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Home , Covering space, CW complex, Hawaiian earring, Retraction (topology), Simply connected space, Subspace topology

Algebraic F18

David Altizio January 12, 2019

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

Contents

1 August 26 4

2 August 28 4 2.1 Introduction ...... 4 2.2 Topology Crash Course, Part I ...... 4

3 August 31 7

4 September 5 10 4.1 Opening Remarks ...... 10 4.2 Quotient Maps and Examples ...... 10 4.3 CW Complexes ...... 12

5 September 7 13 5.1 More on CW complexes ...... 13 5.2 Retractions and ...... 14

6 September 10 16 6.1 The Extension Property ...... 16

7 September 12 19 7.1 Introducing the ...... 19

8 September 14 21 8.1 Basepoint Independence ...... 21 8.2 The Fundamental Group of the Circle ...... 22

9 September 17 23 9.1 Generalizations of IVT ...... 23 9.2 Other Important Theorems ...... 23

10 September 19 25

11 September 21 26

12 September 24 27

13 September 26 29 13.1 Covering Spaces ...... 29

14 September 28 32 14.1 Simply Connected Covering Spaces ...... 32

1 David Altizio 21-752 Lecture Notes

15 October 1 34 15.1 More Simply Connected Covering Spaces ...... 34 15.2 Some More Group Theory ...... 34 15.3 Uniqueness of covering spaces ...... 35 15.4 The punchline ...... 35

16 October 3 36 16.1 Covering Spaces for S1 S1 ...... 36 16.2 Deck Transformations∨ ...... 36

17 October 5 38 17.1 Examples of Deck Transformations ...... 38 17.2 Group Actions ...... 38

18 October 8 40 18.1 Some “Last” Words on Covering Spaces ...... 40 18.2 Van Kampen’s Theorem ...... 40 18.3 Knots ...... 42

19 October 10 43 19.1 The , Revisited ...... 43 19.2 Knots ...... 43 19.3 Gluing 2-Cells ...... 44 19.4 Fundamental Groups of Orientable Surfaces ...... 45

20 October 12 46 20.1 Links ...... 46 20.2 Wirtinger Presentations ...... 46 20.3 Some Parting Examples ...... 47

21 October 15 48 21.1 Building the Intuition for ...... 48

22 October 17 51 22.1 Simplicial Homology ...... 51

23 October 19 53

24 October 22 53 24.1 ...... 53 24.2 Connecting our Two Homologies ...... 53

25 October 24 55

26 October 26 55

27 October 29 55

28 October 31 57 28.1 Relative Homology Groups ...... 57

29 November 2 60 29.1 Excision ...... 60

30 November 5 63 30.1 Good Pairs ...... 63 30.2 Generators of Homology Groups on ∆ -complexes ...... 64

31 November 7 66 31.1 The Home Stretch ...... 66

2 David Altizio 21-752 Lecture Notes

32 November 9 68 32.1 Mayer-Vietoris sequences ...... 68

33 November 12 70 33.1 Degree ...... 70 33.2 for CW Complexes ...... 71

34 November 14 72 34.1 Cellular Homology ...... 72

35 November 16 75 35.1 ...... 75

36 November 26 77 36.1 Homology with coefficients ...... 77 36.2 A Tiny Introduction to ...... 78

37 November 28 79 37.1 A Less Tiny Introduction to Cohomology ...... 79

38 November 30 82 38.1 Free Resolutions ...... 82 38.2 Why cohomology? ...... 83

39 December 3 84 39.1 Mayer-Vietoris ...... 84 39.2 Cup Product ...... 84

40 December 5 87 40.1 ...... 88

41 December 7 90

3 David Altizio 21-752 Lecture Notes

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 spaces with morphisms being continuous functions. Recall that a f : X Y between two metric spaces is continuous if for all x X and for all ε > 0 there exists→δ < 0 such that ∈ dX (x, y) < δ implies dY (f(x), f(y)) < ε. Naturally, the isomorphisms between metric spaces are those continuous functions which also have inverses - that is to say, . 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 are the 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 if f −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 centered at x with radius r.) By the definition⊆ of continuity, there exists δ > 0 such that −1 −1 y Bδ(x) f(y) Bε(f(x)) U. Thus Bδ(x) f (U), and so f (U) is open. ∈ ⇒ ∈ ⊆ ⊆ ( ) Suppose f −1(U) is open for every open U Y . Let x X and ε > 0 be arbitrary. • ⇐ ⊆ −1∈ Then in particular V := Bε(f(x)) is open, and thus so is f (V ) x. Hence we can find −1 3 δ > 0 such that Bδ(x) f (V ), meaning that for any y X with dX (x, y) < δ we have ⊆ ∈ f(y) V = Bε(f(x)) dY (f(x), f(y)) < ε. Since x and ε were arbitrary, we deduce that f is continuous.∈ ⇒

Thus, while continuous functions do not preserve distances, they do preserve open sets. This means 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 is a set X with a collection of (called a topology) such that F ∅ and X ; • ∈ F ∈ F if U and V , then U V ; • ∈ F ∈ F ∩ ∈ F S If Uα for all α in some index set A, then Uα . • ∈ F α∈A ∈ F In other words, is closed under both finite intersections and arbitrary unions. F Definition 2. Let (X, τX ) and (Y, τY ) be two topological spaces. A function f : X Y is −1 −1 → continuous if f (U) τX for all U τY . If additionally f : Y X exists and is continuous, then f is called a ∈ . ∈ →

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

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

4 David Altizio 21-752 Lecture Notes

Definition 4. A topological space X is Hausdorff if for all x, y X with x = y, there exist open ∈ 6 sets U and V in τX such that x U, y V , and U V = ∅. In essence, we can “separate” any two 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 lies in the following theorem. Theorem 1 (Urysohn’s Metrization Theorem). Let X be a topological space which is Hausdorff; • regular, meaning that for E X and any x X E there exist open sets U and • ⊆ ∈ \ V in τX with x U, E V , and U V = ∅; and ∈ ⊆ ∩ second-countable, meaning that there exists a countable collection Ui i∈ τX of subsets in • S { } N ⊆ τX such that for all V τX there exists I N such that V = Ui. ∈ ⊆ i∈I Then the topology on X is induced by some metric.

Proof. Omitted.

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

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

Remark. Note that the Yi are topological spaces, so in particular they already have topologies rigged on them; we thus need the topology on A to be “compatible” with all the topologies on the Yi. 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 f : 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 means U 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 ! [ [ [ Ui = (Xi A) = Xi A, ∩ ∩ i∈I i∈I i∈I S and so Ui τA. i∈I ∈ 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 τA ⊇ ∈ is 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 thus ∈ deduce τ τA. Combining this with the above work yields the desired. ⊇ The next example is important enough to warrant its own definition.

5 David Altizio 21-752 Lecture Notes

Q Definition 6. Given topological spaces Xi for i I, the on Y := Xi is the ∈ i∈I initial topology with respect to projections on the factors Xi. In other words, it is the coarsest topology on Y such that each of the functions πi : Y Xi is continuous. → Remark. The initial topology can be classified by the following not-quite-universal property: for any topological space Z, a function g : Z X is continuous if and only if the function fi g : Z Yi is continuous for each i. → ◦ →

fi X Yi

g fi◦g

Z

6 David Altizio 21-752 Lecture Notes

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 and A◦ = U C⊇A U⊆A C closed U open to be the and 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 open with 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 complements in∪ [0, 1], U and V are also closed. ∩ WLOG let 1 V , and set ∈ u = sup x : x U . { ∈ } We claim that u is in both U and V , contradicting the fact that U V = ∅. First note that because U is closed, the supremum is actually a maximum, so in particular∩ u U. Now define 1 ∈ un := u + for all n N, and remark that since u < 1, un V for all sufficiently large n. Then n ∈ ∈ un u as n , but V is closed, so by the sequential characterization of closed sets we have u →V as well.→ Done. ∞ ∈ We now define something which is stronger than .

Definition 9. A topological space X is -connected if for any x, y X, there exists some 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 disjoint open 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)∈ = x and γ(1) = y. Remark that γ is the continuous of a connected set→ and is thus connected. But now Uγ := γ([0, 1]) U and Vγ := γ([0, 1]) V ∩ ∩ are two open sets with respect to the on γ([0, 1]) which it but are disjoint, contradiction.

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

γ := x, sin 1  : x [0, 1] = x, sin 1  : x [0, 1] (0, x): x [ 1, 1] . x ∈ x ∈ ∪ { ∈ − } 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, as any{ 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 Q 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

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

7 David Altizio 21-752 Lecture Notes with the product topology. Then X is compact but not sequentially compact. (Proof of this via {0,1}N math.stackexchange: write R = 0, 1 N, and define a sequence fn [0, 1] 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 a standard diagonalization argument, since it so happens that the product topology is the topology of 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, and ∈ let fi : Yi X be functions. The final topology on X is the finest topology that makes all the fi continuous.→ Remark. The final topology can be classified by the following not-quite-universal property: for any topological space Z, a function g : X Z is continuous if and only if the function fi g : Yi Z is continuous for each i. → ◦ →

fi Yi X

fi◦g g

Z

As with the initial topology, we introduce some examples - the first is a small example, while the second is important enough to warrant a definition. F Example 3. Suppose Yi (i I) are topological spaces, and X = Yi. The correct topology ∈ F i∈I on X is the final topology induced by the inclusions Yi , Yi. → i∈I 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 from the 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 a continuous 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 then so is f(U) Y , which in turn is equivalent 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 an open cover⊆ of X, extract a finite subcover, and remove everything you don’t want). Thus, since f is 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 x and Vx y which are disjoint. Now S ∈ 3 3 x∈f(C) Ux is an open cover of f(C), so by compactness we may find points x1, . . . , xk such that

k [ f(C) Ux . ⊆ k i=1

Sk c Let V = i=1 Vxk . Then V is an open set containing y which lies exclusively in f(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 in ' R2). To prove this, define g : [0, 1] S1 via θ e2πiθ; note that g(0) = g(1) and that g is → 7→

8 David Altizio 21-752 Lecture Notes continuous. Denote by q the quotient map from [0, 1] to X. From g(0) = g(1) we know that there exists a well-defined map p : X S1. Then we have the following commutative diagram; in particular, since X is rigged with a final→ topology, g continuous p continuous. Note further that p is bijective. ⇒

q [0, 1] [0, 1]/(0 1) ∼

g p

S1

It remains to prove that p is actually a homeomorphism. But this is a consequence of the previous lemma, 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: suppose X is a topological space, and A X. If we say “A is compact”, what does that mean? That A with respect to the subspace topology⊆ is compact, or that any open cover of A in X has a finite subcover? The answer is that there is no difference. Such properties are called “intrinsic”. As another example, connectedness is intrinsic, but only if you define connectedness of a subspace in the right way. (Cue five minutes of confusion and counterexamples that Florian claims will be cleared up Wednesday.)

9 David Altizio 21-752 Lecture Notes

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 V where U and V are open. We say that A X is connected if A with the subspace topology∪ is connected. 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 being connected under the subspace topology. Proposition 3. A space X is connected iff any continuous map f : X D, where D is a , 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 for → {−1 } any set Y 1, 2 , f (Y ) is either ∅, U, V , or X, and all of these are in τX . It follows that 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 consider • the⇒ set U := f −1( y →). Remark that U is nonempty by definition. Furthermore,∈ y is both open 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 spaces in more detail. Recall that a quotient map is the final topology on some surjective map. Most of the rest of this lecture will be spent discussing tons of examples. (Note: some of the finer details are 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 the equivalences (x, 0) (x, 1) and (0, y) (1∼, y) for all x, y [0, 1]. Then ∼ ∼ ∈ 2 1 1 [0, 1] / = S S . ∼ ∼ × 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 can proceed as in example 4; the bijection (x, y) ((cos x, sin x), (cos y, sin y)) suffices.× To embed the 7→ torus in R3, we instead use (for example) the bijection

(x, y) (2 cos x + cos x cos y, 2 sin x + sin x cos y, sin y), 7→ where here we assume that radii of the two circles making up the torus are 2 and 1.

Figure 1: A torus.

10 David Altizio 21-752 Lecture Notes

Example 6. Let D2 = (x, y) R2 : x2 + y2 1 be the closed disk in R2 of radius 1 centered at the origin. I claim that { ∈ ≤ } 2 D 2 S . ∂D2 ' (Here X/A is equivalent to modding X by the relation on A for which all points are equivalent to each other; such a space is called an .) To prove this, note that again it suffices to find a continuous bijection from D2/∂D2 to S2. The explicit bijection is difficult to write down, so what follows is intuition. Imagine an ant starting at the center of the disk D2 and traveling to the 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 are mapped 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 x ∼ ∼ − for all x ∂D2. We define D2/ to be the real RP 2.2 ∈ ∼ Let us first figure out what RP 2 is intuitively. Consider the partition of D2 into light gray and dark 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 dark gray 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 S1 D2 - two filled-in donuts. We’ll × t × analyze one way to glue these donuts together. Consider the function f : S1 S1 S1 D2 such that f(x, y) = (y, x). We write the quotient map of X with respect to the× equivalence→ × relation 2 1 1 2 3 generated by f(x) x as the space Y := D S f S D . Now we claim that Y S , the ∼ × ∪ × ' 3- in R4. For the geometric intuition behind this, consider two interlinked donuts such that one torus passes through the hole of the other torus and vice versa. Now fill in the spaces between these holes. Then what’s left is essentially a round object with no holes in its interior, i.e. a 3-sphere. Here’s the algebraic reasoning (i.e. more formal explanation) for this. Recall that

3 4 2 2 2 2 S = (u, v, x, y) R : u + v + x + y = 1 . { ∈ } 2 2 2 Many other resources seem to define RP as the quotient space of S under this antipodal equivalence relation 2 instead of D ; these seem to be equivalent by the previous example.

11 David Altizio 21-752 Lecture Notes

Consider the surface

4 2 2 1 2 2 1 1 T = (u, v, x, y) R : u + v = = x + y = S S . { ∈ 2 } ∼ × Now T cuts S3 into two components, and it is easy to see these two spaces are homeomorphic (in particular x2 + y2 < 1 u2 + v2 > 1 ). These spaces are both filled-in tori, ergo they are both 2 ⇔ 2 homeomorphic to D2 S1. × 4.3 CW Complexes

We finish today by introducing a fundamental topological idea.

n 0 Definition 13. Define a sequence of topological spaces X n∈N as follows. Let X be a discrete space. Proceeding inductively, assuming Xn−1 has been{ defined,} let Xn be obtained from Xn−1 n−1 n n−1 as a quotient of X α Dα (i.e. a disjoint union of X with some number of n-dimensional t n−1 n−1 n n−1 disks) via gluing maps fα : S X . Note that it is possible for X = X . Then the topological space → [ X := Xn n∈N 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 z2. The point of this mapping is • that it goes around the boundary of the circle twice; hence7→ antipodal points on the circle are mapped to the same point on the disk. Remark. We will usually only study spaces that are CW complexes; even Hatcher states at a certain point that all spaces henceforth will be CW complexes unless otherwise specified.

12 David Altizio 21-752 Lecture Notes

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 alternate between 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 to look 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 many • cells of lower . Note that by “finitely many cells” we mean cells belonging to the same complex. This leads to some strange anomalies that seem unintuitive at first. For example, a 2-cell is not actually a CW complex since its boundary consists of infinitely many 0-cells (we can’t cover it with a single 1-cell because we don’t have any of those to work with!). 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 “”. This means that A X is closed iff A Xn is closed • in Xn for all n . Remark that this makes sense because⊆ of the definition X∩ = S Xn, N n∈N so despite the fact∈ that each space is built by a gluing process all of our sets can be assumed to live in the same ambient space. Now that we know what the letters stand for, we can formulate a sensible condition to determine when 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 collection • of all cells⊆ with dimension ⊆ n, ≤

13 David Altizio 21-752 Lecture Notes 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 the ideas in the discipline with the following example.

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

2 2 Proof. Suppose FSOC that there exists a homeomorphism f : R R . Then R 0 ∼= R f(0) as well. But the LHS is a disconnected topological space while the→ RHS is a connected\{ } topological\{ } space, contradiction.

We might suspect that this same technique can be used to prove that an m-cell is not the same as an n-cell for any m = n. However, while this type of argument can be made to work, it turns 6 out to be a lot more complicated in general. As an example, suppose we wanted to prove R2 = R3. 6∼ Then we could in theory take a line ` in R2 and write 2 3 R ` = R f(`) \ ∼ \ where f is some homeomorphism from R2 to R3. However, this isn’t so simple, as we don’t quite know yet what f(`) is. (For all we know, it could be a space-filling curve!) We will thus need more powerful machinery to prove this result, and this is one of the problems that algebraic topology is able to answer. The above result also suggests homeomorphism might be too strong in the following way. Using the 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 two spaces are still kinda similar - after all, S1 can be thought of as an annulus with an infinitely thin interior. We are thus led to our first fundamental algebraic topology concept. Definition 15. Let X be a topological space and A X.A deformation from X to A is 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 movie which 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 condition. In particular, consider the two spaces shown below; while it is clear that we can deform one to look like the other, this transformation cannot be a deformation retraction. We thus consider a more general definition.

14 David Altizio 21-752 Lecture Notes

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, to show symmetry, suppose γ0 is homotopy equivalent to γ1, so that there exists a continuous function f such that f( , 0) = γ0 and f( , 1) = γ1. Then the function g given by g(x, t) = f(x, 1 t) yields · · − a homotopy from γ1 to γ0. (Think of this as playing a movie in reverse.) We write γ0 γ1 to denote 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 homotopy equivalent (written X Y ) if there exist maps p : X Y and q : Y X such that q p idX ' → → ◦ ' and p q idY . ◦ ' Definition 18. We say a space X is contractible if there exists a point x0 X such that X x0 . ∈ '{ }

15 David Altizio 21-752 Lecture Notes

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 for ⊆ any homotopy F : A [0, 1] Y and for any map fe0 : X Y with fe0 A = F0, there exists a × → → | new homotopy Fe : X Y such that Fe A×[0,1] = F . Then we say that (X,A) has the homotopy extension property (abbreviated→ HEP). |

We’ll later explore spaces which have the HEP as well as spaces which do not, but for now, we show how to use the HEP. Theorem 5. Suppose (X,A) have the HEP and furthermore that A is contractible. Then X and X/A are homotopy equivalent. Remark. The intuition behind this result is the diagram below. Ideally, contracting A to a point should not affect the topology on X, even if X itself is not contractible.

XX

AA ⇒

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 = id extending this contraction f. We will now apply the universal property of quotient→ spaces twice to 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 fet : X/A X/A such that q ft = fet q. This can be seen in the commutative diagram below; in→ particular, within the◦ context◦ of the categorical interpretation of Remark rem:quotient-diagram, take Z = X/A. q X X/A

q◦ft fet

X/A

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

f 1 g

X 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)) = fe1(q(x)) = fe1([x]), ◦ so q g = fe1 fe0 = idX/A. Hence g is a homotopy inverse of q, meaning that indeed X and X/A are homotopy◦ ' equivalent.

16 David Altizio 21-752 Lecture Notes

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 subspace inheriting 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 retraction if ⊆ → 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 “not super 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 A is closed, any two maps X 0 Y and×A { }[0∪, 1] ×Y which agree on A ×0 combine to give a map X 0 A [0, 1]× { Y} →which is continuous.× → By composing this map× { with} the retraction from X ×[0 {, 1]} to ∪ (X× 0 )→ (A [0, 1]), we get an extension X [0, 1] Y of the original pair of 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 of ∈ × every ray through (0, 2) with Dn [0, 1] either lies on the bottom of the (i.e. Dn 0 ) × n ×{ } or the lateral face of the cylinder (i.e. ∂D [0, 1]). In turn, setting rt = tr + (1 t) id gives a × − deformation 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 retraction of Xn [0, 1] onto Xn 0 (Xn−1 An) I, where recall that Xk is the kth step of our inductive construction.× Indeed,×{ remark}∪ that X∪n [0×, 1] is obtained from Xn 0 (Xn−1 An) I by × × { } ∪ ∪ × attaching copies of Dn [0, 1] along Dn 0 ∂Dn [0, 1]. (The Dn 0 part is due to the × × { } ∪ × × { } fact that a copy of Dn must lie in X, while ∂Dn [0, 1] is from our inductive construction of a CW-complex.) This allows us to take each of our deformation× retractions from above (one for each new 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 infinite× concatenation× { } of ∪ homotopies× is a× deformation retraction of X I onto X 0 A I. We are allowed 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, we now give an example of a pair which does not have the HEP property. Example 13. Let X = [0, 1], and set 1 A = 0 : n N . { } ∪ { n ∈ } I claim that (X,A) does not have the HEP. This is actually HW3 P1, and what follows is the proof. Recall that it suffices to show that [0, 1] 0 A [0, 1] is not a retract of X [0, 1]. Suppose for the sake of contradiction that such a retraction× { } ∪ r×: [0, 1]2 [0, 1] 0 A ×[0, 1] exists. Let → × { } ∪ × x0 = (0, 1) A [0, 1], and set ∈ × 1 B := B(x0, ) (A [0, 1]). 2 ∩ ×

17 David Altizio 21-752 Lecture Notes

1 Since r is continuous, there exists δ > 0 such that x B(x0, δ) implies r(x) B(x0, ), but recall ∈ ∈ 2 that B is a retract onto [0, 1] 0 A [0, 1], so in fact x B(x0, δ) implies r(x) B. But 1 × { } ∪ × ∈ 1 ∈ now [0, δ ] 1 is a path which maps under r to a path connecting (0, 1) to ( δ , 1) contained exclusivelyb c in×B { ,} contradiction since B is not path-connected. Therefore our originalb c assumption was false and no such r exists.

18 David Altizio 21-752 Lecture Notes

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 [0, 1]. We can do this using connectedness, but the ideas that we will examine generalize. 6'

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 homotopy × → ∈ tracks loops which always pass through the point x0.

Proposition 5. The loops γ, γ0 : [0, 1] S1 defined by → γ(x) = (1, 0) and γ0(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 y ∼ ∼ ∼ ∼ iff 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 (cos 2πx, sin 2πx). Thus we can consider the maps 7→ 0 γ : [0, 1] R, x 0 and γ : [0, 1] R, x x e → 7→ e → 7→ 0 0 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, then e → ◦ e we will be done: since γt(1) Z for all t, this right endpoint must be constant, which contradicts e ∈ the fact that γe 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 based 0 0 ∈ ∼ at 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 γ based at x0, and so • γ γ. ∼ Symmetry: Suppose γ1 γ2 via some homotopy F : X [0, 1] X. Then the homotopy • ∼ × → G : 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 and • ∼ ∼ × → F2 : X [0, 1] X. Then the homotopy F : X [0, 1] X via × → × → 1 1 F (x, t) = F1(x, 2t) t + F2(x, 2t 1) t { ≤ 2 } − { ≥ 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 set of equivalence classes happens to be a group under concatenation, i.e. “do the first loop, then the 0 second loop”. More formally, given γ, γ : [0, 1] X based at x0, their concatenation is defined as γ γ0 : [0, 1] X via → · → ( γ(2t) t 1 , t ≤ 2 7→ γ0(2t 1) t 1 . − ≥ 2 To show that concatenation actually induces multiplication on equivalence classes of loops, suppose 0 0 γ0 γ0 and γ1 γ1 are four loops based at x0 X. Then given homotopies γt from γ0 to γ1 and 0 ' 0 '0 0 ∈ 0 0 0 0 γt from γ0 to γ1, the homotopy γt γt sends γ0 γ0 γ1 γ1, i.e. γ0 γ0 γ1 γ1. This tells us that our operation is well-defined,· allowing us to· state→ the· following definition.· ' · ·

19 David Altizio 21-752 Lecture Notes

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-called n higher homotopy groups πn(X, x0) for n 2 which keep track of maps S 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 continuous map such that ϕ(0) = 0 and ϕ(1) = 1, and let f be any path. Then→fϕ can be considered a reparametrization of f. Note that reparrametrization preserves its homotopy class since fϕ f via the homotopy ' (x, t) (1 t)ϕ(x) + tx. 7→ − (In other words, adjusting where points in [0, 1] get mapped to without changing the actual path of 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 via • the function whose is shown in the first figure below,· and so f c f. Similarly, by using 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 shows that 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 reparametrizations • of each other via the function whose graph is shown· below.· This· is· because the former path travels along f for the first quarter, g the second quarter, and h the final half, while the second path travels along f for the first half, g for the third quarter, and h for the final quarter. The function below simply adjusts the locations where f transitions into g and g transitions into h.

Figure 10: The reparametrization associated with associativity in π1.

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

20 David Altizio 21-752 Lecture Notes

8 September 14

Last class, we introduced the notion of the fundamental group. This leads us to a slew of interesting questions, 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 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: take two disjoint spaces with different fundamental groups and consider the groups generated by points in 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 path ∈ → from x0 to x1. Define

βh : π1(X, x1) π1(X, x0) via [γ] [h γ h¯]. → 7→ · · 0 This is a homomorphism because given γ, γ π1(X, x1) we may write ∈ 0 0 0 0 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 β¯ [γ] = [γ] = β¯ βh[γ]. · h h · 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 groups → ϕ∗ : 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 homotopy equivalent 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 better { } { } example, X = x , Y = S2.) { } One might think this would be an easy thing to prove: simply add stars to everything and call it a day. Unfortunately, this doesn’t quite work. Indeed, suppose we have two homotopies ϕ : X Y → and 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 isomorphism → ∈ βh : π1(X, ϕ1(x0)) π1(X, ϕ0(x0)) such that (ϕ0)∗ = βh(ϕ1)∗. →

π1(X, ϕ0(x0))

ϕ0

X βh ϕ1

π1(X, ϕ1(X0))

21 David Altizio 21-752 Lecture Notes

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

βhβ¯ [f] = βh[h¯ f h] = [h h¯ f h h¯] = [f]. h · · · · · ·

Similarly βh¯ βh[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 definitions of ϕ and ψ we have the string of morphisms

ϕ∗ ψ∗ π1(X, x0) π1(Y, ϕ(x0)) 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 apply the same argument with the composition reversed to get ϕ∗ψ∗ = βh as well. Thus ϕ∗ and ψ∗ are isomorphic as groups.

8.2 The Fundamental Group of the Circle

This leads us into our first major result.3

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

Proof. Denote by p : R S1 the map s (cos 2πs, sin 2πs). Furthermore, for all n Z set → 7→ ∈ ω˜n : [0, 1] R to be the path which sends s ns for all s [0, 1]. We will first show that any → 1 7→ ∈ element of π1(S ) is homotopy equivalent to someω ˜n, and then show that eachω ˜n is unique up to homotopy. 1 1 Let γ : [0, 1] S be a loop at (1, 0) representing some element of π1(S ). This loop lifts to a → pathγ ˜ : [0, 1] R from 0 to some n Z. Now the homotopy → ∈

t (1 t)˜γ + tω˜n 7→ − shows thatγ ˜ ω˜n; composing this homotopy with p shows that [γ] = [ωn]. ' Now let m and n be integers, and suppose ft is some homotopy fromω ˜m toω ˜n. This lifts to a homotopy f˜t with f˜0(1) = m and f˜1(1) = n. But recall that the map s fs(1) both is 7→ continuous 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.

22 David Altizio 21-752 Lecture Notes

9 September 17

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

9.1 Generalizations of IVT

We start by generalizing the Intermediate Value Theorem. Note that IVT is equivalent to the fact that there does not exist a retraction f : [0, 1] 0, 1 (which recall is true because [0, 1] is connected). 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 living in (the boundary of) D2. Now find a homotopy sending γ to a point, and compose it with the retraction. This reveals the existence of a homotopy within S1 sending γ to a point, which means 1 that all loops are homotopic to the trivial loop. This is a contradiction of π1(S ) = Z, and so such a retraction does not exist.

Our second result is another generalization of the Intermediate Value Theorem. Remark that IVT implies that any function f : [0, 1] [0, 1] must have a fixed point. This is because if → we define g(x) := f(x) x, then g(0) 0 but g(1) 0, so there must exist an x0 for which − ≥ ≤ g(x0) = f(x0) x0 = 0. Looking at IVT in this way leads us to an extremely powerful theorem in analysis, 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 function → r : D2 S1 as follows. Let x D2 be arbitrary. Since x and f(x) are distinct, we may draw a ray with endpoint→ f(x) passing through∈ x in a well-defined manner; let r(x) denote the intersection of this ray with S1. Since f is continuous, so is r (check this for yourself), so we have established the existence 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 show that in fact p is the constant polynomial.→ 1 Remark that for every radius r > 0 we have the existence of a map γr : [0, 1] S given by → p(re2πix)/p(r) x . 7→ p(re2πix)/p(r) | | 1 The division by p(r) sends 0 1, i.e. γ0(x) = 1 for all x. Thus [γr] is trivial in π1(S , 1) for all r > 0. → Now write n n−1 p(z) = z + an−1z + + a0. ··· For sufficiently large z , we have | | n n−1 z > an−1z + + a1z + a0 . | | | ··· | With this in mind, set n n−1 Pt(z) := z + t(an−1z + + a0). ··· n Then P1 = p and P0(z) = z ; furthermore, Pt(z) = 0 for large z. As a result, the map 6 2πix Pt(re )/Pt(r) x 2πix 7→ Pt(re )/Pt(r) | | n for varying t is a homotopy from the loop γr to the loop z z , i.e. the loop which wraps around n 7→ itself n times. But [γr] = 0, so [z z ] = 0 as well, i.e. n = 0. Thus indeed p is constant. 7→

23 David Altizio 21-752 Lecture Notes

Theorem 10 (Borsuk-Ulam). Let f : S2 R be continuous. Then there exists x S2 for which f(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 → f(x) f( x) g(x) = − − . f(x) f( x) | − − | As g( x) = g(x), we may write − − 1 1 h(s + 2 ) = g(η(s + 2 )) = g(cos(2πs + π), sin(2πs + π), 0) = g( (cos 2πs, sin 2πs, 0)) = h(s). − − 1 Now recall that as shown in the calculation of π1(S ) we may the loop h to a path eh : [0, 1] R. The equation h(s + 1 ) = h(s) implies that → 2 − 1 q eh(s + 2 ) = eh(s) + 2 for some odd integer q. (Note that because q depends continuously on s [0, 1 ] and is always an integer, so it must be ∈ 2 constant.) Hence eh(1) = eh(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 nullhomotopic in S1 as well. Contradiction. Corollary 1. Suppose 2 S = A1 A2 A3 ∪ ∪ 2 where each Ai is a closed set. Then some Ai contains two antipodal points on S .

Proof. Consider the function f : S2 R2 via →

x (dist(x, A1), dist(x, A2)). 7→ This function is continuous, so by Borsuk-Ulam there exists x S2 and (r, s) R2 such that f(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 = 0 and s = 0, then both x and x lie strictly outside both A1 and A2, ergo they must • 6 6 − lie in A3. We’re done.

24 David Altizio 21-752 Lecture Notes

10 September 19

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

n Theorem 11. For n 2, π1(S ) = 0. ≥ To prove this, we cite a lemma. S Lemma 4. Let X be a topological space and A be some index set. Suppose X = α∈A Aα, where

each Aα is open and path-connected, • S there exists x0 X such that x0 Aα, and • ∈ ∈ α∈A 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 Aα Aβ

x0 x0

With this, we may prove Theorem 11.

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

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

Then A1 and A2 are open path-connected spaces, and furthermore A1 A2 is path-connected from n ∩ n 2. Now let x0 S be any base point in A1 A2 (the exact base point doesn’t matter due to ≥ ∈ ∩ path-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 of n n stereographic projection A1 and A2 are both homeomorphic to R ; since R is simply connected for n 2, so must A1 and A2. It follows that γi is null-homotopic for each i, meaning that γ is ≥ 1 1 null-homotopic. Since γ was arbitrary, we deduce that π1(S ) = π1(S , x0) = 0. Corollary 2. For any n = 2, Rn = R2. 6 6∼ Proof. Assume n 3, as n = 1 has already been addressed. Suppose for the sake of contradiction ≥ that f : R2 Rn is a homeomorphism. Then f : R2 0 Rn f(0) is also a homotopy. But → n n−1 \{n } → \{ } 1 remark that R x = S for any n N and x R . This gives a contradiction, as π1(S ) = Z n−1 \{ } ∼ ∈ ∈ but π1(S ) = 0.

25 David Altizio 21-752 Lecture Notes

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 to topological spaces with distinguished basepoints as pointed spaces.)∈ We say that∈ the of X and Y , denoted X Y , is the quotient space of the disjoint union of X and Y by the identification ∨ x0 y0, i.e. ∼ X Y = (X Y )/(x0 y0). ∨ t ∼ Definition 25. Let X be a topological space. The CX of X is defined as the topological space CX := (X [0, 1])/(X 0 ). × × { } Intuitively, this construction turns X into a cylinder and collapses one end of the cylinder to a point. Remark. The space CX is always contractible, since there exists a deformation retraction from CX 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 the trivial group; take for example convex sets in . We also know of spaces with fundamental groups that are isomorphic to Z, e.g. S1 or (re the in-class quiz) S1 S2.4 What about other groups? Are they fundamental groups of spaces as well? ∨ To answer this question, we’ll need a lemma.

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

Proof. Note that γ0 γ1 is a loop, so there exists a homotopy sending it to cx (the · 0 at x0). Call this homotopy Ft. Now consider the homotopy γt via γt = Ft γ1. Observe that e e ·

γ0 = F0 γ1 = γ0 γ1 γ1 γ0 e · · · ' and that γ1 = F1 γ1 γ1. So this γt is the homotopy which works. e · ' e With this in mind, we may prove

2 Proposition 9. π1(RP ) ∼= Z2.

Proof. Let q : S2 RP 2 via x [x] be the quotient map. Denote by N and S the two lifts of → 2 7→ 2 2 the chosen basepoint x0 or RP . The map q exhibits RP as evenly covered by S , i.e. for every 2 −1 x RP , there exists U x such that q (U) = U1 U2 and q Ui : Ui U is a homeomorphism. (We’ll∈ explore this idea more3 fully later.) t | → The loop γ : [0, 1] RP 2, where p : D2 RP 2 is the quotient map and t p(2t 1, 0), is → → 7→ − nontrivial. Loop γ lifts to a path γe connecting the north pole N and the south pole S. If there were a homotopy from γ to cx0 (the constant path at x0), it would lift to a homotopy γe fixing endpoints from γe to cN . This is not possible by IVT: at some point in time the endpoint at S has to “jump” to the endpoint at N. 2 Putting this together, we know that π1(RP ) ∼= Z2.

We now have the ability to answer 9 : having fundamental group G means there exists a simply connected quotient by G. Let’s see some examples of this. 1 1 S S : The fundamental group of this space is Z Z. (In general, π1(A B) = π1(A) π1(B), • × 0 00 × 0 00 × × since if γt and γt are homotopies of loops in A and B, then (γt, γt ) is a homotopy of a loop in A B, and vice versa because projections are continuous.) We can use S1 S1 to cover × × R R = R2 in the following way. ×

Figure 11: Tiling R2 with copies of the torus.

4 1 2 1 2 Basic idea: retract S ∨ S onto S by retracting S to a point; this induces an injective homomorphism in the fundamental groups.

27 David Altizio 21-752 Lecture Notes

2 2 RP RP : The fundamental group of this space is Z2 Z2. Analogously, every loop in • × × RP 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 intuit • what∨ it is by drawing diagrams. Consider S1 S1 as the wedge product of two loops a and b. This space is far from simply connected, but∨ we can try to make it simply connected by unraveling the b loop. This has the side effect of copying the a loop infinitely many times and results in the diagram shown below.

Figure 12: The first step in the process.

The next logical step is to unravel all the a-loops. Unraveling each a-loop creates infinitely many copies of Figure 12. This in turn forces us to stretch each of the new a-loops. Repeating this infinitely many times results in the figure below, which just so happens to be the Cayley 1 1 graph of the on two generators Z Z. Thus, we suspect π1(S S ) = Z Z. (This ∗ ∨ ∼ ∗ makes sense intuitively, as any loop in S1 S1 based at the wedge point can freely move between the loops a and b without restrictions.)∨

Figure 13: The almighty of Z Z! ∗ In this way we see that our diagram gives the Cayley graph of the free group on two generators 1 1 Z Z, and so we suspect that π1(S S ) = Z Z. ∗ ∨ ∼ ∗ There are a few things to note about the Cayley graph in particular, namely that it’s a 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∨ the future.

28 David Altizio 21-752 Lecture Notes

13 September 26

13.1 Covering Spaces

We begin with a definition.

Definition 26. Let X be a space. A of X is a space Xe and a map p : Xe X such that for every x X, there is a neighborhood U x such that → ∈ 3 −1 F p (U) = Uα, where Uα is open in X; and • α for each such α, p U : Uα U is a homeomorphism. • | α →

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 consider the case of there being no αs at all), but for the sake of simplicity we will restrict our attention to surjective p.

Example 14. Consider the map from p : S1 S1 given by z z2. I claim that this map covers → 7→ S1. Indeed, for any x S1 and for sufficiently small ε > 0, there exists δ > 0 such that ∈ −1 1 p (B(x, ε) S ) = B(x, δ) B( x, δ). ∩ ∪ − The restriction of p to each of these open sets does yield a homeomorphism.

p−1[U] 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 Klein bottle K; this won’t be a formal proof, but it will lay some good intuition. The key is that R2 is a covering space for K, as shown by the following diagram. Here each square looks like one copy of 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 the map B : R2 R2 given by (x, y) (1 x, y 1), the same thing happens with the horizontal stripes. → 7→ − − 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

29 David Altizio 21-752 Lecture Notes

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

γ R2, we may use 4 to reduce to examining loops γ completely located within one of ⊆ the squares of R2. Now remark that the completely hollow boundary of K is homomorphic to two single loops (visualize this!) while the inside of the square deformation retracts to a point (and thereby does not affect the fundamental group). So indeed a and b generate π1(K) under the relation aba = b. Note that we may apply the series of operations • A B A (x, y) (x + 1, y) (1 (x + 1), y + 1) = ( x, y + 1) (1 x, y + 1) 7→ 7→ − − 7→ − 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 2 Z. h | i ∗ Z

Proposition 10. If p : Xe X is a covering space, ft : Y X is a homotopy, and fe0 : Y Xe → → → a map lifting f0, then there exists a unique homotopy fet : Y Xe of fe0 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 starting • path. Proposition 11. Let p : Xe X be a covering space. Then the induced map on fundamental → groups p∗ : π1(X,e x0) π1(X, x0) is injective. f →

Proof. Let γ be a loop based at x0 such that γ = p γ is null-homotopic (i.e. p∗([γ]) = 0). Denote e f ◦ e e by γt the homotopy sending γ to a point. Lift the homotopy to Xe to see that γe is null-homotopic (i.e. [γe] = 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 has a lift which is a loop. For example, recall the map p : S1 S1 defined by z 2z. Then 1 1 → 7→ p∗ : π1(S ) π1(S ) is defined via n 2n. The image of this p∗ is precisely the set of loops which wrap around→ an even number of times.7→ Our next goal (which will extend into the subsequent two lectures) is to find a covering space p : Xe X such that Xe is simply connected. → Remark. In general, such a construction is not necessarily possible. To see why, suppose p : Xe X is such a map, and let x X be arbitrary. By the covering space property, there exists some U→ x ∈ 3 such that U has a lift Ue, i.e. p : Ue U is a homeomorphism. Let γ be a loop contained entirely → in U which is based at X. Then γ = (p )−1 γ is a loop in Ue. If Xe is simply connected, then e |Ue ◦ there exists a homotopy γet sending γe to a constant loop in Xe. Now project this loop down to X to get that any loop in U is trivial. Thus every X has a neighborhood which is trivial in X.

30 David Altizio 21-752 Lecture Notes

The previous remark warrants a definition. Definition 27. We say that a topological space X is semi-locally simply connected (SLSC for short) if for every x X there exists some neighborhood U x such that every loop in U is null-homotopic in X. ∈ 3 Example 16. The H (i.e. the union of the circles in the Euclidean plan with 1 1 centers ( n , 0) and radii n for all n N) is an example of a topological space which is not SLSC. Indeed, let U (0, 0) be any open neighborhood∈ containing (0, 0), and suppose δ > 0 is such that B((0, 0), δ) 3U. Then any circle with radius strictly smaller than δ is completely contained in U but 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 those in the previous remark.

31 David Altizio 21-752 Lecture Notes

14 September 28

14.1 Simply Connected Covering Spaces

We proceed with constructing simply-connected covering spaces. We will assume below that X is locally 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 only ' ∼ ∼ − ∈ given S1 and we wanted to construct R. We could do this by thinking of R as the collection of homotopy classes of paths starting at (1, 0) which fix endpoints. For example, given x = (1, 0) and y = ( 1, 0), the homotopy class assigned to the path which goes around the semicircle directly − from 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 ∈

Xe := [γ]: γ is a path starting at x0 . { } Furthermore, consider p : Xe X defined via [γ] γ(1). This turns out to be the correct (X,e p) → 7→ pair we need; it remains to put a topology on Xe. A natural first idea is to simply rig the initial topology on Xe with respect to p, but this doesn’t work. As an example, going back to our R S1 1 → case, rigging such a topology on R would not generate the set (0, 2 ), 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 Xe. 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 τ such that for 0 S B ⊆ all U τ, there exists a subcollection such that U = 0 B. ∈ B ⊆ B B∈B The following equivalent definition was mentioned in class. Proposition 12. Equivalently, is a basis of a topology τ if B S B = X, and • B∈B for all B1,B2 and x B1 B2, there exists B3 such that x B3 and B3 B1 B2. • ∈ B ∈ ∩ ∈ B ∈ ⊆ ∩ With this in mind, set

:= U X open and path connected : π1(U) π1(X) is trivial . U { ⊆ → } Lemma 6. The collection of sets is a basis of X. U Proof. Suppose U , and let V U be open and path-connected. Then there exists an inclusion ∈ U ⊆ map from π1(V ) to π1(U). As the map from π1(U) π1(X) is trivial, so must the map from → π1(V ) π1(X). Thus V . → ∈ U From here we can see that forms a basis for X. First remark that since is path-connected, if U U the inclusion π1(U) π1(X) is trivial for one choice of basepoint x0, then it is true for all choices of basepoints. Thus→ every element in X is contained inside at least one such set, and so the union of all such U equals X. Secondly, suppose U1 and U2 are two open and path-connected subsets of X such that the inclusions π1(Ui) π1(X) for i = 1, 2 are trivial. Take V := U1 U2. Then V is → ∩ a path-connected and open subset of X, and furthermore by the previous proposition e.g. V U1, and so V . Both of these properties show that is a basis of X by Proposition 12. ⊆ ∈ U U Now we will take this basis of X and lift it to a basis of Xe. Given U and a path γ in X ∈ U from x0 to some point in U, let

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

Lemma 7. The map p U , where p is defined above, is bijective. | [γ] 5See here to see why both of these assumptions are needed.

32 David Altizio 21-752 Lecture Notes

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 exists • a 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 the endpoints γ(1) and y, and so [γ1] = [γ2].

0 0 Lemma 8. Suppose γ is a path with [γ ] U . Then U = U 0 . ∈ [γ] [γ] [γ ] Proof. Let γ00 be a path contained in U such that [γ0] = [γ γ00]. We will use this γ00 to prove the result using double-containment. ·

( ) Let [µ] U 0 be arbitrary; then • ⊇ ∈ [γ ] [µ] = [γ0 µ0] = [γ γ00 µ] · · · 0 00 for some path µ contained entirely in U. Since γ µ is contained entirely in U,[µ] U[γ]. ·0 0 ∈ ( ) Let [µ] U[γ] be arbitrary, so that [µ] = [γ µ ] for some path µ contained entirely in • U⊆. Since ∈ · [γ µ0] = [γ γ00 γ00 µ0] = [γ0 γ00 µ0], · · · · · · 00 0 and since γ µ is contained entirely inside U,[µ] U 0 . · ∈ [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.

00 Proof. We will use the criterion described in Propsition 12. Let U , V 0 , and [γ ] U V 0 [γ] [γ ] ∈ [γ] ∩ [γ ] be arbitrary. By Lemma 8, we have U[γ] = U[γ00] and V[γ0] = V[γ00]. Now let W be such that 00 ∈ U W U V and γ (1) W , which we know exists since is a basis in X. Then W[γ00] U[γ00] V[γ00] ⊆ 00∩ ∈ U ⊆ ∩ and [γ ] W 00 , and so we have found our desired set. ∈ [γ ] Thus after all of this work, we have rigged Xe with a topology, and we have our function p : Xe X with [γ] γ(1). There are a few more things we need to check. → 7→ 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 to ⊆ U bijects subsets of V 0 U to subsets V U; this implies that p is continuous (check this!). [γ] [γ ] ⊆ [γ] ⊆ In one direction, for any such V , p(V 0 ) = V . In the other direction, p(V ) U = V 0 , since [γ ] ∩ [γ] [γ ] V 0 U 0 = U and V 0 maps onto V by the bijection p. [γ ] ⊂ [γ ] [γ] [γ ] We defer the remaining things to check to the next lecture.

33 David Altizio 21-752 Lecture Notes

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 Xe and p as before.

Proposition 15. The space Xe is path-connected.

Proof. Let [γ] Xe be arbitrary; we want to construct a path in Xe from [cx0 ] [γ]. To do this, ∈ → remark that γ is a path in X sending x0 to γ(1). Now set ( γ(s) if s t, αt(s) = ≤ 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 Xe is simply connected, i.e. π1(X,e [cx0 ]) = 0.

Proof. Note that our map p : Xe X given by [γ] γ(1) lifts to a map p∗ : π1(X,e [cx0 ]) → 7→ → π1(X, x0). It suffices to show that p∗ is trivial, i.e. the image of π1(X,e [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 Xe. The α described 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 abstract objects. 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. Then as seen below, R2 is a covering space of X. However, this is not R2 with the standard Euclidean metric, as it is not possible to tile R2 with regular octagons. Instead, a hyperbolic metric is necessary.

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 a of π1(X, x0). Then there exists a covering space p : XH X such that → p∗(π1(XH , xe0)) = H for some suitably chosen basepoint xH .

Proof. Define a relation on X via the following criterion: [γ] [γ0] if and only if γ(1) = γ0(1) 0 ∼ ∼ and [γ γ ] H. It’s not hard to check that this is an equivalence relation. Now set XH = X/e , · ∈ ∼ and define p : XH X via [γ] γ(1). This works. Indeed, for any loop γ based at x0, its lift → → to Xe starting at the constant loop [cx0 ] ends at [γ]; thus, the image of this lifted path in XH is a loop iff [γ] [cx ], which is true iff [γ] H by the definition of . ∼ 0 ∈ ∼

34 David Altizio 21-752 Lecture Notes

15.3 Uniqueness of covering spaces

Definition 29. Let p1 : Xe1 X and p2 : Xe2 X be covering spaces. We say that p1 and p2 are → → isomorphic if there exists a homeomorhism f : Xe1 Xe2 such that p1 = p2 f. → ◦ Proposition 18. Let X be path-connected and locally-path-connected, and let p1 : Xe1 X and → p2 : Xe2 X be covering spaces. The following are equivalent. → −1 The spaces p1 and p2 are isomorphic via an isomorphism f : Xe1 Xe2 taking x1 p1 (x0) • −1 → e ∈ to x2 p (x0); e ∈ 2 We have the equality • (p1)∗(π1(X,e xe1)) = (p2)∗(π1(X,e xe2)). Proof. The “ ” direction is easy, as isomorphic covering spaces induce isomorphic maps on the fundamental groups.⇒ For the “ ” direction, we need to be a bit more careful and establish a lifting criterion. ⇐

Let p :(X,e x0) (X, x0) be a covering space and suppose f :(Y, y0) (X, x0), where Y is e → → path-connected and locally-path-connected. I claim that a lift fe :(Y, y0) (X,e x0) of f exists if → e and only if f∗(π1(Y, y0)) p∗(π1(X,e x0)). The fact that this is necessary is intuitive. To define ⊆ e 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 fe(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 there exists a bijection between

basepoint-preserving isomorphism classes of path-connected covering spaces { } and subgroups of π1(X, x0) { } explictly given by (X,e x0) p∗(π1(X,e x0)). e ↔

35 David Altizio 21-752 Lecture Notes

16 October 3

There is only one (up to isomorphism) simply connected covering space for a space X which satisfies the usual properties; this is called the universal cover. But that doesn’t mean there aren’t other covering 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 finite case, 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 have • an 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 one could 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 the free group on 2 generators, which is somewhat counterintuitive. Remark. In the first two examples, there exists a symmetry swapping vertices; e.g. in the first example we may change the basepoint and relabel the loops without affecting the presentation of the group. In example 3, this is not the case, as the leftmost and rightmost vertices behave differently 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 from its covering space Xe. Definition 30. Let p : Xe X be a covering space. An isomorphism from p to p is called a deck → transformation. The deck transformations of p form a group G(Xe).

36 David Altizio 21-752 Lecture Notes

Example 21. The universal cover of S1 is R, as has been shown previously. Recall that if p1 : Xe1 X and p2 : Xe2 X are two covering spaces, an isomorphism between → → them is a homeomorphism f : Xe1 Xe2 such that p1 = p2 f. The only such homeomorphisms are those of the form x x + n.→ (Indeed, scaling either removes◦ or adds pairs which differ by integers, and multiplication→ by 1 does not work because 1 and 1 do not project down to the − 10 − 10 same point in S1.) Thus,

1 G(R S ) = ( R R, x x + n, ) = Z = π1(S). → { → 7→ ·} ∼ Definition 31. Let p : Xe X be a covering space. We say p is normal if for all x X and for → ∈ all lifts Xe1 and Xe2 of X, there exists a deck transformation sending Xe1 Xe2. → We are now ready for the punchline.

Theorem 13. Let p :(X,e xe0) (X, x0) be a path-connected covering space of a path-connected, locally-path-connected space X.→ Set

H = p∗(π1(X,e x0)) π1(X, x0). e ⊆ Then the following hold.

The covering space p is normal if and only if H is a of π1(X, x0). • Recalling that N(H) = g G : gH = Hg is the normalizer of G, we have • { ∈ }

G(Xe) ∼= N(H)/H.

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

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

Hi := p∗(π1(X,e xi)) for i 0, 1 . e ∈ { } Then note the following observations. ¯ −1 Let fe be a loop in Xe based at xe0. Then γe fe γe is a loop based at xe1, and so g H0g H1. • · −1· −1 ⊆ The exact same argument shows that gH1g H0 H1 g H0g. • ⊆ ⇒ ⊆ −1 Hence in fact H1 = g H0g, and so [γ] N(H) if and only if p∗(π1(X,e xe0)) = p∗(π1(X,e xe1)). This along 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(Xe) defined via [γ] τ, where τ is the deck → 7→ transformation taking xe0 xe1 via the notation above. Note that ϕ is a group homomorphism, 0 →0 0 0 0 because if [γ ] maps to the τ taking xe0 to xe1, then γ γ lifts to γe τ(γe ), which is a path from xe0 0 0 · · to τ(xe1) = ττ (xe0). To finish, remark that by the previous paragraph ϕ is surjective, and furthermore ker ϕ consists of the deck transformations τ which fix xe0, which are precisely the elements of p∗(π1(X,e xe0)) = H. Thus by the First Isomorphism Theorem,

N(H) N(H) = ϕ(N(H)) = = G(Xe). ker ϕ ∼ ⇒ H ∼ Huzzah.

37 David Altizio 21-752 Lecture Notes

17 October 5

17.1 Examples of Deck Transformations

Recall the following fact: a map h : Y X has a lift eh : Y Xe if and only if h∗(π1(Y )) → → ⊆ p∗(π1(Xe)). Example 22. Consider the map p : S1 S1 via z zn. Then any basepoint lifts to n points evenly spaced out under the preimage. Hence,→ the deck7→ transformations correspond to by 2π multiples of n . (Note that reflections do not work because they don’t commute with the projection operator analogously to Example 21.) Thus, this covering space is normal and the group of deck transformations is Z/nZ. Note however that this does not satisfy the requirements of Theorem 13 because 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 map d d d q : S RP is the universal cover of RP . Given a fixed basepoint x0, there are two inverse →−1 d images q (x0) in S , which means the only possible deck transformations are the identity and the d one which swaps the two given basepoints. This implies π1(RP ) ∼= Z2. Example 24. Recall that 3×3 T T SO(3) = A R : AA = I = A A, det A = 1 { ∈ } 3 = rotations of R fixing the origin . { } Note that every is determined by the axis of rotation plus an angle. In particular, there exists a map from D3 to SO(3) given by the following: an element x D3 determines the direction of the rotation, while π x determines the angle of the rotation. This∈ is a quotient map identifying | | 3 antipodal points on the boundary, so by the same ideas as the previous example SO(3) ∼= RP 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 times it 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 homomorphism from 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 such that ∈ g1, g2 G, if g1(U) g2(U) = ∅, then g1 = g2. ∀ ∈ ∩ 6 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 quotient space Y/ , where y y0 if and only if y = g y0 for some g G. The points of Y/G are the orbits G 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 with swapping the vertices of T T in a cyclical manner (so that G = Z/3Z). Then this is not a properly discontinuous group action on because the centroid O6 of remains fixed under the action g sending i to i + 1 modulo 3, implyingT g O = e O. T · · 6ouch

38 David Altizio 21-752 Lecture Notes

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

Example 27. Consider S3 as the unit sphere in the H. The restriction to unit quaternions 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 of C2. Let p and q be relatively prime positive integers, and consider the action

2πi/p 2πi/q (z1, z2) (e z1, e z2). 7→ · · 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 of this was as follows.

Example 29. Let X = RP 2 RP 2. On the homework we showed that the below space is the ∨ universal cover Xe of X. Here the alternate between covering the first copy of RP 2 and the second 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 Xe? It turns out there are two: either translating by an even number of spheres or flipping the entire figure in the center of some sphere. (The first is valid because of the alternating parity of the spheres, while the second is valid because the covering map from S2 to RP 2 identifies antipodal points.) As a result, we obtain 2 2 2 2 π1(RP RP ) = D∞ = a, b a , b = Z2 Z2. ∨ ∼ h | i ∗ 2 Note that π1(RP ) = Z2, so we expect that the homotopies on either side don’t affect much. In other words, the fact that the wedge product on topological spaces translates to the free product on 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, S and write X = α Aα, where

each Aα is open and path-connected; • T there exists some point x0 Aα; and • ∈ α for all α = β, Aα Aβ is path-connected. • 6 ∩ We thus have the existence of inclusion maps jα : Aα , X and iα,β : Aα Aβ , Aα. Note further that these maps satisfy the following commutative→ diagram. ∩ →

X jα jβ

Aα Aβ

iα,β iβ,α

Aα Aβ ∩

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

−1 (iα,β)∗(ω)(iβ,α)∗(ω) 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 this enough? ∩

40 David Altizio 21-752 Lecture Notes

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, and hence α π1(Aα) π1(X) = ∗ . ∼ N We won’t prove this in class (the proof is in Hatcher). W Example 30. We’ll now make the intuition of Example 29 precise. Suppose X = α Xα, where each Xα is path-connected.

Xα

Uα

Uβ Uγ Xβ Xγ

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

Suppose we also introduce the technical restriction that each base point x0 Xα has an open ∈ neighborhood Uα Xα which deformation retracts to a point. Then we can take ⊆ ! _ Aα = Xα Uβ ∨ β6=α as our open covering of X. W Furthermore, note that for each α = β, the intersection Aα Aβ = γ∈{ / α,β} Uγ is path-connected and contractible, and so each relation6 from equation 18.1 is the∩ trivial relation. We may now appeal to 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 denote the 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 obtain \{ } π1(X) ∼= Z Z Z. But in reality, π1(X) ∼= Z Z, because X is homomorphic to the CW complex consisting 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. FN 1 3 2.A link is an embedding from n=1 S into R , where here N is a positive integer. 3 3 3. Two knots α1 and α2 are said to be equivalent if there is a map H : R [0, 1] R such that × → for all t [0, 1], the map x H(x, t) is a homeomorphism; • ∈ 7→ H(x, 0) = x for all x R3; • ∈ 1 1 the image of α1 under H is α2, i.e. H(α1(S ), 1) = α2(S ). • 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? 3 2 1 3 Let A be the unknot. Then in fact R A S S as shown below, so π1(R A) ∼= Z. Thus, in order to determine whether some other knot\ 'K cannot∨ be untied, we can compute\ the fundamental group of R3 K and see if it does not equal Z; since fundamental groups are preserved through homotopies,\ this tells us such an untying is not possible. We’ll do this when K is the trefoil knot next 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 a bit. a

A1

b A2 b

a

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). ∩ 1 1 ∗ But A1 is homeomorphic to S S from our previous example and A2 is simply-connected, so in ∨ fact π1(K) is isomorphic to a quotient of a, b . h i 1 To determine exactly which quotient this is, identify A1 A2 S , so π1(A1 A2) = Z. In ∩ ' ∩ ∼ particular, there exists some loop γ generating A1 A2. In A2 this loop is the trivial loop, while −1 ∩ in A1 its equal to abab . As a result, Van Kampen implies

−1 π1(K) = a, b abab = e . ∼ h | i 19.2 Torus Knots

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

1 1 1 n m Definition 36. For positive integers m and n, define the map Tm,n : S S S via z (z , z ). → × 7→ For this to be an embedding, we must have gcd(m, n) = 1. Now view this as living in R3 and put your 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 to 3 3 compute π1(R K) ∼= π1(S K). Our reasons for considering the latter topological space stem from 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 to transcribe in lecture notes. I would be very interested in improvements to this section that don’t involve copy and pasting Hatcher. Consider any cross section x ∂D2 of WLOG the first . The knot K intersects every such cross section three times{ at}× equally spaced points, as shown below. (These three points rotate continuously as the cross section varies.) Thus, as also shown in the diagram, the complement of 1 T3,2 in the first solid torus deformation retracts to a space X3 S , as each of the three strands ' can be pushed in and what’s left is a circle. In other words, π1(X3) = Z. Similarly, the complement 1 ∼ of K in the other torus deformation retracts to a space X2 S . ' We’ll now apply Van Kampen on the union X3 X2, or more specifically open neighborhoods of ∪ these two spaces which deformation retract onto themselves. The intersection X3 X2 is a circle ∩

43 David Altizio 21-752 Lecture Notes

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 times a generator and in X2 to a loop representing 2 times a generator. Thus, by Van Kampen, we may conclude that 3 2 π1(X1 X2) = a, b a = b . ∪ h | i One can now check that this is not isomorphic to Z, and so the trefoil is not equivalent to the unknot.

m n Remark. By applying this to general torus knots, we see that π1(Tm,n) = a, b a = b . h | i 19.3 Gluing 2-Cells

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

2 1 Theorem 16. Let X be a path-connected space. Attach 2-cells eα to X via maps ϕα : S X to → obtain a new space Y . Choose some basepoint x0, and paths γα from x0 to some ϕα(sα), where 1 sα S for all α. Then the map π1(X) π1(Y ) induced by inclusion is surjective and its ∈ → is the smallest subgroup containing all equivalence classes of paths [γαϕαγ¯α].

Proof. Slightly enlarge the space Y by thickening the paths γα as shown below to form a new space 2 Z, and let yα e be arbitrary for each α. ∈ α

Figure 25: Hatcher makes really good diagrams.

Observe now the two facts [ Z Y and Z yα X. ' \ α { }' S In turn, cover Z by the two spaces A and B, where A = Z X and B = Z α yα . Now A is contractible and \ \ { } [ B = Z yα X, \ α { }' so Van Kampen implies π1(Y ) = π1(Z) is a quotient of π1(X) by generators of the form [γαϕαγ¯α].

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

W 2 Proof. Consider the wedge sum X := α eα 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 changes to π1(X) = α Z are that the loops corresponding to the relations of G become trivial. ∼ ∗

44 David Altizio 21-752 Lecture Notes

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, we actually know a bunch of these already: the fundamental group of the sphere M0 is the identity, −1 −1 the fundamental group of the torus M1 is a, b aba b , and the fundamental group of the h | −1 −1 i −1 −1 two-holed torus M2 is, as shown below, a, b, c, d aba b cdc d . h | i

Figure 26: A two-holed orientable surface.

In general,

−1 −1 −1 −1 π1(Mg) = a1, . . . , ag, b1, . . . , bg a1b1a b a2b2a b = [a1, b1][a2, b2] , h | 1 1 2 2 · · · i ··· where [a, b] denotes the commutator of a and b. One can check that the abelianizations of these fundamental groups all have different rank, and so any two distinct orientable surfaces are not homotopy 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 by computing the fundamental groups of their complements.

Example 34. Let K1 denote the link consisting of two unlinked unknots and K2 denote the link consisting of two interlinked unknots. I claim that K1 and K2 are not equivalent (i.e. the interlinked knots are actually linked). Indeed, the complement of K1 is homotopy equivalent to the wedge sum of two S2 S1s (i.e. the wedge sum of two complements of a single unknot), while ∨ 2 the complement of K2 is homotopy equivalent to the wedge sum of S and a torus as shown below. Hence 3 3 π1(R K1) = Z Z = Z Z = π1(R K2), \ ∼ ∗ 6∼ × ∼ \ 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 of the 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) knot K. Construct a knot diagram of K, which consists of drawing K in such a way that K is contained • in 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 knot • \ K 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 opposite ends 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, −1 while each square S` induces the relation RiRjRi = 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 3 −1 −1 −1 π1(R T3,2) = x1, x2, x3 x1x3x1 = x2, x2x1x2 = x3, x3x2x3 = x3 . \ ∼ h | i The fact that this is isomorphic to the group a, b a3 = b2 is very nontrivial. h | i

46 David Altizio 21-752 Lecture Notes

Figure 29: A visual representation of the Wirtinger construction.

20.3 Some Parting Examples

We’ll finish our foray into with a few examples. Example 37. On a previous homework assignment, we examined the dunce hat topological space

+ 2 D := (x, y) (R ) : x + y 1 / , { ∈ ≤ } ∼ where is the equivalence relation which pairs (0, x) (x, 0) (x, 1 x) for all x [0, 1]. On that homework∼ assignment, we showed that D [(0, 0)] deformation∼ ∼ retracted− onto a∈ one-dimensional CW complex by explicitly defining the bijection.\{ Now,} we’ll show that D itself is simply-connected by appealing to Van Kampen. Let a denote the common loop around all three sides of the triangle D. As we did with the Klein bottle, partition D into its boundary and its interior (thickened of course). Then the boundary part 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. ∼ h | i Example 38. Start with S1, and let X denote the topological space discovered by gluing in a 3 2-cell along its boundary via the map z z . Then π1(X) = Z/3Z. 7→ ∼ A simple way to see this is to go around S1 three times and lift to the universal cover; this lift is a loop in the covering space and is hence trivial.

47 David Altizio 21-752 Lecture Notes

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 more algebraic perspective. Warning. This section is very hand-wavy, in the sense that we will not define anything precisely or prove any results rigorously (that will start on Wednesday). We start by asking the following question: what are all homotopy types of one-dimensional CW complexes? 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 previous examples, 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 the number of faces of each dimension? But this turns out to be false! Recall our representations of the torus, Klein bottle, and as the quotient space of a square with opposite edges glued together in various ways. In this case, all three spaces have the same number of vertices, edges, and faces, but the different 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 with vertex 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 larger vertex. 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 the kernel of ∂. This logic extends− to any cycle in G, so the kernel of ∂ is spanned− by the cycles of G. Hence g := dim(ker ∂) is equal to the number of cycles in G, which coincides well with the fact Wg 1 that G S . ' i=1 Now we’ll extend this to the two-dimensional case. Consider the case where we have a triangle E V with vertices v1, v2, and v3. Define ∂1 : R R 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 introduce T E − a second linear transformation ∂2 : R R (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.

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

48 David Altizio 21-752 Lecture Notes

v1 v2

t1

t2

v4 v3

Figure 31: A square decomposed into two triangles.

E V T E As before, we have our linear maps ∂1 : R R and ∂2 : R R . 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 basis T elements for R , 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 our topological 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 e1 v1

t1

e2 e2 e3 t2

v1 v1 e1

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 new cycles (namely e1, e2, and e3), but the gluing of faces kills two of them, so there is one nontrivial cycle in the Klein bottle. But in Example 15, we showed that

−1 π1(K) = a, b abab = Z 2 Z, ∼ h | i ∼ ∗ 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, 3 ker ∂1 ∼= Z “as before”, but now instead of quotienting out by e1, only 2e1 is trivial. This implies

ker ∂1 = Z Z2, im ∂2 ∼ ⊕ which(a) solves our generator mismatch problem and (b) happens to be the abelianization of Z 2 Z. ∗ Z

49 David Altizio 21-752 Lecture Notes

v1 e1 v1

t1

e2 e2 e3 t2

v1 v1 e1

Figure 33: The torus decomposed into two triangles.

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

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

( 2 ker ∂1 R over an R-vector space, ∼= 2 1 1 im ∂2 Z = π1(S S ) 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 extending this to three-dimensional tetrahedra before formalizing everything done today.

50 David Altizio 21-752 Lecture Notes

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 iff i < 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 22.4 yields 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 negative terms 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 type of homology has the advantage of being more computable than singular homology, but it’s not obvious from the start that this can be extended to deal with other types of spaces in a nice and well-defined way. Definition 37. Some important definitions for simplicial homology. 1. For n N, the n- ∆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.

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

nα ◦ a) For all α, the restriction of σα to (∆ ) is injective, and furthermore each point of X is in exactly one of these images. n−1 b) Any map σα restricted to a face of an n-simplex is one of the maps σβ : ∆ X in an order-preserving way. (This rules out, for example, the triangle with a cyclic orientation→ of the edges.) c) Any subset A X is open if and only if σ−1(A) is open in ∆n. ⊆ α 3. We define ( )

X ∆n(X) = njσnj nj Z , ∈ j≥1

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, → X i ∂n(σ) = ( 1) σ . − |[v0,...,vˆi,...,vn] i≥1

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

We proceed with an important lemma. Lemma 9. For each n, the composition

∂n ∂n−1 ∆n(X) ∆n−1(X) ∆n−2(X) −→ −→ is zero.

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

X i j ∂n−1∂n(σ) = ( 1) ( 1) σ − − |[v0,...,vˆj ,...,vˆi,...,vn] j

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

∂n ∂n−1 ∂n−2 Cn Cn−1 Cn−2 C0 0 · · · −→ −→ −→ −→ · · · −→ −→

is a 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 of this 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 not depend on the precise homology complex endowed on our topological space. The next several lectures will be devoted to resolving this hole.

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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 . It turns out the conventionally easiest 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 singular homology and is defined for any topological space. Instead of looking at a particular ∆-complex on a topological space, we’ll instead look at the collection of continuous maps σ : ∆n X (called singular n-simplices). Then define →  n  X n  Cn(X) = njσj nj Z, σj : ∆ X ∈ → j=1  to be the collection of linear combinations of these continuous maps. These are called singular n-chains. Furthermore, define ∂ : Cn(x) Cn−1(x) as before, meaning on a basis element σ, → X ∂(σ) = ( 1)jσ . − |[v0,...,vˆj ,...,vn] j 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 at of n- into X instead. To see why, let’s think about S1. Suppose we defined 1 the same theory except our maps are embeddings now. For S , ker ∂1 consists of 1-cycles, i.e. finite subdivisions 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-simplex into X. But for every n, there’s{∗} only one such map: the map σn : ∆n X via x . As a result, → 7→ ∗ the groups Cn(X) are free abelian groups generated by precisely one element, and so Cn(X) ∼= Z for each X. This yields the chain complex

∂3 ∂2 ∂1 ∂0 Z Z Z Z 0. ··· −→ −→ −→ −→ −→ What are the homology groups? Well, note that the map ∂n is defined on a single basis element σ via ( X j X j 0 if n is even, ∂n(σ) = ( 1) σn = ( 1) σn−1 = − |[ˆvj ] − 1≤j≤n 1≤j≤n σn−1 if n is odd. This means that the maps alternate between being the zero map and being the “multiplication by 1” map. From here one can case on the parity of n to deduce that H0(X) ∼= Z and Hn(X) = 0 for n 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 invariant and one in which we can’t compute but is trivially homeomorphism invariant. We’d like to show that these two forms of homology are in fact the same.

53 David Altizio 21-752 Lecture Notes

Definition 39. We define the Hen(X) to be the homology of the chain complex ε Cn(X) C0(X) Z 0, · −→ · · · −→ −→ −→ where ε : C0(X) Z is the map which satisfies → k ! k X X ε niσi = ni. i=1 i=1

In other words, Hen(X) = Hn(X) for n > 0, while He0(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 He0(X) = 0.

Proof. Remember He0(X) = ker ε/ im ∂1. We claim they’re the same group. For one direction, if 1 σ : ∆ X is a singular 1-simplex, then ∂(σ) = σ [v ] σ [v ], so ε(∂(σ)) = 1 + ( 1) = 0. For the → P | 1 − | 0 − other 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 X  X X X ∂ niτi = niσi nix0 = niσi − P 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. Then → f 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) ··· ··· f# f# f# ∂ ∂ ∂ ∂ Cn+1(Y ) Cn(Y ) Cn−1(Y ) ··· ···

Proof. For σ Cn(X), write ∈ ! X j X j f#∂(σ) = f# ( 1) σ = ( 1) f σ = ∂f#(σ). − |[ˆxj ] − ◦ |[ˆxj ] j j 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 also want 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 a boundary.

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 homomorphism → f∗ = 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.

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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 a boundary. ∈ ⊆

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 the collection 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 → X σ ( 1)iF (σ id) , 7→ − ◦ × |[v0,...,vi,wi,...,wn] i where F : X [0, 1] Y is the homotopy from f to g. I claim that ∂P = g# f# P ∂. To prove this, write × → − − X ∂P (σ) = ( 1)i( 1)jF (σ id) − − ◦ × |[v0,...,vˆj ,...,vi,wi,...,wn] j≤i X + ( 1)i( 1)j+1F (σ id) . − − ◦ × |[v0,...,vˆi,wi,...,wˆj ,...,wn] j≥i

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

F (σ id) = g σ = g#(σ) ◦ × |[ˆv0,w0,...,wn] ◦ and F (σ id) = f σ = f#(σ). − ◦ × |[v0,...,vn,wˆn] − ◦ − Furthermore, the i = j terms all sum to P ∂(σ) since 6 − X P ∂(σ) = ( 1)i( 1)jF (σ id) − − ◦ ◦ |[v0,...,vi,wi,...,wˆj ,...,wn] ij 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# if → f# 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 it in the case of a point (and thus in any ). In order to proceed further, we have to understand the relation between Hn(A) and Hn(X) if A X. ⊆ Definition 42. A (perhaps infinite) sequence of homomorphisms

αn+1 αn αn−1 An+1 An An−1 · · · −→ −→ −→ −→ · · · is called exact if ker αj = im αj+1 for all j. In other words, the homology of a chain complex measures how far from exact the chain complex is. Definition 43. An

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 from B to C is surjective. Definition 44. If A, B, C are abelian groups, then

g 0 A B 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 deformation retraction of some neighborhood in X. Then there is⊆ an exact sequence

i∗ j∗ ∂ Hen(A) Hen(X) Hen(X/A) Hen−1(A) He0(X/A). · · · −→ −→ −→ −→ −→ · · · −→

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

n n Corollary 6. We have Hen(S ) = Z and Hei(S ) = 0 for i = n. ∼ 6 n n−1 n n Proof. For n > 0, take X = D , A = S = ∂X, so that X/A ∼= S . Note that Hen(D ) = 0 since Dn is contractible. So using the previous theorem, we know the existence of an exact sequence

n n n−1 n 0 = Hei(D ) Hei(S ) Hei−1(S ) Hei−1(D ) = 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. By → letting i : ∂Dn Dn be the canonical inclusion map, we see that the following diagram commutes. → ∂Dn ∂Dn

i f

Dn

n As a result, we obtain an analogous commutative diagram on homologies. But Hn(∂D ) = n n ∼ Hn(S ) ∼= Z but Hn(D ) = 0, contradiction. Example 46. Let A S2 be a two-point space. We’ll compute the homology of S2/A. Recall by Theorem 17 the sequence⊆

2 2 He2(A) He2(S ) He2(S /A) −→ −→ 2 2 He1(A) He1(S ) He1(S /A) −→ −→ −→ 2 2 He0(A) He0(S ) He0(S /A) 0 −→ −→ −→ −→ is exact. We now proceed in steps. 2 2 2 By Corollary 6, we know that He2(S ) = Z and He1(S ) = He0(S ) = 0. • ∼ By the definition of reduced homology, He2(A) = He1(A) = 0 and He0(A) = Z. • ∼ Recall that (after relabeling to accommodate the previous steps) the subsequence • ϕ 2 0 Z He2(S /A) 0 −→ −→ −→ is exact. In particular, by looking at the left three terms we see that ϕ is injective and by looking at the right three terms we see that ϕ is surjective. Thus ϕ is actually a bijection, 2 and He2(S /A) ∼= Z. Similarly, by looking at the exact subsequence • 2 0 He1(S /A) Z 0 −→ −→ −→ 2 we see that He1(S /A) ∼= Z. 2 Finally, He0(S /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 use them 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 chain complex → 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 supported in A. Note that this yields the following commutative diagram.

i j 0 Cn(A) Cn(X) Cn(X,A) 0

∂ ∂ ∂ i j 0 Cn−1(A) Cn−1(X) Cn−1(X,A) 0

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

0 0 0

∂ ∂ ∂ An+1 An An−1 ··· ··· i i i ∂ ∂ ∂ Bn+1 Bn Bn−1 ··· ··· j j j ∂ ∂ ∂ Cn+1 Cn Cn−1 ··· ···

0 0 0

Let c Cn be a cycle. Since j is surjective, there exists b Bn such that c = j(b). Now push ∈ ∈ forward to ∂b Bn−1. I claim ∂b ker j: indeed, j(∂b) = ∂j(b) = 0, so ∂b im i. Thus there ∈ ∈ ∈ exists 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 which well-definedness might be a problem. Note that a is uniquely determined by ∂b since i is injective. • If we chose a different b0 instead of b with j(b0) = c = j(b), then b b0 ker j = im i, so • 0 0 0 0 0 0 − ∈ 0 b b = i(a ) for some a An b = b + i(a ). So b b changes a a + ∂a because − ∈ ⇒ → → i(a + ∂a0) = i(a) + i∂a0 = ∂b + ∂i(a0) = ∂(b + ia0) = ∂b0. 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 c + ∂c0 for • 0 0 0 0 7→ some c Cn+1. Find b Bn+1 such that c = j(b ), so ∈ ∈ c + ∂c0 = c + ∂j(b0) = c + j(∂b0) = j(b + ∂b0).

58 David Altizio 21-752 Lecture Notes

so we replace b by b + ∂b0. Now when we push b forward, the ∂ term becomes ∂∂b0 = 0, so ∂(b + ∂b0) = ∂b.

We are now led to the final piece of the puzzle. Theorem 18. The sequence

i∗ j∗ ∂ i∗ j∗ Hn(A) Hn(B) Hn(C) Hn−1(A) Hn−1(B) · · · −→ −→ −→ −→ −→ −→ · · · is exact.

Proof. First remark that im i∗ ker j∗ since ji = 0 j∗i∗ = 0. ⊆ ⇒ 0 0 Now we’ll prove ker j∗ im i∗. Let b Bn with ∂b = 0 and j(b) = ∂c for some c Cn+1. (The ⊆ ∈ ∈ former condition arises from the fact that Hn(B) is a quotient of ker ∂, while the latter is due to 0 0 0 our assumption that b ker j∗.) Since j is surjective, c = j(b ) for some b Bn+1. Now ∈ ∈ j(b ∂b0) = j(b) j(∂b0) = j(b) ∂j(b0) = 0 − − − 0 0 0 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 ∂b0) = ∂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 ⊆

i∗ j∗ ∂ Hn(A) Hn(X) 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 are → homotopic 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 operator → also 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 Z cover 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 = Uj j be a collection of sets such that X = jUj . U U { } P ∪ Denote by C (X) Cn(X) to be the collection of linear combinations niσi, where each σi has n ⊆ its image in one set Uj. This gives a chain complex

∂ ∂ ∂ CU (X) CU (X) CU (X) , · · · −→ n −→ n−1 −→ n−2 −→ · · · U with associated homology groups denoted by Hn (X). U Proposition 24. The map i : Cn (X) , Cn(X) is a chain homotopy equivalence, i.e. there is a U → map ρ : Cn(X) Cn (X) such that iρ and ρi are chain homotopic to the identity. Thus, i induces →U isomorphisms Hn (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 1-1 correspondence with chains (total orders) of faces.

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

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

60 David Altizio 21-752 Lecture Notes

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

(IH) ∂S(σ) = ∂b(S∂σ) = S∂σ b∂(S∂σ) = 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 → − T σ = b(σ T ∂σ). − We can check that this map is indeed the correct map via induction, because ∂T σ = ∂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 all m 1, Sm (i.e. S with iterated barycentric subdivisions m times) is chain homotopic to the identity≥ map via the map m−1 X i Dm = TS . i=0 Indeed, checking this is a simple calculation, as

m−1 m−1 X i i X i i ∂Dm + Dm∂ = (∂T S + TS ∂) = ∂T S + T ∂S i=0 i=0 m−1 m−1 m X X X = (∂T + T ∂)Si = (id S)Si = (Si Si+1) = id Sm. − − − i=0 i=0 i=0

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 Lebesgue 8 n −1 ◦ n number of the cover of ∆ by the open sets σ (Uj ). As a result, for each σ : ∆ X there is m U → m = m(σ) such that S (σ) is in C (X), which means we may define D : Cn(X) Cn+1(X) via n → Dσ = Dm(σ)σ.

After all this work, we are almost done. Recall that we originally set out to find ρ : Cn(X) U → Cn(X) with image in Cn (X) such that ∂D + D∂ = id ρ. The easiest way to do this is to consider this equation as defining ρ and set ρ = id ∂D D∂.− Note that − − ∂ρ(σ) = ∂σ ∂2Dσ ∂D∂σ = ∂σ ∂D∂σ − − − and ρ(∂σ) = ∂ρ ρDρσ D∂2ρ = ∂σ ∂D∂σ, − − − U so indeed ρ is a chain map. Finally, to check that ρ takes Cn(X) to Cn (X), we compute ρ(σ) more explicitly and write ρ(σ) = σ ∂Dσ D(∂σ) − − = σ ∂D σ D(∂σ) − m(σ) − = Sm(σ)σ + D (∂σ) D(∂σ), m(σ) − 8See here for a detailed explanation of the Lebesgue Number Lemma.

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m m(σ) U where the last equality is due to ∂Dm + Dm∂ = id S . Remark that S σ Cn (X) by − ∈ i definition of m(σ). Furthermore, Dm(σ)(∂σ) D(∂σ) consists of only the terms TS (σj) with − U i m(σj) (because the subtraction cancels out all other terms!), and these terms lie in Cn (X) ≥ U U since T takes Cn−1(X) to Cn (X). All in all, we have discovered the existence of a map D such that ∂D + D∂ = id iρ (where U − i : Cn (X) , Cn(X) is the inclusion map). Combining this with ∂i = id shows that ρ is a chain homotopy inverse→ for i, and so we are done.

62 David Altizio 21-752 Lecture Notes

30 November 5

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

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

CU (X) C (X) n , n Cn(A) → Cn(A) induces an isomorphism on homology. Furthermore, the map

U Cn(B) Cn (X) Cn(A B) → Cn(A) ∩ induced by inclusion also induces an isomorphism on homology, since both quotient groups are greely 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 = ∅. The pair (X,A) is a good pair if A has a neighborhood in X that deformation retracts to A.6 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, ) = Hen(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. To prove this, consider the long exact sequence

ϕ1 ϕ2 ϕ3 ϕ4 ϕ5 ϕ6 H1( ) H1(X) H1(X, ) H0( ) H0(X) H0(X, ) 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 α • with boundary contained entirely in and outputs precisely the boundary of α. Here, the boundary of any such 1-cycle is {∗} = 0 (i.e. the two endpoints of the 1-cycle coincide {∗} − {∗} but have opposite orientations). Thus the map ϕ3 is the trivial map. As a result,

im ϕ2 = ker ϕ3 = H1(X, ), {∗}

so ϕ2 is surjective.

Similarly, because ϕ1 is the zero map, ker ϕ2 = im ϕ1 = 0, so ϕ2 is injective. Thus in fact ϕ2 • is an isomorphism, and H1(X, ) = H1(X) = He1(X). {∗} ∼ ∼ Finally, note that the map ϕ6 is the zero map. Since im ϕ5 = ker ϕ6 = H0(X, ), we get • {∗} that ϕ5 is surjective. Thus by Definition 44,

H0(X) = Z H0(X, ) = He0(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. Then the 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 following commutative 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 into steps. 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 a • hole”. In particular, remark that

X A (X/A) (A/A), \ ' \ and similarly V A (V/A) (A/A); these homeomorphisms induce an isomorphism on relative homology.\ ' \ Done.

30.2 Generators of Homology Groups on ∆-complexes

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

n n Proposition 26. For all n 0, the group Hn(∆ , ∂∆ ) = Z is generated by the identity map n n ≥ ∼ in : ∆ ∆ . → Proof. We proceed by induction. The base case of n = 0 is obvious, so let n 1 be arbitrary. Let Λ ∆n be the union of all but one of the (n 1)-faces. I claim that ≥ ⊆ − (1) (2) n n n n−1 n−1 Hn(∆ , ∂∆ ) ∼= Hn−1(∂∆ , Λ) ∼= Hn−1(∆ , ∂∆ ).

To prove (1) we examine the long exact sequence for the triple Λ ∂∆n ∆n: this is ⊆ ⊆ n n n ϕ n n Hn(∆ , Λ) Hn(∆ , ∂∆ ) Hn−1(∂∆ , Λ) Hn−1(∆ , Λ) . · · · −→ −→ −→ −→ −→ · · ·

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n n n But Hn(∆ , Λ) and Hn−1(∆ , Λ) are both zero since ∆ deformation retracts onto Λ, so ϕ : n n n Hn(∆ , ∂∆ ) Hn−1(∂∆ , 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. First remark that ≥ ∆n−1 ∂∆n = . ∂∆n−1 ∼ Λ Indeed, one can check that both spaces are homeomorphic to Sn (this is purely geometrical). This homeomorphism implies an isomorphism on the relative homology groups. n−1 n Now consider the map jn induced by ∆ , ∂∆ 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 exactly n−1 n−1 what happens to the map in−1 when viewed in the quotient space ∆ /∂∆ , and so jn is a generator as well. We’re done.

In the previous proof, we examined many homology groups of the form Hn(X,X ). This warrants 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 x. 3 Remark. These groups are not always trivial, surprisingly. Take for X = Rn the long exact sequence

n n n n n n Hk(R x ) Hk(R ) Hk(R , R x ) Hk−1(R x ) Hk−1(R ). \{ } −→ −→ \{ } −→ \{ } −→ n n So Hk(R , R x ) is zero if k = n 1 and Z if k = n 1. In the latter case its isomorphic to \{ } 6 − − k k−1 Hk−1(R x ) = Hk−1(S ) = 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) ∼= Hen(X/A); • n n For all n, Hn(∆ , ∂∆ ) ∼= Z, and furthermore this group is generated by the canonical • identity 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

j A i B C k D ` E

α β γ δ ε

0 j0 0 0 A0 i B0 C0 k D0 ` E0

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 c0 C0. Then k0(c0) = δ(d) for some d D due to surjectivity. I claim d ker `; indeed, ∈ ∈ ∈ ε(`(d)) = `0(δ(d)) = `0(k0(c0)) = 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)) = k0(c0); our goal is to show that γ(c) = c0. To compute this, remark that

k0(c0 γ(c)) = k0(c0) k0(γ(c0)) = k0(c0) δ(k(c)) = 0, − − − so c0 γ(c) ker k0 = im j0. Thus there exists b0 B0 such that j0(b0) = c0 γ(c); by surjectivity of β there− exists∈ b B such that β(b) = b0. ∈ − ∈ But now we have our punchline: write

γ(c + j(b)) = γ(c) + γ(j(b)) = γ(c) + j0(β(b)) = γ(c) + j0(b0) = c0.

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 is injective. The proof is similar and will be omitted.

In order to show that simplicial and singular homology are isomorphic, we need to distinguish the notation. Definition 49. Let H∆(X,A) be the relative simplicial homology groups, where here A X is a n ⊆ 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 Hn (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. Now consider the commutative diagram ≤

∆ k k−1 ∆ k−1 ∆ k ∆ k k−1 ∆ k−1 Hn+1(X ,X ) Hn (X ) Hn (X ) Hn (X ,X ) Hn−1(X )

k k−1 k−1 k k k−1 k−1 Hn+1(X ,X ) Hn(X ) Hn(X ) Hn(X ,X ) Hn−1(X )

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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; see Example 6). As a result,

( b ∆ k k−1 Z for n = k and b the number of k-simplices, H (X ,X ) = n ∼ 0 otherwise.

F k k k k−1 Furthermore, if we take Φ : α(∆α, ∂∆α) (X ,X ) via the expected characteristic maps, there is a homeomorphism of quotient spaces→

F ∆k Xk α α . F k ∼= k−1 α ∂∆α X 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 open ∆ simplices. Using this, we claim the inclusion map Hn (X) Hn(X) is surjective. Indeed, if z is a singular n-cycle, it can be written as a finite linear combination→ of singular n-simplices; each is contained in finite number of simplices, and so z must be contained in Xk for some k. But the ∆ k k finite case implies Hn (X ) Hn(X ) is surjective, so z is homologous to a simplicial n cycle. A similar “compactness” argument→ shows injectivity as well. Finally, to pass to the more general case, consider the commutative diagram

∆ ∆ ∆ ∆ ∆ Hn (A) Hn (X) Hn (X,A) Hn−1(A) Hn−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 column must 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 our labor. 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

i j {A,B} 0 Cn(A B) Cn(A) Cn(B) C (X) 0 −→ ∩ −→ ⊕ −→ n −→ is a short exact sequence. Here

The map i : Cn(A B) Cn(A) Cn(B) is the inclusion map given by x (x, x) (i.e. • we view x as living∩ in A→and B but⊕ flip the boundary of the part in B), while7→ − {A,B} the map j : Cn(A) Cn(B) Cn (X) is given by (x, y) x + y (i.e. we combine the • ⊕ → 7→ 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 CU . • n

Upon iterating this and using the same techniques as in Theorem 18 we obtain a long exact sequence

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 part contained in A. Example 47. Let X = S1 S1 be the torus. Split the torus into two sides A and B (thickened a × little bit). Note that A S1 and B S1, while A B S1 S1. Now consider the Mayer-Vietoris sequence ' ' ∩ ' t

j2 H2(A B) H2(A) H2(B) H2(X) ∩ −→ ⊕ −→ ∂1 i1 j1 H1(A B) H1(A) H1(B) H1(X) −→ ∩ −→ ⊕ −→ ∂0 i0 j0 H0(A B) H0(A) H0(B) H0(X) 0. −→ ∩ −→ ⊕ −→ −→ We now proceed in steps. We know from previous ventures that • ( Z for i = 0, 1, Hi(A) Hi(B) = Hi(A B) = ⊕ ∼ ∩ ∼ 0 for i = 2.

The map j0 : H0(A) H0(B) H0(X) takes singular 0-chains in the former space and • includes them into the⊕ latter space→ via addition. It is easy to see that the image of this map is 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 maps • ∩ → ⊕ them 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 fact that 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 is omitted. 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

∂1 i1 0 H2(K) H1(A B) H1(A) H1(B) −→ −→ ∩ −→ ⊕ j1 ∂0 i0 H1(K) H0(A B) 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) M¨obiusstrip, 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 of • ∩ → ⊕ H0(A B) to (1, 1). Hence i0 is injective, which after some exactness chasing implies j1 is surjective.∩ This will− be useful later.

Consider the map i1 : H1(A B) H1(A) H1(B). This map sends the generator 1 • ∩ → ⊕ of H1(A B) to (2, 2) (recall that the generator of e.g. H1(A) is the central circle, and ∩ − the boundary circle wraps around twice). This makes the computation for H1(K) more interesting, 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(A) H1(B) Z Z H1(K) = im j1 = ⊕ = ⊕ = Z Z2. ∼ ker j1 ∼ (2, 2) ∼ ⊕ h − i 1 1 Remark. Note that we computed H2(S S ) ∼= Z, which means that we can orient every triangle in such a way that the orientations all match× up on each edge (i.e. the kernel of the boundary map is 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 be embedded in R3. If we could, then we could use the outward normal vector to induce an orientation of the edges, but this is not allowed because the kernel is trivial.

In general, a d-dimensional 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 from Example 7, where M is the M¨obiusstrip and D the outside disk. This yields the sequence

2 i 2 H2(M) H2(D) H2(RP ) H1(M D) H1(M) H1(D) H1(RP ) 0. ⊕ −→ −→ ∩ −→ ⊕ −→ −→ We now proceed in steps, except this time I won’t bullet the individual steps. First note that we have previously computed H2(M) = 0, H1(M) = Z, H1(M D) = Z, and Hi(D) = 0 for all i; in ∼ ∩ ∼ 2 turn, H2(M) H2(D) = 0 and H1(M) H1(D) ∼= Z. Thus all that is left is to compute Hi(RP ) for i 1, 2 .⊕ ⊕ ∈ { } The key is to consider the map i : H1(M D) H1(M) H1(D). Recall that H1(M D) is ∩ → ⊕ ∩ generated by a single element 1. This element maps to 2 times the generator of H1(M), and so i(1) = 2. This has two consequences. First, this map is injective, so the mapping from H2(M) 2 2 ⊕ H2(D) H2(RP ) is surjective, and hence H2(RP ) = 0. Second, by the First Isomorphism Theorem,→ 2 H1(M) H1(D) Z H1(RP ) = ⊕ = = Z2. ∼ im i ∼ (2) ∼ Huzzah.

33.1 Degree

We’ll now start exploring applications of homology. n n n n If we have a map f : S S it generates a map on homology f∗ : Hn(S ) Hn(S ). Both of → → these 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 of • ◦ · f∗ 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 a • − nice ∆-structure for Sn where identical ∆-complexes are oppositely orientated with respect to this hyperplane; then f reverses the orientations of all the ∆-complexes, and so f∗(1) = 1. 6 deg( id) = ( 1)n+1, where id denotes reflection about the origin of Sn. Indeed, note • − − − that when embedding Sn into Rn+1, id corresponds to reflection about each of the n + 1 − hyperplanes 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 exists • a homotopy from f to id; the result follows− by the homotopy invariance of degrees. This homotopy is given specifically− by the map (1 t)f(x) tx t − − ; 7→ (1 t)f(x) tx | − − | 1 in particular, “the no fixed points” condition implies this map is well-defined for t = 2 . 2n Theorem 24. The only non-trivial group G that acts properly discontinuously on S is G = Z2.

Proof. There is a map d : G Z via g deg g. The image of d is contained in 1, 1 , so d induces a group homomorphism→ G 7→1, 1 . Now every nontrivial element g {−G satisfies} d(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) = 0. Without loss of → 6 generality (since v(x) = 0) let v(x) = 1. Consider the map f : Sn Sn given by 6 | | → t (sin t) x + (cos t) v(x). 7→ · ·

70 David Altizio 21-752 Lecture Notes

2 2 This is a homotopy on Sn because f(t) = sin 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 will start with this today and continue on Wednesday. Lemma 12. Let X be a CW complex. Then n n−1 Hk(X ,X ) = 0 if k = n and free abelian for k = n with basis the n-cells of X. • n 6 Hk(X ) = 0 if k > n. • n Hk(X ) 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

n n−1 n n−1 Hk(X ,X ) ∼= Hk(X /X ). But recall that Xn/Xn−1 is homeomorphic to a wedge sum of n-dimensional spheres (one for each n-dimensional face); this proves the first bullet point. For the rest of the proof, consider the exact sequence

n n−1 n−1 n n n−1 Hk+1(X ,X ) Hk(X ) Hk(X ) Hk(X ,X ) −→ −→ −→ n n−1 n n−1 n−1 If k / n 1, n , then Hk+1(X ,X ) = Hk(X ,X ) = 0 by the first part, and so Hk(X ) = ∈n { − } n n−1 n ∼ Hk(X ). If k = n, then at the very least Hk+1(X ,X ) = 0, and so the mapping Hk(X ) n−1 0 → Hk(X ) is a surjection. These isomorphisms plus Hk(X ) = 0 for k 1 imply part 2 and 3 in the 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 to construct a long exact sequence on homology

n+1 n dn+1 n n−1 dn n−1 n−2 Hn+1(X ,X ) Hn(X ,X ) Hn−1(X ,X ) . −→ −→ −→ · · · The key observation is that these groups fit into several short exact sequences we already know of, and so we will construct the desired maps dk by snaking between these established sequences. Behold the following diagram!

0

n+1 0 Hn(X ) ∼= Hn(X)

n Hn(X )

∂n+1 jn

n+1 n dn+1 n n−1 dn n−1 n−2 Hn+1(X ,X ) Hn(X ,X ) Hn−1(X ,X )

∂n

jn−1 n−1 Hn−1(X )

0

CW This is a chain complex, and its homology is called cellular homology, denoted by Hn (X). Fortunately, we won’t need to carry the CW notation for long.

CW Theorem 25. For any CW complex X, Hn (X) ∼= Hn(X). n Proof. We know Hn(X) ∼= Hn(X )/ im ∂n+1 via the leftmost diagonal chain. Now due to exactness of the middle diagonal chain, jn is injective, which implies jn(im ∂n+1) = im dn+1. Further, n Hn(X ) gets mapped under jn to im jn = ker ∂n. But jn−1 is injective, so ker ∂n = ker dn. It n follows that jn induces an isomorphism of the quotient ∼= Hn(X )/ 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 cellular homology is often easier to compute than simplicial homology. Below, we see an example of how to 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, −1 and one 2-cell σ attached by the word ab aba. This implies d2(σ) = a b + a + b + a = 3a. So − H1 is generated by a and b, where 3a is trivial.

a b

σ a a

b

Figure 36: A CW complex.

Remark. In general, for σ an n-cell, X dn(σ) = dβ eβ, · where dβ is the degree of the map ∆β defined by the composition

n−1 n−1 n−1 X n−1 S X n−1 = Sβ . → → X eβ ∼ \ 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 the orientable surface Mg. This structure consists of one (common) vertex, four edges, and one 2-cell −1 −1 attached by the word [a1, b1][a2, b2] [ag, bg], where [a, b] = aba b . The chain complex for Mg is thus ··· d2 2g d1 0 Z Z Z 0. −→ −→ −→ −→ The map d1 is zero since there is only one vertex in the complex structure and so the only element n−1 of all X eβ is a point which is contractible. Furthermore, the map d2 is zero: in our attaching \ map 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) = 2g ∼ Z ∼= H2(Mg) and H1(Mg) ∼= Z .

Example 52 (Nonorientable surfaces). Consider the closed nonorientientable surfaces Ng of 2 g. For example, N1 ∼= RP , while N2 is congruent to the Klein bottle. The surface Ng is represented by a CW complex structure with one 0-cell, g 1-cells, and one 2-cell attached by the word a2a2 a2. 1 2 ··· g The chain complex structure of Ng is thus

d2 g d1 0 Z Z Z 0. −→ −→ −→ −→

The map d1 is zero for the same reason as in the previous example. However, this time the 2 1 g generator 1 of H2(X ,X ) is mapped to (2, 2,..., 2) Z ; this is because the attaching map goes ∈ 2 along each one cell twice, and so each ∆β is homomprhic to the map z z , which has degree 2. 7→ 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)

g g−1 of Z , we see that H1(Ng) = Z Z2. ∼ ⊕ Example 53. The following example is especially illustrative. Start with S1 S1, with two 1-cells a 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

2 d2 2 d1 0 Z Z Z 0. −→ −→ −→ −→

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5 −2  The map d1 is still zero. The map d2 is given by the matrix M = −3 1 . The determinant of M is 1, so the linear map x Mx is an isomorphism, which implies H2(X) = 0. Additionally this 7→ map is surjective, so H1(X) = 0. Hence Hei(X) = 0 for all i. But now comes the punchline: the fundamental group of X is, by Van Kampen, equal to

5 −3 3 −2 π1(X) = a, b a b , b (ab) ; h | i this group is nontrivial. (Indeed, there is a nontrivial homomorphism sending this group to the group G of rotational symmetries of a regular dodecahdron: send a to the rotation ρa through angle 2π/5 about the axis through the center of a pentagonal face, and send b to the rotation ρb through 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 using Theorem 14, we see that there exists a closed 3d manifold that has the homology of a sphere but is not a sphere.9 Remark. One thing to notice here is that homology doesn’t care about order to some extent. In particular, the main reason that this space X from above has the properties it does is that the attaching maps only care about the total number of wrappings around the two 1-cells, whereas the fundamental group depends more crucially on the exact nature of the attaching map. With cellular homology, we have greatly expanded the range of possible computations, almost to 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 show that the higher homology groups do as well, but it is possible.

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35 November 16

Example 54. As a quotient, RP n has a CW-complex structure with one cell in each dimension k n. The chain complex thus looks like ≤ dn dn−1 d1 0 Z Z Z Z 0. −→ −→ −→ −→ · · · −→ −→ The difficulty lies in computing the maps dn between these spaces. By the Cellular Boundary n n−1 formula, dn(e ) = kne , where kn is the degree of the composition k−1 n−1 k−1 RP k−1 S RP = S . −→ −→ RP k−2 ∼ This map is a homeomoprhism when restricted to each component of Sk−1 Sk−2. One of these \ components maps identically onto RP k−1 (and thus has degree 1), while the other components maps antipodally onto RP k−1 (and thus has degree ( 1)k). As a result, the degree of the total composition 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.  Z if k = 0 or k = n odd, n  Hk(RP ) ∼= Z2 if k < n is odd, 0 if k n is even. ≤ Hence RP n is orientable iff its dimension is odd.

35.1 Euler Characteristic

Definition 51. For a CW complex X, the Euler characteristic of X is

X n χ(X) = ( 1) cn, − n≥0 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 characteristic is 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 is 8 12 + 6 = 2. − Hmm. Theorem 26. The equality X n χ(X) = ( 1) rank Hn(X). − n≥0 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. th Recall that the n homology group is a quotient of n-boundaries by n-cycles, and so rank Hn = rank Zn rank Bn. Additionally, by “rank-nullity”, rank Cn = rank Zn + rank Bn−1. Combining these two− facts yields rank Cn = rank Bn + rank Hn + rank Bn−1, so X n χ(X) = ( 1) rank Cn − n≥0 X n X n = ( 1) (rank Bn + rank Hn + rank Bn−1) = ( 1) rank Hn, − − n≥0 n≥0 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,

2 X n 2 χ(X) = χ(S ) = ( 1) rank Hn(S ) = 1 0 + 1 = 2. − − n≥0

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 faces we see that χ(X) χ(X/G) = . G | | k k This shows if Z2 acts freely on S , then S has even Euler characteristic; in turn, k is even.

Corollary 11. The complete graph on five vertices K5 is not planar.

2 2 Proof. Suppose there was an embedding K5 , S . This defines a CW-complex X on S . But X has 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; it has 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 a vertex (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 bit of 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 triangles per 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 every closed surface S, Z 2πχ(S) = κ(~x) d~x, S where κ(~x) is the curvature at ~x. This fact is called the Gauβ-Bonnet Theorem, and it’s terribly false for higher .

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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 with coefficients in the abelian group G instead of with Z-coefficients. n n Example 59. Let’s compute Hk(RP , Z2). Recall that the CW complex model of RP consists of 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 ( n Z2, k n, Hk(RP , Z2) = ≤ 0 k > n.

Remark. If X is a CW complex, then the chain group Ck(X; Z2) consists of linear combinations of subsets of the k-dimensional faces of X. More formally, this chain group can be written as a tensor product Ck(X; Z2) = Ck(X) Z2. ⊗Z The above remark warrants a definition. Definition 52. Let G be a group. Then the functor Tor( ,G): Ab Ab measures to which extent 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 via • multiplication 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 • { } ! M M Tor Gi,H ∼= Tor(Gi,H). i∈I i∈I

This allows one to compute Tor explicitly.

Theorem 27 (Universal coefficient theorem). The homology groups with Z-coefficients and G determine 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 the computations. Consider the sequence of chain complexes

n τ n q# n 0 C(RP ; Z2) C(S , Z2) C(RP ; Z2) 0. −→ −→ −→ −→

Here τ is the transfer map τ(σ) =σ ˜1 +σ ˜2 (i.e. the sum of the two lifts), while q# is (as you may recall) the map on chain complexes induced by the quotient map q : Sn RP n. Check that this is a short exact sequence. This induces a long exact sequence in homology→

n n n 0 Hn(RP ; Z2) Hn(S ; Z2) Hn(RP ; Z2) −→ −→ n −→ n n Hn−1(RP ; Z2) Hn−1(S ; Z2) Hn−1(RP ; Z2) −→ −→ −→ −→ · · · n n Recall that Hn(S ; Z2) ∼= Z2 and Hk(S ; Z2) = 0 otherwise. This combined with exactness tells us that n n n Hn−1(S ; Z2) = Hn−2(RP ; Z2) = = H0(RP ; Z2) = Z2. ∼ ∼ ··· ∼ ∼ n n Furthermore, the top level shows Hn(RP ; Z2) ∼= Hn(S ; Z2) ∼= Z2.

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Theorem 28 (Borsuk-Ulam in higher dimensions). A map f : Sn Sn with f( x) = f(x) for all x has odd degree. → − −

Proof. The map f induces a map g : RP n RP n. Now consider the commutative diagram → n ∂ n Hk(RP ; Z2) Hk−1(RP ; Z2)

g# g#

n ∂ n Hk(RP ; Z2) Hk−1(RP ; Z2) induced by g#, where k N is arbitrary. Now g# is an isomorphism in zero-homology (the zeroth ∈ homology group is generated by the path-components of RP n, of which there is only one!), and so th inducting upward yields that g# is an isomorphism in k -homology for all k. Now consider the commutative diagram

n τ∗ n Hk(RP ; Z2) Hk−1(S ; Z2)

g# f#

n τ∗ n Hk(RP ; Z2) Hk−1(S ; Z2)

We have established g# is an isomorphism, and from the previous example we see that τ∗ is an n n isomorphism as well. Hence f# : Hk−1(S ; Z2) Hk−1(S ; Z2) is an isomorphism, so f has odd degree. →

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 so everything 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 Cn = hom(Cn, Z); this reverses all the arrows. Then compute the homology of this “dual” chain complex instead; this is cohomology. But why would you do this? In homology, if you take a linear combination of k-dimensional faces 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 get functions 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 some function 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 ∆1(X; G) H1(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, a solution 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 fact that any such function extends to a homomorphism on k-chains and vice versa.) k k+1 2. Define δk : ∆ (X; G) ∆ (X; G) by ϕ ϕ ∂. This can be considered the “dual map” of the boundary map.→ → ◦ 3. Remark that δ δ ∆2(X; G) 1 ∆1(X; G) 0 ∆0(X; G) 0 · · · ←− ←− ←− ←− th is a chain complex, since δkδk+1 = 0 for all k. Thus, for all n, let the n cohomology group Hn(X; G) of X be the homology of this chain cocomplex. Remark. This is a standard construction in , so it might help (for myself!) to explain 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 do this is to take an element ϕ hom(H; Z) and compose it with f on the right; this changes the input 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 might recognize the analogies with constructing the dual of a vector space.)

n Despite the fact that all we did was flip the directions of the arrows, it is not true that H (C; G) ∼= hom(Hn(C; G)).

Example 61. Let’s look at the cohomology of RP 3. Recall from Example 54 that the associated chain complex was 0 2 0 0 Z Z Z Z 0. −→ −→ −→ −→ −→ ∗ Now let’s dualize with G = Z. Note that Z ∼= Z since each homomorphism is uniquely determined by 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 ϕ 0 = 0, meaning • that f ∗ is also→ the zero map. 7→ ◦

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Suppose f : Z Z is given by the “multiplication by two” map, i.e. f(x) = 2x. Then f ∗ is • defined 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 and 2 3 3 3 H (RP ) ∼= Z2, but H1(RP ) ∼= Z2 and H2(RP ) = 0. Remark. What does this tell us? In the case of homology, it implies the existence of a nontrivial cycle which is made trivial if looped twice. In the case of cohomology, the fact that H1(RP 3) = 0 implies that for any values we write down at the edges we can always assign values to the vertices which 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 the integration problem on the edges of RP 3. Fun fact: the homology groups uniquely determine the cohomology groups. With this fact, it may seem as if cohomology is useless. However, it turns out that cohomology has more structure than 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 of chain complexes by fitting it into the bigger picture below, where each downward arrow represents an application of the ∂ map.

∂ 0 Zn+1 Cn+1 Bn 0

0 ∂ 0 ∂ 0 Zn Cn Bn−1 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 Zn+1 Cn+1 Bn 0

0 δ 0 ∗ ∗ ∗ 0 Zn Cn Bn−1 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← original← ← sequence splits, we can. In our case, its split because Bn is a subgroup of a Cn−1 and is hence free abelian. As a result, this induces a long exact sequence in cohomology

B∗ Z∗ Hn(C; G) B∗ Z∗ . ←− n ←− n ←− ←− n−1 ←− n−1 ←− · · · As always, in order for this LES to be useful, we need to figure out precisely what the connecting ∗ map between sheets is. But in our case, it’s simple: the map is in, where in : Bn Zn is the inclusion map. →

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Exactness at Hn(C; G) means

0 ker i∗ Hn(C; G) coker i∗ 0. (37.1) ←− n ←− ←− n−1 ←− ∗ ∗ is split and exact. What’s ker i ? Well, if ψ : Zn G lies in ker i , then n → n ∗ 0 = i (ψ)(x) = (ψ in)(x) = ψ(in(x)). n ◦ ∗ Hence ker in consists of precisely those homomorphisms ϕ which vanish on the subgroup Bn. These in turn induce homomorphisms ϕ¯ : Zn/Bn G. → ∗ But now recall that Zn/Bn = Hn(C); as a result, ker in = hom(Hn(C),G). This answers our ∗ ∼ n original question: if coker in−1 is zero, then in fact we have an isomorphism between H (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

∗ n 0 coker i H (C; G) hom(Hn(C),G) 0. (37.2) −→ n−1 −→ −→ −→ To get a better sense of this sequence, recall the existence of another exact sequence

in−1 0 Bn−1 Zn−1 Hn−1(C) 0. −→ −→ −→ −→ Now the dual of this short exact sequence is

i∗ ∗ n−1 ∗ ∗ B Z Hn−1(C) 0. (37.3) n−1 ←− n−1 ←− ←− ∗ If Hn−1(C) is free, then in fact the sequence in Equation 37.3 is short exact. Thus in−1 is surjective, ∗ n implying coker in−1 = 0 H (C; G) ∼= hom(Hn(C),G). This means that the only thing that prevents this from happening⇒ is torsion. Thus, we have to measure to what extent the sequence 37.3 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 in. Recall further that we have a short exact sequence 0 Bn−1 Zn−1 Hn−1(C) 0, −→ −→ −→ −→ which we dualized to get

∗ ∗ ∗ 0 B Z Hn−1(C) 0. ←− n−1 ←− n−1 ←− ←− 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 by relations, everything else is zero. The dual of this (vis a vis applying hom( ,G) is) · F ∗ F ∗ F ∗ 0. · · · ←− 2 ←− 1 ←− 0 ←− Recall that this isn’t always exact. But recall that homology tells you how far you are away from being exact, since ∗ n ker fn+1 H (Fn; G) = ∗ . im fn Lemma 13. Given free resolutions F and F 0 of abelian groups H and H0, every homomorphism α : H H0 can be extended to a chain map →

F2 F1 F0 H 0

α α α α

0 0 0 0 F2 F1 F0 H 0 and any two such extensions are chain homotopic. In particular, for any two free resolutions F 0 n n 0 and F of H, there is an isomorphism of groups H (F ; G) ∼= H (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 free resolution, 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 has n homology groups Hn(C), then the cohomology groups H (C; G) are determined by the split exact sequence n 0 Ext(Hn−1(C),G) H (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

n 0 Z 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 are finitely generated Abelian groups. Proposition 27. We have Ext(H H0,G) = Ext(H,G) Ext(H0,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 ⊆ n Hn H (C; Z) = Tn−1. ∼ Tn ⊕

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 that is 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 this × → × down 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 by maps 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. For example, relative homology works similarly as before. We show this now. Recall that 0 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, which means we can construct the same long exact sequence on homology from a short exact sequence of cochain complexes. Definition 56. Define Hn(X,A; G) via δ : Cn(X,A; G) Cn+1(X,A; G) by naturally restricting δ : Cn(X; G) Cn+1(X; G). Then this creates a short→ exact sequence of chain complexes with LES 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, which happens 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 exact sequence 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

{A,B} 0 Cn(A B) Cn(A) Cn(B) C (X) 0; −→ ∩ −→ ⊕ −→ n −→ note in particular this is the same short exact sequence as was from the Mayer-Vietoris sequence for homology.

Remark. Recall the dunce cap, consisting of three vertices v0, v1, and v2 in that ordering, and set G = Z. If α is a 1-chain, what does δα = 0 mean? Well, letting σ be the inner triangle, the condition 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 some spanning tree; then, the relation in Equation 39.1 uniquely determines the values of the edges not contained in this spanning tree. Two examples of cocycles are shown below.

1 1 7 5 3 2

6 4 2 1

2 1

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 groups 1 2 1 are free, and so H (S ; Z) ∼= hom(H1(S ; Z) = 0. This means that the above cocycle is also a coboundary, meaning that we can assign values to the vertices such that the number assigned to each edge is the difference between the numbers along the vertices. Such a construction is easy in this 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 homology does. 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’ll show 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 the 2 vertices on each triangle. For an arbitrary 2-cell e , suppose the vertices are v0, v1, and v2 in this 2 order. Then the value assigned to e in the product α ^ β is the number assigned to [v0, v1] in

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6 4

4 2

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 a value 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 in Equation 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 cup product ϕ ^ ψ Ck+`(X; R) via ∈ ∈ ∈ (ϕ ^ ψ)(σ) = ϕ(σ )ψ(σ ) |[v0,...,vk] |[vk,...,v`] for every (k + `)-cocycle σ. This is essentially rigorizing/generalizing the multiplication process from 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 the • cup 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 Equation • 39.2 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 that 4 H0(M) = Z, H1(M) = Z , and H2(M) = Z. Since all these groups are free, the cohomology ∼j ∼ ∼ groups H (X) precisely isomorphic to hom(Hj(X); Z). This yields 0 1 4 2 H (M; Z) ∼= Z,H (M; Z) ∼= Z , and H (M; Z) ∼= Z. Our main goal is to determine how the cup product H1(M) H1(M) H2(M) behaves. Recall × → that in Z coefficients, a basis for H1(M) is formed by the edges ai and bi for i 1, 2 . A basis ∈ { } for H1(M) determines a dual basis for hom(H1(M),Z); thus, dual to ai is the cohomology class αi assigning the value 1 to ai and 0 to the other basis elements. The basis elements βi dual to bi are constructed similarly.

In order to represent αi by a simplicial cocyle ϕi, we need to choose values for ϕi on the edges radiating out from the central vertex in such a way that δϕi = 0. Simply setting these to zero does not work, as the cocycle condition is no longer satisfied. Instead, consider the dashed arc αi in 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 ψi for 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 to compute the cup product, on each 2-simplex [v0, v1, v2] we multiply the value assigned by ϕ1 to the edge [v0, v1] by the value assigned by ϕ1 to the edge [v1, v2]. One can see that this assigns 0 to each 2-simplex in M1, and so ϕ1 ^ ϕ1 (and hence α1 ^ α1) equals 0. Similarly, α1 ^ α2 = 0 and α1 ^ β2 = 0. However, α1 ^ β1 is nontrivial (consider the positively-oriented triangle with 2 outer edge b1); in fact, a little work shows that α1 ^ β1 is precisely the generator of H (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 and β1 will never cease to intersect.

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40 December 5

We being with another example demonstrating what the cup product is. 2 2 Example 66. Let X = RP ; we’ll work over the group Z2 for simplicity. Recall Hk(RP ; Z2) = Z2; k 2 ∼ since torsion doesn’t exist in this restricted setting, H (RP ; Z2) = Z2 for 0 k 2. ∼ ≤ ≤ How does the cup product behave on X? Consider the structure for RP 2 given below. e

e1 e2 ww v

e

Figure 40: Another visual for RP 2.

1 2 Since H (RP ; Z2) ∼= Z2, we have the existence of a nontrivial element in cohomology. A bit of experimentation yields that the cocycle β with β(e1) = β(e) = 1 and β(e2) = 0 does the trick, as any 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 is also 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 σ is 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 some definitions. Definition 58. Let A be a ring. We say that A is a graded ring if there exists a decomposition M A = Ak k≥0 where each Ak is an additive subgroup of A, such that multiplication takes Ak A` to Ak+`. × Definition 59. The graded ring M H∗(X; R) := Hk(X; R) k≥0 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. ∗ 2 3 Proposition 29. We have the equality H (RP ; Z2) ∼= Z2[β]/(β = 0). 2 i 2 Proof. Let 1, β, and β be the generators for H (RP ; Z2), where i = 0, 1, 2 respectively. Then (in suggestive 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) results from the fact that all the nth cohomology groups are trivial for n 3.) ≥ Remark. In general, ∗ n n+1 H (RP ; Z2) ∼= Z2[β]/(β = 0).

87 David Altizio 21-752 Lecture Notes

Note: one can show α ^ β = ( 1)k`(β α), where α Hk and β H`. Thus a sort of strange − ∪ ∈ ∈ k` anti-commutativity occurs, but in Z2 this just turns into commutativity because ( 1) = 1 regardless of what k and ` are. − ∗ 4 5 ∗ 5 Remark. We can use this to show that H (RP S ; Z2) = H (RP ; Z2), even though these two spaces have the same homology groups. Consider∧ the (characteristic6 cochains of the) two- dimensional and three-dimensional faces α and β in RP n for n 4, 5 ; their product α ^ β is ∈ { } the (characteristic cochain of the) five-dimensional face in RP 5 but is completely zero in RP 4 and hence trivial in RP 4 S5. ∨ 40.1 Duality

One might get the sense that homology and cohomology are “opposites” in some sense and wonder whether 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 C0 by connecting the centers of these squares to form another grid structure.

Figure 41: A CW complex and its dual.

Note that there are several between faces of C and faces of C0. In particular, Every vertex of C is contained in a unique square of C0, and vice versa; and • Every edge of C intersects a unique edge of C0. • These bijections can be used to establish parallels between homology in C and cohomology in C0. Observe the following. Let σ be a two-dimensional face of C. Then the boundary map ∂, when applied to σ, returns • a sum of the edges which are faces of σ, i.e. a linear combination of the four sides of the square σ. Now let σ0 be the vertex of C0 in the middle of σ. Then the coboundary map δ, when applied • to σ0, returns the a sum of edges for which σ0 is a face, i.e. a linear combination of the four edges eminating from σ0. These two sets of edges are in bijective correspondence with each other according to the bijections mentioned previously! This suggests homology in C and cohomology in C0 are actually the same thing and thus, from the fact that C and C0 are identical, we guess that the second homology group of C equals the first cohomology group of C0. Of course, this doesn’t always hold - in particular, the bijection breaks down if this grid of squares ends at some point. We now give some definitions that give light to exactly where this symmetry 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 µx such that the map Hn−1(M,M U) Hn(M,M 7→ \ → \ y ) gives µy and similarly for µx. Here U is a neighborhood around x and y homeomorphic to { } Rn. (“fancy language is sheaf”)

We now present our first of two important theorems. Unfortunately, the proof is omitted, but you 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 )σ . |[v0,...,v`] |[v`,...,vk]

` It is easily checked that this induces a map Hk(X; R) H (X; R) Hk−`(X; R). Indeed, one can 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 (Poincar´eduality). If M is a closed orientable n-manifold with fundamental class k [M] Hn(M; Z) then the map D : H (M; Z) Hn−k(M; Z) which takes a cocycle α and maps it to [M∈] _ α is an isomorphism for all k. The→ same is true for (not not necessarily) orientable M with 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 have no connections to topology. Let G be a graph. How many colors are required to color the vertices of G such that adjacent vertices get different colors? This problem is not inherently topological, but we may rephrase this question in terms of graph homomorphisms. Definition 65. Let G and H be graphs. A between G and H is a map f : V (G) V (H) such that if (v, w) E(G), then (f(v), f(w)) E(H). (Notice that this is not an iff statement!)→ ∈ ∈ Proposition 30. A graph G can be colored in n colors iff there exists a graph homomorphism from G to Kn.

Proof. First suppose a graph G can be colored in n colors, labeled 1, 2, . . . n. Consider the function f : V (G) V (Kn) which sends all vertices with color i to vertex i V (Kn). Note that this is a graph homomorphism→ because any two adjacent vertices in G have different∈ colors and are thus sent to different (and hence adjacent) vertices in Kn.

Conversely, given a graph homomorphism from G to Kn, we can reverse the above proof to establish a coloring on the vertices.

Unfortunately, graphs are strange because sometimes (often times!) there is no homomorphism between 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 of precisely 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.∩ ∈ 2 1

1

3 23 GH Figure 42: Two graphs G and H.

Note that this definition forces a partial ordering on the elements of hom(G, H), where two such elements η and η0 satisfy η η0 iff η(v) η0(v) for all v V (G). This is great, because any partially ordered set P gives≤ rise to a simplicial⊆ complex ∆∈ in a natural way: elements of ∆ are vertices 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 H0 induces a continuous homomorphism 0 → f∗ : hom(G, H) hom(G, H ). → V (H0) Proof. For arbitrary η hom(G, H), define f∗η : V (G) 2 ∅ by f∗η = η f. This works. ∈ → \ ◦

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 Example { } { } 67, 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 { − − } ei , . . . , ei , ej ,..., ej with (ip, jq) E(H) for all 1 p ik, 1 q j`. { 1 k − 1 − ` } ∈ ≤ ≤ ≤ ≤

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This is about as far as we can go in the general case, but when looking at specific H some interesting situations can arise. For instance, set H = K3. Then a quick computation tells us there are 12 elements of hom(K2,K3), and the relevant partially ordered set has Hasse diagram shown below. As a topological space, this is homotopic to S1! In fact, it is generally true that n−2 hom(K2,Kn) S . '

Figure 43: The Hasse diagram for hom(K2,K3). The six subsets with only two elements are in the bottom row, while the six subsets with three elements are in the top row.

Notice from the previous example that hom(K2,H) has a properly discontinuous Z2 symmetry that flips K2: for all j, send ej ej. ↔ − With this, we may finally bridge the gap between topology and our coloring problem. Suppose a graph homomorphism f : H Kn exists, where n 1. We may then use Proposition 31 to induce a homomorphism → ≥

n−2 f∗ : hom(K2,H) 2 hom(K2,Kn) S . →Z '

(Here A Z2 B indicates a map which respects the Z2 actions on A and B.) Suppose further that → n−1 n−1 we may also map S Z2 hom(K2,H), where the Z2 action on S is the natural antipodal map. Then we obtain a→ contradiction, since Borsuk-Ulam implies their composition cannot be properly discontinuous.

Proposition 32. Let X be a space with free (i.e. properly discontinuous) Z2 action. Then n−1 S 2 X exists if X is simply connected and Hk(X; Z) vanish for k n 2. →Z ≤ − Proof. The essential idea is to map it dimension by dimension. Let x Sn−1 be arbitrary, and ∈ map f(x) X; then the Z2 condition implies f( x) is forced. Now “arbitrarily” map one half of ∈ − a great circle passing through x and x in Sn−1 to X; then the image of the second half of this − circle is forced from the Z2 action. Rinse and repeat. The Hk vanishing condition ensures that we can 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, conn K is the largest n such that the first n homology groups vanish.

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. In 1955, Martin∈ Kneser∩ conjectured that

χ(G) = n 2(k 1). − − This was proven by Lovasz in 1978 using topological methods. A purely combinatorial proof was not found until 2004 by Jiˇr´ıMatouˇsek. Remark. One might wonder whether these ideas can be used to prove results about hypergraph colorings. The answer is ‘yes’, but one needs to generalize Borsuk-Ulam to non-Z2 actions, and so far we only know how to do this when n is a power of a prime.

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