations. Other textbooks like Munkres, Hatcher, and Massey, prove itin greater generality.

Corollary 7.9. If P is a plane model for a surface S with boundary word w(x1, . . . , xn), thena presentation for π1(S) is given by:

〈x1, . . . , xn | w = 1〉

Exercise 7.10. Prove the corollary assuming the Seifert-Van Kampen Theorem.

Exercise 7.11. Assuming the full Classification Theorem, prove that a closed orientable surfaceof genus g has fundamental group with presentation:

〈a1, b1, . . . , ag, bg | [a1, b1] · · · [ag, bg] = 1〉

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8 Lecture 8: Jordan Curve Theorem

Our proof of the Jordan Curve Theorem will show us some deep connections with elementarycombinatorics, in the form of graph theory. We will closely follow the treatment of CarstenThomassen’s excellent (and relatively short) paper The Jordan-Schoenflies Theorem and theClassification of Surfaces.

8.1 Graph Theory Cheat Sheet

• An arc is a path which is one-to-one. (Exercise: Prove that a space is path-connected ifand only if it is arc-connected.)

• A polygonal arc is an arc which is the union of a finite number of (straight) line segments.

• A region of an open set in the plane is a maximal path-connected subset.

• A graph Γ is a pair (V,E,φ) where V is a FINITE set of vertices, E is a FINITE set ofedges, and φ is a function from E to (unordered) pairs of vertices. We usually leave outthe φ from the notation and when convenient denote an edge e by vw if φ(e) = {v, w}.In this case we say that e is incident to v and w. DO WE ALLOW LOOPS?

• If there exist edges e1 #= e2 such that φ(e1) = φ(e2), then we say that Γ has multipleedges.

• A path is a graph (V = {v1, . . . , vn}, E = {v1v2, v2v3, . . . vn−1vn}, where all the vi aredistinct.

• If we add the edge v1vn to a path we get a cycle. We denote both the path and the cycleby listing the vertices and hope that it’s clear from the context what we mean.

• If A ⊂ V , then Γ− V is the graph obtained by deleting all

• A graph is connected if every two vertices are joined by a path.

• A graph Γ is 2-connected if it is connected and if for every v ∈ V , Γ− {v} is connected.

• If X is a space, we say that Γ = (V,E) can be embedded in X if there is a one-to-onefunction from V to X such that each edge e = vw ∈ E can be represented by a simplearc in X joining v to w.

• If Γ can be embedded in the plane R2, we say that Γ is a planar graph.

• If Γ actually is embedded in the plane R2, we say that Γ is a plane graph.

• A subdivision of Γ is a graph obtained from Γ by inserting vertices on edges in the obviousway.

Exercise 8.1. Define an isomorphism of graphs. Write down a precise definition of “subdivi-sion”.

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8.2 Polygonal Jordan Curve Theorem

Lemma 8.2 (Polygonal Approximations of Paths in Open Sets). If A is an open path-connectedset in R2, then any two points in A are joined by a simple polygonal arc in A.

Proof. Let p, q ∈ A. Let f be any path in A from p to q. Consider the set

S = {t ∈ [0, 1] | A contains a simple polygonal arc from p to t}.

Then S is nonempty since 0 ∈ S, and it’s clearly bounded above by 1. As we’ve seen before,it’s an easy exercise to show that the least upper bound s of S is actually in S, and moreoverthat s = 1.

Lemma 8.3. If Γ is a planar graph, then Γ admits an embedding in R2 where all the edges aresimple polygonal arcs.

Proof. Let’s abuse notation and think of Γ as an actual plane graph. For every vertex p ∈ Γ, wechoose a closed disk D(p) which is small enough so that it does not intersect any edges whichare not incident to p. We also choose our disks so that D(p) ∩D(q) = ∅ for p #= q. If pq is anedge from p to q, we can consider 3 subintervals of pq determined by D(p), D(q). Then we canuse the previous lemma (carefully) to replace pq with a simple polygonal arc.

Theorem 8.4 (Jordan Curve Theorem: Polygonal Case). If C is a simple closed polygonalcurve in R2, then R2 − C has precisely two regions, each of whose boundaries is C.

Proof. Suppose that p1, p2, p3 belong to distinct regions of R2 − C. Choose any disk D suchthat D ∩ C is a straight line segment. For each i, we can choose a polygonal arc from pi toD (not intersecting C). Since we’re working with an actual disk with a straight line runningthrough it, we conclude that we can join at least two of the three points by a simple polygonalarc, which is a contradiction.

We’ve proved that R2 −C contains at most 2 regions. Now we’ll show it contains at least 2regions, i.e., that R2−C is not path-connected. For each point q ∈ R2−C, we take a (straight)ray R starting at q. Then R ∩ C is a finite collection of intervals and points. Let Q be aninterval (for simplicity, we’ll consider points to be intervals as well). Since we’re in R2 we cantalk about “sides” of lines and intervals and we can distinguish in the obvious way between Ctouching R (entering and leaving Q on the same side of R) and crossing R (entering and leavingon different sides). Let k be the number of crossing points of R with C. Note that the parityof k is independent of the choice of R and depends only on q and C. Now it is easy to find twopoints with different parity by finding a ray that intersects C precisely once, and one that doesnot intersect C at all (which exists since C is a bounded subset of R2.

More Terminology. Now that we understand polygonal closed curves in the plane, we can talkabout the interior of a polygonal closed curve C (the bounded component of R2 −C), denotedint(C). and similarly the exterior of a polygonal closed curve C (the unbounded component),denoted ext(C). We will also denote the closure of int(C) = int(C) ∪ C in the usual way:¯int(C), and similarly for the closure of the exterior...

Exercise 8.5. Let C be a simple closed polygonal curve. Let P be a simple polygonal arcproperly embedded in ¯int(C), i.e., such that P joins two points p, q on C but the interior ofP does not intersect C. Let P1, P2 be the two subarcs of C with endpoints p, q. Prove that

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R2 − (C ∪ P ) has precisely three regions, and their boundaries are C,P1 ∪ P , and P2 ∪ P ,respectively. (HINT: This exercise is a sort of scholium of the Polygonal JCT.) Conclude thatif r, s are two points on the interiors of P1, P2, respectively, then it is not possible to join themby a simple polygonal arc in int(C) without intersecting P .

8.3 General Jordan Curve Theorem

The work we did in the previous subsection was mainly to prove the following fact, which mostof us figured out as kids without actually being motivated to actually prove it. (Or was thatjust me?)

Theorem 8.6. The complete bipartite graph K3,3, also known as the Utilities Graph, is notplanar.

Please note that this is NOT the same as the famous theorem of Kuratowski, which saysthat a graph is nonplanar if and only if it contains a finite list of subgraphs, one of which is theUtilities Graph. That is harder to prove.

Proof that the Utilities Graph is Nonplanar. There is a cycle C of length 6 in K3,3 and we canlabel its vertices x1x2x3x4x5x6. We can think of C as a polygonal simple closed curve withtwo “chords” x1x4, x2x5, x3x6. If K3,3 were planar, then at least two of the three chords wouldhave to be either both in int(C) or both in ext(C). But this would contradict the Exerciseabove.

Exercise 8.7. Prove that the intersection of any straight line in R2 with a simple closed curveis compact.

Proposition 8.8. Let C be a simple closed curve in R2. Then R2 − C is not path-connected.

Proof. If we project C to the x-axis, we will get a closed subinterval [a1, a2] of R. Let L1, L2

denote the vertical lines x = a1,x = a2, respectively, and let L3 be a vertical straight linesituated somewhere in between them. For i = 1, 2, let pi denote the highest point in theintersection Li ∩ C. Let P1, P2 be the two arcs on C with end points p1, p2. Then Pi ∩ L3 arecompact and disjoint. So L3 contains an interval L4 joining P1 to P2 only intersecting C in itsendpoints.

Now, let L5 be an arc in ext(C) joining p1 to p2 consisting of segments of L1, L2, and ahorizontal straight line segment above C. If L4 is in ¯ext(C) then there is a simple polygonalarc in ¯ext(C) from L4 to L5.

But, then C ∪ L4 ∪ L5 ∪ L6 is a plane graph isomorphic to K3,3, which is a contradiction.Hence L4 does not lie in ¯ext(C), so int(C) #= ∅.

We are assuming all our surfaces are closed and connected, in other words, compact withno boundary.

Lemma 8.9 (Key Lemma). If G is a 2-connected graph, H is a 2-connected subgraph of G,then G can be obtained from H by successively adding paths such that each of these paths jointstwo distinct vertices in the current graph and has all other vertices outside the current graph.

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Proof. We induct on N = the number of edges in the set EG\EH . If N = 0 then G = H andwe’re done. Otherwise, if H is not a maximal 2-connected subgraph of G, then we can findH ′ which is a maximal 2-connected subgraph of G containing H , and then apply the inductionhypothesis to go from H to H ′, and then from H ′ to G.

So now assume H is a maximal 2-connected subgraph of G. Since G is connected, thereexists an edge x1x2 /∈ H with x1 ∈ H . Further, since G− x1 is connected, there exists a pathx2x3 . . . xk where xk ∈ H but x2, . . . , xk−1 /∈ H . (We allow the possibility that k = 2.) Butthen joining the edge x1x2 and this path x2x3 . . . xk to H , we get a 2-connected graph, whichmust therefore be all of G, and the lemma is proved.

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9 Lecture 9: Triangulability of Surfaces

9.1 Finishing up Jordan Curve Theorem

More Terminology. The term faces is also used for the regions of the complement of a planegraph Γ in R2. In particular, the unbounded region is often called the outer face. If Γ is2-connected, then the boundary of the outer face is the outer cycle.

Exercise 9.1. If Γ = (V,E) is a plane 2-connected graph with |V | ≥ 3, all of whose edges arerepresented as simple polygonal arcs, prove that R2 −Γ has |E|− |V |+2 regions, each of whichhas a cycle of Γ as boundary.

Define the union of two plane graphs in the obvious way: subdivide edges such that eachedge is actually straight, not just polygonal. If two edges cross, or if a vertex of one graph lieson an edge of another, subdivide further. then the following lemma is clear.

Exercise 9.2. Prove that if Γ1,Γ2 are two plane graphs such that each edge is a simple polygonalarc, then the union of Γ1,Γ2 is a graph Γ3.

Note that if Γ1,Γ2 are 2-connected then their union Γ3 is too, as long as Γ1,Γ2 share atleast two points in common.

Lemma 9.3. Let Γ1,Γ2, . . . ,Γk be plane 2-connected graphs all of whose edges are simplepolygonal arcs such that Γi has at least two points in common with Γi−1,Γi+1, but sharesno points in common with any other Γj. Also assume that the abstract graphs Γi are pairwisedisjoint (i.e., we’re not embedding any subgraphs separately in the plane). Then any point whichis in the outer face of Γ1 ∪ Γ2,Γ2 ∪ Γ3, . . . ,Γk−1 ∪ Γk is also in the outer face of Γ1 ∪ · · · ∪ Γk.

Proof sketch. Take a point p in a bounded face of Γ1 ∪ · · ·∪ Γk. Find a cycle C which containsp in its interior. Choose C minimally in that it is contained in Γi ∪ Γi+1 · · · ∪ Γj with j − i assmall as possible. Argue that j − i = 1.

Proposition 9.4. If A is a simple arc in the plane, then R2 −A is path-connected.

Proof sketch. Let p, q be two points in R2 − A. Choose d such that p, q are both at least adistance of 3d from the arc A. Now, A is the image of a uniformly continuous map of I intoR2. So we can partition A into segments Pi with end points p = p1, p2, . . . , pn such that anypoint on the path in A from pi to pi+1 is at most distance d away from pi. Then let d′ be theminimum distance between all the Pi, Pj , with |i − j| ≥ 2. So d′ ≤ d. Keep going like this,further subdividing each Pi into segments Pi,j . The idea is to cover A with boxes of a sizedesigned to ensure that graphs formed from the union of the boundaries of the boxes satisfiesthe hypotheses of the previous lemma, thus guaranteeing that p, q are in the outer face of thisunion, and the arc A is completely disjoint from this outer face. Thus p, q can be joined by asimple polygonal arc in the complement of P .

Theorem 9.5 (General Jordan Curve Theorem for R2). If C is a simple closed curve in R2,then R2 − C has precisely two regions, each of which has C as boundary.

Proof. Suppose that p1, p2, p3 are distinct points contained in three distinct regions R1, R2, R3,respectively, of R2−C. Let Q1, Q2, Q3 be three pairwise disjoint segments of C. It follows fromthe preceding lemma (plus a little thought) that we can find 9 polygonal arcs Aij satisfying:

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• Aij is a a simple polygonal arc from pi to Qj .

• Aij is contained in Ri

• Aij ∩ Aik = pi if j #= k

• Aij ∩ Ai′j′ = ∅

But then union of the Aij can be extended to form a plane graph isomorphic to K3,3. thisis a contradiction. So by Propositon 8.8, R2 − C has precisely two components. Finally, byProposition 9.4, we know that C is the boundary of each.

Exercise 9.6. State and prove an analogue of the Jordan Curve Theorem for the sphere S2.

Theorem 9.7 (Jordan-Schoenflies). If f is a homeomorphism of one simple closed curve inthe plane onto another, then f can be extended to a homeomorphism of the whole plane.

Prove the following generalizations of Exercises 8.5 and 9.1, respectively, which are usefulin proving the Jordan-Schoenflies Theorem.

Exercise 9.8. Let C be a simple closed curve and let P be a simple polygonal arc in int(C)such that P joins two points p, q on C but does not otherwise intersect C. Let P1, P2 be the twoarcs on C with endpoints at p, q. Then R2\(C ∪P ) has precisely three regions whose boundariesare C,P1 ∪ P , and P2 ∪ P , respectively.

Exercise 9.9. If Γ = (V,E) is a plane 2-connected graph containing a cycle C (which is asimple closed curve) such that all edges in Γ\C are simple polygonal arcs in ¯int(C), then R2\Γhas |E|− |V |+ 2 regions, each of which has a cycle of Γ as boundary.

If we assume the Seifert-van Kampen Theorem, then we can prove the Jordan Curve The-orem as an application of fundamental groups. We sketch this approach as follows.

Let C be a simple closed curve in R2. Choose a homeomorphism h : R2 → S2 − {(0, 0, 1}.Choose a point p ∈ C, and then choose a second homeomorphism k : R2 → S2 − {p}. LetL = k−1(C) − {p}, so that L is some kind of infinite “line” in R2.

Exercise 9.10. Use L to form an appropriate decomposition of R2 − L such that Seifert-vanKampen will tell us that R2 − L is simply connected.

Exercise 9.11. Prove that there is a homeomorphism of R3 taking L to the z-axis.

Exercise 9.12. Assuming any results you may need about extending continuous functions (e.g.,Tietze Extension Theorem; see, e.g., Armstrong), finish the proof of the Jordan Curve Theorem.

9.2 Corollaries of Jordan-Schoenflies

Let F be a closed set in the plane. We say that a point p ∈ F is curve accessible if for eachpoint 1 /∈ F , there is a simple arc A from p to q such that A ∩ F = {p}.

Corollary 9.13. If F is a closed set in the plane with at least three curve-accessible points,then R2\F has at most two regions.

Exercise 9.14. Prove Corollary 9.13 assuming the Jordan-Schoenflies Theorem.

Corollary 9.15. Let Γ,Γ′ be 2-connected plane graphs such that g is a homeomorphism andplane-isomorphism of Γ onto Γ′. Then g can be extended to a homeomorphism of all of R2.

Exercise 9.16. Prove Corollary 9.15. You will need to use the Key Lemma from previouslecture.

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9.3 Proof of Triangulability, assuming Jordan-Schoenflies.

Theorem 9.17. Every surface S is homeomorphic to a triangulated surface.

Proof. As we’ve observed before, it is sufficient to prove that S has a cellular structure.

Setup. For each point p on S, choose a disc neighborhood D(p), which we think of as an actualdisc in the plane, disjoint from all others. InsideD(p) we draw two quadrilateralsQ1(p) ⊂ Q2(p)containing p (interiors nested). Since S is compact, we can take a finite cover of S by int(Q1(pi).The idea of the proof is basically to modify the choices of the Q1’s so that eventually they givea cell structure on S.

The Painful Step. The main difficulty will be arranging things so that the Q1(pi) so that haveonly a finite number of points of intersection on S. The idea is to build up starting with Q1(p1),and show how to choose Q1(p2) to ensure that the two only have a finite number of points ofintersection. More precisely, we will assume inductively that Q1(p1), Q1(p2), . . . Q1(pk−1) havebeen chosen to achieve finitely many intersection points.

Now, look at Q2(pk), the “outer square” for pk. Choose Q3(pk) to be some other squarenested between Q1(pk) and Q2(pk). We are going to use Q3(pk) to find a “new” Q1(pk) whichwill complete the inductive step. Consider intersections of Q1(pj) with Q2(pk). We say asegment P of Q1(pj) is bad if it joins two points of Q2(pk) and is otherwise contained in theinterior of Q2(pk). We say that such a P is very bad if it also intersects Q3(pk).

CLAIM 1: There are a priori infinitely many bad segments in Q2(pk) but only finitely manyvery bad ones.

Exercise 9.18. Prove the claim *carefully* using an argument similar to that used to proveProposition 1.14 on p. 35 of Hatcher’s Algebraic Topology.

Since there are only finitely many very bad segments, their union together with Q2(pk) itselfforms a 2-connected graph which we will call Γ. Using Lemma 8.3, we can replace Γ with aplane-isomorphic graph Γ′ whose edges are all simple polygonal arcs. Corollary 9.15 allows usto extend the isomorphism from Γ to Γ′ to a homeomorphism of the interior of Q2(pk) which isfixed on Q2(pk) itself. Let Q′

1, Q′3 be the images of Q1(pk), Q3(pk) under this homeomorphism.

CLAIM 2: We can now choose a simple closed polygonal curve Q′′3 in the interior of Q2(pk)

such that Q′1 ⊆ int(Q′′

3) and such that Q′′3 intersects only the very bad segments (i.e., disjoint

from any “merely” bad segments).

Exercise 9.19. Prove Claim 2.

The main thing we’ve achieved now is that the very bad segments are now simple polygonalarcs.

Now, Γ′ ∪ Q′′3 is a 2-connected plane graph. If we use Corollary 9.15, we can now take

Q′′3 to be a quadrilateral whose interior contains Q′

1. So we can use Q′′3 as our “new” Q1(pk),

and we have finitely many intersection points with Q1(p1), Q1(p2), . . . , Q1(pk−1). Also, ourconstruction ensures that the Q1(pi) still cover S.

Final Step. The union of the Q1(pi)’s now gives a nice graph Γ on the surface S. TheJordan-Schoenflies theorem now gives the desired homeomorphism to a cellular surface.

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More precisely, each Q2(pi) contains only finitely many very bad segments, which are allsimply polygonal arcs which form a 2-connected graph. Thus each region of the complementof Γ = ∪Q1(pi) is bounded by a cycle C which lives inside some Q2(pi). And For each such C,draw a new convex polygon C′ of, say, unit sidelength, whose vertices are the same as those ofC. Taking the union, this gives a new graph Γ′ which is plane-isomorphic to Γ. So now we usethe Jordan-Schoenflies to extend this to a homeomorphism on the interiors of the C’s.

9.4 Well-definedness of connected sum

Let’s redefine the operation of connected sum as follows. We now know that any surface carriesa cell structure with a single face F , which we think of as sitting in the plane R2. Take twocongruent disjoint triangles T1, T2 in the interior of F , and delete their interiors. Each triangleTi inherits an orientation from R2, which just means a cyclical ordering of vertices – we’llchoose, say, a clockwise ordering. (The choice doesn’t matter, but we just have to make one.)We now identify T1 with T2 in such a way that their orientations disagree. This gives us a newsurface S′. It is an easy exercise to show that we can continuously map the interior of F toitself taking any triangle to any other, so this operation is well defined.

Exercise 9.20. Prove that the surface S′ which results from this operation is homeomorphicto the result of doing a connected sum of S with a torus T 2.

Exercise 9.21. Prove that we can relax this a little bit by allowing the triangles to be takenfrom different faces.

9.5 Finishing Classification

Exercise 9.22. So...where are we? Over the weekend, take stock. Try to write down an outlineof the proof of classification theorem, including any missing pieces if any. What have we proved?What remains to finish off our proof of classification of surfaces, if anything?

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10 Lecture 10: Proof of Triangulability of Surfaces

10.1 Well-definedness of connected sum

Let’s redefine the operation of connected sum as follows. We now know that any surface carriesa cell structure with a single face F , which we think of as sitting in the plane R2. Take twocongruent disjoint triangles T1, T2 in the interior of F , and delete their interiors. Each triangleTi inherits an orientation from R2, which just means a cyclical ordering of vertices – we’llchoose, say, a clockwise ordering. (The choice doesn’t matter, but we just have to make one.)We now identify T1 with T2 in such a way that their orientations disagree. This gives us a newsurface S′. It is an easy exercise to show that we can continuously map the interior of F toitself taking any triangle to any other, so this operation is well defined.

Exercise 10.1. Prove that the surface S′ which results from this operation is homeomorphicto the result of doing a connected sum of S with a torus T 2.

Exercise 10.2. Prove that we can relax this a little bit by allowing the triangles to be takenfrom different faces.

10.2 Finishing Classification

N.B.: I am purposely not writing a lot of details in this particular section of the lecture notes.

Things we have proved, more or less:

1. Every surface is triangulable, using the Jordan Curve Theorem and the Jordan-SchoenfliesTheorem.

2. Every triangulated surface can be represented by a “plane model” or “polygonal model”,that is, a polygon with its sides identified in pairs.

3. Every orientable triangulated surface is homeomorphic to a connected sum of the sphereS2 with some number of tori. This number of tori is the genus of the surface. A priori,the genus of a surface is not even well-defined for a particular surface, let alone for asurface up to homeomophism.

4. Just a note: our argument for showing that every orientable triangulated surface is home-omorphic to a connected sum of S2 with some number of tori can be modified withoutmuch trouble to show how to change any given plane model into the “standard” 4g-gon.(Exercise?)

5. Connected sum is well-defined. In other words, the connected sum of g tori (done inone particular way) is homeomorphic to the connected sum of g tori (done in some otherspecified way).

6. The Euler characteristic of any triangulation of any plane model representing S2 is 2.

7. The Euler characteristic of any triangulation of any plane model representing a (closedorientable) surface which is homeomorphic to the connected sum of S2 with some nonzeronumber g of tori is 2− 2g.

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8. By a connectedness argument, we know that any connected sum involving tori is nothomeomorphic to S2.

To sum up: what we have so far is a complete infinite list of possibilities, but we don’t knowyet whether it’s possible for a surface to be homeomorphic to more than one thing on the list.For example, how do we know that there is no homeomorphism from the connected sum of gtori to the connected sum of h tori for g #= h?

Two ways to finish off the proof, which we haven’t yet proved:

1. Euler characteristic is an invariant of the choice of triangulation.

2. The abelianization of the fundamental group of a surface of genus g is Z2g (and Zk ! Zl

for k #= l, but this part isn’t so hard).

If we were going to try to use fundamental groups as an invariant of surfaces, we would haveto give a full proof of Seifert - van Kampen. If we don’t care about obtaining the most generalpossible statement of SVK, then this can be done for surface groups using something called theSimplicial Approximation Theorem and the technique of “barycentric subdivision” of a giventriangulation. These techniques are similar at least in spirit to those we have seen developed inThomassen’s graph theoretic approach and are completely accessible at this point, though theproof is quite long and involved.

So, we will proceed by showing the former.

Theorem 10.3. Euler characteristic is an invariant of the choice of triangulation and henceinvariant under homeomorphism.

Proof. Let T, T ′ be two different triangulations of S. Then we can represent them each (dis-jointly) as polygonal plane graphs Γ,Γ′, respectively, with some sort of homeomorphism takingΓ′ to the polygon for Γ. Now, the image of Γ′ under this map may not be polygonal, but usingLemma 8.3 we can arrange that it is. Thus, we can form the graph G = Γ ∪ Γ′. Now, each ofΓ,Γ′ can be related to G (and hence to each other) by a finite sequence of moves of the typedescribed in Exercise 4.5. So we’re done.

Theorem 10.4. The homeomorphism type of a compact orientable surface is completely deter-mined by its genus and the number of boundary components.

10.3 Classification of simple closed curves on surfaces

We have seen how important it is to understand just the case of a simple closed curve in theplane, since the entire classification of surfaces rests mainly on the strength of the Jordan CurveTheorem. So we need to understand all the possible ways a simple closed curve can be embeddedin an arbitrary surface. Just like with the Jordan Curve Theorem, what we really need to dois describe the complement of the curve. Hence given a simple closed curve C and a surface S(with C ⊂ S), we define the surface SC to be a surface with two boundary components c1, c2such that the result of gluing c1 to c2 is the surface S, and such that the image of c1, c2 in thisquotient surface is the curve C.

From here on out, curve means simple closed curve.A curve C ⊂ S is nonseparating in S if SC is path-connected, otherwise it is separating.

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Proposition 10.5. Up to homeomorphism, there is a unique nonseparating curve on a givensurface. More precisely, if α,β are two nonseparating curves on a surface S, then there existsa homeomorphism h : S → S such that h(α) = β.

Proof. By Theorem 10.4, there exists a homeomorphism φ : Sα → Sβ . Moreover, we canchoose φ so that it “correctly” matches up boundary components. The map φ induces ahomeomorphism of S taking α to β.

A similar argument shows that if α,β are two separating curves on S, then there is ahomeomorphism of S taking α to β if and only if the corresponding “pieces” of Sα, Sβ arehomeomorphic. If S is closed, we often say that the genus of a separating curve C is theminimum of the genus of the components of SC , although this term is not always consistentlyused, and saying that a curve C has genus g may simply mean that one of the components ofSC has genus g.

Exercise 10.6. Use the Jordan Curve Theorem to prove that SC has at most 2 components.

Exercise 10.7. Classify all topological types of curves on a closed surface of genus g.

If α,β are two curves on a surface S, we define the geometric intersection number of α andβ to be |α∩β|. We denote this by i(α,β). A priori, the set α∩β could have infinite cardinality.Using something called transversality, it turns out that in fact one can always homotope afinite collection of curves to have finitely many points of intersection which are “transverse”, asopposed to “tangencies”. Unfortunately, we don’t have the time in this course to develop thenecessary tools to express this rigorously.

Exercise 10.8. Let α be a nonseparating curve on a surface S. Prove each of the following.

1. If the genus of S is at least 1, then there exists a curve β such that i(α,β) = 1

2. If S is genus two, prove that there exists a curve β such that α and β fill S, that is,S\(α ∪ β) is a disjoint union of disks.

3. Show that there exists a curve α′ which is disjoint from α and not isotopic to α so thatthe union α∪α′ separates S. (For this exercise you will need to assume that S has genusat least 3.

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11 Lecture 11: Basics of curves on surfaces

Acknowledgement: following the Primer by Farb-MargalitLet S = Sg,b,n denote a surface with genus g, with b boundary components and n punctures

(points removed). Recall that throughout, “curve” means “simple closed curve”.

11.1 Essential Curves

A curve is essential on S if it is NOT homotopic to a point or a boundary component or apuncture.

11.2 Algebraic intersection

We defined geometric intersection of two curves in the previous lecture. To define a notion ofalgebraic intersection, we require some notion of an orientation of a surface. Roughly speaking,an orientation is a consistent choice of ordered basis in a neighborhood locally homeomorphic toR2 (usually we choose {(1, 0), (0, 1)}). We need the machinery of smooth manifolds to explainprecisely what we mean by a “consistent choice”, but the idea is that we need to have a well-defined sense of “left”, “right”, “up” and “down”. Note that we don’t need any of this to talkabout an orientation of just a curve, since that can be given by the usual parametrization.

Let α,β be two oriented, transverse curves on an oriented surface. Then i(α,β) is the sumof all signed intersections, where we assign the value +1 to an intersection which “agrees” withour orientation and −1 otherwise. Clearly then we have:

i(α,β) = −i(β,α)

Exercise 11.1. Prove the following:

1. If α is separating, then for all transverse curves β, we have i(α,β) ∈ 2Z, and i(α,β) = 0.

2. In general, algebraic intersection and geometric intersection of α,β agree mod two.

3. Assuming the bigon criterion, for a = [α], b = [β], then i(a, b) is well defined.

4. i(a, a) = i(a, a) = 0.

11.3 Geodesic representatives

Here S is a hyperbolic surface.Recall that there is a bijective correspondence between the set of all conjugacy classes in

the fundamental group of a surface S and the set of all free homotopy classes of oriented curvesin S. So we will often abuse notation and use α to denote a curve in S and also to denote anelement of π1(S), even though that is only well defined up to conjugacy.

A geodesic in a hyperbolic surface S is the image of a geodesic in H2.

Proposition 11.2. Suppose S is a hyperbolic surface (i.e., χ(S) < 0). Let α ∈ π1(S) be a(not necessarily simple) curve in S which is not homotopic to a puncture. Then there exists aunique geodesic γ in the homotopy class of α. Moreover, if α is simple, then so is γ.

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Before we prove the proposition, recall that individual elements in the conjugacy class ofα #= 0 in π1(S) are in one-to-one correspondence with the set of all lifts to H2 of α. The groupπ1(S) acts on the set of lifts, and we can interpret this action as π1(S) acting on cosets of thecyclic group 〈α〉 ≤ π1(S). The stabilizer of the coset γ〈α〉 is the subgroup generated by γα0γwhere α0 is a primitive root of α.

Exercise 11.3. Explain why this correspondence fails to be true for T 2 (with the usual coveringby R2).

Proof of Proposition. We will not prove the last statement of the proposition here. And we’llsimplify things by assuming here that α is primitive in π1(S). Choose a lift α ∈ H2. Let φbe the generator of the stabilizer of α. Then φ is some hyperbolic isometry of H2. Let A beits axis. By the straight-line homotopy, we see that the image of A under the covering mapH2 → S is a geodesic in [α].

Now, S1 × I is compact, and hence distances of lifted homotopies in H2 are bounded. Soany other homotopic lift has the same endpoints at infinity, so we get the same geodesic A inH2. Hence the geodesic representative of [α] is unique.

Exercise 11.4. Finish the proof of the proposition in the case when α is not primitive.

Exercise 11.5. Let α be a simple closed curve. Then the conjugacy class of α in π1(S) consistsof primitive elements.

11.4 Torus example

Here is a basic problem: given a surface S, classify all homotopy classes of simple closed curveson S. This is not really possible in general, at least not in any meaningful sense, but on thetorus we can give a complete answer.

Consider the usual covering map R2 → T 2. Then we have

Aut(R2, p) ∼= Z2 ∼= 〈(1, 0), (0, 1)〉

Then we can represent (p, q) ∈ π(T 2) by the projection to the unit square of the line y = qpx.

Conversely, any lift of an element in π1(S) ends ,at some (p, q), and then we can just use thestraight line homotopy.

Exercise 11.6. Prove the following. In each case, you might want to try first proving the case(p, q) = (1, 0).

• i((p, q), (p′, q′)) = |pq′ − p′q|

• i((p, q), (p′, q′)) = pq′ − p′q

11.5 Bigon criterion

Lemma 11.7. Let χ(S) < 0. Let α,β be transverse curves in S. If α,β form no bigons, thenany pair of lifts α, β in the universal cover (H2)intersect at most in one point.

Proof sketch. Suppose that there are two (or more) points of intersection; then some segmentsof α, β form a disk upstairs. By an innermost disk argument, this disk must project via thecovering map to a disk downstairs, contradicting the assumption that there are no bigons.

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Theorem 11.8 (Bigon Criterion). Two transverse simple closed curves α,β in a surface S arein minimal position if and only if α,β do not form a bigon.

The Bigon Criterion can be proved as follows:

Exercise 11.9. Prove that any two lifts α, β intersect in at most one point, and that theircorresponding axes have distinct endpoints. Conclude that there cannot be any intersection-reducing homotopy of α or β.

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12 Lecture 12: Mapping Class Groups I

12.1 Punctures versus Boundary Components

In Mod(S), punctures are fixed setwise and boundary components are fixed pointwise, notjust by the maps but also during isotopies. We can think of a puncture as being obtained byremoving a closed disk from a surface, whereas a boundary component is obtained by removingan open disk. Thus we can also think of a puncture as a boundary component that is allowedto be twisted and/or swapped with other boundary components. Finally, it is equivalent tothink of punctures as marked points on the surface (rather than deleted points), as long as wemake the condition that isotopies don’t allowed non-marked points to wander over the markedpoints.

12.2 Finite order examples

1. The hyperelliptic involution (kebab embedding, rotated by π)

2. Element of order g (flower embedding)

Exercise 12.1. Find examples of the following finite order elements in the mapping class groupsof the following surfaces:

1. An element of order 4g + 2, for any Sg

2. Another involution besides the one described above. (Make sure to give a convincing reasonwhy it’s different!)

3. An element of order 5 in Mod(S2).

12.3 Some warmup examples

Notation: Sg,b,n has genus g, with b boundary components and n punctures.

Proposition 12.2 (Alexander Lemma). The mapping class group of a disk Mod(D2) is trivial.

So in the triple-index notation, D2 = S0,1,0, or just S0,1.

Proof. Given φ : D2 → D2 with φ = id∂D2 , we define:

F (x, t) =

(1− t)φ( x1−t

) for 0 ≤ |x| ≤ 1− t

x for 1− t ≤ |x| ≤ 1

Exercise 12.3. The mapping class group is also trivial for the following surfaces: S0,1,1, S0,0,1, S2,that is, the disk with one puncture, the sphere with one puncture, and the sphere.

The next calculation gives some idea of what happens with punctures.

Exercise 12.4. If α,β are simple arcs with the same endpoints, then α and β are homotopicrel endpoints.

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Proposition 12.5. The mapping class group Mod(S0,0,3) ∼= Σ3, where Σ3 denotes the sym-metric group on three letters.

Proof sketch: The map is the obvious one induced by what happens to the punctures under amapping class. It’s clearly a homomorphism and clearly surjective. To see that it is injective,consider an element in the kernel. Since endpoints are preserved, it takes any arc to anotherarc with the same endpoints, which is isotopic to the original arc by the previous exercise. Weinvoke an isotopy extension theorem and assume WLOG that this arc is in fact fixed. Thenwe cut open along the arc to obtain a disk with one puncture, which has trivial mapping classgroup.

Remark: The surface S0,0,3 is sometimes called an “open pair of pants”. A pair of pants isthe surface S0,3,0. (It turns out that Mod(S0,3,0 is isomorphic to Z3.)

Exercise 12.6. Prove that Mod(S0,0,2) ∼= Z/2Z.

Now consider the annulus A = S0,2,0.

Proposition 12.7. The mapping class group of the annulus Mod(A) ∼= Z.

Proof sketch. The annulus A is covered in the obvious way by R × [0, 1], which we think ofas embedded in the upper-half plane in the obvious way. Let [φ] ∈ Mod(A), and let f be aproperly embedded arc in A, say, corresponding to {(r, t) | t = 0 and 1 ≤ r ≤ 2}. Think of

f as a path, then we have a unique lift ˜φ(f) to R × [0, 1] based at the origin. The endpointf(1) = (n, 1) for some integer n. Then we define a map

Mod(A) → Z

[φ] .→ n

Well definedness follows from homotopy lifting, and it’s also clear that the map is a homo-morphism.

If we think of the annulus as those points in R2 in terms of polar coordinates with radius1 ≤ r ≤ 2, then we can define a map TA by

(r, θ) .→ (r, θ + πr)

Then T nA .→ n as described above, so the map is surjective.

We can prove injectivity using the straight-line homotopy.

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13 Lecture 13: Mapping Class Groups II

13.1 Dehn twists

The map TA defined on an annulus A in the previous section is an example of a Dehn twist.More generally, if α ⊂ S is a simple closed curve, we can take an annular neighborhood A of α,do TA on A, and then extend by the identity to S. We want to call this a “twist about α”. Apriori, this depends on the choice of the neighborhood A, and on a particular parametrizationof A, but up to isotopy, these choices don’t matter. So if a denotes the isotopy class of α, thenwe can define Ta to be the Dehn twist about a. Note that this does not require any choice oforientation for a, but instead only relies on an orientation of the surface S.

Exercise 13.1. The Dehn twist Ta is trivial in Mod(S) if and only if a is null-homotopic orparallel to a puncture.

Proposition 13.2. Let a, b be isotopy classes of simple closed curves with k ∈ Z. Then

i(T ka (b), b) = |k|i(a, b)

Proof. You can draw a quick picture to find a representation of T ka (b) and b which realizes the

intersection number given in the proposition. I’ll explain this in lecture

Exercise 13.3. Use the Bigon Criterion to show that this picture actually realizes minimalposition, hence completing the proof of the proposition.

Exercise 13.4. Verify the proposition in each of the following cases:

1. Any example with i(a, b) = 1, and k = 1, 2, 3.

2. Any example with a separating and i(a, b) = 2, and k = 1, 2 (you can also do k = 3 ifyou’re really brave).

The “calculus” of Dehn twists is very beautiful. Below we list several Facts/Exercises, andyou should attempt them all. Some are easy, some are hard, but it’s worth attempting themall to appreciate the difference.

Exercise 13.5 (Facts about Dehn twists). Here a, b are isotopy classes of curves and f is anarbitrary mapping class.

1. Ta = Tb if and only if a = b

2. fTaf−1 = Tf(a)

3. f commutes with Ta if and only if f(a) = a

4. If a, b are both nonseparating, then Ta is conjugate to Tb

5. TaTb = TbTa if and only if i(a, b) = 0

6. TaTbTa = TbTaTb if and only if i(a, b) = 1

7. Ta, Tb generated a free group (of rank two) if and only if i(a, b) ≥ 2.

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Here are two specific results we’ll need later, so even if you didn’t make through all of theabove, make sure to do these:

Exercise 13.6. If i(a, b) = 1, then TaTbTa(b) = a.

Exercise 13.7. If i(a, b) = 1, then TbT 2aTbTa reverses the orientation of a.

13.2 Torus, again

Chris proved in his lecture that Mod(T 2) ∼= SL(2,Z). It is known classically that SL(2,Z) canbe generated with two elements; these correspond to two Dehn twists about a meridian anda longitude on T 2 (the curves corresponding to the projected images of the x− and y− axes,respectively.

13.3 Exact sequences

A sequence of maps between groups is exact if the image of the first is precisely equal to thekernel of the next.

Exercise 13.8. Suppose you have the following short exact sequence of groups:

1 → K → G → Q → 1

In particular, this is one way of saying that Q ∼= G/K. Prove that G can be generated by a setof generators for K together with a lift of a set of generators for Q.

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14 Lecture 14: Mapping Class Groups III

14.1 Generating sets

1. Dehn twist generators

2. Finite order generators

3. Bounded generating sets

14.2 The curve complex

Let χ(S) < 0. We define a simplicial complex C(S) corresponding to S as follows:

1. Vertices: C0(S) ↔ isotopy classes of essential simple closed curves on S.

2. Edges: C1(S) ↔ disjoint representatives.

3. k-simplices: Ck(S) : [a1], . . . , [ak+1] span a k-simplex if and only if there exist pairwisedisjoint representatives.

The simplicial complex C(S) is usually called the curve complex (as opposed to “a” curvecomplex). The condition on k-simplices means that C(S) is completely determined by its 1-skeleton C1(S). In other words, C(S) is a flag complex. Despite the simplicity and elegance ofits definition, the curve complex remains an active area of research.

CHANGE OF NOTATION: Sg,n = Sg,b,n

Theorem 14.1 (Harvey, Lickorish). If 3g + n ≥ 5, then C(Sg,n) is connected.

Proof. Given a, b ∈ (S), we must find a sequence c1 = a, c2, . . . , ck = b such that i(ci, ci+1) = 0.We will prove this by induction; we will actually need two base cases: if i(a, b) = 0, the result isobvious. If i(a, b) = 1, then take the boundary γ of a regular neighborhood of a∪ b. The curveγ must be essential in S otherwise we violate the assumption that 3g + n ≥ 5, Then a, [γ], b isthe desired sequence.

Now, we are ready to induct on i(a, b) ≥ 2. Consider transverse (oriented) representativesα,β of a, b, respectively. There are two cases to consider based on whether two consecutiveintersections of α with β have the same or opposite signs. In either case we can do some kindof “surgery” on α,β to reduce intersection. [FILL IN YOUR OWN PICTURES!]

In the case of the torus, we have shown that no two distinct essential curves can be realizeddisjointly. In this case, we use the minimum possible intersection to determine where to putedges, which in the case of the torus is 1, and we get what’s known as the Farey complex.

14.3 Variations on the curve complex

Let N(S) denote the subcomplex of C(S) spanned by (isotopy classes of) nonseparating curves.

Proposition 14.2. For g ≥ 2, the complex N(Sg,n) is connected.

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Proof. Let a, b ∈ N(S). We can connect a, b in C(S), possibly using some separating curves:a = c1, . . . , ck = b. Suppose ci is separating. WLOG, we can assume that ci−1, ci+1 are bothnonseparating, since if, say, ci−1 is also separating, then ci−1, ci cobound a subsurface of genusat least 1, so there are infinitely many choices for a nonseparating curve d that we can insertinto the sequence: . . . , ci−1, d, ci, . . ..

So, we assume that our separating curve ci has two nonseparating neighbors in the sequenceci−1, ci+1. Then ci has two sides, and ci−1, ci+1 are either both on the same side, or else areon different sides. If they’re on different sides, we can simply delete ci from the sequence. Ifthey’re on the same side, then we choose any nonseparating curve from the other side, insert itinto the sequence in place of ci, and we’re done.

One more variation: we define a complex N(S) with the same vertex set as N(S), but weput edges down whenever i(a, b) = 1 instead of i(a, b) = 0. This complex is 1-dimensional.

Exercise 14.3. Use connectedness of N(S) to prove that N(S) is also connected for g ≥ 2(n =0).

The result of the exercise also holds for S1,n, when n ≥ 0. This can be proved by inductionusing the Farey graph as the base case.

Note that Mod(S) acts on N(S), a path-connected space. Variations of the following lemmaappear all over the place in Geometric Group Theory. This version has basically been cookedup for this example.

Lemma 14.4. Suppose G acts by simplicial automorphisms on a connected 1-dimensionalsimplicial complex X. Suppose G acts transitively on the vertex set X0 and on ordered pairs ofvertices connected by an edge. Let v, w be two vertices which share an edge in X, and let h ∈ Gtake w to v. Then the set {h} ∪ Stab (v) generates G.

Note that a priori, the lemma only guarantees a potentially infinite generating set. Muchdepends on what we can say about the stabilizer of a vertex. When you see these kinds ofproofs, remember that everything is cooked up to work out – the hard part is thinking of theright complex for your group to act on in the first place!

Proof. Let g ∈ G. Connect v to g(v) in X : v = v0, v1, . . . , vk = g(v). Now G acts transitivelyon X0, so choose gi such that gi(v) = vi. (We choose g0 = e, and gk = g.) We will use inductionto show gi is in the subgroup H generated by the element h and the elements of Stab(v). Thebase case is g0 = e.

Now, if ei is the edge between vi and vi+1, then g−1i (ei) is an edge between v and g−1

i gi+1(v).Since G acts transitively on ordered pairs of vertices connected by an edge, there is an elementr ∈ G such that r · (v, g−1

i gi+1(v)) = (v, w). So in particular, r ∈ Stab(v). But then we have

rg−1i gi+1(v) = w = h−1(v)

So hrg−1i gi+1 ∈ Stab(v), which implies that gi+1 ∈ H (assuming by induction that gi ∈ H).

14.4 Generation of Mod(S)

If S = Sg,n is a surface of genus g with n punctures, then PMod(S) means the version of themapping class group where the set of punctures is fixed pointwise. The “P” stands for “pure”.

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Theorem 14.5. The pure mapping class group PMod(S0,n) is finitely generated by Dehn twists(about separating curves).

We can use the previous theorem, together with the following short exact sequence, to workout Mod(S0,n)

1 → PMod(S0,n → Mod(S0,n) → Σn → 1

What we’re going to prove more carefully is the following theorem.

Theorem 14.6. The group PMod(Sg,n) is generated by finitely many Dehn twists about non-separating curves for g ≥ 1, n ≥ 0.

Proof. The proof is by double induction on g and n. Chris provided one base case: Mod(T 2),and actually a similar proof goes through for Mod(S1,1.

The key ingredient for this proof is the Birman Exact Sequence (BES), which holds for thevalues of g, n as above.

1 → π1(Sg,n → PMod(Sg,n+1 → PMod(Sg,n) → 1

The first map is the “Push” map, where you “push a marked point along a curve”. Thatthis is well defined is not at all clear, and unfortunately the proof uses the “long exact sequenceof a fibration” and is beyond the scope of this course. The key points for our purposes are:

1. The group π1(Sg,n) is finitely generated by nonseparating loops.

2. If γ is a simple representative of a homotopy class in π1(S), and if a regular neighborhoodof γ has boundary components a, b, then Push(γ) = TaT

−1b (or the inverse).

So we’re done with induction on punctures, and now need to induct on genus. We applyLemma 14.4 to Mod(S) on N(S). By Exercise 13.6, we can take h in the lemma to be theproduct TaTbTa, if a, b are the vertices playing the role of v, w. So we just need to understandStab(a) = Mod(S, a).

There is another short exact sequence which, together with Exercise 13.7, tells us we canassume that the curve a is fixed not only setwise, but with orientation preserved.

1 → Mod(Sg,-a) → Mod(Sg, a) → Z/2Z → 1

Here Mod(Sg,-a) means the subgroup of Mod(Sg) consisting of elements which fix the ori-ented curve -a.

The final piece is to consider the “Delete map” from Mod(Sg,-a) to PMod(S − α), whereα ∈ [a].

Exercise 14.7. Prove that the following sequence is exact, if the “right-hand” map is the Deletemap:

1 → 〈Ta〉 → Mod(Sg,-a) → PMod(S − α) → 1

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