Math 757 Homology theory

Transcription

1 Math 757 Homology theory March 3, 2011

2 (for spaces). Given spaces X and Y we wish to show that we have a natural exact sequence 0 i H i (X ) H n i (Y ) H n (X Y ) i Tor(H i (X ), H n i 1 (Y )) 0 By Eilenberg-Zilber Theorem we have a chain homotopy C(X Y ) C(X ) C(Y ) Hence H n(x Y ) = H n(c(x Y )) = H n(c(x ) C(Y ))

3 (continued). By the for chain complexes we have natural exact 0 i H i (C(X )) H n i (C(Y )) H n(c(x ) C(Y )) i Applying Eilenberg-Zilber we get Tor(H i (C(X )), H n i 1 (C(Y ))) 0 0 i H i (C(X )) H n i (C(Y )) H n(c(x Y )) i Tor(H i (C(X )), H n i 1 (C(Y ))) 0 Finally by definition of sing. simp. homology we conclude 0 i H i (X ) H n i (Y ) H n(x Y ) i Tor(H i (X ), H n i 1 (Y ) 0

4 We must prove the Eilenberg-Zilber Theorem which says that for spaces X and Y there is a chain homotopy Notation for product spaces C(X Y ) C(X ) C(Y ) For a product space X Y we have continuous projection maps p 0 : X Y X and p 1 : X Y Y and given continuous f : W X and g : W Y let f g : W X Y be the continuous map f g(w) = (f (w), g(w)) Hence for any u : W X Y we have u = (p 0 u) (p 1 u)

5 Proof of Eilenberg-Zilber. First we will apply the Acyclic Model Theorem to get a chain map ϕ : C(X Y ) C(X ) C(Y ) Let C be the category Top Top of ordered pairs of spaces with morphisms from (X, Y ) to (Z, W ) given by pairs of maps (f, g) with f : X Z and g : Y W continuous. Let T : Top Top Chain be the functor T (X, Y ) = C(X Y ) and given a morphism (f, g) : (X, Y ) (Z, W ) and a singular simplex σ : n X Y define T (f, g) : C(X Y ) C(Z W ) T (f, g)(σ) = (fp 0 σ) (gp 1 σ)

6 Proof of Eilenberg-Zilber (continued). Let S : Top Top Chain be the functor S(X, Y ) = C(X ) C(Y ) and given a morphism (f, g) : (X, Y ) (Z, W ) and a singular simplices σ : p X and η : q Y define S(f, g) : C(X ) C(Y ) C(Z) C(W ) S(f, g)(σ η) = (f σ) (gη)

7 Proof of Eilenberg-Zilber (continued). Set M n = ( n, n ) Choose models for C the singleton sets M n = {M n } Choose e n T n (M n ) = C n ( n n ) to be the n-chain 1 1 : n n n This is the diagonal map for n. Note that p 0 e n = p 1 e n = 1 n

8 Proof of Eilenberg-Zilber (continued). Claim that the functor T n is free with models M n In other words, we must show that T (X, Y ) = C(X Y ) has free basis { } B = T n(f, g)(e n) (f, g) : ( n, n ) (X, Y ) Given a singular simplex σ : n X Y T n(p 0σ, p 1σ)(e n) = (p 0σp 0e n) (p 1σp 1e n) = (p 0σ1 n) (p 1σ1 n) = (p 0σ) (p 1σ) = σ Hence B contains the singular n-simplices S n(x Y ) T n(f, g)(e n) = (fp 0e n) (gp 1e n) = f g is a map from n to X Y so S n(x Y ) contains B Finally if (f, g) (f, g ) then T n(f, g)(e n) = f g f g = T n(f, g )(e n)

9 Proof of Eilenberg-Zilber (continued). Claim that each M n is S-acyclic In other words, we must show that for k > 0 we have H k (S(M n)) = 0 H k (S(M n)) = H k (S( n, n )) = H k (C( n ) C( n )) = H i (C( n )) H k i (C( n )) i Tor(H i (C( n )), H k i 1 (C( n ))) i = H i ( n ) H k i ( n ) i i Tor(H i ( n ), H k i 1 ( n )) = { 0, k 0 Z, k = 0

10 Proof of Eilenberg-Zilber (continued). Claim there is a natural isomorphism Φ : H 0 (T ) H 0 (S) In other words for each (X, Y ) we have Φ (X,Y ) : H 0(T (X, Y )) H 0(S(X, Y )) H 0(S(X, Y )) = H 0(C(X ) C(Y )) = H i (X ) H i (Y ) i i Tor(H i (X ), H i 1 (Y )) = H 0(X ) H 0(Y ) H 0(T (X, Y )) = H 0(C(X Y )) = H 0(X Y ) Path components of X Y are product of path components of X with path components of Y

11 Proof of Eilenberg-Zilber (continued). Now apply Acyclic Models Theorem to get chain homotopy ϕ : T S inducing an isomorphism H 0 (T ) = H 0 (S). Now we need a futher chain homotopy γ : S T inducing the reverse isomorphism H 0 (S) = H 0 (T ). Here we need larger set of models { } M S n = ( p, q ) p + q = n Let ep,q S S n ( p, q ) = i C i( p ) C n i ( q ) be e S p,q = 1 p 1 q S n is free with models M S n

12 Proof of Eilenberg-Zilber (continued). Claim that each ( p, q ) M S n is T -acyclic In other words, we must show that for k > 0 we have H k (T ( p, q )) = 0 H k (T ( p, q )) = H k (C( p q )) = H k ( p q ) = H k ({ }) { 0, k 0 = Z, k = 0 This concludes the proof of the (for spaces).

13 More applications of Acyclic Model Theorem We could have used Acyclic Model Theorem to prove homotopy invariance of H as follows Let T : Top Chain be Let S : Top Chain be T (X ) = C(X ) S(X ) = C(X I ) M n = { n } with e n = 1 n C n( n ) T n is free with models M n (observe definitions) n is S-acyclic (cone each sing. simplex to fixed p n I to show H k ( n I ) = 0 for k > 0). Thus there is a unique chain homotopy type of chain maps γ : C(X ) C(X I ) inducing the natural isomorphism H 0(X ) = H 0(X I ).

14 More applications of Acyclic Model Theorem Let h 0, h 1 : X X I be the maps h 0(x) = (x, 0) h 1(x) = (x, 1) h 0 and h 1 induce h 0, h 1 : C(X ) C(X I ) which in turn induce the natural isomorphism H 0(X ) = H 1(X I ) Hence h 0 and h 1 are chain homotopic. No suppose f 0, f 1 : X Y are homotopic Have homotopy F : X I Y from f 0 to f 1 with f 0 = F h 0 and f 1 = F h 1 Hence f 0 = F h 0 F h 1 = f 1

15 Definition 172 ( for pairs) (X, A) and (Y, B) topological pairs f, g : (X, A) (Y, B) are homotopic (resp. homotopic rel X X ) if there is F : (X I, A I ) (Y, B) with F (x, 0) = f (x) and F (x, 1) = g(x). (resp. further F X = 1 X ) Denote the class of f : (X, A) (Y, B) by [f ]. (resp. [f ] X ) Let and { } [(X, A); (Y, B)] = [f ] f : (X, A) (Y, B) { } [(X, A); (Y, B)] X = [f ] X f : (X, A) (Y, B)

16 I 0 = {0} I 0 = I n = [0, 1] n R n { } I n = (x 1,, x n ) there is i s.t. x i = 0 or x i = 1 { } I n 1 = (x 1,, x n ) x n = 0 J n 1 = I n int(i n 1 ) Definition 173 ( ) (X, x 0 ) a pointed space and n 0 π n (X, x 0 ) = [(I n, I n ); (X, x 0 )] (X, A, x 0 ) pointed pair of spaces and n 1 π n (X, A, x 0 ) = [(I n, I n, J n 1 ); (X, A, x 0 )]

17 Definition 174 (Group structure of homotopy ) [f ], [g] π n (X, x 0 ) with n 1 or [f ], [g] π n (X, A, x 0 ) with n 2 { f (2x1, x 2,, x n ), x (f g)(x 1,, x n ) = g(2x 1 1, x 2,, x n ), x Setting [f ] [g] = [f g] gives π n a group structure. π 0 (X, x 0 ) and π 1 (X, A, x 0 ) are sets with distinguished identity element [c x0 ] π 1 (X, x 0 ) and π 2 (X, A, x 0 ) are nonabelian (in general) π n (X, x 0 ) and π n+1 (X, A, x 0 ) are abelian for n 2

18 } D n = {x R n x 1 } S n 1 = D n = {x R n x = 1 p = (1, 0,, 0) Definition 175 ( (alt. def.)) (X, x 0 ) a pointed space and n 0 π n (X, x 0 ) = [(S n, p); (X, x 0 )] (X, A, x 0 ) pointed pair of spaces and n 1 π n (X, A, x 0 ) = [(D n, S n 1, p); (X, A, x 0 )]

19 Theorem 176 (Long exact sequence of homotopy ) (X, A) a pair of spaces. There is a natural long exact sequence π n+1 (X, A, x 0 ) π n (A, x 0 ) i π n (X, x 0 ) j i π 1 (X, x 0 ) j π 1 (X, A, x 0 ) π 0 (A, x 0 ) i π 0 (X, x 0 ) For last three maps ker means the preimage of [c x0 ]. i and j are obvious inclusion maps Given f : (I n, I n, J n 1 ) (X, A, x 0 ) [f ] = [f I n 1]

20 Theorem 177 (Long exact homotopy sequence of triple) (X, A, B) a triple of spaces. There is a natural long exact sequence π n+1 (X, A, x 0 ) π n (A, B, x 0 ) i π n (X, B, x 0 ) j j π 2 (X, A, x 0 ) π 1 (A, B, x 0 ) i π 1 (X, B, x 0 ) j π 1 (X, A, x 0 ) Are homotopy a homology theory? Exactness Axiom holds (by above) Axiom holds Dimension Axiom fails (but not badly) Additivity Axiom fails Excision Axiom fails (big problem)

21 Definition 178 (n-connectivity) A space X is n-connected if π k (X, x 0 ) = 0 for all k n A topological pair (X, A) is n-connected if π k (X, A, x 0 ) = 0 for all k n A space is 0-connected iff it is path connected A space is 1-connected iff it is simply connected S n is (n 1)-connected but not n-connected

22 Let X be a space and G = π 1 (X, x 0 ) Then we have a G-action on π n (X, x 0 ) For n 2 this makes π n (X, x 0 ) a G-module Definition 179 (G-module) A right or left G-module M is a abelian group with a right or left Z[G]-module structure. Definition 180 (G-coinvaraiants) Given a G-module M, the G-coinvariants of M are M M G = gm m g G, m M or equivalently M G = M Z[G] Z where Z is the trivial G-module.

23 Basic idea of Hurewicz Homorphism Given [α] π n (X ) want h([α]) H n (X ) have map α : (S n, p) (X, X 0 ) Choose a generator z n H n (S n, p) define h([α]) = α z n To get naturality of h must choose z n compatibly for all n Some work to show h is a homomorphism.

24 Goal: define Hurewicz homorphism Must inductively choose generators of H n (I n, I n ) for all n. Choose z 1 H 1 (I, I ) to be unique element such that z 1 = [1] [0] H 0 ( I = {0, 1}) By Excision have an isomorphism j : H n 1 (I n 1, I n 1 ) = H n 1 ( I n, J n 1 ) Given z n 1 H n 1 (I n 1, I n 1 ) choose z n H n (I n, I n ) so that z n = jz n 1

25 Definition 181 ( ) map is the function h : π n (X, A, x 0 ) H n (X, A) where given [α] π n (X, A, x 0 ) we let h([α]) = α z n H n (X, A) Letting A = x 0 and assuming n 1 we get h : π n (X, A, x 0 ) H n (X, x 0 ) = H n (X ) = H n (X ) Claim: h is well-defined Given α, β : (I n, I n ) (X, A) if α β then by Axiom α = β So α z n = β z n

26 Homology theory Theorem 182 ( ) For (X, A) a topological pair and n 1 the Hurewicz map h : π n (X, A, x 0 ) H(X, A) defined above satisfies the following: 1 h is a homomorphism. 2 h is natural in that we have commutative i π 2 (X, x 0 ) j π 2 (X, A, x 0 ) π 1 (A, x 0 ) i π 1 (X, x 0 ) h i H 2 (X ) h j H 2 (X, A) h H 1 (A) h i H 1 (X ) 3 h is also natural in that for f : (X, A, x 0 ) (Y, B, y 0 ) we have commutative π n (X, A, x 0 ) h H n (X, A) f π n (Y, B, y 0 ) h f H n (Y, B)

27 Homology theory Proof of Theorem Claim: Given f : (X, A, x 0 ) (Y, B, y 0 ) we have commutative f π n (X, A, x 0 ) π n (Y, B, y 0 ) h H n (X, A) Given α : (I n, I n ) (X, A) h f H n (Y, B) h(f [α]) = h([f α]) = (f α) z n = f (α z n) = f (h([α])) Note : We have established commutativity of many squares of LES

28 Homology theory Proof of Theorem 182 (continued). 2 Claim: We have commutative π n (X, A, x 0 ) h H n (X, A) π n 1 (A, x 0 ) h H n 1 (A) Given α : (I n, I n ) (X, A) Let α : (I n 1, I n 1 ) (A, x 0) satisfy α = α I n 1 Then [α] = [α ] If j : (I n 1, I n 1 ) ( I n, J n 1 ) is inclusion we have α = α I n j h( [α]) = h([α ]) = α z n 1 = (α I n) j z n 1 = (α I n) z n = α z n = α z n = h([α])

29 Proof of Theorem 182 (continued). 1 Claim: h : π n (X, A, x 0 ) H n (X, A) is a homomorphism { } Let L I n be (x 1,, x n) x { } Let U I n be (x 1,, x n) x Let L = L int L and U = U int U Consider inclusions Get isomorphism i L : (L, L) (I n, L U) i U : (U, U) (I n, L U) i L i U : H n(l, L) H n(u, U) H n(i n, L U) Consider squeeze maps defined by s L : (I n, I n ) (L, L) s L (x 1,, x n) = ( x 12,, x n ) s U : (I n, I n ) (U, U) s U (x 1,, x n) = ( x ,, x n )

30 Proof of Theorem 182 (continued). 1 h : π n (X, A, x 0 ) H n (X, A) is a homomorphism (continued) We have inculsion i : (I n, I n ) (I n, L U) For z n H n(i n, I n ) i z n = i L s L z n + i U s U z n Now given α, β : (I n, I n ) (X, A) h([α] [β]) = h([α β]) = (α β) z n = ( (α β) i ) zn = (α β) i z n = (α β) (i L s L + i U s U )z n = αz n + βz n = h([α]) + h([β])

31 Theorem 183 Let G = π 1 (X, x 0 ) then for n 1 h : π n (X, x 0 ) H n (X ) factors through the quotient map q : π n (X, x 0 ) π n (X, x 0 ) G to π n (X, x 0 ) G the G-coinvariants of π n (X, x 0 ) Theorem 184 (X, A) a topological pair. Let G = π 1 (A, x 0 ) then for n 2 h : π n (X, A, x 0 ) H n (X ) factors through the quotient map q : π n (X, A, x 0 ) π n (X, A, x 0 ) G to π n (X, A, x 0 ) G the G-coinvariants of π n (X, A, x 0 )

32 Proof of Theorems 183 and 184. [f ] π 1 (X, x 0 ) [α] π n (X, x 0 ) Standard argument gives explicit homotopy between α and f α

33 Exercise 185 (HW9 - Problem 1) Decide whether or not for all spaces X and all n 1 the Hurewicz homomorphism h : π n (X, x 0 ) H n (X ) is surjective.

34 Exercise 186 (HW9 - Problem 2) Hatcher pg : 10, 11, 19, 20 You may use any results from Section 4.2 for these problems including the Hurewicz Theorem and Whitehead s Theorem