0, otherwise Furthermore, H i (X) is free for all i, so Ext(H i 1 (X), G) = 0. Thus we conclude. n i x i. i i

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1 Cohomology of Spaces (continued) Let X = {point}. From UCT, we have H{ i (X; G) = Hom(H i (X), G) Ext(H i 1 (X), G). { Z, i = 0 G, i = 0 And since H i (X; G) =, we have Hom(H i(x); G) = Furthermore, H i (X) is free for all i, so Ext(H i 1 (X), G) = 0. Thus we conclude H i (X; G) = { G, i = 0 { Z, i = 0 or n Let X = S n. Then we have H i (X; G) = Thus again we can compute Hom(H i (X), G) and Ext(H i 1 (X), G) = 0. Hence by universal coefficient theorem, H i (X; G) = Hom(H i (X), G) Ext(H i 1 (X), G) = Reduced Cohomology H i (X; G) We start with the augmented singular chain complex for X: { G, i = 0 or n C 1 (X) C 0 (X) n i x i i ɛ Z 0 i n i Dualizing this (i.e., apply Hom( ; G)), we get δ C 1 δ (X; G) C 0 (X; G) G 0 And since ɛδ = 0, by dualizing, we get δɛ = 0. The homology of this augmented cochain complex is the reduced cohomology of X, denoted by H i (X; G). ɛ Remark We have H i (X; G) = H i (X; G), if i > 0. Also, by UCT, we have H 0 (X; G) = Hom( H 0 (X), G). Relative Cohomology Groups of a Pair (X, A) We start with a ses. of chain complexes: 1

2 0 C (A) C (X) C (X, A) 0 (1) C (X)/C (A) Now let C n (X, A; G) = Hom(C n (X, A), G) be the cochain complex, and let the coboundary maps δ : C n C n+1 be defined by δ = ( ). We see that C n (X, A; G) = Hom(C n (X, A), G) is really the functions on n-simplices of X which vanishes on simplices in A, thus we get a natural inclusion map C n (X, A; G) C n (X; G) Also, δ : C n (X, A; G) C n+1 (X, A; G) is a restriction of δ : C n (X; G) C n+1 (X; G). By dualizing (1), we get the ses. 0 C n (A; G) C n (X; G) C n (X, A; G) 0 i Note that the exactness at C n (A; G) follows by extending a cochain in A by zero, as illustrated below: We start with a ψ C n (A; G), which is a function ψ : C n (A) G by definition. We then define ˆψ : C n (X) G by having ˆψ(ρ) = { ψ(ρ), if ρ Cn (A) 0, if Im(ρ) A =. This can be done since C n(x)has a basis made of simplices contained in A and those contained in X \ A. So now, we get a ses. at each level n. Since i and j commute with, we see that i and j also commute with δ. Thus we really have a ses. of cochain complexes: 0 C (A; G) C (X; G) C (X, A; G) 0 By taking the associated les. of the previous ses. (as homology groups), we get the les. for the cohomology of the pair (X, A): j H n (X, A; G) j H n (X; G) i H n (A; G) δ H n+1 (X, A; G) In fact, we can also start with the augmented chain complexes on X and A, and get a les. for the reduced cohomology groups, with H n (X, A; G) = H n (X, A; G): H n (X, A; G) Hn (X; G) Hn (A; G) H n+1 (X, A; G) In particular, if A = x 0 is a point in X, since a point has trivial reduced cohomology, we get H n (X; G) = H n (X, x 0 ; G). Induced Homomorphisms Suppose f : X Y is a continuous map, then we have f # : C n (X) C n (Y ) ρ f ρ 2

3 and f # = f #. Dualizing f #, we get maps f # : C n (Y ; G) C n (X; G), with f # (ψ) = ψ(f # ) and δf # = f # δ (which is the dualized version of f # = f # ). Thus, we get induced homomorphisms on the cohomology: f : H n (Y, G) H n (X, G). In fact, we can do this for maps of pairs, say f : (X, A) (Y, B). And note that UCT also works for pairs because C n (X, A) = C n (X)/C n (A) is free abelian (whose proof is also by the basis element argument). So, by naturality, we can get a commutative diagram for f : (X, A) (Y, B): 0 Ext(H n 1 (X, A), G) H n (X, A; G) (f ) f 0 Ext(H n 1 (Y, B), G) H n (Y, B; G) h Hom(H n (X, A), G) 0 h (f ) Hom(H n (Y, B), G) 0 Homotopy Invariance Theorem If f g : (X, A) (Y, B), i.e., f is homotopic to g, then f = g : H n (Y, B; G) H n (X, A; G). Corollary If f : X Y is a homotopy equivalence, then f : H n (Y ; G) H n (X; G) is an isomorphism. Proof (of the theorem) Recall from homology that we had the prism operator with P : C n (X, A) C n+1 (Y, B) f # g # = P + P (2) i.e., we have the difference of the middle maps equals to the sum of the two side paths in the following diagram: C n+1 (Y, B) P C n (X, A) f # g # C n (Y, B) P C n (X, A) 3

4 For instance, if we have f, g : X Y, since they re homotopic, we have a homotopy F : X I Y, where F (x, 0) = f(x) and F (x, 1) = g(x). Then the prism operator is defined, cubically, to be P : C n (X) C n+1 (Y ) (ρ : I n X) (F (ρ id) : I n I X I Y ) Or, the prism operator can also be defined, simplicially, by considering F (ρ id) : n I Y, and then break n I into n+1 s. Either way, it follows that f = g. Now we can dualize the prism operators, and get P : C n+1 (Y, B; G) C n (X, A; G) so that after dualizing, (2) becomes f # g # = δp + P δ. Then it follows that f = g on H. Excision Theorem Suppose that Z A X, and that cl(z) int(a). Then i : (X Z, A Z) (X, A) induces isomorphisms i : H n (X, A; G) H n (X Z, A Z; G) for all n. Or, equivalently, if X = int(a) int(b), then (B, A B) (X, A) induces isomorphisms in cohomology. Proof By the naturality of UCT, we have 0 Ext(H n 1(X, A), G) H n (X, A; G) Hom(H n(x, A), G) 0 (i ) i (i ) 0 Ext(H n 1(X Z, A Z), G) H n (X Z, A Z; G) Hom(H n(x Z, A Z), G) 0 Now by excision in homology, we know that i, and hence (i ), are isomorphisms. Thus by 5-lemma, it follows that i is also an isomorphism. Mayer-Vietoris Sequence Theorem Suppose X = int(a) int(b), then there is a les.: H n (X; G) ψ H n (A; G) H n (B; G) φ H n (A B; G) H n+1 (X; G) Remark It s irrelevant which kind of cohomology we use here. Indeed, singular, simplicial, or cellular cohomologies really give the same cohomology. Furthermore, in differential manifold, even the De Rham cohomology is no different than these cohomologies. Proof There is a ses. of cochain complexes: 0 C n (A + B; G) ψ C n (A; G) C n (B; G) φ C n (A B; G) 0 (3) Hom(C n (A + B); G) 4

5 Here, C n (A + B) is the set of simplices in X made of sum of simplices in A and B, and the maps are defined by ψ(η) = (η Cn(A), η Cn(B)) and φ(α, β) = α Cn(A B β Cn(A B). Recall that since C (A + B) i C (X) is a chain homotopy, we have H (A+B) = H (X). It also follows that C (A+B; G) and C (X; G) are chain homotopic, and thus H (A+B; G) = H (X; G). Now the les. associated to (3) gives the Mayer-Vietoris sequence in the cohomology. A Few s { G, i = 0 H i (R n ; G) = This follows immediately by the homotopy invariant theorem. Consider S n. We can cover S n by two open sets A = S n \ N and B = S n \ S, where the N and S are the south pole and north pole, respectively. Then we have A B = S n 1 and A = B = R n. Thus by Mayer-Vietoris sequence (of reduced cohomology) and induction, we have: H i (A B; G) = H i 1 (A B; G) = = H i n (S 0 ; G) = { G, i = n Cellular Cohomology Definition Let X be a CW-complex. Then we define the cellular cochain complex of X to be C (X; G) = {C n = H n (X n, X n 1 ; G), d n = δ n j n }, i.e.: (with all the coefficient group G ignored here) j n 1 H n 1 (X n 1, X n 2 ) H n 1 (X n 1 ) d n 1 δ n 1 H n (X n, X n 1 ) d n H n+1 (X n+1, X n ) j n H n (X n ) δ n Note that j n δ n 1 = 0, since they re two consecutive maps in a les., and so d n d n 1 = δ n j n δ n 1 j n 1 = 0. Thus Im(d n 1 ) Ker(d n ), and this is indeed a cochain complex. Theorem H n (X; G) = H n (C (X; G)). Moreover, C (X; G) is exactly the dual (with respect to G) of the 5

6 cellular chain complex (C (X), d ). Proof Recall from the cellular chain complex, we have C n = H n (X n, X n 1 ) # of n-cells = Z and H i (X n, X n 1 ) = 0 whenever i n. So by UCT, C n = H n (X n, X n 1 ; G) = Hom(C n, G), since the Ext terms vanishes. Also, from UCT again, H i (X n, X n 1 ; G) = 0, whenever i n. So if we consider the leq. of the pair (X n, X n 1 ), we get: H k (X n, X n 1 ) H k (X n ) H k (X n 1 ) H k+1 (X n, X n 1 ) And if k n, n 1, we get isomorphisms H k (X n ) = H k (X n 1 ). Thus if k > n, we get H k (X n ) = H k (X n 1 ) = H k (X n 2 ) = = H k (X 0 ) = 0 Next we show that H n (X n+1 ; G) = H n (X; G). First note X n+1 i X induces isomorphisms on H k, for k < n + 1. So by the naturality of UCT, we get the following diagram: 0 Ext(H n 1 (X, G) H n (X; G) h Hom(H n (X), G) 0 = (i ) i 0 Ext(H n 1 (X n+1, G) H n (X n+1 ; G) = (i ) h Hom(H n (X n+1, G) 0 Then, the 5-lemma tells us that the middle map i is also an isomorphism. Now we have the diagram: (where the diagonal are les. of the pairs) H n 1 (X n 2 ) = 0 j n 1 H n 1 (X n 1, X n 2 ) H n 1 (X n 1 ) δ n 1 d n 1 H n (X n, X n 1 ) d n H n+1 (X n+1, X n ) j n H n (X n ) δ n α H n (X) = H n (X n+1 ) H n (X n 1 ) = 0 So remembering that d n = δ n j n and noting that j n 1 is onto, we calculate: H n (X; G) = H n (X n+1 ; G) = Im(α) = Ker(δ n ) = Ker(d n )/Ker(j n ) = Ker(d n )/Im(δ n 1 ) = Ker(d n )/Im(δ n 1 j n 1 ) = Ker(d n )/Im(d n 1 ) 6

7 Now all that s left is the show that d n = (d n+1 ). By definition, we have d n : H n (X n, X n 1 j ) n H n (X n δ ) n H n+1 (X n+1, X n ) H n (X n, X n 1 ) H n (X n ) H n+1 (X n+1, X n ) : d n+1 Now dualizing d n, we get the arrows reversed, and thus the diagram: d n : H n (X n, X n 1 ) j n H n (X n ) δ n H n+1 (X n+1, X n ) = h (d n+1) : Hom(H n(x n, X n 1 ), G) (jn) Hom(H n(x n ), G) ( n+1) Hom(H n+1(x n+1, X n ), G) The extreme vertical arrows labelled h are really isomorphisms by UCT, since the Ext terms vanishes. The left square is commutative by the naturality of UCT for the inclusion map (X n, ) (X n, X n 1 ), and the right square is commutative by diagram hunting. Thus, the squares commute, and we see that d n = (d n+1 ), and so the theorem follows. h = h Let X = RP 2. Note that X has 1 cell in each dimension 0, 1, and 2, and the cellular chain complex of X is: Z 2 Z 0 Z 0 Now to get H (X; Z), we apply Hom(, Z), and get: Z 2 Z 0 Z 0 (Indeed, H n (X n, X n 1 ; G) = Z # of n-cells.) Thus, we have H i (RP 2 ; Z) = Z, i = 0 Z/2, i = 2. Also, to get H (X; Z/2), we apply Hom(, Z/2) to get: And, 0 Z/2 0 Z/2 0 Z/2 0 H i (RP 2 ; Z/2) = { Z/2, i = 0, 1, or 2 7

8 Let K be the Klein bottle. And recall that the cellular chain complex is given by: Now using coefficient Z/3, we get: 0 Z (2,0) Z Z 0 Z 0 0 Z/3 (2,0) Z/3 Z/3 0 Z/3 0 Note that the map (2, 0) : Z/3 Z/3 Z/3 is really an isomorphism into the first component, so we get the homology: { Z/3, i = 0 or 1 H i (K; Z/3) = Now in cohomology, we dualize to get: And finally, we get (2,0) 0 Z/3 Z/3 Z/3 0 Z/3 0 H i (K; Z/3) = { Z/3, i = 0 or 1 8