algebraic topology f18 · algebraic topology f18 david altizio january 12, 2019 the following notes...

Algebraic Topology F18

David Altizio

January 12, 2019

The following notes are for the course 21-752 Algebraic Topology, taught duringthe Fall 2018 semester by Florian Frick. If you find any errors in these notes, feel freeto contact me at [email protected].

Contents

1 August 26 4

2 August 28 42.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 Topology Crash Course, Part I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

3 August 31 7

4 September 5 104.1 Opening Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104.2 Quotient Maps and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104.3 CW Complexes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

5 September 7 135.1 More on CW complexes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135.2 Retractions and Homotopies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

6 September 10 166.1 The Homotopy Extension Property . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

7 September 12 197.1 Introducing the Fundamental Group . . . . . . . . . . . . . . . . . . . . . . . . . . 19

8 September 14 218.1 Basepoint Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218.2 The Fundamental Group of the Circle . . . . . . . . . . . . . . . . . . . . . . . . . 22

9 September 17 239.1 Generalizations of IVT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239.2 Other Important Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

10 September 19 25

11 September 21 26

12 September 24 27

13 September 26 2913.1 Covering Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

14 September 28 3214.1 Simply Connected Covering Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

1

David Altizio 21-752 Lecture Notes

15 October 1 3415.1 More Simply Connected Covering Spaces . . . . . . . . . . . . . . . . . . . . . . . 3415.2 Some More Group Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3415.3 Uniqueness of covering spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3515.4 The punchline . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

16 October 3 3616.1 Covering Spaces for S1 ∨ S1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3616.2 Deck Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

17 October 5 3817.1 Examples of Deck Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . 3817.2 Group Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

18 October 8 4018.1 Some “Last” Words on Covering Spaces . . . . . . . . . . . . . . . . . . . . . . . . 4018.2 Van Kampen’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4018.3 Knots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

19 October 10 4319.1 The Klein Bottle, Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4319.2 Torus Knots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4319.3 Gluing 2-Cells . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4419.4 Fundamental Groups of Orientable Surfaces . . . . . . . . . . . . . . . . . . . . . . 45

20 October 12 4620.1 Links . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4620.2 Wirtinger Presentations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4620.3 Some Parting Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

21 October 15 4821.1 Building the Intuition for Homology . . . . . . . . . . . . . . . . . . . . . . . . . . 48

22 October 17 5122.1 Simplicial Homology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

23 October 19 53

24 October 22 5324.1 Singular Homology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5324.2 Connecting our Two Homologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

25 October 24 55

26 October 26 55

27 October 29 55

28 October 31 5728.1 Relative Homology Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

29 November 2 6029.1 Excision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

30 November 5 6330.1 Good Pairs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6330.2 Generators of Homology Groups on ∆ -complexes . . . . . . . . . . . . . . . . . . . 64

31 November 7 6631.1 The Home Stretch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

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32 November 9 6832.1 Mayer-Vietoris sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

33 November 12 7033.1 Degree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7033.2 Cellular Homology for CW Complexes . . . . . . . . . . . . . . . . . . . . . . . . . 71

34 November 14 7234.1 Cellular Homology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

35 November 16 7535.1 Euler Characteristic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

36 November 26 7736.1 Homology with coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7736.2 A Tiny Introduction to Cohomology . . . . . . . . . . . . . . . . . . . . . . . . . . 78

37 November 28 7937.1 A Less Tiny Introduction to Cohomology . . . . . . . . . . . . . . . . . . . . . . . 79

38 November 30 8238.1 Free Resolutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8238.2 Why cohomology? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

39 December 3 8439.1 Mayer-Vietoris . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8439.2 Cup Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

40 December 5 8740.1 Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

41 December 7 90

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1 August 26

Just logistical bookkeeping, e.g. introductions, course syllabus, and a beginning survey.

2 August 28

2.1 Introduction

We can motivate the study of topology as follows. Let’s consider the category of metric spaceswith morphisms being continuous functions. Recall that a function f : X → Y between twometric spaces is continuous if for all x ∈ X and for all ε > 0 there exists δ < 0 such thatdX(x, y) < δ implies dY (f(x), f(y)) < ε. Naturally, the isomorphisms between metric spaces arethose continuous functions which also have inverses - that is to say, homeomorphisms. However,this is a bad notion of isomorphism - ideally we would like to imagine isomorphisms as “renamings”of elements in our spaces, yet homeomorphisms do not preserve distances.

This tells us that metric spaces are not quite the right structures to study. But then what arethe right structures? The answer lies in the following proposition.

Proposition 1. Let X and Y be two metric spaces. Then f : X → Y is continuous if and only iff−1(U) ⊆ X is open for every open U ⊆ Y .

Proof. This proposition was not proven in class, but here it is for my own sake.

• (⇒) Suppose f is continuous, and let U ⊆ Y be open. Let x ∈ f−1(U) be arbitrary;then there exists ε > 0 such that Bε(f(x)) ⊆ f(U). (Here Br(x) denotes the open ballcentered at x with radius r.) By the definition of continuity, there exists δ > 0 such thaty ∈ Bδ(x)⇒ f(y) ∈ Bε(f(x)) ⊆ U . Thus Bδ(x) ⊆ f−1(U), and so f−1(U) is open.

• (⇐) Suppose f−1(U) is open for every open U ⊆ Y . Let x ∈ X and ε > 0 be arbitrary.Then in particular V := Bε(f(x)) is open, and thus so is f−1(V ) 3 x. Hence we can findδ > 0 such that Bδ(x) ⊆ f−1(V ), meaning that for any y ∈ X with dX(x, y) < δ we havef(y) ∈ V = Bε(f(x))⇒ dY (f(x), f(y)) < ε. Since x and ε were arbitrary, we deduce that fis continuous.

Thus, while continuous functions do not preserve distances, they do preserve open sets. Thismeans it is the open sets that we should be studying.

2.2 Topology Crash Course, Part I

What follows is a crash course on point-set topology and, in particular, a series of definitions.

Definition 1. A topological space is a set X with a collection of subsets F (called a topology) suchthat

• ∅ ∈ F and X ∈ F ;

• if U ∈ F and V ∈ F , then U ∩ V ∈ F ;

• If Uα ∈ F for all α in some index set A, then⋃α∈A Uα ∈ F .

In other words, F is closed under both finite intersections and arbitrary unions.

Definition 2. Let (X, τX) and (Y, τY ) be two topological spaces. A function f : X → Y iscontinuous if f−1(U) ∈ τX for all U ∈ τY . If additionally f−1 : Y → X exists and is continuous,then f is called a homeomorphism.

Definition 3. A sequence (xn)n ⊆ X converges to X if for all U ⊆ X open with x ∈ U , onlyfinitely many xn are not in U . (Note that by taking the open sets U to be open balls surroundingx we recover the metric form of continuity.)

We now recover one of the most fundamental concepts in point-set topology.

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Definition 4. A topological space X is Hausdorff if for all x, y ∈ X with x 6= y, there exist opensets U and V in τX such that x ∈ U , y ∈ V , and U ∩ V = ∅. In essence, we can “separate” anytwo unequal points.

In the future, almost all of the spaces we will be working with are Hausdorff.

Note that all metric spaces are topological spaces trivially (after all, we introduced them specif-ically as generalizations). But when is a topological space induced by a metric? The answer liesin the following theorem.

Theorem 1 (Urysohn’s Metrization Theorem). Let X be a topological space which is

• Hausdorff;

• regular, meaning that for closed set E ⊆ X and any x ∈ X \ E there exist open sets U andV in τX with x ∈ U , E ⊆ V , and U ∩ V = ∅; and

• second-countable, meaning that there exists a countable collection {Ui}i∈N ⊆ τX of subsets inτX such that for all V ∈ τX there exists I ⊆ N such that V =

⋃i∈I Ui.

Then the topology on X is induced by some metric.

Proof. Omitted.

We now introduce a way to generate new topologies from old ones.

Definition 5. Given maps fi : A → Yi, the initial topology on A is the coarsest topology thatmakes all of the fi continuous.

Remark. Note that the Yi are topological spaces, so in particular they already have topologiesrigged on them; we thus need the topology on A to be “compatible” with all the topologies on theYi.

This definition is a bit abstract, so here are a few examples.

Example 1. Let (X, τX) be a topological space, and let A ⊆ X. Define the inclusion mapf : A→ X via f(x) = x for all x ∈ A. We claim that the topology

τA = {A ∩ U : U ∈ τX}

is the initial topology on A with respect to f .

Again, this was merely stated in class, but here is a proof. We first claim τA is a topology on A.

• Since ∅ = ∅ ∩A and A = X ∩A, both ∅ and A are in τA.

• Suppose U ∈ τA and V ∈ τA are arbitrary. Then there exist sets XU and XV in τX such that

U = XU ∩A and V = XV ∩A.

This meansU ∩ V = (XU ∩A) ∩ (XV ∩A) = (XU ∩XV ) ∩A,

and so U ∩ V ∈ τA.

• Suppose Ui ∈ τA for all i ∈ I. Then there exist sets Xi for all i ∈ I such that Ui = Xi ∩ A.Then ⋃

i∈IUi =

⋃i∈I

(Xi ∩A) =

(⋃i∈I

Xi

)∩A,

and so⋃i∈I Ui ∈ τA.

Thus indeed τA is a topology.

Now we show that if τ is a topology which makes f continuous, then τ ⊇ τA. Indeed, if U ∈ τAis arbitrary, then there exists some set V ∈ τX such that U = V ∩A. Now remark that

f−1(V ) = {x ∈ A : f(x) ∈ V } = {x ∈ A : x ∈ V } = V ∩A = U.

By the definition of continuity, we therefore must have U ∈ τ ; since U was arbitrary, we thusdeduce τ ⊇ τA. Combining this with the above work yields the desired.

The next example is important enough to warrant its own definition.

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Definition 6. Given topological spaces Xi for i ∈ I, the product topology on Y :=∏i∈I Xi is the

initial topology with respect to projections on the factors Xi. In other words, it is the coarsesttopology on Y such that each of the projection functions πi : Y → Xi is continuous.

Remark. The initial topology can be classified by the following not-quite-universal property: forany topological space Z, a function g : Z → X is continuous if and only if the function fi◦g : Z → Yiis continuous for each i.

X Yi

Z

fi

g fi◦g

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3 August 31

We continue with our crash course on point-set topology.

Definition 7. If A ⊆ X for X a topological space, then we define

A =⋂C⊇A

C closed

C and A◦ =⋃U⊆AU open

U

to be the closure and interior of A respectively. Remark that C is closed while U is open.

Definition 8. Let X be a topological space. We say that X is connected if for all U and V openwith X = U ∪ V and U ∩ V = ∅, either U or V is empty.

Example 2. We claim that [0, 1] is connected under the usual topology on R. Suppose [0, 1] =U ∪ V , where U and V are open with U ∩ V = ∅. Note that because V and U are complementsin [0, 1], U and V are also closed.

WLOG let 1 ∈ V , and setu = sup{x : x ∈ U}.

We claim that u is in both U and V , contradicting the fact that U ∩ V = ∅. First note thatbecause U is closed, the supremum is actually a maximum, so in particular u ∈ U . Now defineun := u + 1

n for all n ∈ N, and remark that since u < 1, un ∈ V for all sufficiently large n. Thenun → u as n → ∞, but V is closed, so by the sequential characterization of closed sets we haveu ∈ V as well. Done.

We now define something which is stronger than connectedness.

Definition 9. A topological space X is path-connected if for any x, y ∈ X, there exists somecontinuous function γ : [0, 1]→ X such that γ(0) = x and γ(1) = y.

Proposition 2. If X is path-connected, then X is connected.

Proof. This proof is slightly different from the one given in class. Suppose X = U ∪ V for disjointopen sets U and V , and assume FSOC that neither U nor V is empty. Let x ∈ U and y ∈ V .Then by path-connectedness there exists a continuous function γ : [0, 1] → X such that γ(0) = xand γ(1) = y. Remark that γ is the continuous image of a connected set and is thus connected.But now

Uγ := γ([0, 1]) ∩ U and Vγ := γ([0, 1]) ∩ Vare two open sets with respect to the subspace topology on γ([0, 1]) which cover it but are disjoint,contradiction.

The opposite implication, however, is not true. Consider the topologists sine curve

γ :={(x, sin 1

x

): x ∈ [0, 1]

}={(x, sin 1

x

): x ∈ [0, 1]

}∪ {(0, x) : x ∈ [−1, 1]}.

Note that γ is connected, since sin is continuous and any open ball surrounding any point in{(0, x) : x ∈ [−1, 1]} must also intersect γ at least a second time. But γ is not path-connected, asany path from (0, 0) to (1, sin 1) cannot be of finite length.

We now proceed with another definition.

Definition 10. A topological space X is compact if every open cover of X has a finite subcover.

Theorem 2 (Tychonoff). If the topological space Xi is compact for all i in some index set I, then∏i∈I Xi is a compact topological space when equipped with the product topology.

Proof. Omitted.

Remark. Tychonoff allows us to construct topological spaces which are compact but not sequen-tially compact1. Take X = [0, 1]R, i.e. the set of all functions from R to [0, 1] when equipped

1i.e. any sequence {xi} ⊆ X has a convergent subsequence which converges to something in X

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with the product topology. Then X is compact but not sequentially compact. (Proof of this via

math.stackexchange: write R ∼= {0, 1}N, and define a sequence fn ∈ [0, 1]{0,1}N

via

fn(ω) = ωn for all ω ∈ {0, 1}N.

This sequence of functions does not have a convergent subsequence in the product topology via astandard diagonalization argument, since it so happens that the product topology is the topologyof pointwise convergence.)

Last class, we discussed the concept of an initial topology. Now we discuss its “dual”.

Definition 11. Let X be a set. For some index set I, let Yi (i ∈ I) be topological spaces, andlet fi : Yi → X be functions. The final topology on X is the finest topology that makes all the ficontinuous.

Remark. The final topology can be classified by the following not-quite-universal property: for anytopological space Z, a function g : X → Z is continuous if and only if the function fi ◦ g : Yi → Zis continuous for each i.

Yi X

Z

fi◦g

fi

g

As with the initial topology, we introduce some examples - the first is a small example, whilethe second is important enough to warrant a definition.

Example 3. Suppose Yi (i ∈ I) are topological spaces, and X =⊔i∈I Yi. The correct topology

on X is the final topology induced by the inclusions Yi ↪→⊔i∈I Yi.

Definition 12 (Quotients). Let Y be a topological space, X a set, and f : Y → X be surjective.The final topology on X with respect to f is called the quotient topology. The name comes fromthe fact that we can set e.g. X = Y/ ∼ for some equivalence relation ∼ on X.

We will introduce the above idea with a more sophisticated example. To set up this example,we will need a lemma commonly found in point-set topology courses.

Lemma 1. Let X and Y be topological spaces with X compact and Y Hausdorff. Suppose f is acontinuous bijection from X to Y . Then f is a homeomorphism.

Proof. Since f is bijective, its inverse f−1 : Y → X exists. It suffices to show that f−1 is continuous;this is equivalent to proving that if U ⊆ X is an open set then so is f(U) ⊆ Y , which in turn isequivalent to proving that if C ⊆ X is closed then so is f(C) ⊆ Y (take complements).

Let C ⊆ X be closed. Note that since X is compact, so is C (extend an open cover of C to anopen cover of X, extract a finite subcover, and remove everything you don’t want). Thus, since fis continuous, f(C) ⊆ Y is compact.

Now we claim that in fact f(C) is closed; this will finish the proof. To prove this, let y ∈ f(C)c

be arbitrary. For any x ∈ f(C) we may find open sets Ux 3 x and Vx 3 y which are disjoint. Now⋃x∈f(C) Ux is an open cover of f(C), so by compactness we may find points x1, . . . , xk such that

f(C) ⊆k⋃i=1

Uxk .

Let V =⋃ki=1 Vxk . Then V is an open set containing y which lies exclusively in f(C)c. Since y

was arbitrary, we deduce that f(C)c is open ⇒ f(C) is closed.

Now we turn to the sophisticated example.

Example 4. Consider the equivalence relation ∼ on [0, 1] generated by 0 ∼ 1. Let X = [0, 1]/(0 ∼1) equipped with the aforementioned quotient topology. I claim that X ' S1 (the unit circle inR2). To prove this, define g : [0, 1] → S1 via θ 7→ e2πiθ; note that g(0) = g(1) and that g is

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continuous. Denote by q the quotient map from [0, 1] to X. From g(0) = g(1) we know thatthere exists a well-defined map p : X → S1. Then we have the following commutative diagram; inparticular, since X is rigged with a final topology, g continuous⇒ p continuous. Note further thatp is bijective.

[0, 1] [0, 1]/(0 ∼ 1)

S1

g

q

p

It remains to prove that p is actually a homeomorphism. But this is a consequence of the previouslemma, since X is compact (due to q being continuous and [0, 1] being compact) and S1 is Hausdorff.

We have pretty much finished our crash course on point-set topology. One minor thing: supposeX is a topological space, and A ⊆ X. If we say “A is compact”, what does that mean? That Awith respect to the subspace topology is compact, or that any open cover of A in X has a finitesubcover? The answer is that there is no difference. Such properties are called “intrinsic”. Asanother example, connectedness is intrinsic, but only if you define connectedness of a subspacein the right way. (Cue five minutes of confusion and counterexamples that Florian claims will becleared up Wednesday.)

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4 September 5

4.1 Opening Remarks

We briefly clear up confusion from last time.

Remark. We can consider a space X to be disconnected if there exists a partition X = U ∪ Vwhere U and V are open. We say that A ⊆ X is connected if A with the subspace topology isconnected. As a warning, this is NOT

A ⊆ U ∪ V, U, V open in X and disjoint, U ∩A = V ∩A = ∅.

In particular, two sets U and V which intersect outside of A contradicts the notion of beingconnected under the subspace topology.

Proposition 3. A space X is connected iff any continuous map f : X → D, where D is a discretespace, is constant.

Proof. We proceed to show both directions.

• (⇐) Suppose X is disconnected, so that X = U ∪ V where U and V are open and disjoint.Now define f : X → {0, 1} via f(U) = 0 and f(V ) = 1. Note that f is continuous since forany set Y ⊆ {1, 2}, f−1(Y ) is either ∅, U , V , or X, and all of these are in τX . It followsthat we can always construct such a function, and now we take the contrapositive.

• (⇒) Suppose f : X → D is a continuous map. Pick an arbitrary y ∈ f(X), and considerthe set U := f−1({y}). Remark that U is nonempty by definition. Furthermore, {y} is bothopen and closed in the discrete space, and so U is also open and closed. But X is connected,so U = X, implying that f is constant.

4.2 Quotient Maps and Examples

We now slowly trudge our way into the realm of algebraic topology by discussing quotient spacesin more detail. Recall that a quotient map is the final topology on some surjective map. Most ofthe rest of this lecture will be spent discussing tons of examples. (Note: some of the finer detailsare expanded upon from class to be more rigorous.)

Example 5. Let X := [0, 1]2, and let ∼ be the equivalence relation on X generated by theequivalences (x, 0) ∼ (x, 1) and (0, y) ∼ (1, y) for all x, y ∈ [0, 1]. Then

[0, 1]2/∼ ∼= S1 × S1.

This is canonically the torus.

To prove this, it suffices to find a continuous bijection from [0, 1]2 to S1 × S1, for then we canproceed as in example 4; the bijection (x, y) 7→ ((cosx, sinx), (cos y, sin y)) suffices. To embed thetorus in R3, we instead use (for example) the bijection

(x, y) 7→ (2 cosx+ cosx cos y, 2 sinx+ sinx cos y, sin y),

where here we assume that radii of the two circles making up the torus are 2 and 1.

Figure 1: A torus.

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Example 6. Let D2 = {(x, y) ∈ R2 : x2 + y2 ≤ 1} be the closed disk in R2 of radius 1 centered atthe origin. I claim that

D2

∂D2' S2.

(Here X/A is equivalent to modding X by the relation on A for which all points are equivalent toeach other; such a space is called an adjunction space.) To prove this, note that again it sufficesto find a continuous bijection from D2/∂D2 to S2.

The explicit bijection is difficult to write down, so what follows is intuition. Imagine an antstarting at the center of the disk D2 and traveling to the boundary of the disk radially outward.The analogous trip on S2 would then be the trip which starts at the north pole and travels downward(radially?) to the south pole. This works because now all points on the boundary of the disk aremapped to the south pole.

⇒

Figure 2: A visual representation of the quotient map.

Example 7. Consider the disk D2 and let ∼ be the equivalence relation on ∂D2 such that x ∼ −xfor all x ∈ ∂D2. We define D2/ ∼ to be the real projective plane RP 2.2

Let us first figure out what RP 2 is intuitively. Consider the partition of D2 into light gray anddark gray parts below.

Figure 3: A partitioning of D2 into two parts.

The light gray part is homeomorphic to D2 (simply glue the two parts together!), while the darkgray part is homeomorphic to a Mobius strip due to the nature of ∼. Thus, one can obtain RP 2

by gluing the boundary of a disk to the nonorientable boundary of the Mobius strip.

Example 8. Consider the disjoint union X := D2 × S1 t S1 × D2 - two filled-in donuts. We’llanalyze one way to glue these donuts together. Consider the function f : S1 × S1 → S1 × D2 suchthat f(x, y) = (y, x). We write the quotient map of X with respect to the equivalence relationgenerated by f(x) ∼ x as the space Y := D2 × S1 ∪f S1 × D2. Now we claim that Y ' S3, the3-sphere in R4. For the geometric intuition behind this, consider two interlinked donuts such thatone torus passes through the hole of the other torus and vice versa. Now fill in the spaces betweenthese holes. Then what’s left is essentially a round object with no holes in its interior, i.e. a3-sphere.

Here’s the algebraic reasoning (i.e. more formal explanation) for this. Recall that

S3 = {(u, v, x, y) ∈ R4 : u2 + v2 + x2 + y2 = 1}.2Many other resources seem to define RP 2 as the quotient space of S2 under this antipodal equivalence relation

instead of D2; these seem to be equivalent by the previous example.

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Consider the surface

T = {(u, v, x, y) ∈ R4 : u2 + v2 = 12 = x2 + y2} ∼= S1 × S1.

Now T cuts S3 into two components, and it is easy to see these two spaces are homeomorphic (inparticular x2 + y2 < 1

2 ⇔ u2 + v2 > 12 ). These spaces are both filled-in tori, ergo they are both

homeomorphic to D2 × S1.

4.3 CW Complexes

We finish today by introducing a fundamental topological idea.

Definition 13. Define a sequence of topological spaces {Xn}n∈N as follows. Let X0 be a discretespace. Proceeding inductively, assuming Xn−1 has been defined, let Xn be obtained from Xn−1

as a quotient of Xn−1 tα Dnα (i.e. a disjoint union of Xn−1 with some number of n-dimensionaldisks) via gluing maps fα : Sn−1 → Xn−1. Note that it is possible for Xn = Xn−1. Then thetopological space

X :=⋃n∈N

Xn

is said to be a CW complex when rigged with the topology of nested union.

Example 9. I claim RP 2 is a CW complex. To see this, perform the following iterative construc-tion:

• Start with a single point.

• Take a line segment and glue both endpoints to this point; this yields a circle.

• Glue this to the boundary of D2 via the mapping z 7→ z2. The point of this mapping isthat it goes around the boundary of the circle twice; hence antipodal points on the circle aremapped to the same point on the disk.

Remark. We will usually only study spaces that are CW complexes; even Hatcher states at acertain point that all spaces henceforth will be CW complexes unless otherwise specified.

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5 September 7

5.1 More on CW complexes

We expand on the idea of CW complexes introduces last lecture.

Definition 14. For simplicity we will henceforth refer to Dn as an n-cell.

Example 10. Last time we showed that RP 2 is a CW complex. What about the torus S1 × S1?

1. Start off with a single point.

2. Glue two 1-cells to this point; this forms a figure-eight.

3. Glue the boundary of a square (homeomorphic to a 2-cell) by having the four sides alternatebetween being glued to the first 1-cell and being glued to the second 2-cell.

This construction indeed generates S1 × S1, and so the torus is indeed a CW complex.

Figure 4: The construction showing that the torus is a CW complex. Picture taken from Hatcher.

Example 11. The following construction is also a CW complex. These structures do not need tolook nice.

Figure 5: An unattractive CW-complex.

Remark. What do the letters C and W in the name “CW-complex” stand for?

• The “C” stands for “closure-finite”, i.e. the boundary of any cell is covered by finitely manycells of lower dimension. Note that by “finitely many cells” we mean cells belonging to thesame complex. This leads to some strange anomalies that seem unintuitive at first. Forexample, a 2-cell is not actually a CW complex since its boundary consists of infinitely many0-cells (we can’t cover it with a single 1-cell because we don’t have any of those to workwith!). However, a 2-cell is indeed homeomorphic to a CW-complex; take a single 0-cell,1-cell, and 2-cell, and glue them together in the obvious way.

• The “W” stands for “weak topology”. This means that A ⊆ X is closed iff A ∩Xn is closedin Xn for all n ∈ N. Remark that this makes sense because of the definition X =

⋃n∈NX

n,so despite the fact that each space is built by a gluing process all of our sets can be assumedto live in the same ambient space.

Now that we know what the letters stand for, we can formulate a sensible condition to determinewhen a topological space is a CW-complex.

Theorem 3. If X is a Hausdorff topological space and is partitioned into open cells such that

• (C) each cell has closure covered by finitely many of the open cells, and

• (W) A ⊆ X is closed iff A ⊆ Xn is closed in Xn for all n, where Xn denotes the collectionof all cells with dimension ≤ n,

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then the cells are a CW structure on X.

Proof. Omitted.

5.2 Retractions and Homotopies

We are now ready to start seriously diving into the realm of algebraic topology. We introduce theideas in the discipline with the following example.

Theorem 4. A 1-cell is not the same as a 2-cell. In other words, R 6∼= R2, where both sets arerigged with their standard topologies.

Proof. Suppose FSOC that there exists a homeomorphism f : R→ R2. Then R\{0} ∼= R2 \{f(0)}as well. But the LHS is a disconnected topological space while the RHS is a connected topologicalspace, contradiction.

We might suspect that this same technique can be used to prove that an m-cell is not the sameas an n-cell for any m 6= n. However, while this type of argument can be made to work, it turnsout to be a lot more complicated in general. As an example, suppose we wanted to prove R2 6∼= R3.Then we could in theory take a line ` in R2 and write

R2 \ ` ∼= R3 \ f(`)

where f is some homeomorphism from R2 to R3. However, this isn’t so simple, as we don’t quiteknow yet what f(`) is. (For all we know, it could be a space-filling curve!) We will thus need morepowerful machinery to prove this result, and this is one of the problems that algebraic topology isable to answer.

The above result also suggests homeomorphism might be too strong in the following way. Usingthe same technique, we can also show that the spaces S1 × [0, 1] and S1 are not homeomorphic(removing a point disconnects the latter but not the former). However, in some sense the twospaces are still kinda similar - after all, S1 can be thought of as an annulus with an infinitely thininterior. We are thus led to our first fundamental algebraic topology concept.

Definition 15. Let X be a topological space and A ⊆ X. A deformation retraction from X to Ais a continuous map F : X × [0, 1]→ X such that

• F (x, 1) ∈ A for all x ∈ X, and

• F (a, t) = a for all a ∈ A and t ∈ [0, 1].

Remark. An intuitive way to visualize a deformation retraction is to imagine watching a moviewhich shows the deformation of X. Then for 0 ≤ t ≤ 1, f(·, t) shows the image of this transforma-tion at time t.

Example 12. The function f : (S1× [0, 1])× [0, 1]→ S1× [0, 1] given by f((z, s), t) = (z, s(1− t))is a deformation retraction from S1 × [0, 1] to S1.

Figure 6: The deformation retraction from S1 × [0, 1]→ S1.

While deformation retractions are nice, they are in some sense a bit restrictive due to the subsetcondition. In particular, consider the two spaces shown below; while it is clear that we can deformone to look like the other, this transformation cannot be a deformation retraction. We thus considera more general definition.

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Figure 7: Two spaces which are not subspaces of each other.

Definition 16. Let X and Y be topological spaces. A homotopy is a continuous map F : X ×[0, 1]→ Y . We say that F is a homotopy from f(·, 0) to f(·, 1).

Remark. Homotopy equivalence is, as the name suggests, an equivalence relation. For example, toshow symmetry, suppose γ0 is homotopy equivalent to γ1, so that there exists a continuous functionf such that f(·, 0) = γ0 and f(·, 1) = γ1. Then the function g given by g(x, t) = f(x, 1− t) yieldsa homotopy from γ1 to γ0. (Think of this as playing a movie in reverse.) We write γ0 ' γ1 todenote that the two curves are homotopy equivalent.

We finish with a few small definitions.

Definition 17. Let X and Y be two topological spaces. We say that X and Y are homotopyequivalent (written X ' Y ) if there exist maps p : X → Y and q : Y → X such that q ◦ p ' idXand p ◦ q ' idY .

Definition 18. We say a space X is contractible if there exists a point x0 ∈ X such that X ' {x0}.

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6 September 10

6.1 The Homotopy Extension Property

We introduce the idea of extending a homotopy.

Definition 19. Let X and Y be topological spaces and A ⊆ X a subspace. Suppose that forany homotopy F : A × [0, 1] → Y and for any map f0 : X → Y with f0|A = F0, there exists a

new homotopy F : X → Y such that F |A×[0,1] = F . Then we say that (X,A) has the homotopyextension property (abbreviated HEP).

We’ll later explore spaces which have the HEP as well as spaces which do not, but for now, weshow how to use the HEP.

Theorem 5. Suppose (X,A) have the HEP and furthermore that A is contractible. Then X andX/A are homotopy equivalent.

Remark. The intuition behind this result is the diagram below. Ideally, contracting A to a pointshould not affect the topology on X, even if X itself is not contractible.

X X

A A⇒

Figure 8: A cartoon showcasing why removing contractible subspaces shouldn’t affect topology.

Proof. Let f be a contraction of A. By the HEP, we can find a homotopy ft : X → X with f0 = idextending this contraction f . We will now apply the universal property of quotient spaces twiceto discover two new continuous maps.

• Note that ft(A) ⊆ A, so q ◦ ft maps A to some singleton. We thus have the existence of

a map ft : X/A → X/A such that q ◦ ft = ft ◦ q. This can be seen in the commutativediagram below; in particular, within the context of the categorical interpretation of Remarkrem:quotient-diagram, take Z = X/A.

X X/A

X/A

q◦ft

q

ft

• Note that ft is a contraction, so f1 maps A to some singleton. We thus have the existenceof a map g : X/A → X such that g ◦ q = f1. This can be seen in the commutative diagrambelow; in particular, within the context of the categorical interpretation of Remark 11, takeZ = X.

X X/A

X

f1

q

g

We now claim that g is the desired function which shows that X and X/A are homotopy equivalent.To prove this, first remark that note g ◦ q = f1 ' f0 = idX . Furthermore, for any equivalence class[x] ∈ X/A,

q ◦ g([x]) = q(g(q(x))) = q(f1(x)) = f1(q(x)) = f1([x]),

so q ◦ g = f1 ' f0 = idX/A. Hence g is a homotopy inverse of q, meaning that indeed X and X/Aare homotopy equivalent.

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We now explore examples of space-subspace pairs which have the HEP.

Definition 20. Let X be a CW complex. A subspace A ⊆ X is a subcomplex if it is a subspaceinheriting the CW-complex structure on X.

Proposition 4. Let X be a CW complex an A ⊆ X a subcomplex. Then (X,A) has the HEP.

To prove this result, we first make a definition.

Definition 21. Let X and A ⊆ X be topological spaces. A map r : X → X is called a retractionif

• r(a) = a for all a ∈ A, and

• r(x) ∈ A for all x ∈ X.

If there exists a retraction from X to A, we say that A is a retract of X.

Lemma 2. The pair (X,A) has the HEP iff (X × {0}) ∪ (A× [0, 1]) is a retract of X × [0, 1].

Proof. We’ll show the result when A is closed in X; proving the result when A is not closed is “notsuper hard but requires a lot of analysis”.

First assume that (X,A) has the homotopy extension property. Then the identity map X ×{0}∪A× [0, 1]→ X×{0}∪A× [0, 1] extends to a map X× [0, 1]→ X×{0}∪A× [0, 1], so indeed(X × {0}) ∪ (A× [0, 1]) is a retract of X × [0, 1].

In the other direction, assume that (X × {0}) ∪ (A× [0, 1]) is a retract of X × [0, 1]. Because Ais closed, any two maps X ×{0} → Y and A× [0, 1]→ Y which agree on A×{0} combine to givea map X × {0} ∪A× [0, 1]→ Y which is continuous. By composing this map with the retractionfrom X × [0, 1] to (X × {0}) ∪ (A× [0, 1]), we get an extension X × [0, 1]→ Y of the original pairof maps; thus (X,A) has the HEP.

We are now ready to prove Proposition 4.

Proof. The following proof is in Hatcher and differs from the one sketched in class.

First remark that there exists a retraction r : Dn× [0, 1]→ Dn×{0}∪∂Dn× [0, 1]. For example,consider radial projection from the point (0, 2) ∈ Dn × R; then the second intersection point ofevery ray through (0, 2) with Dn×[0, 1] either lies on the bottom base of the cylinder (i.e. Dn×{0})or the lateral face of the cylinder (i.e. ∂Dn × [0, 1]). In turn, setting rt = tr + (1 − t) id gives adeformation retraction of Dn × [0, 1] onto Dn × {0} ∪ ∂Dn × [0, 1].

The crucial step is that the deformation retraction above gives rise to a deformation retractionof Xn× [0, 1] onto Xn×{0}∪(Xn−1∪An)×I, where recall that Xk is the kth step of our inductiveconstruction. Indeed, remark that Xn × [0, 1] is obtained from Xn × {0} ∪ (Xn−1 ∪ An) × I byattaching copies of Dn × [0, 1] along Dn × {0} ∪ ∂Dn × [0, 1]. (The Dn × {0} part is due to thefact that a copy of Dn must lie in X, while ∂Dn × [0, 1] is from our inductive construction of aCW-complex.) This allows us to take each of our deformation retractions from above (one for eachnew cell we add) and mesh them together in a well-defined way to get the desired result.

Finally, we connect everything together by performing the deformation retraction of Xn × [0, 1]onto Xn × {0} ∪ (Xn−1 × An) × I during the time interval [1/2n+1, 1/2n]. Then this infiniteconcatenation of homotopies is a deformation retraction of X × I onto X × {0} ∪ A × I. We areallowed to perform such an infinite concatenation because CW-complexes have the weak topology,i.e. a map is continuous iff its restriction to each skeleton is continuous.

Now that we have discussed examples of pairs of spaces which do have the HEP property, wenow give an example of a pair which does not have the HEP property.

Example 13. Let X = [0, 1], and set

A = {0} ∪ { 1n : n ∈ N}.

I claim that (X,A) does not have the HEP. This is actually HW3 P1, and what follows is theproof.

Recall that it suffices to show that [0, 1]× {0} ∪A× [0, 1] is not a retract of X × [0, 1]. Supposefor the sake of contradiction that such a retraction r : [0, 1]2 → [0, 1]× {0} ∪A× [0, 1] exists. Letx0 = (0, 1) ∈ A× [0, 1], and set

B := B(x0,12 ) ∩ (A× [0, 1]).

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Since r is continuous, there exists δ > 0 such that x ∈ B(x0, δ) implies r(x) ∈ B(x0,12 ), but recall

that B is a retract onto [0, 1] × {0} ∪ A × [0, 1], so in fact x ∈ B(x0, δ) implies r(x) ∈ B. Butnow [0, b 1

δ c]× {1} is a path which maps under r to a path connecting (0, 1) to (b 1δ c, 1) contained

exclusively in B, contradiction since B is not path-connected. Therefore our original assumptionwas false and no such r exists.

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7 September 12

We now begin the “algebra” section of Algebraic Topology.

7.1 Introducing the Fundamental Group

Our goal for today is motivated by the following proof that S1 6' [0, 1]. We can do this usingconnectedness, but the ideas that we will examine generalize.

Definition 22. Let X be a topological space. A homotopy based at x0 ∈ X is a map F :[0, 1] × [0, 1] → X such that F (0, t) = F (1, t) = x0 for all t ∈ [0, 1]. Intuitively, this homotopytracks loops which always pass through the point x0.

Proposition 5. The loops γ, γ′ : [0, 1]→ S1 defined by

γ(x) = (1, 0) and γ′(x) = (cos 2πx, sin 2πx)

are different, i.e. there is no homotopy based at (1, 0) between them.

The key observation is that S1 ∼= R/ ∼, where ∼ is the equivalence relation on R such that x ∼ yiff x− y is an integer. This is true for the same reason that S1 ∼= [0, 1]/(0 ∼ 1).

We’ll prove the result by lifting our loops up to R. Remark that the quotient map from R to S1

is defined by x 7→ (cos 2πx, sin 2πx). Thus we can consider the maps

γ : [0, 1]→ R, x 7→ 0 and γ′ : [0, 1]→ R, x 7→ x

as the lifted versions of γ and γ′ respectively. Now suppose that γ ' γ′, say via some homotopy γt.If we can show that we can find some other continuous map γt : [0, 1] → R with p ◦ γt = γt, thenwe will be done: since γt(1) ∈ Z for all t, this right endpoint must be constant, which contradictsthe fact that γ is a single point.

This motivates our further studies.

Proposition 6. Let X be a space and x0 ∈ X a base point. Define the relation ∼ on loops basedat x0 such that γ ∼ γ′ iff γ and γ′ are homotopy equivalent. Then ∼ is an equivalence relation.

Proof. We show the three properties of equivalence relations.

• Reflexivity: The identity homotopy sends γ to itself for any loop γ based at x0, and soγ ∼ γ.

• Symmetry: Suppose γ1 ∼ γ2 via some homotopy F : X × [0, 1]→ X. Then the homotopyG : X × [0, 1]→ X via G(x, t) = F (x, 1− t) sends γ2 to γ1, and so γ2 ∼ γ1.

• Transitivity: Suppose γ1 ∼ γ2 and γ2 ∼ γ3 via homotopies F1 : X × [0, 1] → X andF2 : X × [0, 1]→ X. Then the homotopy F : X × [0, 1]→ X via

F (x, t) = F1(x, 2t){t ≤ 12}+ F2(x, 2t− 1){t ≥ 1

2}

sends γ1 to γ3, and so γ1 ∼ γ3.

Thus ∼ is an equivalence relation.

Now that we have an equivalence relation on a set, let’s consider its equivalence classes. The setof equivalence classes happens to be a group under concatenation, i.e. “do the first loop, then thesecond loop”. More formally, given γ, γ′ : [0, 1]→ X based at x0, their concatenation is defined asγ · γ′ : [0, 1]→ X via

t 7→{γ(2t) t ≤ 1

2 ,

γ′(2t− 1) t ≥ 12 .

To show that concatenation actually induces multiplication on equivalence classes of loops, supposeγ0 ' γ′0 and γ1 ' γ′1 are four loops based at x0 ∈ X. Then given homotopies γt from γ0 to γ1 andγ′t from γ′0 to γ′1, the homotopy γt · γ′t sends γ0 · γ′0 → γ1 · γ′1, i.e. γ0 · γ′0 ' γ1 · γ′1. This tells usthat our · operation is well-defined, allowing us to state the following definition.

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Definition 23. The set of equivalence classes of loops at x0 in the space X equipped with con-catenation is a group denoted the fundamental group π1(X,x0) of X at the basepoint x0.

Remark. The subscript 1 in the notation for fundamental group is used because there are so-calledhigher homotopy groups πn(X,x0) for n ≥ 2 which keep track of maps Sn → X.

Theorem 6. It’s actually a group.

Proof. We first make an observation which will save time. Let ϕ : [0, 1]→ [0, 1] be any continuousmap such that ϕ(0) = 0 and ϕ(1) = 1, and let f be any path. Then fϕ can be considered areparametrization of f . Note that reparrametrization preserves its homotopy class since fϕ ' fvia the homotopy

(x, t) 7→ (1− t)ϕ(x) + tx.

(In other words, adjusting where points in [0, 1] get mapped to without changing the actual pathof f does not affect f topologically.)

We can use this to prove all three requirements of a group. Let f, g, h ∈ π1(X,x0) be arbitrary.

• Identity: Let c denote the constant path at x0. Then f · c is a reparametrization of f viathe function whose graph is shown in the first figure below, and so f · c ' f . Similarly, byusing the function whose graph is shown in the second figure below, c · f ' f .

Figure 9: The reparametrization associated with the identity of π1.

• Inverse: Define the inverse f : [0, 1]→ X of f to be the function defined via f(s) = f(1−s)(i.e. “going backwards”). Then f · f is homotopic to c via the homotopy

(f · f)t := ft · ft,

where ft is the path which equals f on the interval [0, 1− t]. An analogous argument showsthat f · f ' c, and so f is a two-sided inverse for f .

• Associativity: For arbitrary f, g, h, the paths (f ·g) ·h and f · (g ·h) are reparametrizationsof each other via the function whose graph is shown below. This is because the former pathtravels along f for the first quarter, g the second quarter, and h the final half, while thesecond path travels along f for the first half, g for the third quarter, and h for the finalquarter. The function below simply adjusts the locations where f transitions into g and gtransitions into h.

Figure 10: The reparametrization associated with associativity in π1.

Thus indeed we have a group on our hands. This might end up being useful some day.

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8 September 14

Last class, we introduced the notion of the fundamental group. This leads us to a slew of interestingquestions, including but not limited to the following:

1 Can π1 be computed efficiently?

2 Is it well-behaved wrt homotopy equivalence?

3 Is every group the fundamental group of some pair (X,x0)?

4 What properties of X are encoded in π1?

5 Does π1 characterize topological spaces up to homotopy? Same for maps?

6 What do subgroups correspond to in this space?

7 Is π1(X,x0) independent of x0?

8 Why do we care?

9 Groups measure symmetries; symmetries of what?

These are all questions we will attempt to answer at some point in this course.

8.1 Basepoint Independence

We will start by addressing 7 . This is clearly false in cases when X is not path-connected: taketwo disjoint spaces with different fundamental groups and consider the groups generated by pointsin the different connected components. But is path-connectedness sufficient?

Proposition 7. Yes.

Proof. Let X be a path-connected topological space and x0, x1 ∈ X. Let h : [0, 1]→ X be a pathfrom x0 to x1. Define

βh : π1(X,x1)→ π1(X,x0) via [γ] 7→ [h · γ · h].

This is a homomorphism because given γ, γ′ ∈ π1(X,x1) we may write

βh[γ · γ′] = [h · γ · γ′ · h] = [h · γ · h · h · γ′ · h] = βh[γ] · βh[γ′].

We can also prove that βh is an isomorphism because we can similarly show

βh · βh[γ] = [γ] = βh · βh[γ].

Thus in fact π1(X,x0) ∼= π1(X,x1), which proves the claim.

In general, any ϕ : X → Y a map with ϕ(x0) = y0 induces a map on the fundamental groupsϕ∗ : π1(X,x0)→ π1(Y, y0) via [γ]→ [ϕγ]. One can check that ϕ∗ is a homeomprhism.

Our goal now is to show the following: if X and Y are path-connected spaces which are homotopyequivalent via some homotopy ϕ : X → Y , then ϕ∗ : π1(X,x0)→ π1(X,ϕ(x0)) is an isomorphism.(Note that the converse is false; take for example X = {a} and Y = {b, c} or, for a slightly betterexample, X = {x}, Y = S2.)

One might think this would be an easy thing to prove: simply add stars to everything and call ita day. Unfortunately, this doesn’t quite work. Indeed, suppose we have two homotopies ϕ : X → Yand ψ : Y → X as well as some base point x0. Then ψ∗ sends π1(Y, ϕ(x0)) to π1(X, (ψ ◦ ϕ(x0)).This is not a priori the same group as π(X,x0)! We thus need to construct a fix.

Lemma 3. Suppose ϕt : X → Y is a homotopy and x0 ∈ X. Then there exists an isomorphismβh : π1(X,ϕ1(x0))→ π1(X,ϕ0(x0)) such that (ϕ0)∗ = βh(ϕ1)∗.

π1(X,ϕ0(x0))

X

π1(X,ϕ1(X0))

ϕ0

ϕ1

βh

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Proof. Let h be a path from ϕ0(x0) to ϕ1(x0) (which we know exists because X is path-connected),and define βh : π1(X,ϕ1(x0))→ π1(X,ϕ0(x0)) via

βh[f ] = [h · f · h].

I claim that this is an isomorphism. To prove this, remark that βh is a homomorphism since

βh[f · g] = [h · f · g · h] = [h · f · h · h · g · h] = βh[f ] · βh[g].

Furthermore, remark that

βhβh[f ] = βh[h · f · h] = [h · h · f · h · h] = [f ].

Similarly βhβh[f ] = [f ] for any f , and so βh is an isomorphism with two-sided inverse βh.

Now we are able to prove the desired result about isomorphisms. Recall that from our definitionsof ϕ and ψ we have the string of morphisms

π1(X,x0)ϕ∗−→ π1(Y, ϕ(x0))

ψ∗−→ π1(X,ψϕ(x0)).

Now consider the homotopy which is ψϕ at time t = 0 and which is a constant path at time t = 1.By the previous lemma, we thus know that there exists a βh such that ψ∗ϕ∗ = βh. Now applythe same argument with the composition reversed to get ϕ∗ψ∗ = βh as well. Thus ϕ∗ and ψ∗ areisomorphic as groups.

8.2 The Fundamental Group of the Circle

This leads us into our first major result.3

Theorem 7. π1(S1) ∼= Z, and furthermore the former group is generated by w(s) = (cos 2πs, sin 2πs).

Proof. Denote by p : R → S1 the map s 7→ (cos 2πs, sin 2πs). Furthermore, for all n ∈ Z setωn : [0, 1] → R to be the path which sends s 7→ ns for all s ∈ [0, 1]. We will first show that anyelement of π1(S1) is homotopy equivalent to some ωn, and then show that each ωn is unique up tohomotopy.

Let γ : [0, 1] → S1 be a loop at (1, 0) representing some element of π1(S1). This loop lifts to apath γ : [0, 1]→ R from 0 to some n ∈ Z. Now the homotopy

t 7→ (1− t)γ + tωn

shows that γ ' ωn; composing this homotopy with p shows that [γ] = [ωn].

Now let m and n be integers, and suppose ft is some homotopy from ωm to ωn. This liftsto a homotopy ft with f0(1) = m and f1(1) = n. But recall that the map s 7→ fs(1) both iscontinuous and has range contained in Z; the only way this can happen is if this map is constant,i.e. m = n.

3Note: this part was reconstructed from Florian’s notes and so may not be entirely accurate.

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9 September 17

The main goal for today is to prove various seemingly-unrelated results using the fact that π1(S1) =Z.

9.1 Generalizations of IVT

We start by generalizing the Intermediate Value Theorem. Note that IVT is equivalent to thefact that there does not exist a retraction f : [0, 1] → {0, 1} (which recall is true because [0, 1] isconnected). This leads us to our first generalization.

Proposition 8. There does not exist a retraction f : D2 → S1.

Proof. Suppose there was such a retraction. Let γ be any loop in S1, and consider it as livingin (the boundary of) D2. Now find a homotopy sending γ to a point, and compose it with theretraction. This reveals the existence of a homotopy within S1 sending γ to a point, which meansthat all loops are homotopic to the trivial loop. This is a contradiction of π1(S1) = Z, and so sucha retraction does not exist.

Our second result is another generalization of the Intermediate Value Theorem. Remark thatIVT implies that any function f : [0, 1] → [0, 1] must have a fixed point. This is because ifwe define g(x) := f(x) − x, then g(0) ≥ 0 but g(1) ≤ 0, so there must exist an x0 for whichg(x0) = f(x0)− x0 = 0. Looking at IVT in this way leads us to an extremely powerful theorem inanalysis, and in fact the tools we have developed so far lead to a quick proof of this theorem.

Theorem 8 (Brouwer). Any map f : D2 → D2 has a fixed point.

Proof. Suppose not, so that there exists some f : D2 → D2 without fixed points. Define a functionr : D2 → S1 as follows. Let x ∈ D2 be arbitrary. Since x and f(x) are distinct, we may draw a raywith endpoint f(x) passing through x in a well-defined manner; let r(x) denote the intersection ofthis ray with S1. Since f is continuous, so is r (check this for yourself), so we have established theexistence of a retraction from D2 to S1. This contradicts the previous proposition.

9.2 Other Important Theorems

Theorem 9 (Fundamental Theorem of Algebra). Let p : C→ C be a complex-valued polynomial.Then there exists some z ∈ C for which p(z) = 0.

Proof. Suppose note, and let p : C → C be some polynomial without zeros; our goal is to showthat in fact p is the constant polynomial.

Remark that for every radius r > 0 we have the existence of a map γr : [0, 1]→ S1 given by

x 7→ p(re2πix)/p(r)

|p(re2πix)/p(r)| .

The division by p(r) sends 0 → 1, i.e. γ0(x) = 1 for all x. Thus [γr] is trivial in π1(S1, 1) for allr > 0.

Now writep(z) = zn + an−1z

n−1 + · · ·+ a0.

For sufficiently large |z|, we have

|z|n > |an−1zn−1 + · · ·+ a1z + a0|.

With this in mind, setPt(z) := zn + t(an−1z

n−1 + · · ·+ a0).

Then P1 = p and P0(z) = zn; furthermore, Pt(z) 6= 0 for large z. As a result, the map

x 7→ Pt(re2πix)/Pt(r)

|Pt(re2πix)/Pt(r)|for varying t is a homotopy from the loop γr to the loop z 7→ zn, i.e. the loop which wraps arounditself n times. But [γr] = 0, so [z 7→ zn] = 0 as well, i.e. n = 0. Thus indeed p is constant.

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Theorem 10 (Borsuk-Ulam). Let f : S2 → R be continuous. Then there exists x ∈ S2 for whichf(x) = f(−x).

Proof. Suppose for the sake of contradiction that there exists some f : S2 → R for which f(x) 6=f(−x) for every x ∈ S2. Denote by η the loop circling the equation of S2 ⊂ R3 by η(s) =cos(2πs, sin 2πs, 0), and let h : [0, 1]→ S1 be the composed loop gη.

Define g : S2 → S1 via

g(x) =f(x)− f(−x)

|f(x)− f(−x)| .

As g(−x) = −g(x), we may write

h(s+ 12 ) = g(η(s+ 1

2 )) = g(cos(2πs+ π), sin(2πs+ π), 0)

= g(−(cos 2πs, sin 2πs, 0)) = −h(s).

Now recall that as shown in the calculation of π1(S1) we may lift the loop h to a path h : [0, 1]→ R.The equation h(s+ 1

2 ) = −h(s) implies that

h(s+ 12 ) = h(s) + q

2 for some odd integer q.

(Note that because q depends continuously on s ∈ [0, 12 ] and is always an integer, so it must be

constant.) Hence h(1) = h(0) + q, and from q being odd we conclude that h is not nullhomotopic.

But h was the composition of g and η, and η is nullhomotopic in S2, so gη must be nullhomotopicin S1 as well. Contradiction.

Corollary 1. SupposeS2 = A1 ∪A2 ∪A3

where each Ai is a closed set. Then some Ai contains two antipodal points on S2.

Proof. Consider the function f : S2 → R2 via

x 7→ (dist(x,A1),dist(x,A2)).

This function is continuous, so by Borsuk-Ulam there exists x ∈ S2 and (r, s) ∈ R2 such thatf(x) = f(−x) = (r, s). Now we case.

• If r = 0, then dist(x,A1) = 0⇒ x ∈ A1, and similarly −x ∈ A1.

• If s = 0, then similar logic as above gives {x,−x} ⊆ A2.

• If r 6= 0 and s 6= 0, then both x and −x lie strictly outside both A1 and A2, ergo they mustlie in A3.

We’re done.

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10 September 19

Today’s lecture was only 15 minutes long. We will prove the following result.

Theorem 11. For n ≥ 2, π1(Sn) = 0.

To prove this, we cite a lemma.

Lemma 4. Let X be a topological space and A be some index set. Suppose X =⋃α∈AAα, where

• each Aα is open and path-connected,

• there exists x0 ∈ X such that x0 ∈⋃α∈AAα, and

• for all α and β in A, the space Aα ∩Aβ is path-connected.

Then every loop γ in X based at x0 is homotopic to a loop of the form

γ1 · γ2 · . . . · γk,

where for all 1 ≤ i ≤ k there exists some α ∈ A such that γi ⊆ Aα.

Proof. The key idea behind this proof is the diagram below. For a full-fledged proof, see Hatcher.

Aα Aβ

x0

Aα Aβ

x0

With this, we may prove Theorem 11.

Proof. Denote by N and S the north and south poles of Sn. Define

A1 := Sn \ {N} and A2 := Sn \ {S}.

Then A1 and A2 are open path-connected spaces, and furthermore A1∩A2 is path-connected fromn ≥ 2. Now let x0 ∈ Sn be any base point in A1 ∩A2 (the exact base point doesn’t matter due topath-connectedness), and suppose γ is any loop based at x0. By the lemma we may write

γ ' γ1 · γ2 · . . . · γk,

where each γk is contained in either A1 or A2. But now recall that by the standard trick ofstereographic projection A1 and A2 are both homeomorphic to Rn; since Rn is simply connectedfor n ≥ 2, so must A1 and A2. It follows that γi is null-homotopic for each i, meaning that γ isnull-homotopic. Since γ was arbitrary, we deduce that π1(S1) = π1(S1, x0) = 0.

Corollary 2. For any n 6= 2, Rn 6∼= R2.

Proof. Assume n ≥ 3, as n = 1 has already been addressed. Suppose for the sake of contradictionthat f : R2 → Rn is a homeomorphism. Then f : R2 \ {0} → Rn \ {f(0)} is also a homotopy. Butremark that Rn \ {x} ∼= Sn−1 for any n ∈ N and x ∈ Rn. This gives a contradiction, as π1(S1) = Zbut π1(Sn−1) = 0.

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11 September 21

In-class quiz 1. The following ideas were introduced during (the review for) the quiz.

Definition 24. Let X and Y be topological spaces, and let x0 ∈ X and y0 ∈ Y . (We refer totopological spaces with distinguished basepoints as pointed spaces.) We say that the wedge sum ofX and Y , denoted X∨Y , is the quotient space of the disjoint union of X and Y by the identificationx0 ∼ y0, i.e.

X ∨ Y = (X t Y )/(x0 ∼ y0).

Definition 25. Let X be a topological space. The cone CX of X is defined as the topologicalspace

CX := (X × [0, 1])/(X × {0}).Intuitively, this construction turns X into a cylinder and collapses one end of the cylinder to apoint.

Remark. The space CX is always contractible, since there exists a deformation retraction fromCX to a point. (Do you see why?)

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12 September 24

Throughout this course, we’ve seen spaces with fundamental groups that are isomorphic to thetrivial group; take for example convex sets in Euclidean space. We also know of spaces withfundamental groups that are isomorphic to Z, e.g. S1 or (re the in-class quiz) S1 ∨ S2.4 Whatabout other groups? Are they fundamental groups of spaces as well?

To answer this question, we’ll need a lemma.

Lemma 5. In a simply-connected space, any two paths γ0 and γ1 with the same endpoints arealways homotopic relative endpoints.

Proof. Note that γ0 ·γ1 is a loop, so there exists a homotopy sending it to cx0 (the constant functionat x0). Call this homotopy Ft. Now consider the homotopy γt via γt = Ft · γ1. Observe that

γ0 = F0 · γ1 = γ0 · γ1 · γ1 ' γ0

and that γ1 = F1 · γ1 ' γ1. So this γt is the homotopy which works.

With this in mind, we may prove

Proposition 9. π1(RP 2) ∼= Z2.

Proof. Let q : S2 → RP 2 via x 7→ [x] be the quotient map. Denote by N and S the two lifts ofthe chosen basepoint x0 or RP 2. The map q exhibits RP 2 as evenly covered by S2, i.e. for everyx ∈ RP 2, there exists U 3 x such that q−1(U) = U1 t U2 and q|Ui : Ui → U is a homeomorphism.(We’ll explore this idea more fully later.)

The loop γ : [0, 1] → RP 2, where p : D2 → RP 2 is the quotient map and t 7→ p(2t − 1, 0), isnontrivial. Loop γ lifts to a path γ connecting the north pole N and the south pole S. If therewere a homotopy from γ to cx0 (the constant path at x0), it would lift to a homotopy γ fixingendpoints from γ to cN . This is not possible by IVT: at some point in time the endpoint at S hasto “jump” to the endpoint at N .

Putting this together, we know that π1(RP 2) ∼= Z2.

We now have the ability to answer 9 : having fundamental group G means there exists a simplyconnected quotient by G.

Let’s see some examples of this.

• S1×S1: The fundamental group of this space is Z×Z. (In general, π1(A×B) = π1(A)×π1(B),since if γ′t and γ′′t are homotopies of loops in A and B, then (γ′t, γ

′′t ) is a homotopy of a loop

in A × B, and vice versa because projections are continuous.) We can use S1 × S1 to coverR× R = R2 in the following way.

Figure 11: Tiling R2 with copies of the torus.

4Basic idea: retract S1 ∨ S2 onto S1 by retracting S2 to a point; this induces an injective homomorphism in thefundamental groups.

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• RP 2 × RP 2: The fundamental group of this space is Z2 × Z2. Analogously, every loop inRP 2 ×RP 2 lifts to a path in S2 × S2 that ends in one of four points, namely (N,N), (N,S),(S,N), and (S, S).

• S1∨S1: We don’t know what the fundamental group of this space is, but we can try to intuitwhat it is by drawing diagrams.

Consider S1 ∨ S1 as the wedge product of two loops a and b. This space is far from simplyconnected, but we can try to make it simply connected by unraveling the b loop. This hasthe side effect of copying the a loop infinitely many times and results in the diagram shownbelow.

Figure 12: The first step in the process.

The next logical step is to unravel all the a-loops. Unraveling each a-loop creates infinitelymany copies of Figure 12. This in turn forces us to stretch each of the new a-loops. Repeatingthis infinitely many times results in the figure below, which just so happens to be the Cayleygraph of the free group on two generators Z ∗Z. Thus, we suspect π1(S1 ∨S1) ∼= Z ∗Z. (Thismakes sense intuitively, as any loop in S1 ∨ S1 based at the wedge point can freely movebetween the loops a and b without restrictions.)

Figure 13: The almighty Cayley graph of Z ∗ Z!

In this way we see that our diagram gives the Cayley graph of the free group on two generatorsZ ∗ Z, and so we suspect that π1(S1 ∨ S1) ∼= Z ∗ Z.

There are a few things to note about the Cayley graph in particular, namely that it’s a simplyconnected space and that there exists a nice even quotient of the Cayley graph onto S1 ∨ S1

(i.e. all of the circles are evenly spaced out, etc.). These properties will be important in thefuture.

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13 September 26

13.1 Covering Spaces

We begin with a definition.

Definition 26. Let X be a space. A covering space of X is a space X and a map p : X → X suchthat for every x ∈ X, there is a neighborhood U 3 x such that

• p−1(U) =⊔α Uα, where Uα is open in X; and

• for each such α, p|Uα : Uα → U is a homeomorphism.

Figure 14: A visual representation of a covering space. Picture due to Wikipedia.

Remark. This definition doesn’t require that p is surjective (because we can technically considerthe case of there being no αs at all), but for the sake of simplicity we will restrict our attention tosurjective p.

Example 14. Consider the map from p : S1 → S1 given by z 7→ z2. I claim that this map coversS1. Indeed, for any x ∈ S1 and for sufficiently small ε > 0, there exists δ > 0 such that

p−1(B(x, ε) ∩ S1) = B(x, δ) ∪B(−x, δ).

The restriction of p to each of these open sets does yield a homeomorphism.

Up−1[U ]

Figure 15: Why multiple Uα are needed.

Example 15. We’ll use the notion of a covering space to find the fundamental group of the Kleinbottle K; this won’t be a formal proof, but it will lay some good intuition. The key is that R2 isa covering space for K, as shown by the following diagram. Here each square looks like one copyof K.

Now let us examine quotient maps on R2. Under the map A : R2 → R2 defined by (x, y) 7→(x + 1, y), the plane R2 maps to a cylinder, i.e. all vertical stripes become identical. Under themap B : R2 → R2 given by (x, y) 7→ (1 − x, y − 1), the same thing happens with the horizontalstripes.

We now have two different ways of looking at K; let’s connect them together.

• Let a denote the path(s) going along the horizontal lines of K and let b denote the path(s)going along the horizontal lines of K. I claim that a and b generate π1(K). For any loop

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Figure 16: Tiling R2 with copies of the Klein bottle. Notice the subtledifferences between this and Figure 11.

γ ⊆ R2, we may use 4 to reduce to examining loops γ completely located within one ofthe squares of R2. Now remark that the completely hollow boundary of K is homomorphicto two single loops (visualize this!) while the inside of the square deformation retracts toa point (and thereby does not affect the fundamental group). So indeed a and b generateπ1(K) under the relation aba = b.

• Note that we may apply the series of operations

(x, y)A7→ (x+ 1, y)

B7→ (1− (x+ 1), y + 1) = (−x, y + 1)A7→ (1− x, y + 1)

on points (x, y) ∈ R2. This final point is exactly B(x, y), and so ABA = B as functions.

These two approaches strongly suggest that

π1(K) = 〈a, b | aba = b〉 =: Z ∗2Z Z.

Proposition 10. If p : X → X is a covering space, ft : Y → X is a homotopy, and f0 : Y → Xa map lifting f0, then there exists a unique homotopy ft : Y → X of f0 that lifts ft.

Proof. Omitted (it’s Theorem 1.7 in Hatcher).

Remark. Note the following two special cases of interest.

• If Y = {∗}, this proposition says we can uniquely lift any path given a starting point.

• If Y = [0, 1], this proposition says we can uniquely lift homotopies of paths given a startingpath.

Proposition 11. Let p : X → X be a covering space. Then the induced map on fundamentalgroups p∗ : π1(X, x0)→ π1(X,x0) is injective.

Proof. Let γ be a loop based at x0 such that γ = p ◦ γ is null-homotopic (i.e. p∗([γ]) = 0). Denote

by γt the homotopy sending γ to a point. Lift the homotopy to X to see that γ is null-homotopic(i.e. [γ] = 0) and furthermore that our lifted homotopy fixes endpoints from construction.

Remark. The image of p∗ is precisely all the equivalence classes of loops [γ] such that γ hasa lift which is a loop. For example, recall the map p : S1 → S1 defined by z 7→ 2z. Thenp∗ : π1(S1)→ π1(S1) is defined via n 7→ 2n. The image of this p∗ is precisely the set of loops whichwrap around an even number of times.

Our next goal (which will extend into the subsequent two lectures) is to find a covering space

p : X → X such that X is simply connected.

Remark. In general, such a construction is not necessarily possible. To see why, suppose p : X → Xis such a map, and let x ∈ X be arbitrary. By the covering space property, there exists some U 3 xsuch that U has a lift U , i.e. p : U → U is a homeomorphism. Let γ be a loop contained entirelyin U which is based at X. Then γ = (p|U )−1 ◦ γ is a loop in U . If X is simply connected, then

there exists a homotopy γt sending γ to a constant loop in X. Now project this loop down to Xto get that any loop in U is trivial. Thus every X has a neighborhood which is trivial in X.

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The previous remark warrants a definition.

Definition 27. We say that a topological space X is semi-locally simply connected (SLSC forshort) if for every x ∈ X there exists some neighborhood U 3 x such that every loop in U isnull-homotopic in X.

Example 16. The Hawaiian earring H (i.e. the union of the circles in the Euclidean plan withcenters ( 1

n , 0) and radii 1n for all n ∈ N) is an example of a topological space which is not SLSC.

Indeed, let U 3 (0, 0) be any open neighborhood containing (0, 0), and suppose δ > 0 is such thatB((0, 0), δ) ⊆ U . Then any circle with radius strictly smaller than δ is completely contained in Ubut is not null-homotopic in the Hawaiian earring.

Example 17. Consider the cone CH of the Hawaiian earing. Note that because CH is con-tractible, CH is indeed SLSC, but it is not locally simply-connected for reasons analogous to thosein the previous remark.

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14 September 28

14.1 Simply Connected Covering Spaces

We proceed with constructing simply-connected covering spaces. We will assume below that X islocally path-connected, path-connected5, and SLSC.

The motivation for our construction comes from the following example.

Example 18. Recall that S1 ' R/ ∼, where x ∼ y iff x − y ∈ Z. Now suppose we were onlygiven S1 and we wanted to construct R. We could do this by thinking of R as the collection ofhomotopy classes of paths starting at (1, 0) which fix endpoints. For example, given x = (1, 0) andy = (−1, 0), the homotopy class assigned to the path which goes around the semicircle directlyfrom x to y can be assigned the number 0.5, the class assigned to the path which wraps around S1

once before landing on y can be assigned the number 1.5, and so on.

With this in mind, let x0 ∈ X, and define

X := {[γ] : γ is a path starting at x0}.

Furthermore, consider p : X → X defined via [γ] 7→ γ(1). This turns out to be the correct (X, p)

pair we need; it remains to put a topology on X. A natural first idea is to simply rig the initialtopology on X with respect to p, but this doesn’t work. As an example, going back to our R→ S1

case, rigging such a topology on R would not generate the set (0, 12 ), for instance, because any

preimage of an open set in S1 contains countably many intervals at integer distances apart. Thus,we actually have to do some work to find the right topology for X.

To do this, we recall a fundamental concept in point-set topology.

Definition 28. Let τ be a topology. A basis of τ is a collection of open sets B ⊆ τ such that forall U ∈ τ , there exists a subcollection B′ ⊆ B such that U =

⋃B∈B′ B.

The following equivalent definition was mentioned in class.

Proposition 12. Equivalently, B is a basis of a topology τ if

• ⋃B∈B B = X, and

• for all B1, B2 ∈ B and x ∈ B1 ∩B2, there exists B3 ∈ B such that x ∈ B3 and B3 ⊆ B1 ∩B2.

With this in mind, set

U := {U ⊆ X open and path connected : π1(U)→ π1(X) is trivial}.

Lemma 6. The collection of sets U is a basis of X.

Proof. Suppose U ∈ U , and let V ⊆ U be open and path-connected. Then there exists an inclusionmap from π1(V ) to π1(U). As the map from π1(U) → π1(X) is trivial, so must the map fromπ1(V )→ π1(X). Thus V ∈ U .

From here we can see that U forms a basis for X. First remark that since U is path-connected, ifthe inclusion π1(U)→ π1(X) is trivial for one choice of basepoint x0, then it is true for all choicesof basepoints. Thus every element in X is contained inside at least one such set, and so the unionof all such U equals X. Secondly, suppose U1 and U2 are two open and path-connected subsets ofX such that the inclusions π1(Ui)→ π1(X) for i = 1, 2 are trivial. Take V := U1 ∩ U2. Then V isa path-connected and open subset of X, and furthermore by the previous proposition e.g. V ⊆ U1,and so V ∈ U . Both of these properties show that U is a basis of X by Proposition 12.

Now we will take this basis of X and lift it to a basis of X. Given U ∈ U and a path γ in Xfrom x0 to some point in U , let

U[γ] := {[γ · η] : η is a path in U with η(0) = γ(1)} ⊆ X.

Lemma 7. The map p|U[γ], where p is defined above, is bijective.

5See here to see why both of these assumptions are needed.

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https://math.stackexchange.com/questions/135463/path-connectedness-and-locally-path-connected


Proof. In order to prove bijectivity, we must prove injectivity and surjectivity.

• The map p is surjective because U is path-connected: given an arbitrary x ∈ U , there existsa path η contained in U with η(0) = γ(1) and η(1) = x.

• The map p is injective because U is SLSC: if γ1 and γ2 are paths in U with γ1(1) = γ2(1) = y,then the SLSC condition guarantees the existence of a homotopy from γ1 to γ2 fixing theendpoints γ(1) and y, and so [γ1] = [γ2].

Lemma 8. Suppose γ′ is a path with [γ′] ∈ U[γ]. Then U[γ] = U[γ′].

Proof. Let γ′′ be a path contained in U such that [γ′] = [γ · γ′′]. We will use this γ′′ to prove theresult using double-containment.

• (⊇) Let [µ] ∈ U[γ′] be arbitrary; then

[µ] = [γ′ · µ′] = [γ · γ′′ · µ]

for some path µ′ contained entirely in U . Since γ′′ · µ is contained entirely in U , [µ] ∈ U[γ].

• (⊆) Let [µ] ∈ U[γ] be arbitrary, so that [µ] = [γ · µ′] for some path µ′ contained entirely inU . Since

[γ · µ′] = [γ · γ′′ · γ′′ · µ′] = [γ′ · γ′′ · µ′],and since γ′′ · µ′ is contained entirely inside U , [µ] ∈ U[γ′].

We now reach our crucial claim.

Proposition 13. The collection of U[γ] for all possible U and γ forms a basis for a topology.

Proof. We will use the criterion described in Propsition 12. Let U[γ], V[γ′], and [γ′′] ∈ U[γ] ∩ V[γ′]

be arbitrary. By Lemma 8, we have U[γ] = U[γ′′] and V[γ′] = V[γ′′]. Now let W ∈ U be such thatW ⊆ U∩V and γ′′(1) ∈W , which we know exists since U is a basis in X. Then W[γ′′] ⊆ U[γ′′]∩V[γ′′]

and [γ′′] ∈W[γ′′], and so we have found our desired set.

Thus after all of this work, we have rigged X with a topology, and we have our function p : X →X with [γ] 7→ γ(1). There are a few more things we need to check.

Proposition 14. The function p is actually continuous.

Proof. I claim that for any U ⊆ X and for any homotopy class of paths [γ], the restriction of p toU[γ] bijects subsets of V[γ′] ⊆ U[γ] to subsets V ⊆ U ; this implies that p is continuous (check this!).

In one direction, for any such V , p(V[γ′]) = V . In the other direction, p(V ) ∩ U[γ] = V[γ′], sinceV[γ′] ⊂ U[γ′] = U[γ] and V[γ′] maps onto V by the bijection p.

We defer the remaining things to check to the next lecture.

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15 October 1

15.1 More Simply Connected Covering Spaces

As before, X is a path-connected, locally-path-connected, and SLSC topological space. Define Xand p as before.

Proposition 15. The space X is path-connected.

Proof. Let [γ] ∈ X be arbitrary; we want to construct a path in X from [cx0] → [γ]. To do this,

remark that γ is a path in X sending x0 to γ(1). Now set

αt(s) =

{γ(s) if s ≤ t,γ(t) if s > t.

This corresponds to the path which travels t amount of the way from x0 to γ(1) before stopping.This α works.

Proposition 16. The space X is simply connected, i.e. π1(X, [cx0]) = 0.

Proof. Note that our map p : X → X given by [γ] 7→ γ(1) lifts to a map p∗ : π1(X, [cx0 ]) →π1(X,x0). It suffices to show that p∗ is trivial, i.e. the image of π1(X, [cx0

]) is the trivial element.Once we do this, the fact that p∗ is injective implies the desired result.

To do this, recall that if [γ] is a loop in the image of p∗, then γ lifts to a loop in X. The αdescribed in the previous proposition is one such lift, and in particular α is a loop iff [γ] = [cx0 ].This holds if and only if γ is null-homotopoic, and so the image of p∗ is indeed trivial.

Now we have the existence of these covering spaces, but at the moment they are quite abstractobjects. We need to visualize what these spaces actually are. This is hard.

Example 19. Let X denote the topological space which is the sphere with two holes in it. Thenas seen below, R2 is a covering space of X. However, this is not R2 with the standard Euclideanmetric, as it is not possible to tile R2 with regular octagons. Instead, a hyperbolic metric isnecessary.

15.2 Some More Group Theory

We’re now ready to dive into some more sophisticated group arguments.

Proposition 17. Suppose X is path-connected, locally-path-connected, and SLSC. Let H be asubgroup of π1(X,x0). Then there exists a covering space p : XH → X such that

p∗(π1(XH , x0)) = H

for some suitably chosen basepoint xH .

Proof. Define a relation ∼ on X via the following criterion: [γ] ∼ [γ′] if and only if γ(1) = γ′(1)

and [γ · γ′] ∈ H. It’s not hard to check that this is an equivalence relation. Now set XH = X/ ∼,and define p : XH → X via [γ] → γ(1). This works. Indeed, for any loop γ based at x0, its lift

to X starting at the constant loop [cx0] ends at [γ]; thus, the image of this lifted path in XH is a

loop iff [γ] ∼ [cx0], which is true iff [γ] ∈ H by the definition of ∼.

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15.3 Uniqueness of covering spaces

Definition 29. Let p1 : X1 → X and p2 : X2 → X be covering spaces. We say that p1 and p2 areisomorphic if there exists a homeomorhism f : X1 → X2 such that p1 = p2 ◦ f .

Proposition 18. Let X be path-connected and locally-path-connected, and let p1 : X1 → X andp2 : X2 → X be covering spaces. The following are equivalent.

• The spaces p1 and p2 are isomorphic via an isomorphism f : X1 → X2 taking x1 ∈ p−11 (x0)

to x2 ∈ p−12 (x0);

• We have the equality(p1)∗(π1(X, x1)) = (p2)∗(π1(X, x2)).

Proof. The “⇒” direction is easy, as isomorphic covering spaces induce isomorphic maps on thefundamental groups. For the “⇐” direction, we need to be a bit more careful and establish a liftingcriterion.

Let p : (X, x0) → (X,x0) be a covering space and suppose f : (Y, y0) → (X,x0), where Y is

path-connected and locally-path-connected. I claim that a lift f : (Y, y0) → (X, x0) of f exists if

and only if f∗(π1(Y, y0)) ⊆ p∗(π1(X, x0)). The fact that this is necessary is intuitive. To define

such a map, suppose γ is a path from y0 to y. Then f ◦ γ (a path in X) has a unique lift f ◦ γ.

Now let f(y) = f ◦ γ(1); it is an exercise to check that this is well-defined.

15.4 The punchline

We’ve basically proven the following theorem.

Theorem 12. Let X be a path-connected, locally-path-connected, and SLSC space. Then thereexists a bijection between

{basepoint-preserving isomorphism classes of path-connected covering spaces}

and{subgroups of π1(X,x0)}

explictly given by (X, x0)↔ p∗(π1(X, x0)).

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16 October 3

There is only one (up to isomorphism) simply connected covering space for a space X which satisfiesthe usual properties; this is called the universal cover. But that doesn’t mean there aren’t othercovering spaces.

16.1 Covering Spaces for S1 ∨ S1

There exist many covering spaces for the topological space S1 ∨ S1; Hatcher lists 14 of them.

Proposition 19. Let G be a 4-regular graph. Then G is a covering space of S1 ∨ S1.

Proof. We’ll explain how to do this for finite graphs; the infinite case is a bit trickier. In the finitecase, we may perform the following steps.

• Find a Eulerian cycle in G.

• Label the edges of this cycle a, b, a, b . . . alternately in order.

• Orient the edges in the correct way, noting that we can always do this because we never havean instance where four “a” edges or four “b” edges meet in the same vertex.

Example 20. The following three examples (taken from Hatcher) were all described in class.Here, the group presentation was found by reading from the graph the possible closed loops onecould form.

Figure 17: Three different covering spaces of S1 ∨ S1.

A few remarks are in order.

Remark. Note that the first example tells us the free group on 3 generators is a subgroup of thefree group on 2 generators, which is somewhat counterintuitive.

Remark. In the first two examples, there exists a symmetry swapping vertices; e.g. in the firstexample we may change the basepoint and relabel the loops without affecting the presentationof the group. In example 3, this is not the case, as the leftmost and rightmost vertices behavedifferently than the middle one. We will revisit this idea later.

16.2 Deck Transformations

We now seek to answer the question of how to read off the fundamental group of a space X fromits covering space X.

Definition 30. Let p : X → X be a covering space. An isomorphism from p to p is called a decktransformation. The deck transformations of p form a group G(X).

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Example 21. The universal cover of S1 is R, as has been shown previously.

Recall that if p1 : X1 → X and p2 : X2 → X are two covering spaces, an isomorphism betweenthem is a homeomorphism f : X1 → X2 such that p1 = p2 ◦ f . The only such homeomorphismsare those of the form x → x + n. (Indeed, scaling either removes or adds pairs which differ byintegers, and multiplication by −1 does not work because 1

10 and − 110 do not project down to the

same point in S1.) Thus,

G(R→ S1) = ({R→ R, x 7→ x+ n, ·}) ∼= Z = π1(S).

Definition 31. Let p : X → X be a covering space. We say p is normal if for all x ∈ X and forall lifts X1 and X2 of X, there exists a deck transformation sending X1 → X2.

We are now ready for the punchline.

Theorem 13. Let p : (X, x0) → (X,x0) be a path-connected covering space of a path-connected,locally-path-connected space X. Set

H = p∗(π1(X, x0)) ⊆ π1(X,x0).

Then the following hold.

• The covering space p is normal if and only if H is a normal subgroup of π1(X,x0).

• Recalling that N(H) = {g ∈ G : gH = Hg} is the normalizer of G, we have

G(X) ∼= N(H)/H.

In particular, if X is the universal cover of X, then H is the trivial group, and so G(X) ∼=π1(X,x0).

Proof. First let x1 ∈ p−1(x0), and γ a path from x0 to x1. Note that γ projects to a loop γ in X;let G = [γ] ∈ π1(X,x0). Set

Hi := p∗(π1(X, xi)) for i ∈ {0, 1}.

Then note the following observations.

• Let f be a loop in X based at x0. Then ¯γ · f · γ is a loop based at x1, and so g−1H0g ⊆ H1.

• The exact same argument shows that gH1g−1 ⊆ H0 ⇒ H1 ⊆ g−1H0g.

Hence in fact H1 = g−1H0g, and so [γ] ∈ N(H) if and only if p∗(π1(X, x0)) = p∗(π1(X, x1)). Thisalong with the lifting criterion mentioned in the proof of Proposition 18 yields the first part.

To prove the second part, consider ϕ : N(H) → G(X) defined via [γ] 7→ τ , where τ is the decktransformation taking x0 → x1 via the notation above. Note that ϕ is a group homomorphism,because if [γ′] maps to the τ ′ taking x0 to x′1, then γ · γ′ lifts to γ · τ(γ′), which is a path from x0

to τ(x′1) = ττ ′(x0).

To finish, remark that by the previous paragraph ϕ is surjective, and furthermore kerϕ consistsof the deck transformations τ which fix x0, which are precisely the elements of p∗(π1(X, x0)) = H.Thus by the First Isomorphism Theorem,

N(H)

kerϕ∼= ϕ(N(H)) =⇒ N(H)

H∼= G(X).

Huzzah.

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17 October 5

17.1 Examples of Deck Transformations

Recall the following fact: a map h : Y → X has a lift h : Y → X if and only if h∗(π1(Y )) ⊆p∗(π1(X)).

Example 22. Consider the map p : S1 → S1 via z 7→ zn. Then any basepoint lifts to n pointsevenly spaced out under the preimage. Hence, the deck transformations correspond to rotations bymultiples of 2π

n . (Note that reflections do not work because they don’t commute with the projectionoperator analogously to Example 21.) Thus, this covering space is normal and the group of decktransformations is Z/nZ. Note however that this does not satisfy the requirements of Theorem 13because S1 is not simply connected.

Example 23. Consider the space RP d for d ≥ 2. Recall that RP 2 is a quotient space of S1

by the equivalence relation which identifies antipodal points as being identical. Thus, the mapq : Sd → RP d is the universal cover of RP d. Given a fixed basepoint x0, there are two inverseimages q−1(x0) in Sd, which means the only possible deck transformations are the identity and theone which swaps the two given basepoints. This implies π1(RP d) ∼= Z2.

Example 24. Recall that

SO(3) = {A ∈ R3×3 : AAT = I = ATA,detA = 1}= {rotations of R3 fixing the origin}.

Note that every rotation is determined by the axis of rotation plus an angle. In particular, thereexists a map from D3 to SO(3) given by the following: an element x ∈ D3 determines the directionof the rotation, while π|x| determines the angle of the rotation. This is a quotient map identifyingantipodal points on the boundary, so by the same ideas as the previous example SO(3) ∼= RP 3

and π1(SO(3)) ∼= Z2.

Remark. The previous exmaple implies that there exists a nontrivial loop γ such that γ ·γ is trivial.This seems hard to believe at first, but with watching Florian move his hands around a few timesit becomes convincing enough.

17.2 Group Actions

Definition 32. Let G be a group and Y a space. An action of G on Y is a group homomorphismfrom G→ Homeo(Y ), i.e. g1(g2y) = (g1g2)y for all y ∈ Y and g1, g2 ∈ G.

For example, deck transformations act on covering spaces.

Definition 33. An action is properly discontinuous if for all y ∈ Y , y has a neighborhood U suchthat

∀g1, g2 ∈ G, if g1(U) ∩ g2(U) 6= ∅, then g1 = g2.

In other words, every nontrivially element of the group moves some neighborhood off of itself.

Definition 34. Given a group G acting on a topological space Y , define Y/G to be the quotientspace Y/ ∼, where y ∼ y′ if and only if y = g · y′ for some g ∈ G. The points of Y/G are the orbitsG · y = {g · y | y ∈ G}, and Y/G is called the orbit space.

Theorem 14. Suppose G acts on Y properly discontinuously. Then the following hold.

• The quotient map p : Y → Y/G is a normal covering space.

• If Y is path-connected, then G is precisely the group of deck transformations on Y .

Proof. Omitted.

Now we explore a bunch of non-examples before exploring some examples.

Example 25. Consider an equilateral triangle T with group action swapping the vertices of Tin a cyclical manner (so that G = Z/3Z). Then this is not a properly discontinuous group actionon T because the centroid O6 of T remains fixed under the action g sending i to i+ 1 modulo 3,implying g ·O = e ·O.6ouch

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Example 26. Consider a tetrahedron with corners labeled 1 through 4, and let G = Z/4Z acton said tetrahedron by permuting the vertices cyclically. This is also not a properly discontinuousgroup action on the tetrahedron. This is because if g is the action which sends i → i + 1 (whereindeces are taken modulo 4), then g2 sends 1 ↔ 3, and so the midpoint of this segment is fixedunder g2.

Example 27. Consider S3 as the unit sphere in the quaternions H. The restriction to unitquaternions is a group. In particular, the subgroup

O8 = {±1,±i,±j,±k}

acts on S3 properly discontinuously.

Example 28. We can generalize the example above by considering S3 as an embedding of C2.Let p and q be relatively prime positive integers, and consider the action

(z1, z2) 7→ (e2πi/p · z1, e2πi/q · z2).

By letting the group G = Z/pZ×Z/qZ act on C2 under these cyclic actions, we obtain an example.These spaces are called Lens spaces, and the quotient is denoted by L(p, q).

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18 October 8

Before we begin, a fun fact: π1(SO(n)) ∼= Z2 for all n ≥ 3.

18.1 Some “Last” Words on Covering Spaces

During the past homework assignment, we constructed various covering spaces. An example ofthis was as follows.

Example 29. Let X = RP 2 ∨ RP 2. On the homework we showed that the below space is theuniversal cover X of X. Here the spheres alternate between covering the first copy of RP 2 and thesecond copy of RP 2.

V−2 V−1 V0 V1 V2

· · · · · ·

Figure 18: An infinite wedge sum of spheres.

What are all possible deck transformations of X? It turns out there are two: either translatingby an even number of spheres or flipping the entire figure in the center of some sphere. (The first isvalid because of the alternating parity of the spheres, while the second is valid because the coveringmap from S2 to RP 2 identifies antipodal points.) As a result, we obtain

π1(RP 2 ∨ RP 2) ∼= D∞ = 〈a, b | a2, b2〉 = Z2 ∗ Z2.

Note that π1(RP 2) = Z2, so we expect that the homotopies on either side don’t affect much. Inother words, the fact that the wedge product on topological spaces translates to the free producton the fundamental groups here isn’t too surprising.

18.2 Van Kampen’s Theorem

We now construct a second means of determining fundamental groups of spaces. Let X be a space,and write X =

⋃αAα, where

• each Aα is open and path-connected;

• there exists some point x0 ∈⋂αAα; and

• for all α 6= β, Aα ∩Aβ is path-connected.

We thus have the existence of inclusion maps jα : Aα ↪→ X and iα,β : Aα ∩ Aβ ↪→ Aα. Notefurther that these maps satisfy the following commutative diagram.

X

Aα Aβ

Aα ∩Aβ

jα jβ

iα,β iβ,α

By Lemma 4, we already know that π1(X) is a quotient of the free product ∗α π1(Aα), sincegiven such a covering we may always reduce to studying loops strictly within each Aα. Certainlywe must also impose the relations

(iα,β)∗(ω)(iβ,α)∗(ω)−1 for ω ∈ π1(Aα ∩Aβ); (18.1)

this is because given a loop γ in Aα ∩Aβ we may view γ as living in both Aα and Aβ . But is thisenough?

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Theorem 15 (Van Kampen). Let Φ be a surjective group homomorphism from ∗α π1(Aα) →π1(X). Suppose we also impose the restriction that each triple intersection Aα ∩Aβ ∩Aγ is path-connected. Then ker Φ is the normal subgroup N generated by the relations in equation 18.1, andhence

π1(X) ∼= ∗α π1(Aα)

N.

We won’t prove this in class (the proof is in Hatcher).

Example 30. We’ll now make the intuition of Example 29 precise. Suppose X =∨αXα, where

each Xα is path-connected.

Xα

Uα

Uβ

XβUγ

Xγ

Figure 19: A diagram depicting a wedge sum of path-connectedspaces as well as the Uα we’ll use later.

Suppose we also introduce the technical restriction that each base point x0 ∈ Xα has an openneighborhood Uα ⊆ Xα which deformation retracts to a point. Then we can take

Aα = Xα ∨( ∨β 6=α

Uβ

)

as our open covering of X.

Furthermore, note that for each α 6= β, the intersection Aα∩Aβ =∨γ /∈{α,β} Uγ is path-connected

and contractible, and so each relation from equation 18.1 is the trivial relation. We may now appealto Van Kampen to conclude π1(X) = ∗α π1(Xα).

Remark. Notice why we needed to thicken each of the Xα a bit to include neighborhoods surround-ing the basepoint - this is so our intersections are all path-connected.

Example 31. The following example shows that triple intersections are necessary. Let X denotethe space given below, and set Aα = X \ {a}, Aβ = X \ {b}, and Aγ = X \ {c}.

a b c

Figure 20: A counterexample for pairwise intersections.

Note that this space X satisfies all the given conditions except for the fact that Aα ∩ Aβ ∩Aγ = X \ {a, b, c} is not path-connected. If we could apply Van Kampen, then we would obtainπ1(X) ∼= Z ∗ Z ∗ Z. But in reality, π1(X) ∼= Z ∗ Z, because X is homomorphic to the CW complexconsisting of two loops.

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18.3 Knots

We’ll finish this lecture (and continue into the next one) with a digression into knot theory.

Definition 35. Here are some important definitions involving knots.

1. A knot is an embedding from S1 into R3.

2. A link is an embedding from⊔Nn=1 S1 into R3, where here N is a positive integer.

3. Two knots α1 and α2 are said to be equivalent if there is a map H : R3 × [0, 1] → R3 suchthat

• for all t ∈ [0, 1], the map x 7→ H(x, t) is a homeomorphism;

• H(x, 0) = x for all x ∈ R3;

• the image of α1 under H is α2, i.e. H(α1(S1), 1) = α2(S1).

An example of such a knot is the trefoil knot described below.

Figure 21: The trefoil knot.

One of the main questions in knot theory is to determine whether a given knot can be untied,i.e. whether it is equivalent to the unknot. How can we answer this question?

Let A be the unknot. Then in fact R3 \A ' S2∨S1 as shown below, so π1(R3 \A) ∼= Z. Thus, inorder to determine whether some other knot K cannot be untied, we can compute the fundamentalgroup of R3 \ K and see if it does not equal Z; since fundamental groups are preserved throughhomotopies, this tells us such an untying is not possible. We’ll do this when K is the trefoil knotnext time.

Figure 22: The complement of an unknot.

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19 October 10

Today is going to be all about examples of Van Kampen’s Theorem.

19.1 The Klein Bottle, Revisited

Example 32. We’ll compute the fundamental group of the Klein bottle K. Write K = A1 ∪ A2,where A1 is the boundary of the below square thickened a bit and A2 is the interior thickened abit.

A1

A2

a

a

b b

Figure 23: A parition of the Klein bottle.

Since A1∩A2 is path-connected, Van Kampen tell us that π1(K) is a quotient of π1(A1)∗π1(A2).But A1 is homeomorphic to S1 ∨ S1 from our previous example and A2 is simply-connected, so infact π1(K) is isomorphic to a quotient of 〈a, b〉.

To determine exactly which quotient this is, identify A1 ∩ A2 ' S1, so π1(A1 ∩ A2) ∼= Z. Inparticular, there exists some loop γ generating A1 ∩ A2. In A2 this loop is the trivial loop, whilein A1 its equal to abab−1. As a result, Van Kampen implies

π1(K) ∼= 〈a, b | abab−1 = e〉.

19.2 Torus Knots

Let’s get back to talking about knots. As a reminder, we only care about knots which are homotopicto those which are built by finitely many line segments. (Otherwise things get messy.)

Definition 36. For positive integers m and n, define the map Tm,n : S1 → S1×S1 via z 7→ (zn, zm).For this to be an embedding, we must have gcd(m,n) = 1. Now view this as living in R3 and putyour knot in there. We call such a knot a torus knot.

Example 33. Consider the knot K := T3,2, otherwise known as the trefoil. We will attempt tocompute π1(R3 \K) ∼= π1(S3 \K). Our reasons for considering the latter topological space stemfrom Example 8, where in particular the complement of a torus is also a torus.

Warning. The rest of this example relies heavily on geometric intuition, and thus is difficult totranscribe in lecture notes. I would be very interested in improvements to this section that don’tinvolve copy and pasting Hatcher.

Consider any cross section {x}×∂D2 of WLOG the first solid torus. The knot K intersects everysuch cross section three times at equally spaced points, as shown below. (These three points rotatecontinuously as the cross section varies.) Thus, as also shown in the diagram, the complement ofT3,2 in the first solid torus deformation retracts to a space X3 ' S1, as each of the three strandscan be pushed in and what’s left is a circle. In other words, π1(X3) ∼= Z. Similarly, the complementof K in the other torus deformation retracts to a space X2 ' S1.

We’ll now apply Van Kampen on the union X3∪X2, or more specifically open neighborhoods ofthese two spaces which deformation retract onto themselves. The intersection X3 ∩X2 is a circle

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Figure 24: Deformation retracting a torus knot.

and hence generated by a single loop γ. This γ is homotopic in X3 to a loop representing 3 timesa generator and in X2 to a loop representing 2 times a generator. Thus, by Van Kampen, we mayconclude that

π1(X1 ∪X2) = 〈a, b | a3 = b2〉.One can now check that this is not isomorphic to Z, and so the trefoil is not equivalent to theunknot.

Remark. By applying this to general torus knots, we see that π1(Tm,n) = 〈a, b | am = bn〉.

19.3 Gluing 2-Cells

We finish today with a final application of Van Kampen’s Theorem.

Theorem 16. Let X be a path-connected space. Attach 2-cells e2α to X via maps ϕα : S1 → X to

obtain a new space Y . Choose some basepoint x0, and paths γα from x0 to some ϕα(sα), wheresα ∈ S1 for all α. Then the map π1(X)→ π1(Y ) induced by inclusion is surjective and its kernelis the smallest subgroup containing all equivalence classes of paths [γαϕαγα].

Proof. Slightly enlarge the space Y by thickening the paths γα as shown below to form a new spaceZ, and let yα ∈ e2

α be arbitrary for each α.

Figure 25: Hatcher makes really good diagrams.

Observe now the two facts

Z ' Y and Z \⋃α

{yα} ' X.

In turn, cover Z by the two spaces A and B, where A = Z \X and B = Z \⋃α{yα}. Now A iscontractible and

B = Z \⋃α

{yα} ' X,

so Van Kampen implies π1(Y ) = π1(Z) is a quotient of π1(X) by generators of the form [γαϕαγα].

Corollary 3. Every group G is the fundamental group of some 2-dimensional CW complex XG.

Proof. Consider the wedge sum X :=∨α e

2α of loops, where each eα corresponds to a generator of

G. Now glue in disks along the relations of G; then the previous result implies the only changesto π1(X) ∼= ∗α Z are that the loops corresponding to the relations of G become trivial.

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19.4 Fundamental Groups of Orientable Surfaces

We end with a quick remark.

Remark. We can examine the fundamental groups of various orientable surfaces. Indeed, weactually know a bunch of these already: the fundamental group of the sphere M0 is the identity,the fundamental group of the torus M1 is 〈a, b | aba−1b−1〉, and the fundamental group of thetwo-holed torus M2 is, as shown below, 〈a, b, c, d | aba−1b−1cdc−1d−1〉.

Figure 26: A two-holed orientable surface.

In general,

π1(Mg) = 〈a1, . . . , ag, b1, . . . , bg | a1b1a−11 b−1

1 a2b2a−12 b−1

2 · · · 〉 = [a1, b1][a2, b2] · · · ,

where [a, b] denotes the commutator of a and b. One can check that the abelianizations of thesefundamental groups all have different rank, and so any two distinct orientable surfaces are nothomotopy equivalent.

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20 October 12

20.1 Links

We’ll now give some more examples of knots in action. Recall that we can tell knots apart bycomputing the fundamental groups of their complements.

Example 34. Let K1 denote the link consisting of two unlinked unknots and K2 denote thelink consisting of two interlinked unknots. I claim that K1 and K2 are not equivalent (i.e. theinterlinked knots are actually linked). Indeed, the complement of K1 is homotopy equivalent tothe wedge sum of two S2 ∨ S1s (i.e. the wedge sum of two complements of a single unknot), whilethe complement of K2 is homotopy equivalent to the wedge sum of S2 and a torus as shown below.Hence

π1(R3 \K1) ∼= Z ∗ Z 6∼= Z× Z ∼= π1(R3 \K2),

proving the inequivalence.

Figure 27: The complement of two unknots linked together.

Example 35. The Borromean Rings, which are shown below, have the property that any two ofthe three rings are not linked but all three together are.

Figure 28: The Borromean Rings.

20.2 Wirtinger Presentations

Now we give a general algorithm which allows us to compute the knot group of any (tame) knotK.

• Construct a knot diagram of K, which consists of drawing K in such a way that K is containedin a plane except for crossings, of which finitely many occur.

• Construct a deformation retraction of R3 \K in the following way: every portion of the knotK lying inside the plane (henceforth referred to as strands) is covered by a rectangle Ri,and wherever three strands intersect, a square S` covers the intersection so that the oppositeends join together. See the figure below.

• Apply Van Kampen to the resulting deformation retraction. In particular, each rectangle(which creates an extra cylinder) results in an extra copy of Z in the fundamental group,while each square S` induces the relation RiRjR

−1i = Rk, where Ri, Rk, and Rk are labeled

as in the figure.

Example 36. By this same reasoning, we can see that the trefoil T3,2 has fundamental group

π1(R3 \ T3,2) ∼= 〈x1, x2, x3 | x1x3x−11 = x2, x2x1x

−12 = x3, x3x2x

−13 = x3〉.

The fact that this is isomorphic to the group 〈a, b | a3 = b2〉 is very nontrivial.

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Figure 29: A visual representation of the Wirtinger construction.

20.3 Some Parting Examples

We’ll finish our foray into homotopy theory with a few examples.

Example 37. On a previous homework assignment, we examined the dunce hat topological space

D := {(x, y) ∈ (R+)2 : x+ y ≤ 1}/ ∼,

where ∼ is the equivalence relation which pairs (0, x) ∼ (x, 0) ∼ (x, 1−x) for all x ∈ [0, 1]. On thathomework assignment, we showed that D \ {[(0, 0)]} deformation retracted onto a one-dimensionalCW complex by explicitly defining the bijection. Now, we’ll show that D itself is simply-connectedby appealing to Van Kampen.

Let a denote the common loop around all three sides of the triangle D. As we did with the Kleinbottle, partition D into its boundary and its interior (thickened of course). Then the boundarypart is generated by a single loop a, and the common relation this time around is aa−1a = a = e,so

π1(D) ∼= 〈a | a〉 = id.

Example 38. Start with S1, and let X denote the topological space discovered by gluing in a2-cell along its boundary via the map z 7→ z3. Then π1(X) ∼= Z/3Z.

A simple way to see this is to go around S1 three times and lift to the universal cover; this liftis a loop in the covering space and is hence trivial.

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21 October 15

21.1 Building the Intuition for Homology

Today we move on to a new topic: homology. This explores algebraic topology from a morealgebraic perspective.

Warning. This section is very hand-wavy, in the sense that we will not define anything preciselyor prove any results rigorously (that will start on Wednesday).

We start by asking the following question: what are all homotopy types of one-dimensional CWcomplexes?

Example 39. Here are four graphs and their associated homology types (i.e. what “common”topological spaces they are homotopic to).

Figure 30: Some graphs and their associated homology types.

We notice that by counting the number of vertices and edges in the graphs of each of the previousexamples, we can determine the number of cycles in the graphs and hence the homotopy types.(Indeed, the placement of the cycles does not matter; do you see why?)

Motivated by this, we are next led to ask whether the same is true of 2-dimensional CW com-plexes. In other words, are all 2-dimensional CW complexes classified up to homotopy by thenumber of faces of each dimension?

But this turns out to be false! Recall our representations of the torus, Klein bottle, and realprojective plane as the quotient space of a square with opposite edges glued together in variousways. In this case, all three spaces have the same number of vertices, edges, and faces, but thedifferent ways we can glue these edges yield vastly different spaces. So counting faces is not enough;we need to actually keep track of the gluing information.

We’ll first explore this query within the one-dimensional framework. Let G be a graph withvertex set V and edge set E. Define7 the function ∂ : RE → RV via

∂((vi, vj)) = vj − vi whenever i < j.

This is equivalent to orienting all the edges in G to point from the smaller vertex to the largervertex. Now note that

∂((v1, v2)) = v2 − v1, ∂((v2, v3)) = v2 − v3, and ∂((v1, v3)) = v3 − v1.

Thus ∂((v1, v2) + (v2, v3) − (v1, v3)) = 0, which means that (v1, v2) + (v2, v3) − (v1, v3) is in thekernel of ∂. This logic extends to any cycle in G, so the kernel of ∂ is spanned by the cycles ofG. Hence g := dim(ker ∂) is equal to the number of cycles in G, which coincides well with the factthat G ' ∨gi=1 S1.

Now we’ll extend this to the two-dimensional case. Consider the case where we have a trianglewith vertices v1, v2, and v3. Define ∂1 : RE → RV via ∂1((vi, vj)) = vj − vi. Note that as before,the kernel of ϕ1 is spanned by (v1, v2) + (v2, v3) − (v1, v3). The new portion now is to introducea second linear transformation ∂2 : RT → RE (where T is the collection of triangles in our space)defined by

∂2((v1, v2, v3)) = (v1, v2) + (v2, v3)− (v1, v3).

Example 40. Consider the “graph” shown below, with triangles t1 and t2 filled in.

7Here RV denotes the vector space over R whose basis elements are the vertices of G, and RE is defined similarly.

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v1 v2

v3v4

t1

t2

Figure 31: A square decomposed into two triangles.

As before, we have our linear maps ∂1 : RE → RV and ∂2 : RT → RE . I claim that

ker ∂1 = span((v1, v2, v3), (v1, v3, v4)) = im ∂2.

Indeed, the second equality follows from the fact that (v1, v2, v3) and (v1, v3, v4) are the basiselements for RT , while the first equality comes from the fact that cycles are annihilated under ∂1.Intuitively, this makes sense: the gluing of the vertices and edges creates two new cycles in ourtopological space, but adding in the triangles kills both of them.

Example 41. Now consider the following decomposition of the Klein bottle with edges labeled.Let ∂1 and ∂2 be as before.

v1 v1

v1v1

t1

t2

e1

e2e2

e1

e3

Figure 32: The Klein bottle decomposed into two triangles.

Note that∂1(e1) = ∂1(e2) = ∂1(e3) = v1 − v1 = 0,

so ker ∂1 is spanned by e1, e2, and e3. But now

∂2(t1) = e1 + e2 − e3 and ∂2(t2) = e3 + e1 − e2.

Thus, over R, the image of ϕ2 is spanned by e1 and e2 − e3, so the dimension of im ∂2 is 2.

This presents a problem. According to this, the gluing of vertices and edges produces three newcycles (namely e1, e2, and e3), but the gluing of faces kills two of them, so there is one nontrivialcycle in the Klein bottle. But in Example 15, we showed that

π1(K) ∼= 〈a, b | abab−1〉 ∼= Z ∗2Z Z,

which is a group with two generators, not one.

But this is resolved if we work over a Z-vector space rather than an R-vector space. In this case,ker ∂1

∼= Z3 “as before”, but now instead of quotienting out by e1, only 2e1 is trivial. This implies

ker ∂1

im ∂2

∼= Z⊕ Z2,

which(a) solves our generator mismatch problem and (b) happens to be the abelianization of Z∗2ZZ.

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v1 v1

v1v1

t1

t2

e1

e2e2

e1

e3

Figure 33: The torus decomposed into two triangles.

Example 42. For a final example, consider the torus S1 × S1, whose decomposition is shownbelow.

As before, ∂1 ≡ 0. But now

∂2(t1) = ∂2(t2) = e1 + e2 − e3,

so im ∂2 is a one-dimensional vector space. As a result, we obtain

ker ∂1

im ∂2

∼={R2 over an R-vector space,

Z2 ∼= π1(S1 × S1) over a Z-vector space.

We have the same coincidence as before.

Perhaps these crazy ideas might not be so wacky after all. Next time, we’ll explore extendingthis to three-dimensional tetrahedra before formalizing everything done today.

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22 October 17

We explore tetrahedra before rigorously defining what homology is.

Example 43. Consider a tetrahedron with vertices v1 through v4 with each vi pointing to vj iffi < j. Using this, we define the equalities

∂[v1, v2, v3] = [v1, v2] + [v2, v3]− [v1, v3], (22.1)

∂[v1, v2, v4] = [v1, v2] + [v2, v4]− [v1, v4], (22.2)

∂[v1, v3, v4] = [v1, v3] + [v3, v4]− [v1, v4], (22.3)

∂[v2, v3, v4] = [v2, v3] + [v3, v4]− [v2, v4]. (22.4)

Now notice after staring at this for a bit that adding 22.1 and 22.2 and subtracting 22.3 and 22.4yields a 0 on the right hand side, and so it is sensible to define

∂[v1, v2, v3, v4] = [v1, v2, v3] + [v1, v3, v4]− [v1, v2, v4]− [v2, v3, v4].

Remark that the positive terms don’t contain the even labeled vertices v2 and v4 while the negativeterms don’t contain the odd labeled vertices v1 and v3. This is not a coincidence.

22.1 Simplicial Homology

We first define a notion of homology on slightly simpler structures called ∆-complexes. This typeof homology has the advantage of being more computable than singular homology, but it’s notobvious from the start that this can be extended to deal with other types of spaces in a nice andwell-defined way.

Definition 37. Some important definitions for simplicial homology.

1. For n ∈ N, the n-simplex ∆n is the convex hull of n + 1 points in general position in Rn.Here, general position means that no three points are collinear, no four points are coplanar,and so on.

2. A ∆-complex on a space X is a collection of maps σα : ∆nα → X such that the followinghold.

a) For all α, the restriction of σα to (∆nα)◦ is injective, and furthermore each point of Xis in exactly one of these images.

b) Any map σα restricted to a face of an n-simplex is one of the maps σβ : ∆n−1 → X in anorder-preserving way. (This rules out, for example, the triangle with a cyclic orientationof the edges.)

c) Any subset A ⊆ X is open if and only if σ−1α (A) is open in ∆n.

3. We define

∆n(X) =

{∑j≥1

njσnj

∣∣∣nj ∈ Z

},

where each σnj is an n-simplex in our ∆-complex structure X.

4. Set ∂n : ∆n(X)→ ∆n−1(X) such that for all n-simplices σ,

∂n(σ) =∑i≥1

(−1)iσ|[v0,...,vi,...,vn].

Here the notation [v0, . . . , vi, . . . , vn] implies that we skip the vertex vi but otherwise keepall other vertices.

We proceed with an important lemma.

Lemma 9. For each n, the composition

∆n(X)∂n−→ ∆n−1(X)

∂n−1−→ ∆n−2(X)

is zero.

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Proof. For any basis element σ of ∆n(X), write

∂n−1∂n(σ) =∑j<i

(−1)i(−1)jσ|[v0,...,vj ,...,vi,...,vn]

+∑j<i

(−1)i(−1)j+1σ|[v0,...,vi,...,vj ,...,vn].

Here each term cancels out, and so the total sum is zero.

This lemma suggests viewing homology theory in an algebraic context. Now for more definitions.

Definition 38. Let Cn = ∆n(X) for all n ∈ N.

1. A sequence of group homomorphisms

· · · −→ Cn∂n−→ Cn−1

∂n−1−→ Cn−2∂n−2−→ · · · −→ C0 −→ 0

is a chain complex if ∂n−1 ◦ ∂n = 0 for all n.

2. The homology of a chain complex is the collection of groups {Hn}n≥0 defined via

Hn = ker ∂n/ im ∂n+1 for all n ∈ N.

3. The elements of ker ∂n are called n-cycles. The elements of im ∂n+1 are called boundaries.The elements of Hn are called homology classes.

We now proceed with an example.

Example 44. Remove one triangle from a ∆-complex structure on the torus. The boundary ofthis triangle is certainly a 1-cycle, but the complement in the torus is itself a boundary (after all,there’s cancellation everywhere else!), so it is trivial in H1.

As alluded to before, we need to make sure that this theory of simplicial homology does notdepend on the precise homology complex endowed on our topological space. The next severallectures will be devoted to resolving this hole.

52


23 October 19

No class due to mid-semester break.

24 October 22

Reminder: we have a quiz on Wednesday.

24.1 Singular Homology

Today we will start to show that homology is homotopy invariant. It turns out the conventionallyeasiest way to do this is to define a new homology theory which does not make reference to the∆-complex and show that this is actually the exact same theory. This theory is called singularhomology and is defined for any topological space.

Instead of looking at a particular ∆-complex on a topological space, we’ll instead look at thecollection of continuous maps σ : ∆n → X (called singular n-simplices). Then define

Cn(X) =

n∑j=1

njσj

∣∣∣nj ∈ Z, σj : ∆n → X

to be the collection of linear combinations of these continuous maps. These are called singularn-chains. Furthermore, define ∂ : Cn(x)→ Cn−1(x) as before, meaning on a basis element σ,

∂(σ) =∑j

(−1)jσ|[v0,...,vj ,...,vn].

This is a chain complex of abelian groups

· · · −→ Cn∂−→ Cn−1

∂−→ Cn−2 −→ · · · .

Then the collection of groups Hn(X) = ker ∂n/ im ∂n+1 is called the singular homology of X.

Remark. One may ask why we’re defining singular homologies in this way as opposed to looking atembeddings of n-simplexes into X instead. To see why, let’s think about S1. Suppose we definedthe same theory except our maps are embeddings now. For S1, ker ∂1 consists of 1-cycles, i.e. finitesubdivisions of S1. But any two of these would be different if we defined these wrt embeddings(i.e. we can’t embed a 2-simplex into S1).

Example 45. Consider X = {∗}; we’ll compute Hn(X). The objects are maps from the n-simplexinto X. But for every n, there’s only one such map: the map σn : ∆n → X via x 7→ ∗. As a result,the groups Cn(X) are free abelian groups generated by precisely one element, and so Cn(X) ∼= Zfor each X. This yields the chain complex

· · · ∂3−→ Z ∂2−→ Z ∂1−→ Z ∂0−→ Z −→ 0.

What are the homology groups? Well, note that the map ∂n is defined on a single basis element σvia

∂n(σ) =∑

1≤j≤n

(−1)jσn|[vj ] =∑

1≤j≤n

(−1)jσn−1 =

{0 if n is even,

σn−1 if n is odd.

This means that the maps alternate between being the zero map and being the “multiplication by1” map. From here one can case on the parity of n to deduce that H0(X) ∼= Z and Hn(X) = 0 forn ≥ 1.

For now, this is the only example we can compute, because these groups are just so large.

24.2 Connecting our Two Homologies

We have two theories: one in which we can compute but we don’t know is homeomorphism invariantand one in which we can’t compute but is trivially homeomorphism invariant. We’d like to showthat these two forms of homology are in fact the same.

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Definition 39. We define the reduced homology Hn(X) to be the homology of the chain complex

·Cn(X) −→ · · · −→ C0(X)ε−→ Z −→ 0,

where ε : C0(X)→ Z is the map which satisfies

ε

(k∑i=1

niσi

)=

k∑i=1

ni.

In other words, Hn(X) ∼= Hn(X) for n > 0, while H0(X)⊕ Z ∼= H0(X).

The advantage of reduced homology is that the point now has trivial homology. This generalizes.

Proposition 20. If X is non-empty and path-connected, then H0(X) = 0.

Proof. Remember H0(X) = ker ε/ im ∂1. We claim they’re the same group. For one direction, ifσ : ∆1 → X is a singular 1-simplex, then ∂(σ) = σ|[v1] − σ|[v0], so ε(∂(σ)) = 1 + (−1) = 0. For theother direciton, let

∑niσi be in ker ε. Note that each of the σi is a zero-simplex. Let τi be a path

from some basepoint x0 to σi. Thus

∂(∑

niτi

)=∑

niσi −∑

nix0 =∑

niσi

since∑ni = 0 due to the kernel condition.

We’ll now show that reduced homology is homotopy invariant. Let f : X → Y be a map. Thenf induces a homomorphism f# : Cn(X)→ Cn(Y ) via

f#(σ) = f ◦ σ.Proposition 21. The following diagram commutes.

· · · Cn+1(X) Cn(X) Cn−1(X) · · ·

· · · Cn+1(Y ) Cn(Y ) Cn−1(Y ) · · ·

∂ ∂

f#

∂

f#

∂

f#

∂ ∂ ∂ ∂

Proof. For σ ∈ Cn(X), write

f#∂(σ) = f#

(∑j

(−1)jσ|[xj ])

=∑j

(−1)jf ◦ σ|[xj ] = ∂f#(σ).

So each loop in the diagram commutes as desired.

We want a few things from this. First, we want f# to induce a map on homologies. We alsowant it to be functorial (i.e. passing from structures to homomorphisms between structures).

Definition 40. A map ϕn : Cn(X)→ Cn(Y ) is called a chain map if ϕ∂ = ∂ϕ.

Proposition 22. Let ϕn : Cn(X)→ Cn(Y ) be a chain map.

1. The image of a cycle under ϕn is also a cycle.

2. The image of a boundary under ϕn is also a boundary.

Proof. For the first part, suppose α ∈ Cn(X) with ∂α = 0 (i.e. α is a cycle in Cn(X). Then

0 = ϕ(0) = ϕ(∂α) = ∂(ϕα),

so ϕα is also a cycle.

The second part is even easier: ϕ(∂β) = ∂ϕ(β), so the image of the boundary ∂β is also aboundary.

Corollary 4. Any chain map ϕ : Cn(X) → Cn(Y ) induces a map on homology ϕ∗ : Hn(X) →Hn(Y ). Proof: we checked for kernels and images, so yay.

Fact: (fg)∗ = f∗g∗ because the composition of maps is associative. Fact: id∗ = id.

Corollary 5. If two maps f, g : X → Y are homotopic, then they induce the same homomorphismf∗ = g∗ : Hn(X)→ Hn(Y ) on homology. In particular, if X ' Y , then Hn(X) ∼= Hn(Y ) for all n.

We’ll do this next week, it’s not too difficult.

54


25 October 24

Second quiz on both covering spaces and Van Kampen.

26 October 26

No class due to the presidential inauguration.

27 October 29

We first prove the result from last time.

Proof. For every cycle α ∈ ker ∂n ⊆ Cn(X), we have to show that f#(α) and g#(α) differ by aboundary.

We first show that for any n-simplex ∆n, we can split the space ∆n× [0, 1] into (n+1)-simplices.The bottom n-simplex is [v0, . . . , vn], and the top n-simplex is [w0, . . . , wn]. Now consider thecollection of (n+ 1)-simplices given by

[v0, . . . , vn, wn], [v0, . . . , vn−1, wn−1, wn], · · · [v0, w0, . . . , wn−1, wn].

This works.

Define the prism operator P : Cn(X)→ Cn+1(Y ) given by

σ 7→∑i

(−1)iF ◦ (σ × id)|[v0,...,vi,wi,...,wn],

where F : X × [0, 1]→ Y is the homotopy from f to g. I claim that ∂P = g#− f#−P∂. To provethis, write

∂P (σ) =∑j≤i

(−1)i(−1)jF ◦ (σ × id)|[v0,...,vj ,...,vi,wi,...,wn]

+∑j≥i

(−1)i(−1)j+1F ◦ (σ × id)|[v0,...,vi,wi,...,wj ,...,wn].

The i = j terms telescope and thus cancel except for the first and the last, i.e. the terms

F ◦ (σ × id)|[v0,w0,...,wn] = g ◦ σ = g#(σ)

and−F ◦ (σ × id)|[v0,...,vn,wn] = −f ◦ σ = −f#(σ).

Furthermore, the i 6= j terms all sum to −P∂(σ) since

P∂(σ) =∑i<j

(−1)i(−1)jF ◦ (σ ◦ id)|[v0,...,vi,wi,...,wj ,...,wn]

+∑i>j

(−1)i−1(−1)jF ◦ (σ ◦ id)|[v0,...,vj ,...,vi,wi,...,wn],

and all the signs are flipped. So g#− f# = ∂P +P∂, and so if α ∈ Cn(X) is a cycle (i.e. ∂α = 0),then

g#(α)− f#(α) = ∂P (α) + P∂(α) = ∂P (α).

These indeed differ by only a boundary, so [f#(α)] = [g#(α)].

This leads to a definition.

Definition 41. A map P : Cn(X)→ Cn+1(Y ) for all n is called a chain homotopy from f# to g#

iff# − g# = ∂P + P∂.

In this case we say that f# and g# are chain-homotopic.

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We have thus shown that chain homotopic maps induce the same homomorphism on homology.Unfortunately, this homology is somewhat of a monster, and so right now we can only compute itin the case of a point (and thus in any contractible space). In order to proceed further, we haveto understand the relation between Hn(A) and Hn(X) if A ⊆ X.

Definition 42. A (perhaps infinite) sequence of homomorphisms

· · · −→ An+1αn+1−→ An

αn−→ An−1αn−1−→ · · ·

is called exact if kerαj = imαj+1 for all j. In other words, the homology of a chain complexmeasures how far from exact the chain complex is.

Definition 43. An exact sequence

0 −→ A −→ B −→ C −→ 0

is called a short exact sequence. In this case, the map from A to B is injective, and the map fromB to C is surjective.

Definition 44. If A, B, C are abelian groups, then

0 −→ A −→ Bg−→ C −→ 0

is split if g has a section. If it’s split, then B ∼= A⊕ C.

Theorem 17. Let X be a space and suppose A ⊆ X is nonempty, closed, and a deformationretraction of some neighborhood in X. Then there is an exact sequence

· · · −→ Hn(A)i∗−→ Hn(X)

j∗−→ Hn(X/A)∂−→ Hn−1(A) −→ · · · −→ H0(X/A).

Here i∗ is the inclusion map and j∗ is the quotient map.

Corollary 6. We have Hn(Sn) ∼= Z and Hi(Sn) = 0 for i 6= n.

Proof. For n > 0, take X = Dn, A = Sn−1 = ∂X, so that X/A ∼= Sn. Note that Hn(Dn) = 0 sinceDn is contractible. So using the previous theorem, we know the existence of an exact sequence

0 = Hi(Dn) −→ Hi(Sn) −→ Hi−1(Sn−1) −→ Hi−1(Dn) = 0.

for every i ≥ 0 and n ≥ 0. Now use induction.

It remains to prove Theorem 17. We’ll do this on Wednesday.

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28 October 31

We begin with a corollary to Corollary 6.

Corollary 7. For all n ≥ 2, ∂Dn is not a retract of Dn.

Proof. Suppose for the sake of contradiction that such a retraction f : ∂Dn → Dn existed. Byletting i : ∂Dn → Dn be the canonical inclusion map, we see that the following diagram commutes.

∂Dn ∂Dn

Dn

if

As a result, we obtain an analogous commutative diagram on homologies. But Hn(∂Dn) ∼=Hn(Sn) ∼= Z but Hn(Dn) = 0, contradiction.

Example 46. Let A ⊆ S2 be a two-point space. We’ll compute the homology of S2/A. Recall byTheorem 17 the sequence

H2(A) −→ H2(S2) −→ H2(S2/A)

−→ H1(A) −→ H1(S2) −→ H1(S2/A)

−→ H0(A) −→ H0(S2) −→ H0(S2/A) −→ 0

is exact. We now proceed in steps.

• By Corollary 6, we know that H2(S2) ∼= Z and H1(S2) = H0(S2) = 0.

• By the definition of reduced homology, H2(A) = H1(A) = 0 and H0(A) ∼= Z.

• Recall that (after relabeling to accommodate the previous steps) the subsequence

0 −→ Z ϕ−→ H2(S2/A) −→ 0

is exact. In particular, by looking at the left three terms we see that ϕ is injective and bylooking at the right three terms we see that ϕ is surjective. Thus ϕ is actually a bijection,and H2(S2/A) ∼= Z.

• Similarly, by looking at the exact subsequence

0 −→ H1(S2/A) −→ Z −→ 0

we see that H1(S2/A) ∼= Z.

• Finally, H0(S2/A) is located between two 0s in the chain and is thus the trivial group.

28.1 Relative Homology Groups

We’ll look at these instead of quotient spaces a lot of the time because theyre nicer. We’ll usethem to prove the theorem.

Definition 45. Let A ⊆ X.

1. Define the collection of groups {Cn(X,A)}n≥1 via

Cn(X,A) := Cn(X)/Cn(A).

2. Notice that ∂ : Cn(X) → Cn−1(X) sends Cn(A) → Cn−1(A), so it induces a map ∂ :Cn(X,A) → Cn−1(X,A). As before the composition of two yields zero, so we get a chaincomplex

−→ Cn(X,A) −→ Cn−1(X,A) −→ Cn−2(X,A) −→ · · ·The homology of this chain complex is called the relative homology Hn(X,A).

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X

A

Figure 34: An example of a 1-chain which belongs in H1(X,A).

Geometrically, elements of Hn(X,A) are relative cycles, i.e. elements α such that ∂α is supportedin A.

Note that this yields the following commutative diagram.

0 Cn(A) Cn(X) Cn(X,A) 0

0 Cn−1(A) Cn−1(X) Cn−1(X,A) 0

i

∂

j

∂ ∂

i j

This yields a short exact sequence of chain complexes. The key idea is that this allows us to geta long exact sequence of homology groups via snaking around the diagram. To see this, we rotatethe diagram above 90◦. Here An = Cn(A), Bn = Cn(X), and Cn = Cn(X,A). In order to proveTheorem 17, it suffices to construct a map ∂ from Hn(C) to Hn−1(A).

0 0 0

· · · An+1 An An−1 · · ·

· · · Bn+1 Bn Bn−1 · · ·

· · · Cn+1 Cn Cn−1 · · ·

0 0 0

i

∂

i

∂

i

∂

j

∂

j

∂

j

∂

∂ ∂ ∂

Let c ∈ Cn be a cycle. Since j is surjective, there exists b ∈ Bn such that c = j(b). Now pushforward to ∂b ∈ Bn−1. I claim ∂b ∈ ker j: indeed, j(∂b) = ∂j(b) = 0, so ∂b ∈ im i. Thus thereexists a ∈ An−1 for which i(a) = ∂b. Now define ∂[c] = [a].

Proposition 23. This is well-defined.

Proof. We proceed in steps. Each step details some possible portion of the previous proof in whichwell-definedness might be a problem.

• Note that a is uniquely determined by ∂b since i is injective.

• If we chose a different b′ instead of b with j(b′) = c = j(b), then b − b′ ∈ ker j = im i, sob′ − b = i(a′) for some a′ ∈ An ⇒ b′ = b+ i(a′). So b→ b′ changes a→ a+ ∂a′ because

i(a+ ∂a′) = i(a) + i∂a′ = ∂b+ ∂i(a′) = ∂(b+ ia′) = ∂b′.

These only differ by a boundary and are hence in the same equivalence class.

• Now suppose we have a different choice of c within its homology class, i.e. c 7→ c + ∂c′ forsome c′ ∈ Cn+1. Find b′ ∈ Bn+1 such that c′ = j(b′), so

c+ ∂c′ = c+ ∂j(b′) = c+ j(∂b′) = j(b+ ∂b′).

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so we replace b by b + ∂b′. Now when we push b forward, the ∂ term becomes ∂∂b′ = 0, so∂(b+ ∂b′) = ∂b.

We are now led to the final piece of the puzzle.

Theorem 18. The sequence

· · · −→ Hn(A)i∗−→ Hn(B)

j∗−→ Hn(C)∂−→ Hn−1(A)

i∗−→ Hn−1(B)j∗−→ · · ·

is exact.

Proof. First remark that im i∗ ⊆ ker j∗ since ji = 0⇒ j∗i∗ = 0.

Now we’ll prove ker j∗ ⊆ im i∗. Let b ∈ Bn with ∂b = 0 and j(b) = ∂c′ for some c′ ∈ Cn+1. (Theformer condition arises from the fact that Hn(B) is a quotient of ker ∂, while the latter is due toour assumption that b ∈ ker j∗.) Since j is surjective, c′ = j(b′) for some b′ ∈ Bn+1. Now

j(b− ∂b′) = j(b)− j(∂b′) = j(b)− ∂j(b′) = 0

since ∂j(b′) = ∂c′ = j(b). Hence b− ∂b′ ∈ ker j = im i, so b− ∂b′ = i(a) for some a ∈ An. But now

i(∂a) = ∂i(a) = ∂(b− ∂b′) = ∂b = 0,

and since i is injective we deduce that a is a cycle.

The proofs for the relations involving ∂ are similar and are left as an exercise.

From this, we obtain the following result.

Theorem 19 (Long Exact Sequence of a pair). Let X be a space and A ⊆ X. Then the sequence

· · · −→ Hn(A)i∗−→ Hn(X)

j∗−→ Hn(X,A)∂−→ Hn−1(A) −→ · · · −→ H0(X,A) −→ 0

is exact.

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29 November 2

Warning. This lecture is very technical and might be hard to read; for a more thorough exposition,read pages 119-124 in Hatcher.

We’ll start with the following observation: if we have two maps f, g : (X,A)→ (Y,B) which arehomotopic through maps of pairs, then f∗ = g∗ : Hn(X,A)→ Hn(Y,B). Indeed, for A = B = ∅,we showed this by constructing the so-called prism operator Cn(X)→ Cn+1(Y ). But this operatoralso maps Cn(A) to Cn+1(B), so the same proof works.

29.1 Excision

The key point of today’s lecture is the following theorem.

Theorem 20. Let Z ⊆ A ⊆ X such that Z ⊆ A◦. Then

Hn(X,A) ∼= Hn(X \ Z,A \ Z).

Remark. The condition Z ⊆ A◦ is equivalent to saying that the interiors of A and B := X \ Zcover X, and then the theorem above rewrites as

Hn(X,A) ∼= Hn(B,A ∩B).

So in some sense this is analogous to Van Kampen’s theorem.

In order to prove this, we’ll first need the following definition.

Definition 46. Let X be a space and U = {Uj}j be a collection of sets such that X = ∪jU◦j .

Denote by CUn (X) ⊆ Cn(X) to be the collection of linear combinations∑niσi, where each σi has

its image in one set Uj . This gives a chain complex

· · · −→ CUn (X)∂−→ CUn−1(X)

∂−→ CUn−2(X)∂−→ · · · ,

with associated homology groups denoted by HUn (X).

Proposition 24. The map i : CUn (X) ↪→ Cn(X) is a chain homotopy equivalence, i.e. there is amap ρ : Cn(X)→ CUn (X) such that iρ and ρi are chain homotopic to the identity. Thus, i inducesisomorphisms HUn (X) ∼= Hn(X) for all n.

Proof. This proof is the proof given in class with much additional exposition taken from Hatcher.

The main idea is to take barycentric subdivisions of simplices. Here a barycentric subdivi-sion of ∆n is such that vertices are in 1-1-correspondence with faces of ∆n and faces are in 1-1correspondence with chains (total orders) of faces.

Figure 35: Diagrams showing barycentric subdivisions, courtesy of Hatcher.

One can check that the diameter of faces in a bary subdivision is nn+1 times that of diam(∆n).

Our first goal is to extend this barycentric operator to a map S : Cn(X)→ Cn(X) on n-chains.We’ll adopt the notation

b([v0, . . . , vn]) = [b, v0, . . . , vn].

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Now define S inductively via Sσ = b(S∂σ); in other words, to define S on a basis element σ ofCn(X), take its boundary ∂σ (which has dimension n− 1), apply S to it, and finish with the coneoperation b.

Observe that∂b([v0, . . . , vn]) = [v0, . . . , vn]− b∂[v0, . . . , vn];

this follows by writing out the boundary operator and casing on whether the excluded vertex is b.With this in mind, we may prove that ∂S = S∂ by induction. Indeed, write

∂S(σ) = ∂b(S∂σ) = S∂σ − b∂(S∂σ)(IH)= S∂σ − bS(∂∂σ) = S∂σ,

where in the last equality we use the fact that ∂∂ = 0.

Now we show S is chain-homotopic to the identity, i.e. there exists T : Cn(Y )→ Cn+1(Y ) with∂T + T∂ = id−S. We’ll define T inductively on dimension, via

Tσ = b(σ − T∂σ).

We can check that this map is indeed the correct map via induction, because

∂Tσ = ∂b(σ − T∂σ)

= σ − T∂σ − b∂(σ − T∂σ)

= σ − T∂σ − b(∂σ − ∂T (∂σ))

(IH)= σ − T∂σ − b(S(∂σ) + T∂(∂σ))

= σ − T∂σ − Sσ.

With this in hand, we may extend even further to iterated barycentric subdivisions. For allm ≥ 1, Sm (i.e. S with iterated barycentric subdivisions m times) is chain homotopic to theidentity map via the map

Dm =

m−1∑i=0

TSi.

Indeed, checking this is a simple calculation, as

∂Dm +Dm∂ =

m−1∑i=0

(∂TSi + TSi∂) =

m−1∑i=0

∂TSi + T∂Si

=

m−1∑i=0

(∂T + T∂)Si =

m−1∑i=0

(id−S)Si =

m∑i=0

(Si − Si+1) = id−Sm.

Here is where we use the fact that the diameters of our subdivisions become smaller and smaller.In particular, for m large, the diameters of the simplices of Sm(∆n) will be less than a Lebesguenumber8 of the cover of ∆n by the open sets σ−1(U◦j ). As a result, for each σ : ∆n → X there is

m = m(σ) such that Sm(σ) is in CUn (X), which means we may define D : Cn(X)→ Cn+1(X) via

Dσ = Dm(σ)σ.

After all this work, we are almost done. Recall that we originally set out to find ρ : Cn(X) →Cn(X) with image in CUn (X) such that ∂D+D∂ = id−ρ. The easiest way to do this is to considerthis equation as defining ρ and set ρ = id−∂D −D∂. Note that

∂ρ(σ) = ∂σ − ∂2Dσ − ∂D∂σ = ∂σ − ∂D∂σand

ρ(∂σ) = ∂ρ− ρDρσ −D∂2ρ = ∂σ − ∂D∂σ,so indeed ρ is a chain map. Finally, to check that ρ takes Cn(X) to CUn (X), we compute ρ(σ) moreexplicitly and write

ρ(σ) = σ − ∂Dσ −D(∂σ)

= σ − ∂Dm(σ)σ −D(∂σ)

= Sm(σ)σ +Dm(σ)(∂σ)−D(∂σ),

8See here for a detailed explanation of the Lebesgue Number Lemma.

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http://www.cs.bsu.edu/~hfischer/math441/lebesgue.pdf


where the last equality is due to ∂Dm + Dm∂ = id−Sm. Remark that Sm(σ)σ ∈ CUn (X) bydefinition of m(σ). Furthermore, Dm(σ)(∂σ) − D(∂σ) consists of only the terms TSi(σj) withi ≥ m(σj) (because the subtraction cancels out all other terms!), and these terms lie in CUn (X)since T takes CUn−1(X) to CUn (X).

All in all, we have discovered the existence of a map D such that ∂D + D∂ = id−iρ (wherei : CUn (X) ↪→ Cn(X) is the inclusion map). Combining this with ∂i = id shows that ρ is a chainhomotopy inverse for i, and so we are done.

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30 November 5

With the above proposition proven, we may prove Theorem 20.

Proof. Set U = {A,B}. We showed last time the existence of a map D such that ∂D+D∂ = id−iρ.Both of these maps take chains in A to chains in A, and so the inclusion map

CUn (X)

Cn(A)↪→ Cn(X)

Cn(A)

induces an isomorphism on homology. Furthermore, the map

Cn(B)

Cn(A ∩B)→ CUn (X)

Cn(A)

induced by inclusion also induces an isomorphism on homology, since both quotient groups aregreely generated by the singular n-simplices in B which do not lie in A. Composing these isomor-phisms yields Hn(B,A ∩B) ∼= Hn(X,A) as desired.

30.1 Good Pairs

We first start with a definition.

Definition 47. Let A and X be topological spaces with A 6= ∅. The pair (X,A) is a good pair ifA has a neighborhood in X that deformation retracts to A.

Theorem 21. For good pairs (X,A), the quotient map g : (X,A)→ (X/A,A/A) induces isomor-phisms

g∗ : Hn(X,A)→ Hn(X/A,A/A).

To prove this, we will need a lemma.

Lemma 10. For any n ∈ N, Hn(X, {∗}) ∼= Hn(X).

Proof. Consider the long exact sequence

Hn({∗}) −→ Hn(X) −→ Hn(X, {∗}) −→ Hn−1({∗}) −→ · · · .

Recall that the nth homology group of a point is Z if n = 0 and the trivial group otherwise. Thus,for n > 1, we easily obtain the desired isomorphism. It suffices to check the result for n = 1. Toprove this, consider the long exact sequence

H1({∗}) ϕ1−→ H1(X)ϕ2−→ H1(X, {∗}) ϕ3−→ H0({∗}) ϕ4−→ H0(X)

ϕ5−→ H0(X, {∗}) ϕ6−→ 0.

Now we proceed in steps.

• By our previous work, H1({∗}) = 0 and H0({∗}) ∼= Z.

• Consider the map ϕ3. Via the analysis from October 31 we know that ϕ3 takes in a 1-cycle αwith boundary contained entirely in {∗} and outputs precisely the boundary of α. Here, theboundary of any such 1-cycle is {∗} − {∗} = 0 (i.e. the two endpoints of the 1-cycle coincidebut have opposite orientations). Thus the map ϕ3 is the trivial map. As a result,

imϕ2 = kerϕ3 = H1(X, {∗}),

so ϕ2 is surjective.

• Similarly, because ϕ1 is the zero map, kerϕ2 = imϕ1 = 0, so ϕ2 is injective. Thus in fact ϕ2

is an isomorphism, and H1(X, {∗}) ∼= H1(X) ∼= H1(X).

• Finally, note that the map ϕ6 is the zero map. Since imϕ5 = kerϕ6 = H0(X, {∗}), we getthat ϕ5 is surjective. Thus by Definition 44,

H0(X) ∼= Z⊕H0(X, {∗}) ∼= H0(X).

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We’ll also need the following result, which is related to Theorem 19.

Proposition 25 (Long Exact Sequence of a Triple). Let B ⊆ A ⊆ X be topological spaces. Thenthe sequence

· · · −→ Hn(A,B) −→ Hn(X,B) −→ Hn(X,A) −→ Hn−1(A,B) −→ · · ·

is exact.

Proof. This is derived in the same manner as before via the short exact sequence

0 −→ Cn(A,B) −→ Cn(X,B) −→ Cn(X,A) −→ 0.

The details are ommitted.

We may now prove Theorem 21.

Proof. Let V be a neighborhood of A in X that deformation retracts to A. Consider the followingcommutative diagram.

Hn(X,A) Hn(X,V ) Hn(X \A, V \A)

Hn(X/A,A/A) Hn(X/A, V/A) Hn((X/A) \ (A/A), (V/A) \ (A/A))

The crucial claim is that every arrow is an isomorphism. We’ll split the proof of this fact intosteps.

• Since A ⊆ V ⊆ X, start with the long exact sequence

Hn(V,A) −→ Hn(X,A) −→ Hn(X,V ) −→ Hn−1(V,A).

Because V deformation retracts onto A, Hn(V,A) = Hn−1(V,A) = 0, so by exactness the in-ner map is an isomorphism. The same argument holds for the arrow mapping Hn(X/A,A/A)to Hn(X/A, V/A).

• The two left arrows are isomorphisms by Theorem 20.

• Finally, the rightmost downward arrow yields an isomorphism because we’re “blowing up ahole”. In particular, remark that

X \A ' (X/A) \ (A/A),

and similarly V \ A ' (V/A) \ (A/A); these homeomorphisms induce an isomorphism onrelative homology.

Done.

30.2 Generators of Homology Groups on ∆-complexes

Now before moving on we prove a result that should be intuitive.

Proposition 26. For all n ≥ 0, the group Hn(∆n, ∂∆n) ∼= Z is generated by the identity mapin : ∆n → ∆n.

Proof. We proceed by induction. The base case of n = 0 is obvious, so let n ≥ 1 be arbitrary. LetΛ ⊆ ∆n be the union of all but one of the (n− 1)-faces. I claim that

Hn(∆n, ∂∆n)(1)∼= Hn−1(∂∆n,Λ)

(2)∼= Hn−1(∆n−1, ∂∆n−1).

To prove (1) we examine the long exact sequence for the triple Λ ⊆ ∂∆n ⊆ ∆n: this is

· · · −→ Hn(∆n,Λ) −→ Hn(∆n, ∂∆n)ϕ−→ Hn−1(∂∆n,Λ) −→ Hn−1(∆n,Λ) −→ · · · .

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But Hn(∆n,Λ) and Hn−1(∆n,Λ) are both zero since ∆n deformation retracts onto Λ, so ϕ :Hn(∆n, ∂∆n)→ Hn−1(∂∆n,Λ) is an isomorphism. The exact isomorphism: look at the boundary.

To prove (2), assume n ≥ 1, as the n = 1 case is degenerate and can be proven separately. Firstremark that

∆n−1

∂∆n−1∼= ∂∆n

Λ.

Indeed, one can check that both spaces are homeomorphic to Sn (this is purely geometrical). Thishomeomorphism implies an isomorphism on the relative homology groups.

Now consider the map jn induced by ∆n−1 ↪→ ∂∆n which takes an (n − 1)-dimensional ∆-complex and maps it to the face not contained in Λ. This map when viewed in the quotient space∂∆n/Λ becomes homeomorphic to the map sending ∆n−1 to the n-sphere. But this is exactlywhat happens to the map in−1 when viewed in the quotient space ∆n−1/∂∆n−1, and so jn is agenerator as well. We’re done.

In the previous proof, we examined many homology groups of the form Hn(X,X \ {∗}). Thiswarrants a definition.

Definition 48. For x ∈ X, the group Hn(X,X \ {x}) is called a local homology group.

This definition warrants some remarks, as it first seems like these might not be of much use.

Remark. By excision,Hn(X,X \ {x}) ∼= Hn(U,U \ {x})

for sufficiently small open sets U 3 x.

Remark. These groups are not always trivial, surprisingly. Take for X = Rn the long exactsequence

Hk(Rn \ {x}) −→ Hk(Rn) −→ Hk(Rn,Rn \ {x}) −→ Hk−1(Rn \ {x}) −→ Hk−1(Rn).

So Hk(Rn,Rn \ {x}) is zero if k 6= n− 1 and Z if k = n− 1. In the latter case its isomorphic to

Hk−1(Rk \ {x}) ∼= Hk−1(Sk−1) ∼= Z.

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31 November 7

31.1 The Home Stretch

Recall what we’ve shown so far:

• If (X,A) is a good pair, then Hn(X,A) ∼= Hn(X/A);

• For all n, Hn(∆n, ∂∆n) ∼= Z, and furthermore this group is generated by the canonicalidentity map.

We will now use these two facts to show that simplicial and singular homology are the same.The workhorse of this proof is the following algebraic fact.

Lemma 11 (Five Lemma). Suppose we have a commutative diagram

A B C D E

A′ B′ C ′ D′ E′

i

α

j

β

k

γ

`

δ ε

i′ j′ k′ `′

with rows are all exact and α, β, δ, and ε all isomorphisms. Then γ is an isomorphism too.

Proof. First we’ll show that if β and δ are surjective and ε is injective then γ is surjective.

Pick c′ ∈ C ′. Then k′(c′) = δ(d) for some d ∈ D due to surjectivity. I claim d ∈ ker `; indeed,

ε(`(d)) = `′(δ(d)) = `′(k′(c′)) = 0,

so `(d) = 0 due to injectivity of ε. Thus there exists c ∈ C such that d = k(c).

At this point, all we know is that δ(k(c)) = k′(c′); our goal is to show that γ(c) = c′. To computethis, remark that

k′(c′ − γ(c)) = k′(c′)− k′(γ(c′)) = k′(c′)− δ(k(c)) = 0,

so c′ − γ(c) ∈ ker k′ = im j′. Thus there exists b′ ∈ B′ such that j′(b′) = c′ − γ(c); by surjectivityof β there exists b ∈ B such that β(b) = b′.

But now we have our punchline: write

γ(c+ j(b)) = γ(c) + γ(j(b)) = γ(c) + j′(β(b)) = γ(c) + j′(b′) = c′.

So c+ j(b) is the desired element in C.

The second part involves showing that if β and δ are injective and α is surjective then γ isinjective. The proof is similar and will be omitted.

In order to show that simplicial and singular homology are isomorphic, we need to distinguishthe notation.

Definition 49. Let H∆n (X,A) be the relative simplicial homology groups, where here A ⊆ X is a

subcomplex of the ∆-complex X. These are defined in an analogous way as Hn(X,A).

This leads into a theorem.

Theorem 22. The homomorphism H∆n (X,A)→ Hn(X,A) given by the characteristic map is an

isomorphism for all n and every pair (X,A) for X a ∆-complex and A a subcomplex.

Proof. First assume that X is finite-dimensional and A = ∅. We’ll prove this by induction on k.Denote by Xk the k-skeleton of X, that is, the subcomplex of all faces of dimension ≤ k. Nowconsider the commutative diagram

H∆n+1(Xk, Xk−1) H∆

n (Xk−1) H∆n (Xk) H∆

n (Xk, Xk−1) H∆n−1(Xk−1)

Hn+1(Xk, Xk−1) Hn(Xk−1) Hn(Xk) Hn(Xk, Xk−1) Hn−1(Xk−1)

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By our induction hypothesis, we know that the fifth and the second columns are isomorphisms.

For the fourth and the first columns, check that by modding out Xk−1 chains from Xk chains,we’re left with a wedge sum of k-dimensional spheres (as the boundary collapses to a point; seeExample 6). As a result,

H∆n (Xk, Xk−1) ∼=

{Zb for n = k and b the number of k-simplices,

0 otherwise.

Furthermore, if we take Φ :⊔α(∆k

α, ∂∆kα) → (Xk, Xk−1) via the expected characteristic maps,

there is a homeomorphism of quotient spaces⊔α ∆k

α⊔α ∂∆k

α

∼= Xk

Xk−1.

So Φ induces an isomorphism on homology.

To pass to the infinite case, recall that any compact set C in X only meets finitely many opensimplices. Using this, we claim the inclusion map H∆

n (X) → Hn(X) is surjective. Indeed, if z isa singular n-cycle, it can be written as a finite linear combination of singular n-simplices; each iscontained in finite number of simplices, and so z must be contained in Xk for some k. But thefinite case implies H∆

n (Xk)→ Hn(Xk) is surjective, so z is homologous to a simplicial n cycle. Asimilar “compactness” argument shows injectivity as well.

Finally, to pass to the more general case, consider the commutative diagram

H∆n (A) H∆

n (X) H∆n (X,A) H∆

n−1(A) H∆n−1(X)

Hn(A) Hn(X) Hn(X,A) Hn−1(A) Hn−1(X)

We know that the outer four columns signify isomorphisms by Theorem 22, so the middle columnmust as well. Done.

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32 November 9

32.1 Mayer-Vietoris sequences

Now that we’ve slogged through the details once, we are now ready to reap the benefits of ourlabor. We first record a result which is similar in spirit to Theorem 19.

Theorem 23 (Mayer-Vietoris). Suppose X is a space covered by the interiors of A and B. Then

0 −→ Cn(A ∩B)i−→ Cn(A)⊕ Cn(B)

j−→ C{A,B}n (X) −→ 0

is a short exact sequence. Here

• The map i : Cn(A ∩ B) → Cn(A) ⊕ Cn(B) is the inclusion map given by x 7→ (x,−x) (i.e.we view x as living in A and B but flip the boundary of the part in B), while

• the map j : Cn(A) ⊕ Cn(B) → C{A,B}n (X) is given by (x, y) 7→ x + y (i.e. we combine the

two loops x ∈ Cn(A) and y ∈ Cn(B)).

Proof. There are a few things to check, none of which are hard.

• The map i is injective: project onto first coordinate.

• The composition j ◦ i is zero: obvious.

• The kernel of j is the image of i: obvious.

• The map j is surjective: definition of CUn .

Upon iterating this and using the same techniques as in Theorem 18 we obtain a long exactsequence

· · · −→ Hn(A ∩B) −→ Hn(A)⊕Hn(B) −→ Hn(X)

∂−→ Hn−1(A ∩B) −→ Hn−1(A)⊕Hn−1(B) −→ · · · −→ H0(X) −→ 0.

The map ∂ behaves similarly to the boundary map in the relative case: consider a singular n-chain,subdivide it into parts completely contained in either A or B, and take the boundary of the partcontained in A.

Example 47. Let X = S1 × S1 be the torus. Split the torus into two sides A and B (thickened alittle bit). Note that A ' S1 and B ' S1, while A∩B ' S1 tS1. Now consider the Mayer-Vietorissequence

H2(A ∩B) −→ H2(A)⊕H2(B)j2−→ H2(X)

∂1−→ H1(A ∩B)i1−→ H1(A)⊕H1(B)

j1−→ H1(X)

∂0−→ H0(A ∩B)i0−→ H0(A)⊕H0(B)

j0−→ H0(X) −→ 0.

We now proceed in steps.

• We know from previous ventures that

Hi(A)⊕Hi(B) ∼= Hi(A ∩B) ∼={Z for i = 0, 1,

0 for i = 2.

• The map j0 : H0(A) ⊕ H0(B) → H0(X) takes singular 0-chains in the former space andincludes them into the latter space via addition. It is easy to see that the image of this mapis Z, generated by e.g. (1, 0). But exactness tells us j0 is surjective, so H0(X) ∼= Z.

• The map i1 : H1(A ∩ B) → H1(A) ⊕H1(B) takes the generators (1, 0) and (0, 1) and mapsthem both to (1,−1); thus the image of this map is isomorphic to Z. In turn, the kernel of∂1 is also Z. But exactness tells us ∂1 is injective, so H2(X) ∼= Z.

• In fact, we know more: both ker i0 = im ∂0 and ker j1 = im i1 are generated by (1, 0)− (0, 1).These groups both have rank 1, and so at the very least H1(X) has rank 1 + 1 = 2. The factthat H1(X) has no torsion is somewhat intuitive because none of our maps induce torsion;one can derive this result purely algebraically, but this was not done in class and so it isomitted. In the end, H1(X) ∼= Z⊕ Z.

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Example 48. Let A and B be two Mobius strips, and glue them along their common boundary.This yields a Klein bottle K. Consider the long exact sequence

0 −→ H2(K)∂1−→ H1(A ∩B)

i1−→ H1(A)⊕H1(B)

j1−→ H1(K)∂0−→ H0(A ∩B)

i0−→ H0(A)⊕H0(B) −→ H0(K) −→ 0.

We now compute the homology groups in steps.

• Recall that A and B are both homeomorphic to S1 (deformation retract to the central circle),so H0(A) ∼= H0(A) ∼= Z and similarly for B. Furthermore, A ∩ B is the boundary circle of(either) Mobius strip, and so H0(A ∩B) ∼= H1(A ∩B) ∼= Z as well.

• Since K has one connected component, H0(K) ∼= Z.

• Consider the map i0 : H0(A ∩ B) → H0(A) ⊕ H0(B). This map sends the generator 1 ofH0(A ∩B) to (1,−1). Hence i0 is injective, which after some exactness chasing implies j1 issurjective. This will be useful later.

• Consider the map i1 : H1(A ∩ B) → H1(A) ⊕ H1(B). This map sends the generator 1of H1(A ∩ B) to (2,−2) (recall that the generator of e.g. H1(A) is the central circle, andthe boundary circle wraps around twice). This makes the computation for H1(K) moreinteresting, but at least its still injective, which yields H2(K) = 0.

• To compute H1(K), we combine both bullet points above with the First Isomorphism The-orem to yield

H1(K) = im j1 ∼=H1(A)⊕H1(B)

ker j1∼= Z⊕ Z〈(2,−2)〉

∼= Z⊕ Z2.

Remark. Note that we computed H2(S1 × S1) ∼= Z, which means that we can orient every trianglein such a way that the orientations all match up on each edge (i.e. the kernel of the boundary mapis nontrivial). This is what people mean when they say a surface is orientable.

On the other hand, we computed H2(K) = 0. This means that the Klein bottle cannot beembedded in R3. If we could, then we could use the outward normal vector to induce an orientationof the edges, but this is not allowed because the kernel is trivial.

In general, a d-dimensional manifold M is orientable iff Hd(M) is nontrivial.

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33 November 12

We start with another example.

Example 49. What’s the homology group of RP 2?

To compute this, view RP 2 as a quotient of D2, and consider the decomposition of RP 2 fromExample 7, where M is the Mobius strip and D the outside disk. This yields the sequence

H2(M)⊕H2(D) −→ H2(RP 2) −→ H1(M ∩D)i−→ H1(M)⊕H1(D) −→ H1(RP 2) −→ 0.

We now proceed in steps, except this time I won’t bullet the individual steps. First note that wehave previously computed H2(M) = 0, H1(M) ∼= Z, H1(M ∩D) ∼= Z, and Hi(D) = 0 for all i; inturn, H2(M)⊕H2(D) = 0 and H1(M)⊕H1(D) ∼= Z. Thus all that is left is to compute Hi(RP 2)for i ∈ {1, 2}.

The key is to consider the map i : H1(M ∩D) → H1(M) ⊕H1(D). Recall that H1(M ∩D) isgenerated by a single element 1. This element maps to 2 times the generator of H1(M), and soi(1) = 2. This has two consequences. First, this map is injective, so the mapping from H2(M) ⊕H2(D) → H2(RP 2) is surjective, and hence H2(RP 2) = 0. Second, by the First IsomorphismTheorem,

H1(RP 2) ∼= H1(M)⊕H1(D)

im i∼= Z

(2)∼= Z2.

Huzzah.

33.1 Degree

We’ll now start exploring applications of homology.

If we have a map f : Sn → Sn it generates a map on homology f∗ : Hn(Sn)→ Hn(Sn). Both ofthese groups are Z, so they’re generated by single element.

Definition 50. The value of f∗(1) ∈ Z is called the degree of the map f .

We now record some facts about degree.

• For any maps f and g, deg f ◦ g = deg f · deg g; this follows from the fact that the image off∗ is generated by f∗(1). In particular, if f is a homotopy equivalence, then deg f = ±1.

• If f denotes the reflection of Sn about a hyperplane, then deg f = −1. To prove this, pick anice ∆-structure for Sn where identical ∆-complexes are oppositely orientated with respectto this hyperplane; then f reverses the orientations of all the ∆-complexes, and so f∗(1) 6= 1.

• deg(− id) = (−1)n+1, where − id denotes reflection about the origin of Sn. Indeed, notethat when embedding Sn into Rn+1, − id corresponds to reflection about each of the n + 1hyperplanes of the form {xi = 0}; each one of these reflections multiplies the degree by −1.

• If f has no fixed points, then deg f = deg(− id). To prove this, it suffices to show there existsa homotopy from f to − id; the result follows by the homotopy invariance of degrees. Thishomotopy is given specifically by the map

t 7→ (1− t)f(x)− tx|(1− t)f(x)− tx| ;

in particular, “the no fixed points” condition implies this map is well-defined for t = 12 .

Theorem 24. The only non-trivial group G that acts properly discontinuously on S2n is G = Z2.

Proof. There is a map d : G → Z via g 7→ deg g. The image of d is contained in {−1, 1}, sod induces a group homomorphism G → {−1, 1}. Now every nontrivial element g ∈ G satisfiesd(g) = −1 because there are no fixed points, so we fail if there are two nontrivial elements.

Corollary 8 (Hairy Ball Theorem). At every point, Sn has a nonzero tangent vector iff n is odd.

Proof. Let v : Sn → Rn+1 be a continuous tangent vector field with v(x) 6= 0. Without loss ofgenerality (since v(x) 6= 0) let |v(x)| = 1. Consider the map f : Sn → Sn given by

t 7→ (sin t) · x+ (cos t) · v(x).

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This is a homotopy on Sn because |f(t)|2 = sin2 t + cos2 t = 1. Furthermore, it sends id to − id;by our previous work, this implies n is odd.

Conversely, if n = 2k − 1, embed Sn into Rn+1 and write

v(x1, x2, . . . , x2k) = (−x2, x1,−x4, x3, · · · ,−x2k, x2k−1).

33.2 Cellular Homology for CW Complexes

We now seek to take our homology results from before and apply them to CW complexes. We willstart with this today and continue on Wednesday.

Lemma 12. Let X be a CW complex. Then

• Hk(Xn, Xn−1) = 0 if k 6= n and free abelian for k = n with basis the n-cells of X.

• Hk(Xn) = 0 if k > n.

• Hk(Xn)→ Hk(X) is an iso for k < n and surjective for k = n.

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34 November 14

We start with the proof of the result from last time.

Proof. First assume that the dimension of X is finite. Note that (Xn, Xn−1) form a good pair, so

Hk(Xn, Xn−1) ∼= Hk(Xn/Xn−1).

But recall that Xn/Xn−1 is homeomorphic to a wedge sum of n-dimensional spheres (one for eachn-dimensional face); this proves the first bullet point.

For the rest of the proof, consider the exact sequence

Hk+1(Xn, Xn−1) −→ Hk(Xn−1) −→ Hk(Xn) −→ Hk(Xn, Xn−1)

If k /∈ {n−1, n}, then Hk+1(Xn, Xn−1) = Hk(Xn, Xn−1) = 0 by the first part, and so Hk(Xn−1) ∼=Hk(Xn). If k = n, then at the very least Hk+1(Xn, Xn−1) = 0, and so the mapping Hk(Xn) →Hk(Xn−1) is a surjection. These isomorphisms plus Hk(X0) = 0 for k ≥ 1 imply part 2 and 3 inthe finite case. In the infinite case, instead appeal to a compactness argument.

34.1 Cellular Homology

Our goal now is to define a notion of homology for CW complexes; in other words, we want toconstruct a long exact sequence on homology

Hn+1(Xn+1, Xn)dn+1−→ Hn(Xn, Xn−1)

dn−→ Hn−1(Xn−1, Xn−2) −→ · · · .

The key observation is that these groups fit into several short exact sequences we already knowof, and so we will construct the desired maps dk by snaking between these established sequences.Behold the following diagram!

0

0 Hn(Xn+1) ∼= Hn(X)

Hn(Xn)

Hn+1(Xn+1, Xn) Hn(Xn, Xn−1) Hn−1(Xn−1, Xn−2)

Hn−1(Xn−1)

0

jn∂n+1

dn+1

∂n

dn

jn−1

This is a chain complex, and its homology is called cellular homology, denoted by HCWn (X).

Fortunately, we won’t need to carry the CW notation for long.

Theorem 25. For any CW complex X, HCWn (X) ∼= Hn(X).

Proof. We know Hn(X) ∼= Hn(Xn)/ im ∂n+1 via the leftmost diagonal chain. Now due to exactnessof the middle diagonal chain, jn is injective, which implies jn(im ∂n+1) = im dn+1. Further,Hn(Xn) gets mapped under jn to im jn = ker ∂n. But jn−1 is injective, so ker ∂n = ker dn. Itfollows that jn induces an isomorphism of the quotient ∼= Hn(Xn)/ im ∂n+1 onto ker dn/ im dn+1,and so cellular and singular homologies coincide.

This is nicer because CW complexes have fewer variables than ∆-complexes and therefore cellularhomology is often easier to compute than simplicial homology. Below, we see an example of howto compute cellular homology.

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Example 50. Consider the CW complex shown below, which has one 0-cell, two 1-cells a and b,and one 2-cell σ attached by the word ab−1aba. This implies d2(σ) = a − b + a + b + a = 3a. SoH1 is generated by a and b, where 3a is trivial.

b

a

a

b

a

σ

Figure 36: A CW complex.

Remark. In general, for σ an n-cell,

dn(σ) =∑

dβ · eβ ,

where dβ is the degree of the map ∆β defined by the composition

Sn−1 → Xn−1 → Xn−1

Xn−1 \ eβ∼= Sn−1

β .

The proof of this is in Hatcher (Section 2.2, pages 140-141).

Example 51 (Orientable Surfaces). Recall from Remark 19.4 the CW complex structure for theorientable surface Mg. This structure consists of one (common) vertex, four edges, and one 2-cellattached by the word [a1, b1][a2, b2] · · · [ag, bg], where [a, b] = aba−1b−1. The chain complex for Mg

is thus0 −→ Z d2−→ Z2g d1−→ Z −→ 0.

The map d1 is zero since there is only one vertex in the complex structure and so the only elementof all Xn−1 \ eβ is a point which is contractible. Furthermore, the map d2 is zero: in our attachingmap along every edge we glue along both the edge and its inverse, so each dβ equals zero.

As a result, the cellular homology groups are identical to the chain complex groups, so H0(Mg) ∼=Z ∼= H2(Mg) and H1(Mg) ∼= Z2g.

Example 52 (Nonorientable surfaces). Consider the closed nonorientientable surfaces Ng of genusg. For example, N1

∼= RP 2, whileN2 is congruent to the Klein bottle. The surfaceNg is representedby a CW complex structure with one 0-cell, g 1-cells, and one 2-cell attached by the word a2

1a22 · · · a2

g.The chain complex structure of Ng is thus

0 −→ Z d2−→ Zg d1−→ Z −→ 0.

The map d1 is zero for the same reason as in the previous example. However, this time thegenerator 1 of H2(X2, X1) is mapped to (2, 2, . . . , 2) ∈ Zg; this is because the attaching map goesalong each one cell twice, and so each ∆β is homomprhic to the map z 7→ z2, which has degree 2.

It remains to compute the cellular homology groups. Note that d2 is injective, so H2(Ng) = 0.Furthermore, H0(Ng) ∼= Z. Finally, by considering the basis

(1, 0, . . . , 0), (0, 1, 0, . . . , 0), . . . (0, 0, . . . , 1, 0), (1, 1, . . . , 1)

of Zg, we see that H1(Ng) ∼= Zg−1 ⊕ Z2.

Example 53. The following example is especially illustrative. Start with S1 ∨S1, with two 1-cellsa and b; then glue in two 2-cells via the words a5b−3 and b3(ab)−2, and call the resulting space X.We want to compute H2(X) and H1(X).

As we have one 0-cell, two 1-cells, and two 2-cells, the relevant part of our chain complex is

0 −→ Z2 d2−→ Z2 d1−→ Z −→ 0.

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The map d1 is still zero. The map d2 is given by the matrix M =(

5 −2−3 1

). The determinant of M

is 1, so the linear map x 7→ Mx is an isomorphism, which implies H2(X) = 0. Additionally this

map is surjective, so H1(X) = 0. Hence Hi(X) = 0 for all i.

But now comes the punchline: the fundamental group of X is, by Van Kampen, equal to

π1(X) = 〈a, b | a5b−3, b3(ab)−2〉;

this group is nontrivial. (Indeed, there is a nontrivial homomorphism sending this group to thegroup G of rotational symmetries of a regular dodecahdron: send a to the rotation ρa throughangle 2π/5 about the axis through the center of a pentagonal face, and send b to the rotation ρbthrough angle 2π/3 along the axis through a vertex of this face.)

Also, this group acts properly discontinuously on S3, so via quotienting by this action and usingTheorem 14, we see that there exists a closed 3d manifold that has the homology of a sphere butis not a sphere.9

Remark. One thing to notice here is that homology doesn’t care about order to some extent. Inparticular, the main reason that this space X from above has the properties it does is that theattaching maps only care about the total number of wrappings around the two 1-cells, whereas thefundamental group depends more crucially on the exact nature of the attaching map.

With cellular homology, we have greatly expanded the range of possible computations, almostto the point where we can compute the homology groups of any space we want.

9Technically, this only shows that the zeroth and first homology groups coincide; it is a bit more work to showthat the higher homology groups do as well, but it is possible.

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35 November 16

Example 54. As a quotient, RPn has a CW-complex structure with one cell in each dimensionk ≤ n. The chain complex thus looks like

0 −→ Z dn−→ Z dn−1−→ Z −→ · · · d1−→ Z −→ 0.

The difficulty lies in computing the maps dn between these spaces. By the Cellular Boundaryformula, dn(en) = kne

n−1, where kn is the degree of the composition

Sn−1 −→ RP k−1 −→ RP k−1

RP k−2∼= Sk−1.

This map is a homeomoprhism when restricted to each component of Sk−1 \ Sk−2. One of thesecomponents maps identically onto RP k−1 (and thus has degree 1), while the other componentsmaps antipodally onto RP k−1 (and thus has degree (−1)k). As a result, the degree of the totalcomposition is 1 + (−1)k, which is zero if k is even and 2 if k is odd.

From this, it is not hard to see that the homology groups alternate between 0 and Z2, i.e.

Hk(RPn) ∼=

Z if k = 0 or k = n odd,

Z2 if k < n is odd,

0 if k ≤ n is even.

Hence RPn is orientable iff its dimension is odd.

35.1 Euler Characteristic

Definition 51. For a CW complex X, the Euler characteristic of X is

χ(X) =∑n≥0

(−1)ncn,

where cn is the number of n-cells in X.

We proceed with some examples.

Example 55. The tetrahedron consists of 4 vertices, 6 edges, and 4 faces, so its Euler characteristicis 4− 6 + 4 = 2.

Example 56. The octahedron consists of 6 vertices, 12 edges, and 8 faces, so its Euler character-istic is 6− 12 + 8 = 2.

Example 57. The cube consists of 8 vertices, 12 edges, and 6 faces, so its Euler characteristic is8− 12 + 6 = 2.

Hmm.

Theorem 26. The equality

χ(X) =∑n≥0

(−1)n rankHn(X).

holds; in particular χ is a topological invariant.

Proof. Denote by Cn, Bn, and Zn the groups of n-chains, n-cycles, and n-boundaries respectively.Recall that the nth homology group is a quotient of n-boundaries by n-cycles, and so rankHn =rankZn − rankBn. Additionally, by “rank-nullity”, rankCn = rankZn + rankBn−1. Combiningthese two facts yields

rankCn = rankBn + rankHn + rankBn−1,

so

χ(X) =∑n≥0

(−1)n rankCn

=∑n≥0

(−1)n(rankBn + rankHn + rankBn−1) =∑n≥0

(−1)n rankHn,

where the last equality is due to telescoping sums.

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What follows is a series of corollaries.

Corollary 9. If X is a CW complex homeomorphic to the sphere, then χ(X) = 2.

Proof. In this case,

χ(X) = χ(S2) =∑n≥0

(−1)n rankHn(S2) = 1− 0 + 1 = 2.

Corollary 10. The properly discontinuously result follows from this.

Proof. Its true that if G is finite and acting on X properly discontinuously, then by counting faceswe see that

χ(X/G) =χ(X)

|G| .

This shows if Z2 acts freely on Sk, then Sk has even Euler characteristic; in turn, k is even.

Corollary 11. The complete graph on five vertices K5 is not planar.

Proof. Suppose there was an embedding K5 ↪→ S2. This defines a CW-complex X on S2. But Xhas 5 vertices, 10 edges, and at most 6 edges, and 5− 10 + 6 = 1 < 2.

Example 58. The torus has χ = 0. Consider the standard square representation of the torus; ithas one vertex, 2 edges, and one face, and so indeed its Euler characteristic is 1− 2 + 1 = 0.

We can use the previous ideas to prove a combinatorial result regarding triangulations of surfaces.Let X be a 2-dimensional ∆-complex. Denote by δ the average number of triangles incident to avertex (if X is a surface its the same as number of edges). Suppose we have f0 vertices, f1 edges,and f2 triangles. By double counting, we have the relations 2f1 = 3f2 and δf0 = 3f2. Hence a bitof computation yields

2δχ(X) = 2δ(f0 − f1 + f2) = (6− δ)f2.

In turn, the sign of χ(X) is equal to the sign of 6− δ.What this tells us is that any triangulation of the torus must have an average of six triangles

per vertex. This also tells us that any triangulation of the sphere must have less than 6 trianglesper vertex; in fact, this is why there are only a finite number of regular polyhedra.

Before, we go, fun fact: Euler characteristic in dimension 2 detects curvature. Indeed, for everyclosed surface S,

2πχ(S) =

∫S

κ(~x) d~x,

where κ(~x) is the curvature at ~x. This fact is called the Gauβ-Bonnet Theorem, and it’s terriblyfalse for higher dimensions.

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36 November 26

36.1 Homology with coefficients

The main idea of homology with coefficients is simple: generate the singular chain complex withcoefficients in the abelian group G instead of with Z-coefficients.

Example 59. Let’s compute Hk(RPn,Z2). Recall that the CW complex model of RPn consistsof one face of each dimension ≤ n, and so the relevant chain complex is

0 −→ Z2 −→ Z2 −→ Z2 −→ · · · −→ Z2 −→ 0.

When we had Z-coefficients, the maps alternated between being the zero map and the “multipli-cation by two” map. Thus with Z2 coefficients all maps are zero, meaning that

Hk(RPn,Z2) =

{Z2, k ≤ n,0 k > n.

Remark. If X is a CW complex, then the chain group Ck(X;Z2) consists of linear combinationsof subsets of the k-dimensional faces of X. More formally, this chain group can be written as atensor product

Ck(X;Z2) = Ck(X)⊗Z Z2.

The above remark warrants a definition.

Definition 52. Let G be a group. Then the functor Tor(·, G) : Ab → Ab measures to whichextent tensoring with G is nonexact. In particular, Tor satisfies the following properties.

• For any nonnegative integer n, Tor(Zn, G) is the kernel of the map ϕn : G → G defined viamultiplication by n,

• For any groups G and H, Tor(G,H) = Tor(H,G),

• For any groups G and H, Tor(G,H) = 0 if G or H is torsion free, and

• For any sequence of groups {Hα}α∈I , we have

Tor

(⊕i∈I

Gi, H

)∼=⊕i∈I

Tor(Gi, H).

This allows one to compute Tor explicitly.

Theorem 27 (Universal coefficient theorem). The homology groups with Z-coefficients and Gdetermine Hk(X;G). More specifically,

0 −→ Hk(X;Z) −→ Hk(X;G) −→ Tor(Hk−1(X,Z), G) −→ 0

is a short exact sequence.

Example 60. Let’s run through Example 59 again, but this time use Theorem 27 to do thecomputations. Consider the sequence of chain complexes

0 −→ C(RPn;Z2)τ−→ C(Sn,Z2)

q#−→ C(RPn;Z2) −→ 0.

Here τ is the transfer map τ(σ) = σ1 + σ2 (i.e. the sum of the two lifts), while q# is (as you mayrecall) the map on chain complexes induced by the quotient map q : Sn → RPn. Check that thisis a short exact sequence. This induces a long exact sequence in homology

0 −→ Hn(RPn;Z2) −→ Hn(Sn;Z2) −→ Hn(RPn;Z2)

−→ Hn−1(RPn;Z2) −→ Hn−1(Sn;Z2) −→ Hn−1(RPn;Z2) −→ · · ·

Recall that Hn(Sn;Z2) ∼= Z2 and Hk(Sn;Z2) = 0 otherwise. This combined with exactness tells usthat

Hn−1(Sn;Z2) ∼= Hn−2(RPn;Z2) ∼= · · · ∼= H0(RPn;Z2) ∼= Z2.

Furthermore, the top level shows Hn(RPn;Z2) ∼= Hn(Sn;Z2) ∼= Z2.

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Theorem 28 (Borsuk-Ulam in higher dimensions). A map f : Sn → Sn with f(−x) = −f(x) forall x has odd degree.

Proof. The map f induces a map g : RPn → RPn. Now consider the commutative diagram

Hk(RPn;Z2) Hk−1(RPn;Z2)

Hk(RPn;Z2) Hk−1(RPn;Z2)

∂

g# g#

∂

induced by g#, where k ∈ N is arbitrary. Now g# is an isomorphism in zero-homology (the zerothhomology group is generated by the path-components of RPn, of which there is only one!), and soinducting upward yields that g# is an isomorphism in kth-homology for all k.

Now consider the commutative diagram

Hk(RPn;Z2) Hk−1(Sn;Z2)

Hk(RPn;Z2) Hk−1(Sn;Z2)

τ∗

g# f#

τ∗

We have established g# is an isomorphism, and from the previous example we see that τ∗ is anisomorphism as well. Hence f# : Hk−1(Sn;Z2) → Hk−1(Sn;Z2) is an isomorphism, so f has odddegree.

36.2 A Tiny Introduction to Cohomology

Here’s the basic idea behind cohomology; we don’t have the time to go into the details today soeverything will be defined on Wednesday. Recall that we used the chain complex

−→ Cn −→ Cn−1 −→ Cn−2 −→ · · ·

to establish the homology groups. Now consider instead the dual groups C∗n = hom(Cn,Z); thisreverses all the arrows. Then compute the homology of this “dual” chain complex instead; this iscohomology.

But why would you do this? In homology, if you take a linear combination of k-dimensionalfaces and apply boundary, you get a linear combination of (k − 1)-dimensional faces. In contrast,in cohomology, if you take functions on (k − 1)-dimensional faces and apply coboundary, you getfunctions on k-dimensional faces. This sounds like integration.

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37 November 28

37.1 A Less Tiny Introduction to Cohomology

We’ll start with a simple example. Let X be a directed graph and G an abelian group. Define

∆0(X;G) := {f : V (X)→ G},∆1(X;G) := {f : E(X)→ G}.

Consider the map δ : ∆0(X;G) → ∆1(X;G) which inputs ϕ : V (X) → G and outputs somefunction from E(X)→ G defined via

[v, w]→ ϕ(w)− ϕ(v).

Notice the relationship between this map and the boundary map ∂ given by ∂([u, v]) = v − u.

Then set

H1(X;G) :=∆1(X;G)

im δ.

Note that H1 is trivial iff for each ψ ∈ ∆1(X;G) there is a solution ψ = δϕ. This is, in essence, asolution to an integration problem.

Definition 53. Let X be a CW complex.

1. Set∆k(X;G) = {f : {k faces of X} → G} ∼= hom(∆k(X), G),

where ∆k(X) denotes the k-chains in X as before. (The last congruence comes from the factthat any such function extends to a homomorphism on k-chains and vice versa.)

2. Define δk : ∆k(X;G)→ ∆k+1(X;G) by ϕ→ ϕ ◦ ∂. This can be considered the “dual map”of the boundary map.

3. Remark that· · · ←− ∆2(X;G)

δ1←− ∆1(X;G)δ0←− ∆0(X;G)←− 0

is a chain complex, since δkδk+1 = 0 for all k. Thus, for all n, let the nth cohomology groupHn(X;G) of X be the homology of this chain cocomplex.

Remark. This is a standard construction in homological algebra, so it might help (for myself!) toexplain briefly what is going on. Suppose we have two groups G and H and a map f : G → H.We would like to consider maps between e.g. hom(G;Z) and hom(H;Z). The easiest way to dothis is to take an element ϕ ∈ hom(H;Z) and compose it with f on the right; this changes theinput from living in H to living in G. This natural composition thus yields a map

f∗ : hom(H;Z)→ hom(G;Z);

notice in particular that the locations of G and H swap. (Some linear algebra experts mightrecognize the analogies with constructing the dual of a vector space.)

Despite the fact that all we did was flip the directions of the arrows, it is not true thatHn(C;G) ∼=hom(Hn(C;G)).

Example 61. Let’s look at the cohomology of RP 3. Recall from Example 54 that the associatedchain complex was

0 −→ Z 0−→ Z 2−→ Z 0−→ Z −→ 0.

Now let’s dualize with G = Z. Note that Z∗ ∼= Z since each homomorphism is uniquely determinedby its value at 1, so the associated chain cocomplex is

0←− Z←− Z←− Z←− Z←− 0.

The claim now is that the maps between copies of Z are also the same. To check this, we case.

• Suppose f : Z → Z is given by the zero map. Then f∗ is given by ϕ 7→ ϕ ◦ 0 = 0, meaningthat f∗ is also the zero map.

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• Suppose f : Z→ Z is given by the “multiplication by two” map, i.e. f(x) = 2x. Then f∗ isdefined by

(f∗ ◦ ϕ)(x) = (ϕ ◦ f)(x) = ϕ(2x) = 2ϕ(x),

where in the last step we use the fact that ϕ is a homomorphism. Thus f∗ is also the“multiplication by two” map.

This means that the homology and cohomology groups are reversed: note that H1(RP 3) = 0 andH2(RP 3) ∼= Z2, but H1(RP 3) ∼= Z2 and H2(RP 3) = 0.

Remark. What does this tell us? In the case of homology, it implies the existence of a nontrivialcycle which is made trivial if looped twice. In the case of cohomology, the fact that H1(RP 3) = 0implies that for any values we write down at the edges we can always assign values to the verticeswhich solve the integration problem. But we can’t always do the same thing in two dimensions:there may be assignments of values to the triangles of RP 3 that do not yield a solution to theintegration problem on the edges of RP 3.

Fun fact: the homology groups uniquely determine the cohomology groups. With this fact, itmay seem as if cohomology is useless. However, it turns out that cohomology has more structurethan homology that we’ll be able to exploit.

To show this result, let

Zn = {n-cycles in X} = ker ∂n,

Bn = {n-boundaries in X} = im ∂n+1;

note that we first used this notation in the proof of Theorem 26. Recall that

0 −→ Zn+1 −→ Cn+1∂−→ Bn −→ 0

is a short exact sequence of free abelian groups. We can turn this into a short exact sequence ofchain complexes by fitting it into the bigger picture below, where each downward arrow representsan application of the ∂ map.

0 Zn+1 Cn+1 Bn 0

0 Zn Cn Bn−1 0

0

∂

∂ 0

∂

The right downward arrow is actually the zero map because of commutativity and the fact that∂∂ = 0, while the left downward arrow is also the zero map from the fact that all elements in Zn+1

have no boundary by definition.

Now dualize.

0 Z∗n+1 C∗n+1 B∗n 0

0 Z∗n C∗n B∗n−1 0

0 δ 0

Actually, the above picture is a slight lie in general.

Remark. One can show that if the sequence 0→ A→ B → C → 0 is exact, then so is A∗ ← B∗ ←C∗ ← 0. Normally we can’t add a zero to this to form a short exact sequence, but if the originalsequence splits, we can. In our case, its split because Bn is a subgroup of a free abelian groupCn−1 and is hence free abelian.

As a result, this induces a long exact sequence in cohomology

←− B∗n ←− Z∗n ←− Hn(C;G)←− B∗n−1 ←− Z∗n−1 ←− · · · .

As always, in order for this LES to be useful, we need to figure out precisely what the connectingmap between sheets is. But in our case, it’s simple: the map is i∗n, where in : Bn → Zn is theinclusion map.

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Exactness at Hn(C;G) means

0←− ker i∗n ←− Hn(C;G)←− coker i∗n−1 ←− 0. (37.1)

is split and exact. What’s ker i∗n? Well, if ψ : Zn → G lies in ker i∗n, then

0 = i∗n(ψ)(x) = (ψ ◦ in)(x) = ψ(in(x)).

Hence ker i∗n consists of precisely those homomorphisms ϕ which vanish on the subgroup Bn. Thesein turn induce homomorphisms

ϕ : Zn/Bn → G.

But now recall that Zn/Bn = Hn(C); as a result, ker i∗n∼= hom(Hn(C), G). This answers our

original question: if coker i∗n−1 is zero, then in fact we have an isomorphism between Hn(C;G)and hom(Hn(C), G). Hence, it remains to understand the case where this cokernel is nontrivial.

Now rewriting Equation 37.1 gives us the split exact sequence

0 −→ coker i∗n−1 −→ Hn(C;G) −→ hom(Hn(C), G) −→ 0. (37.2)

To get a better sense of this sequence, recall the existence of another exact sequence

0 −→ Bn−1in−1−→ Zn−1 −→ Hn−1(C) −→ 0.

Now the dual of this short exact sequence is

B∗n−1

i∗n−1←− Z∗n−1 ←− Hn−1(C)∗ ←− 0. (37.3)

If Hn−1(C) is free, then in fact the sequence in Equation 37.3 is short exact. Thus i∗n−1 is surjective,implying coker i∗n−1 = 0 ⇒ Hn(C;G) ∼= hom(Hn(C), G). This means that the only thing thatprevents this from happening is torsion. Thus, we have to measure to what extent the sequence37.3 is not exact; this is precisely the difference between homology and cohomology.

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38 November 30

Recall that it remains to determine the structure of coker i∗n. Recall further that we have a shortexact sequence

0 −→ Bn−1 −→ Zn−1 −→ Hn−1(C) −→ 0,

which we dualized to get

0←− B∗n−1 ←− Z∗n−1 ←− Hn−1(C)∗ ←− 0.

38.1 Free Resolutions

Definition 54. A free resolution of an abelian group H is a sequence of the form

· · · −→ F2 −→ F1 −→ F0 −→ H −→ 0,

where Fi are free abelian and the sequence is exact.

One possible free resolution of an abelian group: F0 generated by generators, F1 generated byrelations, everything else is zero.

The dual of this (vis a vis applying hom(·, G) is)

· · · ←− F ∗2 ←− F ∗1 ←− F ∗0 ←− 0.

Recall that this isn’t always exact. But recall that homology tells you how far you are away frombeing exact, since

Hn(Fn;G) =ker f∗n+1

im f∗n.

Lemma 13. Given free resolutions F and F ′ of abelian groups H and H ′, every homomorphismα : H → H ′ can be extended to a chain map

F2 F1 F0 H 0

F ′2 F ′1 F ′0 H ′ 0

α α α α

and any two such extensions are chain homotopic. In particular, for any two free resolutions Fand F ′ of H, there is an isomorphism of groups Hn(F ;G) ∼= Hn(F ′;G).

Proof. Omitted. (something something surjectivity and exactness)

We now understand the cokernel to some extent: it’s the first homology group of some freeresolution, and by the previous result the specific resolution doesn’t matter.

Definition 55. This first homology group is denoted by Ext(H,G).

Theorem 29 (Universal Coefficient Theorem). If a chain complex C of free abelian groups hashomology groups Hn(C), then the cohomology groups Hn(C;G) are determined by the split exactsequence

0 −→ Ext(Hn−1(C), G) −→ Hn(C;G) −→ hom(Hn(C), G) −→ 0.

Let’s look at some examples.

Example 62. What’s Ext(Zn, G)? To compute this, first remark that the free resolution is

0 −→ Z n−→ Z −→ Zn −→ 0.

The dual of this is

0←− Ext(Zn, G)←− hom(Z, G)←− hom(Z, G)←− hom(Zn, G)←− 0.

Now hom(Z, G) ∼= G, and mult by n stays preserved under hom (why?), and so Ext(Zn, G) =G/nG.

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The next few (unproven) propositions detail how to compute Ext(H,G) whenever H and G arefinitely generated Abelian groups.

Proposition 27. We have

• Ext(H ⊕H ′, G) ∼= Ext(H,G)⊕ Ext(H ′, G);

• Ext(H,G) = 0 if H is free;

• Ext(Zn, G) ∼= G/nG.

Proposition 28. If Hn(C) and Hn−1(C) are finitely generated with torsion subgroups Tn−1 ⊆Hn−1 and Tn ⊆ Hn, then

Hn(C;Z) ∼= Hn

Tn⊕ Tn−1.

38.2 Why cohomology?

As stated before, the Universal Coefficient Theorem tells us that the cohomology groups are deter-mined uniquely from the homology groups, but cohomology allows for some extra structure thatis not present in homology. Let’s mention briefly this structure now.

Suppose we wanted to endow the sequence of homology groups Hi(X) with a product structure.It is easy to construct a product going from Hi(X)×Hj(X)→ Hi+j(X ×X), but projecting thisdown to Hi+j(X) is tricky. However, with cohomology, the sequence

0 −→ Hi(X)×Hj(X)→ Hi+j(X ×X)→ Hi+j(X)

is somewhat more natural: because of the nature of cohomology, the latter arrow is induced bymaps from X → X × X, and there is a natural candidate for such a map: the diagonal map∆(x) = (x, x).

As an added bonus, all the things we showed for homology are the same for cohomology. Forexample, relative homology works similarly as before. We show this now.

Recall that0 −→ Cn(A) −→ Cn(X) −→ Cn(X,A) −→ 0

is a short exact sequence of chain complexes. The dual of this is the sequence

Cn(A;G)←− Cn(X;G)←− Cn(X,A;G)←− 0.

One can check that we can add zero to left hand side and the sequence still remains exact, whichmeans we can construct the same long exact sequence on homology from a short exact sequenceof cochain complexes.

Definition 56. Define Hn(X,A;G) via δ : Cn(X,A;G)→ Cn+1(X,A;G) by naturally restrictingδ : Cn(X;G) → Cn+1(X;G). Then this creates a short exact sequence of chain complexes withLES in homology

· · · −→ Hn(X,A;G) −→ Hn(X;G) −→ Hn(A;G) −→ Hn+1(X,A;G) −→ · · · .

Remark. As usual, the main difficulty lies in determining the connecting homomorphism δ :Hn(A;G) → Hn+1(X,A;G). There in fact exists a duality relationship between δ and the con-necting homomorphism ∂ : Hn+1(X,A) → Hn(A); this is realized via the diagram below, whichhappens to be commutative.

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

hom(Hn(A);G) hom(Hn+1(X,A);G)∂∗

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39 December 3

39.1 Mayer-Vietoris

There is an analogue of the Mayer-Vietoris sequence for cohomology.

Theorem 30 (Mayer-Vietoris for Cohomology). If X = A◦ ∪ B◦, then there is a long exactsequence in homology

· · · −→ Hn(X;G) −→ Hn(A;G)⊕Hn(B;G) −→ Hn(A ∩B;G) −→ Hn+1(X;G) −→ · · · .

Proof. This follows immediately from dualizing the short exact sequence of chain complexes

0 −→ Cn(A ∩B) −→ Cn(A)⊕ Cn(B) −→ C{A,B}n (X) −→ 0;

note in particular this is the same short exact sequence as was from the Mayer-Vietoris sequencefor homology.

Remark. Recall the dunce cap, consisting of three vertices v0, v1, and v2 in that ordering, andset G = Z. If α is a 1-chain, what does δα = 0 mean? Well, letting σ be the inner triangle, thecondition rewrites as

0 = δα(σ) = α(∂σ) = α([v0, v1] + [v1, v2]− [v0, v2]) = α[v0, v1] + α[v1, v2]− α[v0, v2]. (39.1)

Thus δα = 0 if this condition on the numbers assigned to the edges of σ is satisfied.

Example 63. Let’s consider the simplicial homology of S2. The ∆-complex of S2 is a tetrahedron.We can construct any cocycle on this ∆-complex by assigning arbitrary values to the edges of somespanning tree; then, the relation in Equation 39.1 uniquely determines the values of the edges notcontained in this spanning tree. Two examples of cocycles are shown below.

2

46

157

1

12

123

Figure 37: A ∆-complex for S2 as well as two different cocyles.

What’s the first cohomology group of the sphere? In this case, recall that the homology groupsare free, and so H1(S2;Z) ∼= hom(H1(S1;Z) = 0. This means that the above cocycle is also acoboundary, meaning that we can assign values to the vertices such that the number assigned toeach edge is the difference between the numbers along the vertices. Such a construction is easy inthis case: assign the lower-left corner vertex a 0, and then all other labels are forced.

39.2 Cup Product

We mentioned previously that cohomology admits a more natural product structure than homologydoes. We explore this in an example before diving into the general formula.

Example 64. Let α and β be the two cocycles in the previous example from left to right. We’llshow how to compute a product of α and β, which is a two-dimensional cocycle we will call α ^ β.(The order here is important: this product is not commutative!)

Here’s how we’ll do it. Note that the arrows in the ∆-complexes above induce orderings of thevertices on each triangle. For an arbitrary 2-cell e2, suppose the vertices are v0, v1, and v2 in thisorder. Then the value assigned to e2 in the product α ^ β is the number assigned to [v0, v1] in

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2

46

4

Figure 38: The product of the two cocycles in Example 63.

α with the number assigned to [v1, v2] in β. For example, the bottom-most triangle is assigned avalue of 2 ·1 = 2. Performing this computation for all four triangles results in the above 2-cochain.

At first glance this seems quite random, but the interesting thing is that it results in a cocycle.In other words, we can assign values to the edges of the tetrahedron above so that the relation inEquation 39.1 holds at every triangle.

We now present the general construction.

Definition 57. Let R be a ring, and set ϕ ∈ Ck(X;R) and ψ ∈ C`(X;R). We’ll define the cupproduct ϕ ^ ψ ∈ Ck+`(X;R) via

(ϕ ^ ψ)(σ) = ϕ(σ|[v0,...,vk])ψ(σ|[vk,...,v`])

for every (k + `)-cocycle σ. This is essentially rigorizing/generalizing the multiplication processfrom above.

Remark. The following things are true.

• For any ϕ ∈ Ck(X;R) and ψ ∈ C`(X;R),

δ(ϕ ^ ψ) = δϕ ^ ψ + (−1)kϕ ^ δψ. (39.2)

Proof: “literally write down both sides”.10

• If δϕ = δψ = 0, then δ(ϕ ^ ψ) = 0. This follows from Equation 39.2 and the fact that thecup product of a trivial cochain with anything else is a trivial cochain.

• The product of a cocycle and a coboundary is a coboundary. Indeed, if δϕ = 0, then Equation39.2 tells us that

ϕ ^ δψ = ±(δ(ϕ ^ ψ)).

So ^ is well-defined on cohomology.

Example 65. Let M = M2 be the orientable surface of genus 2. Recall from Example 51 thatH0(M) ∼= Z, H1(M) ∼= Z4, and H2(M) ∼= Z. Since all these groups are free, the cohomologygroups Hj(X) precisely isomorphic to hom(Hj(X);Z). This yields

H0(M ;Z) ∼= Z, H1(M ;Z) ∼= Z4, and H2(M ;Z) ∼= Z.

Our main goal is to determine how the cup product H1(M)×H1(M)→ H2(M) behaves. Recallthat in Z coefficients, a basis for H1(M) is formed by the edges ai and bi for i ∈ {1, 2}. A basisfor H1(M) determines a dual basis for hom(H1(M), Z); thus, dual to ai is the cohomology classαi assigning the value 1 to ai and 0 to the other basis elements. The basis elements βi dual to biare constructed similarly.

In order to represent αi by a simplicial cocyle ϕi, we need to choose values for ϕi on the edgesradiating out from the central vertex in such a way that δϕi = 0. Simply setting these to zerodoes not work, as the cocycle condition is no longer satisfied. Instead, consider the dashed arc αiin the diagram below; then all edges which intersect this dashed arc get a 1 and all others get a 0.It is easy to check that this assignment of values to the edges is indeed a cocycle. The cochains ψifor the βis are determined similarly.

10This is Lemma 3.6 in Hatcher.

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Figure 39: Generating the characteristic functions on M2.

Now we may proceed with the computation. First consider ϕ1 ^ ϕ1. Recall that in order tocompute the cup product, on each 2-simplex [v0, v1, v2] we multiply the value assigned by ϕ1 tothe edge [v0, v1] by the value assigned by ϕ1 to the edge [v1, v2]. One can see that this assigns 0to each 2-simplex in M1, and so ϕ1 ^ ϕ1 (and hence α1 ^ α1) equals 0. Similarly, α1 ^ α2 = 0and α1 ^ β2 = 0. However, α1 ^ β1 is nontrivial (consider the positively-oriented triangle withouter edge b1); in fact, a little work shows that α1 ^ β1 is precisely the generator of H2(M) ∼= Z.

Remark. Cohomology essentially tests how many things intersect up to homotopy. For example,α1 and β2 do not intersect, and two copies of α1 can be pulled apart to not intersect, but α1 andβ1 will never cease to intersect.

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40 December 5

We being with another example demonstrating what the cup product is.

Example 66. LetX = RP 2; we’ll work over the group Z2 for simplicity. RecallHk(RP 2;Z2) ∼= Z2;since torsion doesn’t exist in this restricted setting, Hk(RP 2;Z2) ∼= Z2 for 0 ≤ k ≤ 2.

How does the cup product behave on X? Consider the structure for RP 2 given below.

e1 e2

e

e

vw w

Figure 40: Another visual for RP 2.

Since H1(RP 2;Z2) ∼= Z2, we have the existence of a nontrivial element in cohomology. A bit ofexperimentation yields that the cocycle β with β(e1) = β(e) = 1 and β(e2) = 0 does the trick, asany assignment of values to the vertices v and w must induce equal values for the edges e1 and e2.(Note that this is essentially unique; although the assignment γ(e) = γ(e2) = 1 and γ(e1) = 0 isalso nontrivial, they differ by an assignment of the edges which is trivial in cohomology.)

Using the cup product formula, we may compute β ^ β(σ) = 0 · 1 = 0, β ^ β(τ) = 1, where σis top face, τ is bottom face. This means that β ^ β is nontrivial in H2.

Since the cup product is associative and distributive, it is natural to try to make it the multipli-cation in a ring structure on the cohomology groups of a space X. To do this, we will need somedefinitions.

Definition 58. Let A be a ring. We say that A is a graded ring if there exists a decomposition

A =⊕k≥0

Ak

where each Ak is an additive subgroup of A, such that multiplication takes Ak ×A` to Ak+`.

Definition 59. The graded ring

H∗(X;R) :=⊕k≥0

Hk(X;R)

equipped with the cup product structure for multiplication is called the cohomology ring of X.

We may immediately apply our definition to obtain an interesting result.

Proposition 29. We have the equality H∗(RP 2;Z2) ∼= Z2[β]/(β3 = 0).

Proof. Let 1, β, and β2 be the generators for Hi(RP 2;Z2), where i = 0, 1, 2 respectively. Then (insuggestive notation) we can see that

1 · 1 = 1, 1 · β = β, 1 · β2 = β2,β · β = β2, β · β2 = 0, β2 · β2 = 0,

from which the desired isomorphism follows. (Note that modding out by the ideal (β3 = 0) resultsfrom the fact that all the nth cohomology groups are trivial for n ≥ 3.)

Remark. In general,H∗(RPn;Z2) ∼= Z2[β]/(βn+1 = 0).

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Note: one can show α ^ β = (−1)k`(β ∪ α), where α ∈ Hk and β ∈ H`. Thus a sort of strangeanti-commutativity occurs, but in Z2 this just turns into commutativity because (−1)k` = 1regardless of what k and ` are.

Remark. We can use this to show that H∗(RP 4 ∧ S5;Z2) 6= H∗(RP 5;Z2), even though thesetwo spaces have the same homology groups. Consider the (characteristic cochains of the) two-dimensional and three-dimensional faces α and β in RPn for n ∈ {4, 5}; their product α ^ β isthe (characteristic cochain of the) five-dimensional face in RP 5 but is completely zero in RP 4 andhence trivial in RP 4 ∨ S5.

40.1 Duality

One might get the sense that homology and cohomology are “opposites” in some sense and wonderwhether this sense can be strengthened. It turns out that in a specific class of topological spaces,it can, and it leads to one of the most important parts of the theory of cohomology.

We motivate this in the following way. Consider a CW complex C consisting of a grid of squares,as shown below. We can construct a “dual” CW complex C ′ by connecting the centers of thesesquares to form another grid structure.

Figure 41: A CW complex and its dual.

Note that there are several bijections between faces of C and faces of C ′. In particular,

• Every vertex of C is contained in a unique square of C ′, and vice versa; and

• Every edge of C intersects a unique edge of C ′.

These bijections can be used to establish parallels between homology in C and cohomology inC ′. Observe the following.

• Let σ be a two-dimensional face of C. Then the boundary map ∂, when applied to σ, returnsa sum of the edges which are faces of σ, i.e. a linear combination of the four sides of thesquare σ.

• Now let σ′ be the vertex of C ′ in the middle of σ. Then the coboundary map δ, when appliedto σ′, returns the a sum of edges for which σ′ is a face, i.e. a linear combination of the fouredges eminating from σ′.

These two sets of edges are in bijective correspondence with each other according to the bijectionsmentioned previously! This suggests homology in C and cohomology in C ′ are actually the samething and thus, from the fact that C and C ′ are identical, we guess that the second homologygroup of C equals the first cohomology group of C ′.

Of course, this doesn’t always hold - in particular, the bijection breaks down if this grid ofsquares ends at some point. We now give some definitions that give light to exactly where thissymmetry exists.

Definition 60. M is an n-manifold if

• M is Hausdorff and second-countable;

• every x ∈M has a neighborhood homeomorphic to Rn.

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Definition 61. A manifold M is a closed manifold if it’s compact.

Definition 62. A local orientation of M at x ∈M is a choice of generator µn of Hn(M,M\{x}) ∼=Z.

Definition 63. An orientation is a map x 7→ µx such that the map Hn−1(M,M \U)→ Hn(M,M \{y}) gives µy and similarly for µx. Here U is a neighborhood around x and y homeomorphic toRn. (“fancy language is sheaf”)

We now present our first of two important theorems. Unfortunately, the proof is omitted, butyou probably know where to find it.

Theorem 31. If M is a closed connected orientable n-dimensional manifold, then the map Hn(M)→Hn(M,M \ {x}) is an isomorphism for all x. In particular, M is orientable iff Hn(M) ∼= Z, oth-erwise Hn(M) = 0. This generator is called fundamental class, denoted [M ] ∈ Hn(M).

We now formalize the intuition from above by constructing a “dual” operator to the cup product.

Definition 64. Let R be a ring. Let _: Ck(X;R)× C`(X;R)→ Ck−`(X;R) via

σ _ ϕ := ϕ(σ|[v0,...,v`])σ|[v`,...,vk].

It is easily checked that this induces a map Hk(X;R) ×H`(X;R) → Hk−`(X;R). Indeed, onecan check that

∂(σ _ ϕ) = (−1)`(∂σ _ ϕ− σ _ δϕ).

All these definitions lead to the main result of this lecture. Its proof is again omitted.

Theorem 32 (Poincare duality). If M is a closed orientable n-manifold with fundamental class[M ] ∈ Hn(M ;Z) then the map D : Hk(M ;Z)→ Hn−k(M ;Z) which takes a cocycle α and maps itto [M ] _ α is an isomorphism for all k. The same is true for (not not necessarily) orientable Mwith Z2 coefficients.

This theorem admits an easy but important corollary.

Corollary 12. Let M be a closed (2n+ 1)-dimensional manifold. Then χ(M) = 0.

Proof. Use Theorem 26, symmetry, and the universal coefficient theorem.

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41 December 7

We’ll finish the semester by exploring applications of topology to problems which seemingly haveno connections to topology.

Let G be a graph. How many colors are required to color the vertices of G such that adjacentvertices get different colors? This problem is not inherently topological, but we may rephrase thisquestion in terms of graph homomorphisms.

Definition 65. Let G and H be graphs. A graph homomorphism between G and H is a mapf : V (G) → V (H) such that if (v, w) ∈ E(G), then (f(v), f(w)) ∈ E(H). (Notice that this is notan iff statement!)

Proposition 30. A graph G can be colored in n colors iff there exists a graph homomorphismfrom G to Kn.

Proof. First suppose a graph G can be colored in n colors, labeled 1, 2, . . . n. Consider the functionf : V (G) → V (Kn) which sends all vertices with color i to vertex i ∈ V (Kn). Note that this isa graph homomorphism because any two adjacent vertices in G have different colors and are thussent to different (and hence adjacent) vertices in Kn.

Conversely, given a graph homomorphism from G to Kn, we can reverse the above proof toestablish a coloring on the vertices.

Unfortunately, graphs are strange because sometimes (often times!) there is no homomorphismbetween two arbitrary graphs, and homomorphisms in general don’t give rise to topological spaces.We thus need a bit of a workaround.

Definition 66. Let G and H be graphs. Define hom(G,H) as the collection of maps η : V (G)→2V (H) \∅ such that if (v, w) ∈ E(G) then η(v)× η(w) ⊆ E(H).

Example 67. Consider the two graphs G and H shown below. Then hom(G,H) consists ofprecisely those maps η for which η(2) and η(3) are disjoint; if η(2) ∩ η(3) contains some vertex v,then the hom condition implies (v, v) ∈ E(H), which is not possible.

1

2

3

1

2 3

G H

Figure 42: Two graphs G and H.

Note that this definition forces a partial ordering on the elements of hom(G,H), where two suchelements η and η′ satisfy η ≤ η′ iff η(v) ⊆ η′(v) for all v ∈ V (G). This is great, because anypartially ordered set P gives rise to a simplicial complex ∆ in a natural way: elements of ∆ arevertices of P, and for all k ≥ 1, (k + 1)-dimensional faces of ∆ are chains in P of length k.

Proposition 31. Any graph homomorphism f : H → H ′ induces a continuous homomorphismf∗ : hom(G,H)→ hom(G,H ′).

Proof. For arbitrary η ∈ hom(G,H), define f∗η : V (G) → 2V (H′) \ ∅ by f∗η = η ◦ f . Thisworks.

Example 68. Let G = K2. Our goal is to determine what hom(K2, H) is for arbitrary graphs H.

Suppose H has vertex set {1, 2, . . . , n} and K2 has vertex set {1, 2}. By following Example67, we see that elements of hom(K2, H) correspond to (distinguishable) disjoint pairs of sub-sets of the vertices of H. To model this, we can pick the vertices from the 2n-element set{e1, . . . , en,−e1, . . . ,−en} and say that elements of hom(K2, H) are subsets of the form

{ei1 , . . . , eik ,−ej1 , . . . ,−ej`} with (ip, jq) ∈ E(H) for all 1 ≤ p ≤ ik, 1 ≤ q ≤ j`.

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This is about as far as we can go in the general case, but when looking at specific H someinteresting situations can arise. For instance, set H = K3. Then a quick computation tells usthere are 12 elements of hom(K2,K3), and the relevant partially ordered set has Hasse diagramshown below. As a topological space, this is homotopic to S1! In fact, it is generally true thathom(K2,Kn) ' Sn−2.

Figure 43: The Hasse diagram for hom(K2,K3). The six subsets withonly two elements are in the bottom row, while the sixsubsets with three elements are in the top row.

Notice from the previous example that hom(K2, H) has a properly discontinuous Z2 symmetrythat flips K2: for all j, send ej ↔ −ej .

With this, we may finally bridge the gap between topology and our coloring problem. Supposea graph homomorphism f : H → Kn exists, where n ≥ 1. We may then use Proposition 31 toinduce a homomorphism

f∗ : hom(K2, H)→Z2 hom(K2,Kn) ' Sn−2.

(Here A→Z2B indicates a map which respects the Z2 actions on A and B.) Suppose further that

we may also map Sn−1 →Z2 hom(K2, H), where the Z2 action on Sn−1 is the natural antipodalmap. Then we obtain a contradiction, since Borsuk-Ulam implies their composition cannot beproperly discontinuous.

Proposition 32. Let X be a space with free (i.e. properly discontinuous) Z2 action. ThenSn−1 →Z2

X exists if X is simply connected and Hk(X;Z) vanish for k ≤ n− 2.

Proof. The essential idea is to map it dimension by dimension. Let x ∈ Sn−1 be arbitrary, andmap f(x) ∈ X; then the Z2 condition implies f(−x) is forced. Now “arbitrarily” map one half ofa great circle passing through x and −x in Sn−1 to X; then the image of the second half of thiscircle is forced from the Z2 action. Rinse and repeat. The Hk vanishing condition ensures that wecan map both halves without difficulty.

We are now ready to state the main theorem.

Theorem 33. Let G be a graph. Then

χ(G) ≥ conn hom(K2, G) + 3,

where for a simply connected space K, connK is the largest n such that the first n homology groupsvanish.

Example 69. Consider the graph G with vertices all k-element subsets of [n], and such that(A,B) ∈ E(G) iff A ∩B = ∅. For example, if n = 5 and k = 2, this yields the Petersen graph. In1955, Martin Kneser conjectured that

χ(G) = n− 2(k − 1).

This was proven by Lovasz in 1978 using topological methods. A purely combinatorial proof wasnot found until 2004 by Jirı Matousek.

Remark. One might wonder whether these ideas can be used to prove results about hypergraphcolorings. The answer is ‘yes’, but one needs to generalize Borsuk-Ulam to non-Z2 actions, and sofar we only know how to do this when n is a power of a prime.

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algebraic topology f18 · algebraic topology f18 david altizio january 12, 2019 the following notes...

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