Math 757 Homology theory March 9, 2011
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 )
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
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. Exercise 186 (HW9 - Problem 2) Hatcher pg. 389-392: 10, 11, 19, 20 You may use any results from Section 4.2 for these problems including the Hurewicz Theorem and Whitehead s Theorem
Definition 187 Let C n (X, A) k x 0 C n (X, A) Be linear combinations of maps σ : n X Where σ takes vertices of n to x 0 and the k-skeleton of n to A. Goal: Prove that if X is path-connected and (X, A) is k-connected for some k 0 then the inclusion is a chain homotopy equivalence. C(X, A) k x 0 C(X, A)
Lemma 188 Suppose X is a space and D C(X ) is a sub-chain complex satisfying 1 D is generated by singular simplices 2 For each singular simplex σ : n X there is P σ : n I X 3 P σ (z, 0) = σ(z) 4 If σ : n I X is given by σ(x) = P σ (z, 1) then σ D 5 P i σ = P σ i σ I Then the inclusion D C(X ) is a chain homotopy equivalence.
Proof of Lemma 188. Let be the inclusion. Let be given by p(σ) = σ p is a chain map since i : D C(X ) p : C(X ) D p( σ) = j ( 1) j P σ j σ {1} = σ = p(σ) p i = 1 D Claim: i p 1 C(X )
Proof of Lemma 188 (continued). Claim: i p 1 C(X ) Let be the maps h 0, h 1 : n n I h 0(z) = (z, 0) h 1(z) = (z, 1) We showed there is a chain homotopy T : C( n ) C( n I ) between h 0 and h 1. Let T X : C(X ) C(X ) be Then T X (σ) = (P σ T )(1 n) ( T X + T X )(σ) = (P σ T )(1 n) + (P σ T )( 1 n) = P σ T (1 n) + P σ T (1 n) = P σ ( T + T )(1 n) = P σ (h 1 h 0 )(1 n) = σ σ = (i p 1 C(X ) )(σ)
Proposition 189 1 X is path connected 2 For some k 0 we have (X, A) is k-connected (π n (X, A, x 0 ) = 0 for n k) Then the inclusion C(X, A) k x 0 C(X ) is a chain homotopy equivalence.
Proof of Proposition 189. Given Lemma 188 above want P σ : n I X for each σ S n (X ) Will define P inducting on n Suppose n = 0 then 0 = {1} Each σ : 0 X is given by a point σ(1) X For each x X choose path γ x : I X with γ x(0) = x and γ x(1) = x 0 For x 0 X choose γ x0 = c x0 Let P σ : 0 I X be P σ(1, t) = γ σ(1) (t) We have P σ defined for all σ S 0(X ) Suppose 0 < n k and P σ defined for all σ S m (X ) with m < n. Given σ S n(x ) If σ C(X, A) k x 0 let P σ : n I X be P σ(z, t) = σ(z)
Proof of Proposition 189 (continued). If σ / C(X, A) k x 0 Define P σ on n {0} P σ(z, 0) = σ(z) On n I inductive assumption gives def of P σ π n(x, A, x 0) = 0 so there is a homotopy which we use to fill in P σ such that Pσ( n 1) A Suppose n > k and P σ defined for all σ S m (X ) with m < n. Given σ S n(x ) k-skeleton of σ contained in union of faces of σ so σ is in C(X, A) k x 0 iff every face of σ is in C(X, A) k x 0 By inductive assumption P σ is defined on n I n {0} Fill in P σ for n {i} with map that follows n [i, 0] n {0} Faces of σ are in C(X, A) k x 0 by inductive assumption so σ is.
Definition 190 ( ) C(X, A) k x n (X, A) = H 0 n C(X, A) k x 0 C(A) H (k,x0) Inclusion of chain complexes induce homomophisms H (2,x0) n (X, A) H n (1,x0) (X, A) H n (0,x0) (X, A) H n (X, A) Corollary 191 1 A is path connected 2 (X, A) k-connected Then for all n above homomorphisms induce isomorphisms H (k,x0) n (X, A) = H n (X, A)
Homology theory Proof of Corollary 191. Claim 1: C(X, A) k x 0 C(A) C(A) is a chain homotopy equivalence A is path-connected so (A, {x 0}) is 0-connected By Proposition 189 we get above chain homotopy equivalence. Claim 2: C(X, A) k x 0 C(X ) is a chain homotopy equivalence (X, A) is k-connected and X path-connected By Proposition 189 we get above chain homotopy equivalence. Have morphism of short exact sequences of chain complexes 0 C(X, A) k x 0 C(A) C(X, A) k x 0 C(X,A) k x 0 C(X,A) k x 0 C(A) 0 C(A) C(X ) C(X, A) 0 induces morphism of long exact sequence with 2/3 maps isomorphisms 5-Lemma gives desired isomorphism 0
X and A path-connected G = π 1 (A, x 0 ) (X, A) n-connected For now let π n (X, A, x 0 ) be homotopy classes of maps α : ( n, n, 0 n ) (X, A, x 0 ) Any such map α is a singular simplex in C(X, A) (n 1) x 0 Hurewicz homomorphism factors as h : π n (X, A, x 0 ) H n (X, A) so the π n (X, A, x 0 ) q π n (X, A, x 0 ) G h H (n 1,x0) n (X, A) H n (X, A) Let h : π n (X, A, x 0 ) G H n (X, A) be the last composition.
Theorem 192 (RHI(n) Relative Hurewicz Isomorphism) (n 2). Let A be path connected, (X, A) be (n 1)-connected and let G = π 1 (A, x 0 ). Then h : π n (X, A, x 0 ) G H n (X, A) is an isomorphism. Theorem 193 (AHI(n) Hurewicz Isomorphism) (n 1). Let X be (n 1)-connected. Then h : π n (X, x 0 ) H n (X, x 0 ) is an isomorphism.
In the (n + 1)-simplex n+1 with (n 1)-skeleton ( n+1 ) n 1 Let λ i n+1 = [e 0,, ê i,, e n ] λ 0 n+1 : ( n, n, e 0 ) ( n+1, ( n+1 ) n 1, e 1 ) λ i n+1 : ( n, n, e 0 ) ( n+1, ( n+1 ) n 1, e 0 ) Let γ : I n+1 be the linear path with γ(t) = (1 t)e 0 + te 1. Let b n π n ( n+1, ( n+1 ) n 1, e 0 ) b 2 = γ [λ 0 3][λ 2 3][λ 1 3] 1 [λ 3 3] 1 b n = γ [λ 0 n+1] + i ( 1) i [λ i n+1] We have the inclusion j : ( n+1, ( n+1 ) n 1, e 0 ) ( n+1, ( n+1 ) n 1, e 0 ) Lemma 194 (SF(n) Homotopy sum of simplicial faces) (n 1). j b n = 0
Homology theory Proof of RHI(n), AHI(n) and SF(n). The proof will follow these steps 1 SF(1) and AHI(1) were proved when we showed π 1 (X ) Ab = H1 (X ) 2 AHI(1), AHI(2),, AHI(n 1) = SF(n) for n 2 3 SF(n) = RHI(n) for n 2 4 RHI(n) = AHI(n) for n 2 Part 1 is shown so we start with part 2 2 Claim: AHI(1), AHI(2),, AHI(n 1) = SF(n) for n 2 Consider the homotopy long exact sequence of the triple ( n+1, n+1, ( n+1 ) n 1 ) and the pair ( n+1, ( n+1 ) n 1 ) π n+1( n+1, n+1, e 0) π n( n+1, ( n+1 ) n 1, e 0) j π n( n+1, ( n+1 ) n 1, e 0) i π n( n+1, e 0) π n 1(( n+1 ) n 1, e 0) Commutativity comes from naturality of homotopy LES and inclusion ( n+1, n+1, e 0) ( n+1, n+1, ( n+1 ) n 1 )
Homology theory Proof of RHI(n), AHI(n) and SF(n) (continued). 2 Claim: AHI(1), AHI(2),, AHI(n 1) = SF(n) for n 2 π n+1( n+1, n+1, e 0) π n( n+1, ( n+1 ) n 1, e 0) j = π n( n+1, ( n+1 ) n 1, e 0) i π n( n+1, e 0) π n 1(( n+1 ) n 1, e 0) n+1 is contractible so in homotopy LES of pair ( n+1, n+1 ) we see is an isomorphism. Hence ker j = Im = Im(i ) = Im i = ker Hence claim SF(n) (j b n = 0) will follow from showing (b n) = 0
Homology theory Proof of RHI(n), AHI(n) and SF(n) (continued). 2 Claim: AHI(1), AHI(2),, AHI(n 1) = SF(n) for n 2 π n+1( n+1, n+1, e 0) π n( n+1, ( n+1 ) n 1, e 0) j = π n( n+1, ( n+1 ) n 1, e 0) i π n( n+1, e 0) π n 1(( n+1 ) n 1, e 0) If n = 2 easy calculation shows b 2 = 1 π 1(( 3 ) 1, e 0) = F 3 If n 3 then ( n+1 ) n 1 contains 2-skeleton of n so ( n+1 ) n 1 has same trival π 1 as n. For k n 2 H k (( n ) n 1, v 0) = H k ( n, v 0) = 0 By AHI(1),, AHI(n 2) we get ( n+1 ) n 1 is (n 2)-connected. By AHI(n 1) we have an isomorphism h : π n 1(( n+1 ) n 1, v 0) = H n 1( n+1 ) n 1, v 0)
Homology theory Proof of RHI(n), AHI(n) and SF(n) (continued). 2 Claim: AHI(1), AHI(2),, AHI(n 1) = SF(n) for n 2 π n+1( n+1, n+1, e 0) π n( n+1, ( n+1 ) n 1, e 0) j = π n( n+1, ( n+1 ) n 1, e 0) i π n( n+1, e 0) π n 1(( n+1 ) n 1, e 0) In π n 1(( n+1 ) n 1, v 0) computation is difficult but in H n 1(( n+1 ) n 1, v 0) Part 2 is complete. h b n = h(b n) so b n = 0 = [ ( 1) i λ i n+1 = [1 n+1] ]
Proof of RHI(n), AHI(n) and SF(n) (continued). 3 Claim: SF(n) = RHI(n) for n 2 The Hurewicz homomorphism h : π n(x, A, x 0) H n(x, A) factors as π n(x, A, x 0) G h H (n 1,x 0) n (X, A) = H n(x, A) Enough to construct inverse w : H (n 1,x 0) n (X, A) π n(x, A, x 0) G of h First define w on chains C n(x, A) n 1 x 0 If σ : ( n, n, e 0) (X, A, x 0) is a sing. simpl. in C n(x, A) n 1 x 0 then [σ] π n(x, A, x 0) G Set w([σ]) = [σ]. By freeness of C n(x, A) n 1 x 0 have a homorphism w : C n(x, A) n 1 x 0 π n(x, A, x 0) G
Proof of RHI(n), AHI(n) and SF(n) (continued). 3 Claim: SF(n) = RHI(n) for n 2 If Im σ A then w[σ] = 0. So w induces a homorphism w : C n(x, A) n 1 x 0 C n(x, A) n 1 x 0 C πn(x, A, x0) G n(a) If w = 0 then w induces map on homology. Given σ : ( n+1, n+1, e 0) (X, A, x 0) Apply SF(n) to get w σ = ( 1) i [ i σ] = σ j ( 1) i [λ i n+1] = σ j b n = 0 w h = 1 so h is an isomorphism. Part 3 is complete.
Proof of RHI(n), AHI(n) and SF(n) (continued). 4 Claim: RHI(n) = AHI(n) for n 2 If X is (n 1)-connected then the pair (X, x 0) is (n 1)-connected. Apply RHI(n) to the pair (X, x 0). Get AHI(n). This completes the proof.