1 Introduction

When are two spaces X,Y homeomorphic, that is, when is there a continuous map f : X → Y with continuous inverse? Let f, g : X → Y be continuous maps. f is homotopic to g if there exists a , H : X × I → Y such that H(x, 0) = f(x) and H(x, 1) = g(x). Put Ht : X → Y by Ht(x) = H(x, t), then H0 = f, H1 = g. Notation: f, g, ft with f0 = f, f1 = g from Hatcher. If f is homotopic to g, then we write f ' g. Similarly for spaces. Let X,Y be two spaces. They have the same homotopy type if there are continuous f : X → Y and g : Y → X such that g◦f : X → X and f ◦g : Y → Y are homotopic to the appropriate identity maps. Examples: In Rn, we have Bn = Dn = {x ∈ Rn : kxk ≤ 1}. Bn has the homotopy type of a point, for example, {0}. Let i : {0} → Bn be the inclusion, and r : Bn → {0} by r(x) = 0. n n n Then r◦i = id{0} and i◦r : B → B is homotopic to idBn by H : B ×I → I by H : Bn × I → Bn, H(x, t) = (1 − t)x. But for n > 1, Sn−1 = {x ∈ Rn : kxk = 1} are not homotopy equivalent to a point.

2 Fundamental

Definition 2.1 (). Let X be a topological . A path in X is f : I → X a continuous map. Two paths from x0 to x1 are homotopic if there exists H : I × I → S such that H(s, 0) = f(s), H(s, 1) = g(s), H(0, t) = x0 and H(1, t) = x1. In (in fact, any convex subspace of Euclidean space) every path is homotopic to a straight line path, by a straight line homotopy. Let f be a path from x0 to x1 and g be a path from x1 to x2, then f · g is a  f(2t) 0 ≤ t ≤ 1/2 new path from x to x defined by (f · g)t) = 0 2 g(2t − 1) 1/2 ≤ t ≤ 1

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

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

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

Example: O(2) has two path components, the base point is x0 = I. 2 What is π1(O(2),I)? SO(2) is the of all rotations of R , and it is home- omorphic to the . We claim that π1(X, x0) is a group, and call it the of X at x0.

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

−1 Proof. [f][g] = [f · g] is the product. Take the identity to be [cx0 ]. For [f] , ¯ ¯ we take [f] where f = f(1 − t). We must check (f · g) · g 'p f · (g · h), ¯ ¯ f · cx0 'p f 'p cx0 · f and f · f 'p f · f 'p cx0 . We can do these using reparameterizations. If f : I → X is any path, and γ : I → I is continuous with γ(0) = 0 and γ(1) = 1, then f(γ(t)) is a path which is homotopic to f. F (s, t) = f((1 − t)γ(s) + ts). Now, see page 27. Lemma 2.2 (Pasting Lemma). Let X be a such that X = A1 ∪ ... ∪ A`, each Ai is closed. Let f : X → Y be a function to another topological space. If ∀i, f|Ai : Ai → Y is continuous, then so is f. Proof. Let C be a in Y . Then f −1(C) = f|−1(C) ∪ ... ∪ f|−1(C). A1 A` Each term in this is closed in the appropriate Ai, and as Ai is closed in X, each term is itself closed in X. As the finite union of closed sets is closed, f −1(C) is closed in X, and so f is continuous. Let f, g, h be paths in X such that (f · g) · h is defined. We want to show that this composition is associative. That is, that (f · g) · h 'p f · (g · h). The right hand side is also equal to [(f · g) · h] ◦ γ, where

 1 1 2 t 0 ≤ t ≤ 2  1 1 3 γ = t − 2 2 ≤ t ≤ 4  3 2t − 1 4 ≤ t ≤ 1  f( 4s ) 0 ≤ s ≤ (t + 1)/4  t+1 And so, we define H(s, t) = g(4s − t − 1) (t + 1)/4 ≤ s ≤ (t + 2)/t  4s−2−t h( 2−t )(t + 2)/4 ≤ s ≤ 1 Change of Base Point Let X be a topological space, and x0, x1 ∈ X such that there is a path h ¯ from x0 to x1. Then ∃βh : π1(X, x1) → π1(X, x0) given by βh([f]) = [h · f · h] which is an of groups. And so, if X is path-connected, we can write π1(X). Definition 2.4 (Simply-Connected). Call X simply connected if X is path connected and π1(X) is trivial. Quotient Topology I = [0, 1], and we want to identify 0 ∼ 1. ≈ So I/ ∼ is a space, and we believe it is homeomorphic (→) the circle S1 ⊂ C. So we define p : I → S1 such that p(t) = e2πit. p(0) = p(1) = 1 ∈ S1 ⊂ C. Definition 2.5 (Quotient Topology). More generally, let X be a topological space and ∼ be an . Let g : X → X/ ∼ take x to [x]. Call V ⊂ X/ ∼ open iff g−1(V ) is open in X. Then we get a topology on X/ ∼ and we get that

2 g ... X ...... X/ ∼ ...... ∀f ...... ∃h Y

1 Proposition 2.3. π1(S , 1) ' Z. 1 Proof. We want to define a Φ : Z → π1(S ) be Φ(n) = [ωn] 2πins where ωn(s) = e . We need to show that this is a homomorphism. Let p : R → S1 by p(s) = 2πis e . Note that ωn = p ◦ ω˜n, whereω ˜n : I → R is given byω ˜n(s) = ns. R ...... ω˜n ...... p ...... ωn ... I ...... S1 ˜ ˜ Φ(n) = [p ◦ f] whenever f is any path in R from 0 to n. Thenω ˜n must be homotopic to f˜. And so, Φ is a homomorphism. Let τm : R → R be a translation, τm : x 7→ x + m. Thenω ˜m · τmω˜n is a path in R from 0 (to m) to n + m. Thus Φ(n + m) = [p ◦ (ω ˜m · τmω˜n] = [p ◦ ω˜m] · [p(τmω˜n)] = Φ(m) · Φ(n). We need to show that Φ is surjective and that ker Φ = {0}. We will need two facts.

1 −1 1. If f : I → S starting at x0 ∈ X and letx ˜0 ∈ p (x0), then ∃ unique ˜ f : I → R starting atx ˜0. 1 This implies that Φ is surjective, as we let [f] ∈ π1(S , 1) and taking ˜ ˜ x0 = 1, x˜0 = 0, we apply it and get f. Then p(f(1)) = f(1) = 1. So ˜ f˜(1) = m ∈ Z, e2πif(1) = 1 so Φ(m) = [p ◦ f˜] = [f]. 1 1 2. Let F : I×I → S be a homotopy of paths, all starting at x0 ∈ S . Choose −1 ˜ somex ˜0 ∈ p (x0). Then, ∃ a unique lifted homotopy F : I × I → R of paths starting atx ˜0.

(R, x˜0) ...... ˜ .. . F ...... p ...... F .... 1 I × I ...... (S , x0) ˜ We note that p ◦ F˜ = F , and so p ◦ ft = ft. The second fact implies that Φ is injective. Suppose that Φ(n) = e = [c1] ∈ (S1, 1). Assume that n 6= 0. Then [ωn] = e = [c1]. Then ωn is path homotopic to the constant map, and so, by the second fact, we can lift this homotopy

3 toω ˜n 'p ω˜0 = c0, but these have different endpoints, and so cannot be path homotopic, which is a contradiction. All that remains is to prove these two facts. 1 Observe the property of p : R → S : there exists an open cover {Uα} 1 −1 of S such that p (Uα) = ∪iVαi where the Vαi are disjoint open sets and p| : V → U is a . Vαi αi α One such cover is the covering of S1 by two open sets with appropriate overlap. We will first prove fact 1: (R, x¯0) ...... ˜ .. . f ...... p ...... f ...... 1 (I, 0) ...... (S , x0) The Lebesgue Covering Lemma states that if X is a compact and {Wα} is an open cover of X, then there exists λ > 0 such that every of X with diameter less than λ lies in some Wα. So, there exists a partition 0 = t0 < t1 < . . . , t` = 1 such that f([ti−1, ti]) ⊂ ˜ some Uα. We construct a lifting f on [0, tk] by induction on k. We start with ˜ ˜ ˜ t0, [0, 0], f(0) =x ˜0 Assume k < `, have f such that p ◦ f = f. We need ˜ ˜ to define f on [tk, tk+1]. We know that pf(tk) = f(tk), f(tk) ∈ Uα for some ˜ ˜ α, so f(tk) ∈ Vαi for some unique i. Put f(t) for tk ≤ t ≤ tk+1, equal to p| f(t) ∈ V ⊂ , so existence is done. Vαi α,i R Now we prove uniqueness. If p : X˜ → X is a covering projection, that is, p is onto and if given a continuous map X,˜ x˜0 ...... h, g ...... p ...... f . ... Y, y0 ...... X, x0 and if g, h :(Y, y0) → (X,˜ x˜0) are two lifting maps of f, then g = h provided that Y is connected, Y is locally connected. We prove this by putting A = {y ∈ Y : g(y) = h(y)} so that y0 ∈ A. Put B = {y ∈ Y : g(y) 6= h(y)}. Both A and B are open. We now must prove fact 2. R, x˜0 ...... ˜ .. . F ...... p ...... F ...... 1 I × I, (0, 0) .... S , x0 Proof by picture: Divide the unit square into a smaller grid. F (lower left square) ⊂ Uα, and so

4 ˜ F (segment I) ⊂ Vαi , and you can extend the map by continuing it from square to adjacent square. Or see Hatcher. Theorem 2.4 (Brouwer Fixed Point Theorem). Any continuous f : D2 → D2 has a fixed point. Usually this is proved with the no theorem, though can be proved the other way. The no retraction theorem states that there is no continuous r : D2 → S1 such that r(x) = x for x ∈ S1. Theorem 2.5 (Borsuk-Ulam Theorem). If f : Sn → Rn is continuous, then there exists x ∈ Sn such that f(x) = f(−x). For n = 1, intermediate value theorem proves it. For n = 2, use the funda- mental group, see page 32. For n ≥ 2, need and transfer or cohomol- ogy. Look at page 38, exercise 9. π1 as a . (X, x0) gets π1(X, x0) the fundamental group. φ φ∗ Also, (X, x0) → (Y, y0) induces a homomorphism of groups π1(X, x0) →

π1(Y, y0) define by φ∗([f]) = [φ ◦ f]. Note φ = idX ⇒ φ∗ = idπ1(X,x0) and φ ψ (X, x0) → (Y, y0) → (Z, z0) gives ψ∗ ◦ φ∗ : π1(X, x0) → π1(Z, z0) equal to (ψ ◦ φ)∗.

Proposition 2.6. π1(X×Y ) ' π1(X)×π1(Y ), and the isomorphism is obtained by the induced maps from the projection maps of X × Y to X,Y . 1 1 n π1(S × ... × S ) ' Z | {z } n times 1 1 1 1 S ∨ S is two intersecting at a point. Then π1(S ∨ S ) = F2. We will need the Seifert-Van Kampen theorem. If we take C \ {−1, 1}, we see that it retracts onto S1 ∨ S1. n Proposition 2.7. π1S = 0 for n > 1. n Corollary 2.8. π1(R \{0}) = 0 if n > 2. n−1 n x S ⊂ R \{0} is a homotopy equivalence by r(x) = kxk , and rt(x) = x tx + (1 − t) kxk . Definition 2.6 (Retraction). If A ⊆ X, then a retraction of X onto A is r : X → A such that r(a) = a for all a ∈ A. If i : A → X be the inclusion, then r ◦ i = idA.

i∗ Choose x0 ∈ A ⊆ X. Then π1(A, x0) → π1(X, x0) and r∗ in the other direction, we now have that r∗ ◦ i∗ = idπ1(A). And so, i∗ is injective. ∃ no retraction r : D2 → S1. Why? If there were one, then there would be 1 2 a homomorphism π1(S ) → π1(D ) which is injective, that is, an injective map Z → 0, which is impossible. Notation: X ≈ Y is used to denote homeomorphic, X ' Y is homotopic. If f ' g :(X, x0) → (Y, y0) then f∗ = g∗ : π1(X, x0) → π1(Y, y0).

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

The reason is that ∃g :(Y, y0) → (X, x0) such that g ◦ f ' idX , f ◦ g ' idY as based maps, and so π1(X, x0) → π(Y, y0) has an inverse homomorphism, g∗, and so they are inverse . Definition 2.7 (Deformation Retraction). Suppose A ⊆ X is a subspace. A deformation retraction of X onto A is a homotopy H : X × I → X such that for all x ∈ X, H(x, 0) = x, H(x, 1) ∈ A and H(a, t) = a for all t if a ∈ A.

Then i : A → X is a homotopy equivalence, and r = H(x, 1) : X → A is a homotopy inverse. n n+1 x n n+1 Note S ⊆ R \{0} shows, by x 7→ tx + (1 − t) kxk that S ' R \{0}

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

Proof. Sn = U ∪V where U, V are homeomorphic to Rn. U ∩V is path connected and homotopic to Sn−1 × (−, ). n We now show that π1(S , x0) = {1}. Claim: If X = U ∪ V , U, V open, x0 ∈ U ∩ V , and U ∩ V path connected, consider: π1(X, x0) ...... i∗ ...... j∗ ......

π1(U, x0) π1(V, x0)

Then each element of π1(X, x0) is a finite product of elements in Im i∗ or Im j∗. Proof of claim: Let f : I → X be a loop at x0, f : [0, 1] → X. Use Lebesgue covering lemma to get partition 0 = t0 < t1 < . . . < t`−1 < t` = 1 such that f([ti−1, ti]) is contained in U or V . Assume that f([tt−1, ti]) ⊆ U, then f([ti, ti+1]) ⊆ V . We call the restriction to each of these fi, and so f 'p f1 · ... · f`, and it is homotopic to (f1 · g¯1) · ... · (g`−1 · f`). f 'p f1 · f2 · f3 'p f1g¯1 · g1 · f2 · g¯2 · g2 · f3, that is, a loop in U, one in V , and then one in U.

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

π1(X, x0) so that each Im(iα∗ ) is a subgroup of π1(X, x0).

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

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

6 Lemma 2.13. Let φt : X → Y be a homotopy, let x0 ∈ X, and let h(t) = φt(x0), a path in Y from φ0(x0) to φ1(x0). Then here’s a commutative triangle π1(Y, φ1(x0)) ...... β ...... h ...... φ ...... 1∗ ...... φ0 ... ∗ π1(X, x0) ...... π1(Y, φ0(x0)) ¯ Proof. Put ht(s) = h(ts). Let f be a loop in X at x0. Inspect ht · (φtf) · ht, a loop at φ0(x0). If t = 0, get φ0∗ ([f]). If t = 1, then get βh(φ1∗ ([f])). Now we prove the prop:

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

We now go back to chapter zero to define CW-Complexes. Definition 2.8 (CW-Complexes). Let X be a Hausdorff Space and assume that X has an increasing filtration by closed subspaces, that is, there are X0 ⊂ X1 ⊂ ... ⊂ Xn ⊂ ..., such that

n n 1. X = ∪n≥0X We call X the n-skeleton 2. X0 is discrete called the vertices or 0-cells.

3. X1 the 1-skeleton adjoins the edges, the 1-cells, gives a . 4. Xn is obtained from Xn−1 by adjoining n-cells as follows: Xn = (Xn−1 t ` n n n−1 n−1 n α Dα)/ ∼ if x ∈ ∂Dα = Sα , then x ∼ φα(x) ∈ X for φα : ∂Dα → Xn−1 continuous.

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

Note: When is A ⊂ Xn closed? True iff A ∩ Xn−1 is closed and under the n n−1 ` n n −1 n characteristic maps Φα : Dα ⊂ X t β Dβ → X , Φα (A) is closed in Dα. n n n n n−1 Also, eα = Φα(int of Dα) ⊂ X is an open n-cell. Φα(Dα,Sα ) → n n−1 n n−1 S n (X ,X ), Φ | n−1 = φ , X = X ∪ e . α Sα α α α n n n n n n−1 Third, eα = Φα(Dα) may be identified with e˙α = eα \ eα = φα(Sα ) Examples

1. Any graph, even infinite.

V 1 2. α Sα for any index set.

7 P i 3. If X is a finite CW-complex, then χ(X) = i≥0(−1) αi(X) where αi(X) 2 is the number of i-cells. For T , α0 = 1, α1 = 2, α2 = 1 so χ = 0.

4. Sn start with S1 the circle in C. χ(S1) = 0 because we get n edges and n vertices. For S2, we see the half are 2-cells, and then notice that we get 2 i-cells for each, which gives χ(S2) = 2. Sn is a CW complex with 2 i-cells for 0 ≤ i ≤ n. More economically,  2 n ∈ 2 Sn = e0 ∪ en, and so χ(Sn) = 1 + (−1)n = Z 0 n∈ / 2Z 5. RPn = Sn/ ∼ where x ∼ −x. It is also the set of lines through the origin in Rn+1. 6. CPn = S2n+1 ⊂ Cn+1 with x ∼ y if x = λy for λ ∈ S1.

RPn = RPn−1 ∪ en? Well, RPn = Sn/ ∼ identifying antipodal points, which is equal to Dn/ ∼, identification of antipodal points on the . φ : Dn → Dn/ ∼ gives us the attaching map, because Sn−1 → Sn−1/ ∼= RPn−1 are contained in Dn. Example: {Sn}, Sn−1 ⊂ Sn. Then S∞ = ∪Sn is contractible. RP∞ = ∪RPn and CP∞ = ∪CPn. Definition 2.9 (Subcomplex). Let X be a CW-complex. Call a subspace A ⊂ X n n if A is closed, and if eα ∩ A 6= ∅ then eα ⊆ A. Put An = A ∩ Xn for all n. Then A is a CW-complex. For example, each Xn is a subcomplex of X. We call (X,A) a CW-pair if X is a CW-complex and A is a subcomplex of X. Then X/A is what you get when you identify all points of A. X/A can be made naturally into a CW-complex. FACT: If (X,A) is a CW-pair and A is contractible to a point, then g : X → X/A is a homotopy equivalence. Homotopy Extension Property: Given A ⊂ X, A closed and f : A → Y , can you extend f to f˜ : X → Y ? Sometimes, given i : A → X and f : A → Y , there exists f ' g such that g does have an extensiong ˜ : X → Y . Want to be able to conclude that f has extension f˜ : X → Y . Definition 2.10 (Homotopy Extension Property: Provisional Def). (X,A), A closed in X, has the HEP iff whenever f ' g : A → X and g has an extension f˜ : X → Y so does g. Definition 2.11 (Homotopy Extension Property). (X,A) has HEP if any map F : A × I ∪ X × 0 → Y has extension F˜ : X × I → Y . When does (X,A) have HEP? Answer is iff A × I ∪ X × 0 is a retract of X × I. Example: (Dn,Sn−1) has HEP. Proof: Dn × I is a solid cylinder, Sn−1 × I ∪ Dn × 0 is the boundary, minus the top. It is a retraction because if we take

8 a point higher up and hold it still, we draw a line through it to any point on the sides or bottom, and everything on that line gets projected to that point. This implies that any CW-pair (X,A) has HEP (prop 0.16). Prop 0.17 says that if (X,A) has HEP and A is contractible, then X → X/A is a homotopy equivalence. Topological Properties of CW-complexes: X is path connected iff X1 is path connected. X is locally contractible. X is normal. X is compactly generated. If K is a compact subset of X, then ∃n such that K ⊆ Xn, even better, K is contained in a finite union of closed cells, best, K is contained in a finite subcomplex. And now, we return to chapter 1. We need an algebraic aside so that we can do VK. Problem: Given groups {Gα}α∈A and φα : Gα → H, H another group, want some group G together with homomorphisms iα : Gα → G such that ∃Φ: G → H unique subject to iα .... Gα ...... G ...... φ ...... α ...... Φ G Then (G, iα) is the or of the Gα’s.

Make G = ∗αGα as all words h1h2 . . . h` with hi ∈ Gαi \{1}. We call it a reduced word if it cannot be simplified further. G is the reduced words under concatenation.

Theorem 2.14 (Seifert-van Kampen). Let X be a topological space, x0 ∈ X such that X = ∪α∈I Aα with Aα open, path-connected and x0 ∈ Aα for all α. If Aα ∩ Aβ ∩ Aγ is always path-connected, then we have a homomorphism Φ: ∗απ1(Aα) → π1(X) which is surjective, and N = ker Φ is the normal −1 subgroup of the free product generated by all elements iαβ∗(ω)iβα∗(ω) where iαβ : Aα ∩ Aβ → Aα is the inclusion. That is, π1(X) ' ∗α(Aα)/N. A universal property holds: G ...... ∃! ...... φ .. .. φβ α ...... π1(X) ...... π1(Aα) π1(Aβ) ...... π1(Aα ∩ Aβ) W Example: π1( α Xα) with xα ∈ Xα and ∀α, xα ∈ Uα ⊂ Xα, Uα open and Uα deformation retracts to xα. Then π1(X) ' ∗π1(Xα), where X = ∨αXα. Also, assume each Aα is path connected, then so is X.

9 W W Proof. Take Aα = Xα ∨( β Uβ) ' Xα. Then for α 6= β, Aα ∩Aβ = γ Uγ ' x0. Now we apply VK Theorem trivially.

1 1 Using this, we find that π1(S ∨S ) = Z∗Z, the on two generators. 1 2 π1(S ∨ S ) ' Z. In fact, if G is any connected graph, then a maximal tree can be contracted to a point, and so has fundamental group free, and so is a bouquet of circles. 2 N 1 π1(R \N pts) =free group on n generators. This space contracts to ∨i=1S . 2 So π1(S \ N) ' free group on n − 1 generators. n n Suppose R \ K is path connected, K is closed and bounded. π1(R \ K) → n π1(S \ K) is an isomorphism when n > 2.

n n Proof. Take U = R \ K and V = S \ BR(0) where K ⊆ BR(0). Note that n U ∩ V = R \ BR(0), which is path connected. (ADD DETAILS LATER) 3 1 3 Let A be a circle in the plane. Then R \ A ' S , so π1(R \ A) = Z. Proposition 2.15. Inspect Sp+q−1 ⊂ Pp+q. It has Sp−1,Sq−1 as subspheres. Then there exists a deformation retraction from Sp+q−1 \Sq−1 onto Sp−1. Thus, 3 1 π1(S \ A) ' π1(S ).

p+q−1 q−1 p+q−1 q−1 Proof. Define ft : S \ S → S \ S by ft(x, y) = ((1 − t)(x, y) + x t( kxk ), 0)/(kTOP k).

R3 \ (A ∪ B) where A, B are unknotted circles contracts to S1 ∨ S1. R3 \ (A ∪ B) where A, B are linked, then R3 \ (A ∪ B) ' S1 × S1, and so the fundamental group is Z × Z. 3 What is π1(R \ (A ∪ B)), the of linked circles? It is isomorphic 3 3 2 2 2 to π1(S \ (A ∪ B)). Now, S = {(z, w) ∈ C : |z| + |w| = 1}. Put Tz = 3 2 3 2 {(z, w) ∈ S : |w| ≤ 1/2} and Tw = {(z, w) ∈ S : |z| ≤ 1/2}. Note that 3 {(z, 0) : |z| = 1} ⊂ Tz and {(0, w): |w| = 1} ⊂ Tw. Now, S = Tz ∪ Tw and 3 2 2 1 1 1 2 Tz ∩ Tw = {(z, w) ∈ S : |z| = |w| + 1/2}' S × S . So Tz ' S × D the solid . 3 Tz ∩ Tw is a deformation retract of S \ (A ∪ B). 2 2 2 iθ 2 2 Tz = {(z, w): |z| + |w| = 1, |w| ≤ 1/2}. z = |z|e and |z| = 1 − |w| so (z, w) 7→ (eiθ = z/|z|, w) ∈ S1 × √1 D2 ' S1 × D2. 2 m −n π1(Km,n) where (m, n) = 1 is the group ha, b|a b i. Attaching 2-cells creates relations by how they are attached. 1 Let X be a path , x0 ∈ X Attach 2-cells via ϕα :(S , s0) → F 2 1 (X, x0) and put Y = (X t α Dα)/ ∼ where ϕα(x) ∼ x for all x ∈ Sα.

Proposition 2.16. Then π1(Y, x0) ' π1(X, x0)/N, N is the smallest subgroup containing all [ϕα] ∈ π1(X, x0) Proof. Case 1: a Single 2-cell. Φ : D2 → e2 ⊂ Y , y = Φ(0). Y = (Y \{y})∪e2 = 2 1 e \{y}' S . VK implies that π1(Y, Φ(d1)) ' π1(X \{y}, Φ(d1))/minimal subgroup generated by [f]. π1(Y, Φ(d1)) ' π1(Y, x0) ' π1(X, x0)/[ϕα].

10 Case 2: Induction for the case of a finite number of 2-cells. Case 3: Any number at all: 1 S 2 Let ϕα : S → X. Y = X ∪ eα, yα = Φα(0). S α∈A Uα = X ∪ β(eβ \ yβ) ∪ Eα = Y = ∪β{yβ} ∪ eα. Uα will be open and path connected . Thus, α 6= β ⇒ Uα ∩ Uβ = Y − ∪β∈A{yβ}' X. Let F be all finite subsets of A. Put for F ∈ F , UF = ∪α∈F Uα ' X ∪ S e . αF α We know π1(UF ) = π1(X)/normal subgroup generated by [ϕα] for α ∈ F .

VK Theorem: Note that UF1 ∩UF2 = UF1∩F2 , and also for triple intersections. Thus, G ...... ∃! ...... π1(Y ) ......

π1(UF1 ) π1(VF2 ) ......

π1(UF1∩F2 ) If F is the set of all finite subsets of A and we order F by inclusion, we obtain an directed ordered set. Given F1,F2 ∈ F , note that F1 ⊂ F1 ∪ F2 and F2 ⊂ F1 ∪ F2. Thus, π1(Y ) is the colimit (or direct limit) of {π1(UF )}F ∈F along with the homomorphisms from inclusions in F .

An algebraic comment: If D is a directed set α ∈ D with Gα a group and 0 0 d ≤ d implies there is a homomorphism id,d0 : Gd → Gd0 and d = d implies it is the identity, with d ≤ d0 ≤ d00 gives .... Gd ...... Gd00 ...... Gd0 Then lim = colim(G , i 0 ) = ∗ G /N where N is the normal subgroup −→ α α,α d α −1 0 generated by {gα iα,α0 (gα)} with gα ∈ Gα and α ≤ α . NEXT: Covering Spaces, read the section Definition 2.12 (). A covering map p : X˜ → X is a map such −1 that X is covered by evenly covered open sets U, that is, p (U) = ∪αVα, Vα’s open and disjoint in X˜ with p : Vα → U a homeomorphism. Remarks: p is a local homeomorphism, meaning for allx ˜ ∈ X˜, there is an open nbhd V ofx ˜ such that p|V : V → p(V ) is a homeomorphism p(V ) open in X. p is an open mapping. p is a quotient mapping, provided p is surjective.

11 Examples: ∅ and any X, id : X → X. The number of sheets is the index of the subgroup that is the fundamental group of the covering space. It is normal if you get the same subgroup (rather than a conjugate one) from a change of basepoint. Lifting Theorems Theorem 2.17 (Homotopy Lifting Property). Given f˜ 0 ... Y ...... X˜ ...... p . inc ...... {ft} ... Y × I ...... X ˜ Then there exist a unique {ft} : Y × I → X˜.

Theorem 2.18. If p : X˜ → X is any covering map and p(x ˜0) = x0 then

1. p∗ : π1(X,˜ x˜0) → π1(X, x0) is injective ˜ ˜ 2. Im p∗ = {[f]: f based at x0 such that the lifting f such that f(0) =x ˜0 is a loop}. ˜ Proof. 1. how that ker p∗ = 1. Suppose that [f] ∈ π1(X,˜ x˜0) such that ˜ ˜ [p ◦ f] = 1 ∈ π1(X, x0). So p ◦ f = f ' cx0 . The HLT says that we can ˜ ˜ lift to get f 'p cx˜0 , and so [f] = 1. 2. ⊇ is clear. ⊆ is also clear.

Proposition 2.19. If X˜ and X are path connected, then |p−1(x)|, x ∈ X is constant and equal to [π1(X, x0): p∗(π1(X,˜ x˜0))]. Theorem 2.20 (Lifting Theorem). Assume that Y is path connected and locally path connected, with f :(Y, y0) → (X, x0) and p :(X,˜ x˜0) → (X, x0) is a ˜ covering map. A lift f : Y → X˜ exists iff f∗(π1(Y, y0)) ⊆ p∗(π1(X,˜ x˜0)). Proof. It f˜ exists, the inclusion is clear. Assume that the inclusion holds. Take y ∈ Y run a path γy = γ from y0 to ˜ y. Then f ◦ γy is a path in X from x0 to f(y). Lift to f ◦ γy which is a path in ˜ ˜ ˜ X˜ fromx ˜0 to a point we shall name f(y). f(y) = f ◦ γy(1) is set to f(y) by p. ˜ Must check the independence of path from y0 to y, that f : Y → X˜ is ˜ continuous and f(y0) =x ˜0, though the last part is trivial. ˜ ˜ So is f ◦ γy(1) = f ◦ δy(1) for δ, γ paths from y0 to y?(f ◦ γy) · f ◦ δy ' ¯ f ◦ (γy · δy) 'p p ◦  where  is a loop atx ˜0 in X˜. Thus f ◦ γy ' (p ◦ ) · (f ◦ δy), ˜ ˜ and so f ◦ γy 'p  · f ◦ δy. Thus, the endpoints are the same. Why must f˜ : Y → X˜ be continuous? Let y ∈ Y , and check continuity at y. ˜ f(y) ∈ U evenly covered, f(g) ∈ Vα, p : Vα → U homeo. Show that there is a

12 ˜ neighborhood W of y such that f(W ) ⊆ Vα. Since U is open and f(y) ∈ U, there is an open nbhd W of y such that f(W ) ⊂ U hypotheses on Y let us assume that ˜ ˜ W is path connected. Recall, f(y) = f ◦ γy(1) ∈ Vα. Let w ∈ W , run a path in ˜ ˜ ˜ W , ζw from y to w. Then f(w) = f ◦ (γy · ζw)(1) = (f ◦ γy) · (f ◦ ζw)(1). The first part ends at f(y), and the second part happens in U. ˜ −1 And so, this equals f ◦ γy ·p (f ·ζw)(1) ∈ Vα. And so, the map is continuous.

Next we construct the universal cover. If H ≤ π1(X, x0) then we construct X˜H → X such that pα(X˜H ) = H. We will also attempt to classify all covering spaces of some ”good” X isomorphism. Graphs For a graph, χ(X) = |V | − |E|, and if X˜ is an n-sheeted cover of X, then χ(X˜) = nχ(X). If we take X → X/T ≈ W S1 is a homotopy equivalence, and so χ(X) = χ(W S1) = 1−k, where k is the number of circles. Thus, χ(X˜) = n(1−k) = 1−` and ` = 1 + (n − 1)n. If X is some nice space, path conn and locally path conn, then an isomor- phism of spaces over X is f : X˜1 → X˜2 such that p2 ◦ f = p1. We define Cov(X) =all isomorphism classes of coverings of X. Y˜ = p−1(X˜) X˜ ...... p . p ...... φ ...... YX...... then Y˜ = {(y, x˜) ∈ Y × X˜ : φ(y) = p(˜x)} maps down to Y by πY . Thus, we get a map φ∗ : Cov(X) → Cov(Y ) Assume that X is as before and also that for each x ∈ X, there is an open U containing x such that every loop in U is homotopic to the constant loop in X. Then we call it SLSC (Semi-Locally Simply Connected). [Losternik-Schrilmann of a Space]: RPn = Sn/(x ∼ −x). Let n n Ui ⊂ RP be given by xi 6= 0. So RP is a union of n + 1 contractible sets. In general, X = V0 ∪ ... ∪ Vn is an open cover such that ιi : Vi → X is homotopic to a constant map. This is similar to SLSC. Theorem 2.21. There exists a simply connected covering space if SLSC holds for X, and conversely.

Let pu : Xu → X be simply connected covering space of X and let X˜ → X be any other covering space, X˜ also path connected. Then the diagram commutes: ...... ˜ Xu, xu . p X, x˜0 ...... puφ ......

X, x0

13 Note: §1.3 Ex 16 says that φ is also a covering map. Can a s.c. X have covering space no isom to id : X → X? p : X˜ → X given, then isomorphic as it must be a one-sheeted covering space. Easy: If 1-sheet, then p is a homeomorphism. Also if s is a section of the −1 covering map (p ◦ s = idX ) and s(x) ∈ p (x) for all x ∈ X, then p is a homeomorphism.

Proposition 2.22. If H is a subgroup of π1(X, x0) then there exists a covering X˜ → X such that the induced map p∗(π1(X,˜ x˜0)) = H. We define an of a covering space to be an isomorphism f : X˜ → X˜. We also call these Deck Transformations. We name the group of these as G(X,˜ p) = G(X/X˜ ). If for any two pointsx ˜1, x˜2 such that p(˜x1) = p(˜x2) ∈ X, there exists φ ∈ G(X,˜ p) such that φ(˜x1) =x ˜2, call p normal.

Proposition 2.23. Let p :(X,˜ x˜0) → (X, x0) be a covering with X,X˜ path conn and X locally path conn. Then

1. p is normal iff p∗π1(X,˜ x˜0) is a normal subgroup of π1(X, x0).

2. Assuming p is normal, there exists an isomorphism G(X,˜ p) ' π1(X, x0)/p∗π1(X,˜ x˜0).

So for the universal covering, G(X,˜ p) ' π1(X, x0) −1 Assume X˜ → X is a universal cover. So G(X,˜ p) acts transitively on p (x0). Define φ : π1(X, x0) → G(X,˜ p) by if [γ] ∈ π1(X, x0) lift γ toγ ˜ a path in X˜ −1 withγ ˜(0) =x ˜0 andγ ˜(1) ∈ p (x0). Then there exists g ∈ G(X,˜ p) such that g(˜x0) =γ ˜(1). Put φ([γ])(˜x0) =γ ˜(1), and φ([γ]) ∈ G(X,˜ p). This element is unique, as if g1, g2 ∈ G(X,˜ p) and ifx ˜0 such that g1x˜0 = g2x˜0 then g1 = g2 by uniqueness of lifting. Then we can show easily that φ is a homomorphism of groups and that φ is surjective. φ is also injective. Change point of view: If Y is a given space and G is a group, G × Y → Y an action such that g : Y → Y is a homeomorphism of Y . Then y ∼ gy for g ∈ G gives an equivalence relation, and Y → Y/G is the space of orbits. n Favorite Example: Y = S , G = {±1}' Z2 by (+1)y = y and (−1)y = −y, n n n the antipodal map. Y/G = S /Z2 ' RP . Also π1(RP ) ' G ' Z2. Let G act on Y . Call the action even if ∀g ∈ Y , ∃ open neighborhood V of y such that g 6= h ⇒ gV ∩ hV = ∅. Lemma 2.24. If G acts evenly on Y , then p : Y → Y/G is a covering map. Proof. We know p is continuous and open since if W ⊆ Y is open then p−1(p(W )) = ∪g∈GgW . In fact, this is a disjoin union. Claim: For V as in definition, then p(V ) is evenly covered. Note that p|gV : gV → V is bijective. [Clearly surjective, for injective, we say gv1, gv2 are in the same orbit, so ∃h ∈ G such that gv1 = hgv2, disjointness implies that hg = g, so h = 1, hence v1 = v2.]

14 We call such a covering space a G-covering. Isomorphism of G-coverings is φ : Y1 → Y2 over Y1/G ≈ X ≈ Y2/G such that φ is a homeo and φ(gy) = gφ(y). Example, Y = S2n−1 ⊂ Cn, G = {z ∈ C : zk = 1}, then G×S2n−1 → S2n−1 by v 7→ zv. Then S2n−1 → S2n−1/G gives us a . Trivial G-covering over a given X is X × G → X by projection where G has the discrete topology on it. h(x, g) = (x, hg). Lemma 2.25. Any G-covering Y → X is locally trivial as a G-covering. ie, X can be covered by open sets U such that gv ∈ p−1(U) → U is isomorphic to U × G → U by projection. Proposition 2.26. Let p : Y → X be a G-covering. Then we have a homo- G → Aut(Y, p) which is injective. The the case where Y is connected, this is an isomorphism.

Proof. Pick y0 ∈ Y , then for ϕ ∈ Aut(Y, p) = G(Y, p), ϕ(Y0) satisfies p(ϕ(y0)) = py0 so ∃g ∈ G such that ϕ(y0) = gy0. So ϕ and y 7→ gy0 are 2 elements of Aut(Y, p) which agree on y0. Uniqueness props for lifting implies that φ sends y → gy. Proposition 2.27. Let p : Y → X be a covering, Y connected, then Aut(Y, p) acts evenly on Y . If Aut(Y, p) acts transitively on a fiber of p, then the covering is a G-covering for G = Aut(Y, p). Chapter 11 and 13 of Fulton. Proposition 2.28. Let G act evenly on a simply connected and locally path connected space Y . Then for X = Y/G, π1X = G.

Proof. We’ve seen that π1(X) ' Aut(Y, p) for p : Y → X and prop above implies that G ' Aut(Y, p), so π1(X) ' Aut(Y, p) ' G. Chapter 14 - The Van Kampen Theorem x0 ∈ X = U ∪V , x0 ∈ U ∩V , U, V, U ∩V path connected, then we can define the fundamental group and a map π1(X) → G which is unique to each G that makes a commutative diagram commute. What is the meaning of a π1(X, x0) → G? Answer: G-coverings of X up to isomorphism preserving basepoints. Suppose given a homomorphism ρ : π1(X, x0) → G. Construct a G-covering pρ : Yρ → X with yρ ∈ Yρ s.t. pρ(yρ) = x0. Give G the discrete topology. Consider X˜ × G and a left action of π1(X, x0) on it by [σ] · (z, g) = ([σ] · z, g · ρ([σ])−1). Then, put Yρ = (X˜ × G)/π1(X).

15 3 Homology

Let X be a topological space. We want to define Hi(X) the homology groups i ≥ 0 with some good properties. We expect it to be functorial and to be useful. ∆-complexes We look at T 2. v a v ...... •...... •...... c ...... a ...... b ...... v...... b v. •...... •. m And calling the inside of the triangles U and L. Let v0, . . . , vn ∈ R , assume that they are affinely independent. That is, no three are collinear, no four coplanar, etc. Equivalently, if c0v0 + ... + cnvn = 0 and c0 + ... + cn = 0 then ci = 0 for all i. So, let v0, . . . , vn be affinely independent. Then define [v0, . . . , vn] = all Pn P i=0 tivi where ti ≥ 0 and ti = 1. 1 P The Barycenter is n+1 vi. n n+1 The standard n-simplex is ∆ = [e0, . . . , en] ⊆ R . n If v0, . . . , vn are affinely independent points, ∃ a homeomorphism ∆ → n [v0, . . . , vn], as any point in ∆ is of the form (t0, . . . , tn), and we can send it to P tivi. We can orient the edges by the edge between vi and vj points to the greater one. We define for a ∆-complex. n We look at X = ∪eα open n-simplexes and this union is disjoint. We have n n n n for each eα σα : ∆ → [v0, . . . , vn], which gives a map from int∆ → eα. We define ∆n(X) to be the free generated by all open n- n n simplexes. We can identify eα ↔ σα. So the elements look like finite sums of simplices or of these maps. We call each of these a simplicial chain. We now attempt to define a boundary operator. ∂([v0, v1]) = v1 − v0, ∂([v0, v1, v2]) = [v0, v1] + [v1, v2] − [v0, v2] = [ˆv0, v1, v2] − [v0, vˆ1, v2] + [v0, v1, vˆ2]. P i In general ∂([v0, . . . , vn]) = (−1) [v0,..., vˆi, . . . , vn]. ∂ And so, we have for ∆-complex X, ∆n(X) → ∆n−1(X). Lemma 3.1. ∂∂ = 0. ∂ We define Zn(C) = ker(∆n(X) → ∆n−1(X)) to be the cycles and Bn(X) = Im(∆n+1(X) → ∆n(X)). Bn ⊆ Zn by the lemma. ∆ We define Hn (X) = Zn(X)/Bn(X). 2 2 For the torus above, ∆2(T ) has basis U, L, ∆1(T ) has basis a, b, c and 2 ∆0(T ) has basis v. ∂(U) = ∂(V ) = a + b − c, ∂(a) = ∂(b) = ∂(c) = 0. H0 = ∆0/0 ' Z. H1 = ∆1/(a + b − c) ' Z ⊕ Z. H2 = Z(U − L)/0 ' Z.

16 [Eilenberg] Let X be any space. A singular n-simplex in X is a continuous map σ : ∆n → X. Put Cn(X) = the generated by all singular n-simplexes, P n so elements look like nσσ where σ is a map ∆ → X where nσ ∈ Z and only finitely many are nonzero. n Pn i ∂ : Cn(X) → Cn−1(X) is defined by, if σ : ∆ → X, ∂σ = i=0(−1) σi n−1 th where σi : ∆ → X is the inclusion of the i face, φi followed by σ. φi(t0, . . . , tn−1) = (t0, . . . , ti−1, 0, ti, . . . , tn−1). Lemma 3.2. ∂2 = 0. ∼ ∼ If X = Y , then Hn(X) = Hn(Y ).  n = 0 Proposition 3.3. If X = {pt}, then H (X) = Z n 0 else

n Proof. For n ≥ 0, σn : ∆ → {pt} is unique. Thus, ∂σn is the alternative sum of σn−1’s, which is 0 if n is odd, 1 if n is even. Thus, we get the complex Z → Z → Z → ... → Z. In each, either everything is a cycle and everything is a boundary, or else none of either is. Except in dimension zero, as everything is a cycle, but nothing is a boundary, so H1(X) = Z. ∼ Proposition 3.4. If X has path components Xα, then Hn(X) = ⊕αHn(Xα). ∼ Proposition 3.5. If X is path connected and nonempty, then H0(X) = Z. P Proof. Choose x0 ∈ X. Consider C0(X) = x∈X nxx. ∂  C1(X) → C0(X) → 0, but we can augment it, so instead we get C0(X) → Z. P P ( nxx) = nx. Note: ∂ = 0 since ∂σ1 = 1 − 1 = 0. Claim: ker  = Im ∂. We know that Im ∂ ⊂ ker . We need to show the converse. Let c = n1x1 +...+n`x` ∈ C0(X) with (c) = 0 so n1 +...+n` = 0. Choose 1 paths fi from x0 to xi (1 ≤ i ≤ `) so fi = σi : ∆ → X with ∂σi = xi − x0 for all i. P P P Then σ( niσi) = σni(xi − x0) = nixi − nix0 = c Definition 3.1 (Reduced Homology). Let X 6= ∅ be out space. We have ... →   H (X) n > 0 C (X) → C (X) → C (X) → . Define H˜ (X) = n 2 1 0 Z n ker / Im ∂ n = 0

Fact: If X is path connected, x0 ∈ X, then there exists a homomorphism π1(X, x0) → H1(X) called the Hurewicz homomorphism which is surjective and has the of π1. Definition 3.2 (). If f : X → Y is a continuous map, n n then f] : Cn(X) → Cn(Y ) by composing f with σ : ∆ → X to get σ : ∆ → Y .

17 For c ∈ Cn(X), ∂f](c) = f]∂(c). Note: f](Zn(X)) ⊆ Zn(Y ) and f](Bn(X)) ⊂ Bn(Y ) induces f∗ : Hn(X) → Hn(Y ). 0 → Bn(X) → Zn(X) → Hn(X) → 0, f] gives an homomorphism of short exact to 0 → Bn(Y ) → Zn(Y ) → Hn(Y ) → 0.

Theorem 3.6. f, g : X → Y and f ' g, then f∗ = g∗ : Hn(X) → Hn(Y ). Proof. Let F : X × I → Y continuous with F (x, 0) = f(x) and F (x, 1) = g(x) for all x ∈ X. We want to show that f∗ = g∗ : Hn(X) → Hn(Y ). n n n Consider ∆ × I. ∆ × {0} = [v0, . . . , vn], ∆ × {1} = [w0, . . . , wn]. We have additional n-simplices [v0, . . . , vi, wi+1, . . . , wn]. n Sn Note: ∆ × I = i=0[v0, . . . , vi, wi, . . . , wn] is an n + 1-simplex. Pn i Define P : Cn(X) → Cn+1(Y ) by σ 7→ i=0(−1) F ◦(σ×id)|[v0,...,vi,wi,...,wn]. Check: ∂ ◦P = g] −f] −P ◦∂ as hom Cn(X) → Cn(Y ). or ∂P +P ∂ = g] −f]. (P is a chain homotopy between f] and g]) Let z ∈ Zn(X). Thenz ¯ ∈ Hn(X). Then f∗z¯ = f]z and g∗z¯ = g]z, so to show that f∗ = g∗, we must show that f]z − g]z is a boundary. f]z − g]z = −(∂P z + P ∂z) = ∂(−P z).

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

k What is Hn(S ) when k, n > 0? ∆ i∗ Want relative homology long exact , → Hn+1(X,A) → Hn(A) → j∗ Hn(X) → Hn(X,A) → Given a space X, A ⊆ X, Z ⊂ A such that Z¯ ⊂int(A). Have (X \ Z,A \ Z) → (X,A).

Theorem 3.8. i∗ : Hn(X \ Z,A \ Z) → Hn(X,A) is an isomorphism. Use the above to determine the homology groups of Sk by induction on k; later (or read)

Definition 3.3 (Relative Homology). We define the relative chain complex of a pair (X,A) where A ⊂ X as Cn(X,A) = Cn(X)/Cn(A). The relative homology is the homology of this chain complex. Proof of Theorem:

Proof. We have, easily, i∗, j∗, we need ∆ : Hn(X,A) → Hn−1(A). Let z ∈ Zn(X,A) withz ¯ ∈ Hn(X,A). Pick x ∈ Cn(X) with x 7→ z. Then there exists a unique a ∈ Cn−1(A) such that a 7→ ∂x. Note that ∂a = 0 so a ∈ Zn−1(A). Put ∆(¯z) =a ¯ ∈ Hn−1(A). Must check that ∆ is well-defined.

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

18 Proof. Must verify exactness at Hn(A), Hn(X), Hn(X,A). 0 At Hn(A): i∗∆¯z = 0 is clear from the definition of ∆¯z. Suppose a¯ ∈ Hn(A). 0 0 0 0 0 a ∈ Zn(A) and i]a¯ = 0 in Hn(X). Then i]a = ∂x , x ∈ Cn+1(X). Exactness follows by diagram chasing. At Hn(X): Clearly j∗i∗ = 0. Suppose givenz ¯ ∈ Hn(X) such thatz ¯ → 0 in Hn(X,A), there exists a ∈ Cn(A) and n ∈ Cn+1X sch that z = ∂n + i∗a check that a is a cycle, i∗n¯ =z ¯. Proposition 3.10. Let f, g :(X,A) → (Y,B) be homotopic as maps of pairs. Then natural f∗ : Hn(X,A) → Hn(Y,B) agrees with g∗.

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

...∆ ...... Hn+1(X,A...... H. )n...... (A)...... H...... n(X.....H..) ...... n(X,A) ...... f∗ . f∗ . f∗ . f∗ . ...∆ ...... Hn+1(Y,B.....H.. )n...... (B)...... H...... n(Y ..)....H...... nY,B commutes. Reformulation of Excision Theorem X ⊃ A, Z ⊂ A put B = X \ X, so A ∩ B = A \ Z. Now, Z = X \ int(X \ Z). So Z ⊂ int(A) means that X \ int(B) ⊂ int(A), thus A = R (A) ∪ int(B). Theorem 3.11. If U = {A, B} covers X so that X = int(A) ∪ int(B), then Hn(B,A ∩ B) → Hn(X,A) is an isomorphism. Application: X is a space SX is a suspension of X,m that is, X × I/ ∼ with (x, 1) ∼ (x0, 1) and (x, 0) ∼ (x0, 0) for all x, x0 ∈ X. Then there exists isomorphisms H˜n+1(SX) → H˜n(X) ˜ n Corollary 3.12. Hk(S ) is Z for k − n, 0 else. Proof. We now prove the suspension theorem. A = SX \{S}, B = SX \{N}. A ∩ B ' X, so 0 → H˜n+1(SX) → Hn+1(SX,A) → 0 exact (thus middle is an isomorphism), and also Hn+1(SX,A) ∆ isomorphic to Hn+1(B,A∩B), and 0 → Hn+1(B,A∩B) → H˜n(A∩B) → 0. Generalized Excision Theorem: X is given. Let U = {Uα} be a cover of X such that {int(Uα)} is a cover U of X. Let Cn (X) ⊂ Cn(X) be a subgroups generated by all U-small singular n n n-complexes σ : ∆ → X meaning σ(∆ ) ⊂ Uα for some α. Observe that σ U-small implies that ∂σ is U-small. U U U Thus, ∂ maps Cn (X) → Cn−1(X). So get ι∗ : Hn (X) → Hn(X) is an isomorphism.

19 U More precisely, ι : Cn (X) → Cn(X) is an inclusion which has chain homo- U topy inverse ρ : Cn(X) → Cn (X), ι and ρ are both chain maps and ρ ◦ ι 'c id and ι ◦ ρ 'c id where 'c is chain homotopy. Exercise: From this, you can get the excision theorem. Proof of the generalized excision theorem

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

1. Subdivide simplexes, with iterated barycentric subdivision. N N If [w0, . . . , wn] is an n-simplex in R and b ∈ R , then b · [w0, . . . , wn] = [b, w0, . . . , wn], an n + 1 simplex.

Aim: Barycentric Subdivision of an n-simplex [v0, . . . , vn]. We define the −1 simplex to be [∅]. We will define this by induction on n. For n = −1, nothing. For n = 0, also no change.

For n > 0, assume that each (n−1)-face of [v0, . . . , vn] has been subdivided into n!(n − 1)-faces, which we will call [w0, . . . , wn−1] 1 Then we let b = n+1 (v0+...+vn) be the barycenter of [v0, . . . , vn] and con- sider all n-simplexes [b, w0, . . . , wn−1] = b · [w0, . . . , wn−1] giving a grand total of n!(n + 1) = (n + 1)! n-simplexes.

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

N 2. Y ⊆ R a . Let LCn(Y ) ⊂ Cn(Y ) be the subgroup generated by all linear maps σ : ∆n → Y . ∂ Note: LCn(Y ) → LCn−1(Y ), so the linear chains give us a chain complex. If b ∈ Y , we get b : LCn → LCn+1 by b · [w0, . . . , wn] = [b, w0, . . . , wn]. Check that ∂b(α) = α−b(∂α) for all α ∈ LCn(Y ). That is, ∂b+b∂ = id −0 on LCn(Y ).

Take α = [w0, . . . , wn], then ∂b[w0, . . . , wn] = ∂[b, w0, . . . , wn] = [w0, . . . , wn]− [b, w1, . . . , wn] + ..., so it works out. n Define S : LCn(Y ) → LCn(Y ) by induction on n, let λ : ∆ → Y , λ = [w0, . . . , wn] and let bλ = the barycenter of [w0, . . . , wn]. Put S([∅]) = [∅] and S([w0]) = [w0]. In general, S(λ) = bλ · (S∂λ) in LCn(Y ).

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

3. We will now subdivide general chains. S : CnX → CnX by Sσ = σ]S · [e0, . . . , en], where S[e0, . . . , en] ∈ LCn. n Check ∂Sσ = S∂σ and putting T σ = σ]T ∆ , then ∂T + T ∂ = id −S.

20 4. Iterated subdivision. Recall that we have S, T such that ∂S = S∂ and ∂T + T ∂ = id −S. This says that Si is a chain map, and that Si is even chain homotopic to id. P i P i m Dm = 0≤i≤m TS = T ( S ), check that ∂Dm + Dm∂ = id −S . n m U For each σ : ∆ → X, there exists m such that S (σ) ∈ Cn (X). Put m(σ) =least m that works. Define D : Cn(X) → Cn+1(X) by Dσ = U Dm(σ)σ, then ∂Dσ + Dσ∂ = σ − ρ(σ). We discover that ρ(σ) ∈ Cn (X). U Check ρ : Cn(X) → Cn (X) is a chain map, ∂ρ = ρ∂, and moreover ∂D + D∂ = id −ι ◦ ρ where ι is the inclusion of CU into C. Note that ρι = id on CU , and so we are done.

Theorem 2.13

Definition 3.4 (Good Pair). Hn(X,A) → H˜n(X/A) isomorphism if A 6= ∅,A closed, A ⊂ V open, A a deformation retract of V . Then we call (X,A) a good pair.

Prop 2.21 U C∗ (X) → C∗(X) the inclusion, p in the opposite direction, with pi 'c id and ip 'c id Relative version, (X, U ), (A, U |A)...... U |A ...... U ...... U ...... 0 .. C∗ (A) .. C∗ (X) .. C∗ (X,A) .. 0 ...... 0 ...... C∗(A) ...... C∗(X) ...... C∗(X,A) ...... 0 U implies that H (X,A) → H∗(X,A) is an isomorphism. ∗ R R Pf. of excision theorem in form X = (A)∪ (B) implies that H∗(B,A∩B) = H∗(X,A).

Proof. Let U = {A, B} ...... U |A ...... U ...... U ...... 0 . C∗ . C∗ (X) . C∗ (X,A) . 0

C∗(A) C∗(A) ⊕ C∗(B) with vertical columns equivalences. U C∗A+C∗B C∗B C∗B Then C (X,A) = ' = = C∗(B,A∩B). ∗ C∗A C∗(A)∩C∗(B) C∗(A∩B)

21 (X,A) a good pair implies that q :(X,A) → (X/A, A/A) induces an isomor- phism H∗(X,A) → H∗(X/A, A/A) → H˜∗(X/A). A closed subset of V open, A deformation retract of V implies that A/A closed subset of V/A, A/A a def retract of V/A. (X, A, B) triple, X ⊃ A ⊃ B, then 0 → C∗(A, B) → C∗(X,B) → C∗(X,A) → 0 is a short exact sequence of chain complexes, and so we get a long exact se- quence on relative homology groups. '... H∗(X,A) ...... H∗(X,V ) ...... q∗ ...... '... H∗(X/A, A/A....)...H...... ∗(X/A, V/A

In fact, we can excise A and get isomorphisms to H∗(X \ A, V \ A) and to H∗(X/A \ A/A, V/A \ A/A). i : A → V is a homotopy equivalence, so Hn(A) ' Hn(V ) → Hn(V,A) = 0 → Hn−1(A) ' Hn−1(V ) Excision holds for CW-pairs. X = A ∪ B, X a CW complex and A, B CW- subcomplexes implies that H∗(B,A ∩ B) ' H∗(X,A). induces isomorphisms ....? W If X = α Xα where xα ∈ Xα for all α,(X, {xα}) is a good pair for all α, then ια : Xα → X the inclusions induce ⊕αiα∗ : ⊕H˜∗(Xα) → H˜∗(X).

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

m m Proof. Let x ∈ U. Excision implies that Hm(U, U \{x}) → Hm(R , R \{x}), ˜ m−1 m m−1 is an isomorphism, as are Hm(S ) to Hm(R ,S ) to it, and so we get m−1 δ`mZ = Hm(S ). induce isomorphisms on homology, so m = n, as δ`mZ = δ`nZ. Theorem 3.15. If X is a ∆-complex, then there exists a natural isomorphism ∆ H∗ ' H∗(X). In fact, if A is a subcomplex of a ∆-complex X, then get isomorphism ∆ H∗ (X,A) = H∗(X,A). Proof. Take dim X < ∞, A = ∅. Let Xk be the union of all simplexes in X of dim less than or equal to k, call this the k-skeleton. Induction on k, ∅ ⊂ X0 ⊂ ... ⊂ Xn−1 ( Xn = X, dim X = n. See diagram on bottom of 128 for (Xk,Xk−1). By the 5-lemma, if the outside maps are isomorphisms, then the inside one is.

Degree of a map f : Sn → Sn

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

22 Here are some basic properties:

1. deg(id) = 1

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

4. deg fg = deg f deg g. 5. R : Sn → Sn a reflection implies deg R = −1. (Why?) 6. deg A = (−1)n+1, where Ax = −x for all x.

7. If f : Sn → Sn has no fixed points, then deg f = (−1)n−1 8. If A ∈ O(n + 1), then deg A = det A.

n 1 9. fn : z 7→ z on S has deg fn = n. 10. f, g : Sn → Sn and deg f = deg g imply that f ' g.

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

n n n −1 12. f : S → S , y ∈ S , f (y) = {x1, . . . , xk}, degx f = ±1, then deg f = P i degxi f. Cellular Homology 0 n S i X a CW-complex, ∅ ⊆ X ⊆ ... ⊆ X ⊆ ..., X = i X . CW Hk(X) ' Hk (X), the cellular homology given by CW d CW d CW 2 → Cn+1 (X) → Cn (X) → Cn−1(X) →, d = 0. CW n n−1 Put Cn (X) = Hn(X ,X ) is free abelian, a generator for each n cell of X. n n−1 n n−1 n n Why? (X ,X ) is a good pair, X \ X = ∪αeα where eα is homeo- morphic to the open ball. n Let xα correspond to the origin in the homeomorphism. Put V = X \{xα} n−1 n n−1 open and there exists a retraction of V onto X . Hence, Hn(X ,X ) ' ˜ n n−1 ˜ n n−1 n Hn(X /X ) = Hn(∨αBα/sα ) ' ⊕Z with generators eα. n n−1 n−1 n−2 We need d : Hn(X ,X ) → Hn−1(X ,X ), we get it by j∗ ◦ ∆ n−1 n−1 n−1 n−2 through Hn−1(X ) where j :(X , ∅) → (X ,X ) is the inclusion but as a map of pairs. 2 n n−1 n−1 n−2 We must check that d = 0, d = ∆◦j∗ : Hn(X ,X ) → Hn−1(X ,X ), CW CW which is true as j∗ ◦ ∆ = 0. Thus H∗(C∗ (X)) = H∗ (X). n n−1 Lemma 3.16. 1. Hk(X ,X ) = 0 if k 6= n

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

n n 3. i : X → X induces isomorphism i∗ : HkX → HkX if k < n. Proof. Proof of b: Use induction on n. For n = 0, clear. Suppose n > 0, n−1 n result is known for smaller skeleton. Then Hk(X ) = 0 → Hk(X ) → n n−1 n Hk(X ,X ) = 0, so Hk(X ) = 0. Proof of c: First, suppose dim X < ∞, X = XN , N = dim X. Xn ⊆ Xn+1 ⊂ ... ⊂ XN = X. If n = N, good and simple. Suppose that n < n+1 n n n+1 N for k < n, we get 0 = Hk+1(X ,X ) → Hk(X ) → Hk(X ) → n+1 n Hk(X ,X ) = 0 so we get isomorphism for all k. Need the diagram on page 139. Proof by diagram chasing. We will need n n+1 N the next result to do this. If k < n, Hk(X ) → Hk(X ) → ... → Hk(X ) isomorphisms, but does not necessarily reach Hk(X). Lemma 3.17. If K is a compact subset of a CW-complex X, then K ⊂ Xn for some n.

n Proof. Suppose not, then for each n, choose xn ∈ K with xn ∈/ X . Then p K0 = {xn} ⊂ K is an infinite set. Note: K0 ∩ X is finite for all p, so K0 ∩ X is closed. Thus, K0 is compact. Note that each subset of K0 is closed, and so K0 has the discrete topology, contradiction. Easiest Applications 1) Suppose there are no n-cells at all for some n. Then Hn(X) = 0. CW k CW ` 2) Suppose there are k n-cells. Then Cn (X) ' Z , so Zn (X) ' Z , ` ≤ k, so Hn(X) has ≤ k generators. In particular, rank Hn(X) ≤ k. 3) Suppose there are n-cells but no n + 1 or n − 1-cells. Then d = 0 in CW CW C∗ (X), so C∗ (X) = Hn(X) ' free abelian group on the n-cells. n 4) H2k(CP ) ' Z for 0 ≤ k ≤ n, others are 0. Application of Degree

2n Theorem 3.18. Z2 is the only nontrivial group that can map freely on S . Proof. Any homeomorphism h : S2n → S2n has degree ±1, so we get a ho- momorphism deg : G → {±1}. If g ∈ G, g 6= 1, then gx 6= x for all x. So we x 7→ gx is homotopic to A : x 7→ −x, so deg g = (−1)2n+1 = −1. Hence, ker deg = {1}, so deg G → {±1} is injective. Thus |G| ≤ 2.

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

n+1 n ∆...... n Hn+1(D ,S )... H˜n(S ) ...... Φ . f . ∗ . ∗ ......

...... n ...... ∆...... n ...... H˜n+1(X) ... Hn+1(X,S ) ... H˜n(S ) ... H˜n(X)

24 With the first map on the bottom injective and the last surjective. So we conclude that Hn(X) ' Z/dZ, Hn+1(X) = 0 and Hi(X) = 0 if i 6= n, n + 1. 1 2 n If f = f2 : S → S is a map of degree two, we can see that RP has ˜ homology Hn = Z/2 iff n = 1 and 0 otherwise. IF X,Y are CW-Complexes and f : X → Y call f cellular if f(Xn) ⊂ Y n for all n ≥ 0. FACT: Cellular Approximation Theorem: Any continuous f : X → Y be- tween CW-complexes is homotopic to a cellular map. Ex: Any f : Sm → Sn with m < n is homotopic to a constant map. n n−1 n n−1 n n−1 Then f :(X ,X ) → (Y ,Y ) for all n, so get a chain map Hn(X ,X ) → n n−1 CW CW Hn(Y ,Y ), so we get Hn (X) → Hn (Y ), thus we get a commutative square ... Hn(Y ) ...... Hn(X) ... f∗ ...... ' . ' ...... CW ... CW H (Y ) ...... H (X) n CW n f∗ Cellular Boundary Formula n n−1 Prelude: Choose gen of H˜n(S ) compatibly under isomorphisms H˜n+1(S ) → ˜ n 0 n n−1 ˜ n−1 n Hn(S ) starting with S . Then also Hn(D ,S ) → Hn−1(S ). Use {eα} CW n P n−1 as gens for Cn (X). Then dn(eα) = β dαβeβ , dαβ ∈ Z the degree of n−1 n−1 Xn−1 n−1 Sα → X → n−1 n−1 ' Sβ . X \eβ Homology with Coefficients If R is a commuative ring with identity (say a field) then we take Cn(X; R) to be the free R-module generated by the singular simplexes, which is, in fact, ˜ R ⊗Z Cn(X), and every works out with Hn(X; R). So for example, Hn(X; R) ' R for X = Sn. Let X be a finite complex, and αi the number of cells in each dimension. Pdim X i Then χ(X) = i=0 (−1) αi. βi Hi(X) = Z + Fi where Fi is finite. βi is called the ith Betti number. [Hi(X) ⊗Z Q has dim βi]. Fact: For finitely generated abelian groups, if 0 → A → B → C → 0 is exact, then rank A − rank B + rank C = 0.

Pdim X i Theorem 3.19. χ(X) = i=0 (−1) βi. CW CW CW CW Proof. 0 → Bi (X) → Zi (X) → Hi (X) → 0 is s.e.s. Hi (X) ' CW CW CW Hi(X), and 0 → Zi (X) → Ci (X) → Bi−1 (X) → 0 s.e.s. P u P i CW P i CW So (−1) rank Hi(X) = (−1) rank Hi (X) = (−1) [rank Zi (X)− CW P i CW CW CW rank Bi (X)] = (−1) (rank Ci (X) − rank Bi−1 (X) − rank Bi (X)) = P i P i CW P i CW P i (−1) αi − (−1) Bi (X) − (−1) rank Bi−1 (X) = (−1) αi = χ(X).

Mayer-Vietoris Sequence

25 (Vietoris) Let A, B ⊂ X, int(A) ∪ int(B) = X. Our aim is a l.e.s. ... → Φ Ψ ∆ Hn(A ∩ B) → Hn(A) ⊕ Hn(B) → Hn(X) → Hn−1(A ∩ B) → ... Recall that Cn(A)+Cn(B) → Cn(X) for all n gives C∗(A)+C∗(B) → C∗(X) which induces an isomorphism on homology groups. Next: 0 → C∗(A ∩ B) → C∗(A) ⊕ C∗(B) → C∗(A) + C∗(B) ⊂ C∗(X) where + is the join.  x  x x 7→ , 7→ x + y. This is a short exact sequence. This gives: −x y ∆ → Hn(A ∩ B) → Hn(A) ⊕ Hn(B) → Hn(C∗(A) + C∗(B)) ' Hn(X) → ... Snake Lemma and some applications

4

Why bother with Cohomology? Because it has a ring structure, given by a cup product Hp(X) ⊗ Hq(X) → Hp+q(X). In fact, it is a graded ring. This is like differential forms (they are a primary example). A cellular chain, if X is a CW-complex, looks like 0 → Cn → Cn−1 → ... → C1 → C1 → 0. f Cochains: If A is an abelian group, so is hom(A, Z). A → B, then we get f ∗ hom(A, Z) ← hom(B, Z), so this is a contravariant functor. Similarly, if we have f g A → B → C, we get a similar sequence of homs. If A = Zk, B = Z`, then f : A → B a function. What is f ∗ : hom(B, Z) → hom(A, Z)? Well hom(Zk, Z) ' Zk, but this is not a natural isomorphism. So a cochain is 0 ← hom(Cn, Z) ← hom(Cn−1, Z) ← ... ← hom(C1, Z) ← ∗ hom(C0, Z) ← 0. Traditionally, we write the map as δ = ∂ and call it the coboundary. We have δδ = 0. k n And so, we define H (X) = ker δ/ Im δ. We often denote hom(Cn, Z) as C . Why did we do this? This comes from differential forms on . A p-form is useful, because it can be integrated over a p-simplex to obtain a real R p number ∆p α. That is, α ∈ C (M, R). For differential forms, there is a map d from p-forms to (p + 1)-forms called the exterior differential. The big property R R p is that ∆ dα = ∂∆ α, called Stokes’ Theorem. If α ∈ C is not necessarily a ∂ α p-form, then what is δα? We have Cp+1 → Cp → R. So δα = α ◦ ∂. That is, P i δα(∆) = (−1) α(∆i). k We hope that H (X) = hom(Hk(X), Z). x2 0 Example: X = RP2. We have 0 → Z → Z → Z → 0, the chain complex C∗(X). x2 0 C∗(X) is 0 ← Z ← Z ← Z ← 0, so H0(X) = Z. H1(X) = 0 and H2(X) = Z2. So our hopes fail...but we did get the same groups, just in the wrong places. 2 ∗ For S , we have C∗ is 0 → Z → 0 → Z → 0, so C is 0 ← Z ← 0 ← Z ← 0, so the cohomology groups are the same as the homology groups. In fact, this is a general phenomenon, for finitely generated abelian groups, the free part stays when switching to cohomology, but the part switches

26 dimensions. Let f∗ : A∗ → B∗ be a chain map between chain complexes. We want to define the algebraic mapping cylinder. Definition 4.1 (Mapping Cylinder). f : X → Y is a continuous map of topo- logical spaces. Take (X × I ∪ Y )/(x, 1) ∼ f(x). We call this M(f). The mapping cylinder has nice properties, like M(f) is homotopy equivalent to Y an that X,→ M(f), and the diagram M(f) ...... ' ...... X ...... Y It gives us a pair (M(f),X) for the map between X and Y , which we can use to obtain long exact sequences on homology.

Definition 4.2 (Algebraic Mapping Cylinder). We define M(f)k = Ak ⊕ Bk ⊕ Ak−1. Then the map M(f)k → M(f)k−1 is given by ∂ on Ak,Bk and to ∂ on Ak−1 to Ak−2, but also f∗ from Ak−1 to Bk−1 and the identity from Ak−1 to Ak−1. That is, with rows summed across: Ak Ak−1 Bk ...... ∂ . -∂ . ∂ ...... a ...... f ...... Ak−1 Ak−2 Bk−1 with the sign change made to make sure that this is a complex. We call this Error:complex Incorrect the algebraic label specification mapping cylinder, Error: Incorrect so ∂(a, a label0, b) = specification (∂a+a0, −∂a0,Error: ∂b+fa Incorrect0). label specification Error: Incorrect label specification

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

Proposition 4.1. Suppose that C∗ and D∗ are chain complexes of free abelian groups and f∗ : C∗ → D∗ is a chain map such that f∗ induces an isomorphism on homology. Then f∗ is a chain homotopy equivalence. That is, there exist g∗ : D∗ → C∗, S : C∗ → C∗+1 and T : D∗ → D∗+1 so that ∂S + S∂ = 1 − g∗f∗ and ∂T + T ∂ = 1 − f∗g∗. This implies that f ∗ : H∗(D∗) → H∗(D∗) is an isomorphism. Proof. S∗δ + δS∗ = 1 − f ∗g∗ and T ∗δ + δT ∗ = 1 − g∗f ∗. Check this!

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

27 0 So we were at f∗ : A∗ → B∗, M(f)k = Ak ⊕ Ak−1 ⊕ Bk with ∂(a, a , b) = (∂a + a0, −∂a0, ∂b − fa0). 0 s : M(f)∗ → M(f)∗+1, s(a, a , b) = (0, a, 0). ∂S + S∂(a, a0, b) = ∂(0, a, 0) + s(∂a + a0, −∂a0, ∂b − fa0) = (a, −∂a, −fa) + (−, ∂a, a0, 0) = (a, a0, −fa) = 1 − r. r(a, a0, b) = (0, 0, b + fa). And so we get B∗ ...... r∗ ...... f∗ ...... 0 ...... A∗ ...... M(f)∗ ...... C∗(f) ...... 0

Get that f∗ induces isos on homology iff H∗(C(f)) = 0. C(f)∗ acyclic implies that f∗ is a chain equiv (later) Definition 4.3 (Free Resolution). Let H be an abelian group. A resolution is a long exact sequence ... → F2 → F1 → F0 → H → 0. Proof. Resolutions are unique up to chain homotopy...... F2 ...... F1 ...... F0 ...... H ...... 0 ...... φ . φ . 0 ...... G2 ...... G1 ...... G0 ...... K ...... 0 Extends to a chain map which is unique up to chain homotopy, as we can suppose that there are φ¯’s such that the diagram also commutes with them. ¯ We want S : F∗ → G∗+1 with ∂S + S∂ = φ − φ. We want s : F0 → G1 with ∂s = φ − φ¯. Want ∂sz = (φ − φ¯)z, this gets hit by ∂ iff ∂(φ − φ¯)z = 0, and that is (φ∂ − φ∂)z = 0. ¯ Want s : F1 → G2 such that ∂s + s∂ = φ − φ, z a generator in F1. Want ∂sz = (φ−φ¯−s∂)z. There is an element sz with the right ∂ iff ∂(φ−φ¯−s∂)z = 0, ¯ ¯ ¯ and this is (∂φ − ∂φ − ∂s∂)z = (φ0∂ − φ0∂ − ∂s∂)z = (φ0 − φ0 − ∂s)∂z.

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

28 may develop homology, and we thus obtain derived . MISSED A LECTURE f∗ Suppose that A∗ → B∗ is such that f∗ induces an isomorphism on Homology. Then H∗(C(f)) = 0 because 0 → A∗ → M(f)∗ → C(f)∗ → 0 is exact. H∗(C(f)) = 0 ⇒ C(f)∗ is chain contractible, as C(f)∗ is a resolution of 0, so chain equivalent to the zero complex by uniqueness of resolution. C(f)∗ is chain contractible implies that f has a chain homotopy inverse, as S : C(f)∗ → C(F )∗+1 is a chain contraction, ∂ : C(f)∗ → C(f)∗01, then  −∂ 0   k `  : A ⊕ B → A ⊕ B . S = : A ⊕ B → −f ∂ ∗−1 ∗ ∗−2 ∗−1 m n ∗−1 ∗ A∗ ⊕ B∗+1. ∂S + S∂ = I, and can be worked out via multiplication. n There is a splitting H (X,G) ' hom(Hn(X),G) ⊕ Ext(Hn(X),G), but it is not natural, so it must be used carefully, if at all. Cup Product (Cohomology is into a commutative ring Λ with identity) Let ϕ ∈ Ck(X, Λ) and ψ ∈ C`(X, Λ), then ϕ∪ψ ∈ Ck+`(X, Λ) by (ϕ∪ψ)σ = ϕ(σ[v0, . . . , vk])ψ(σ[vk, . . . , vk+`]) δ(ϕ ∪ ψ) = δϕ ∪ ψ + (−1)kϕ ∪ δψ. Why is this good? If ϕ, ψ are cocycles then ϕ ∪ ψ is a cocycle, ie, if δϕ = δψ = 0 then δ(ϕ ∪ ψ) = 0. Also (ϕ + δα) ∪ (ψ + δβ) = ϕ ∪ ψ + ϕ ∪ δβ + δα ∪ ψ + δα ∪ δβ = ϕ ∪ ψ + δ(α ∪ ψ) + δ(α ∪ δβ) + (−1)k(ϕ ∪ β), so it is ϕ ∪ ψ + δγ for some γ. Thus, this defines Hk(X) × H`(X) → Hk+`(X). We should check the formula for δ(ϕ ∪ ψ). We take σ = hv0, . . . , vk+`+1i. P i δ(ϕ ∪ ψ)(σ) = (ϕ ∪ ψ)δσ = (ϕ ∪ ψ) (−1) hv0,..., vˆi, . . . , vk+`+1i. P i δϕ∪ψ(σ) = δϕ[v0, . . . , vk+1]ψ[vk+1, . . . , vk+`+1] = (−1) ϕ[v0,..., vˆi, . . . , vk+1]ψ[vk+1, . . . , vk+`+1] P i ϕ ∪ δψ(σ) = ϕ[v0, . . . , vk] (−1) ψ[vk,..., vˆi, . . . , vk+`+1] Sign counting to check that it’s right, see various sources for details. Poincare : For a closed orientable , there is an isomrophism n−k n H (M) → Hk(M ) by a ”cap product”

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