theory Math 757 theory January 27, 2011
theory Definition 90 (Direct sum in Ab) Given abelian groups {A α }, the direct sum is an abelian group A with s γ α : A α A satifying the following universal property of direct sums: Given s f α : A α K there is a unique f : A K s.t. α f α = f γ α Definition 91 (Direct sum in Chain) The direct sum for chain complexes {C α } is the direct sum as abelian groups with the added stipulation that all s must be chain maps
theory Theorem 92 (Sing. simp. homology satisfies Axiom A4) X α disjoint spaces. Then ( ) H n Xα = α Exercise 93 (HW4 - Problem 1) H n (X α ) Verify that the homology functors H n : Chain Ab commute with direct sums. In particular, show that if C α are chain complexes then H n ( α ) C α = H n (C α ) α This may be done by verifying that H n ( α C α ) satisfies the universal property for direct sums.
theory Definition 94 (Good pair) (X, A) is a good pair if there is a neighborhood V of A in X such that V deformation retracts to A. Theorem 95 (Relative homology and quotient spaces) If (X, A) is a good pair then we have an isomorphism q : H n (X, A) H n (X /A) where q is induced by the composition C(X, A) C(X /A, A/A) C(X /A)
theory Proof of Theorem 95. H n(x, A) α H n(x, V ) β H n(x A, V A) q q H n(x /A, A/A) γ H n(x /A, V /A) β and η are isomorphisms by excision. η H n(x /A A/A, V /A A/A) The long exact sequence of the triple (X, V, A) gives H n (V, A) H n (X, A) α H n (X, V ) H n 1 (V, A) Deformation retaction of V to A shows So α is an isomorphism H n (V, A) = H n (A, A) = 0 Same argument for (X /A, V /A, A/A) shows γ is isomorphism. q is an isomorphism since (X A, V A) (X /A A/A, V /A A/A) is a homeo q = γ 1 ηq β 1 α
theory Corollary 96 (Long exact sequence for a good pair) If (X, A) is a good pair then H n (A) H n (X ) H n (X /A) H n 1 (A) is exact. Proof. By Proposition 61 (Week 2) we have a long exact sequence H n (A) H n (X ) H n (X, A) H n 1 (A) By Theorem 95 above H n (X, A) = H(X /A)
theory Proposition 97 ( groups of spheres) Proof. True for S 1 =. True for S 0 = { 1, 1}. Suppose true for S n 1 H k (S n ) = { Z, k = n 0, k n Consider good pair (D n, S n 1 ) H k (D n ) H k (D n /S n 1 ) H k 1 (S n 1 ) H k 1 (D n ) D n /S n 1 = S n H k (S n ) = H k 1 (S n 1 ) = { Z, k 1 = n 1 0, k 1 n 1
theory Exercise 98 (HW4 - Problem 2) Show that for X nonempty H n (X ) = H n+1 (SX ) Exercise 99 (HW4 - Problem 3) Let X be the n-point set. Compute H k (X ) Exercise 100 (HW4 - Problem 4) Let Y = R n = n i=1 S 1 be the rose with n-petals (see Hatcher pg. 10). Compute H k (X )
theory Definition 101 (Local homology groups) x X the local homology of X at x is H n (X, X x) Let V be any neighborhood of x By excision with (X, A, U) = (X, X x, X V ) we have H n (X, X x) = H n (V, V x) Hence local homology depends only on topology near x Can use local homology show X not homeomorphic to Y
theory Example 102 R n is contractible so the long exact sequence of the pair (R n, R n 0) yeilds H k (R n ) H k (R n, R n 0) H k 1 (R n 0) H k 1 (R n ) R n 0 S n 1 so the local homology of R n at 0 is H k (R n, R n 0) = H k 1 (R n 0) = H k 1 (S n 1 ) = { Z, k = n 0, k n So R n R m if n m. Definition 103 An n-dimensional homology manifold is a space X such that for all x X { H k (X, X x) Z, k = n = 0, k n
theory Suppose we have a commutative diagram of chain complexes and chain maps with exact rows. 0 C γ D 0 C D E 0 δ E ε 0 The Snake Lemma gives H n+1 (E) H n (C) H n (D) H n (E) H n 1 (C) ε γ δ ε γ H n+1 (E ) H n (C ) H n (D ) H n (E ) H n 1 (C ) Fact that H n : Chain Ab is a functor shows some squares commute.
theory Theorem 104 (Naturality of ) Given a commutative diagram of chain complexes and chain maps with exact rows. 0 C γ D 0 C D E 0 δ E ε 0 The following diagram commutes: H n+1 (E) H n (C) ε H n+1 (E ) γ H n (C )
theory Proof of Theorem 104. i 0 C n q n n Dn En 0 C n D n E n i n 1 0 C n 1 Dn 1 We must verify that ε = γ. q n 1 En 1 0 By definition [e n ] = [c n 1 ] where i n 1 (c n 1 ) = D n d n for some d n D n with q n (d n ) = e n Note that εe n = εq n (d n ) = q nδ(d n ) And D n δ(d n ) = δ D n (d n ) = δi n 1 (c n 1 ) = i n 1 γ(c n 1) so ε [e n ] = [εe n ] = [γ(c n 1 )] = γ [c n 1 ] = γ [e n ]
theory Note on category theory: Definition 105 (Natural transformation) Given two functors F, G : C D A natural transformation η from F to G assigns a morphism η C : F (C) G(C) for each C Ob(C) such that the following diagram commute for all morphisms f Mor C (C 1, C 2 ) F (C 1 ) F (f ) F (C 2 ) η C1 η C2 G(C 1 ) G(f ) G(C 2 )
theory How does our naturality fit into this picture? Let C = SESChain be the category of short exact sequences of chain complexes C D E with morphisms given by commuting diagrams of chain maps from one short exact sequence to another. Let D = Ab Let F (C D E) = H n (E) Let G(C D E) = H n 1 (C) Let η (C D E) = : H n (E) H n 1 (C) Or Let C = TopPair Let D = Ab Let F (X, A) = H n (X, A) Let G(X, A) = H n 1 (A) Let η (X,A) = : H n (X, A) H n 1 (A)
theory Exercise 106 (HW4 - Problem 5) Show that if P is the set of path components of X then H 0 (X ) is isomorphic to the free abelian group on P and that if X is nonempty then H 0 (X ) is the free abelian group on P {c 0 } where c 0 is a path component of X.
theory Definition 107 (Chain complex for cellular homology) Let X be a CW complex (see Hatcher pg. 5). Let X 1 = and { Hn (X Cn CW (X ) = n, X n 1 ), n 0 0, n < 0 Let be the CW : Cn CW (X ) Cn 1(X CW ) H n (X n, X n 1 ) H n 1 (X n 1, X n 2 ) from the long exact sequence of the triple (X n, X n 1, X n 2 ) Proposition 108 C CW (X ) is a chain complex
theory Proof of Proposition 108. We must verify that ( CW ) 2 = 0. We have the following commutative diagram of chain maps 0 C(X n ) C(X n+1 ) C(X n+1, X n ) 0 j 0 C(X n, X n 1 ) C(X n+1, X n 1 ) C(X n+1, X n ) 0 1 says the following square from the long exact sequence of homology commutes: H n+1 (X n+1, X n ) H n (X n ) 1 H n+1 (X n+1, X n ) CW H n (X n, X n 1 ) j So CW = j
theory Proof of Proposition 108 (continued). We have commutative H n (X n ) j H n+1 (X n+1, X n ) CW H n (X n, X n 1 ) CW H n 1 (X n 1, X n 2 ) j H n 1 (X n 1 ) Vertical composition H n (X n ) j H n (X n, X n 1 ) H n 1 (X n 1 ) is part of long exact sequence of the pair (X n, X n 1 ) So ( CW ) 2 = (j ) (j ) = j ( j ) = j 0 = 0
theory Definition 109 ( ) If X is a CW complex then the nth cellular homology group of X is H CW n (X ) = H n (C CW (X )) For a CW complex X how do Hn CW (X ) and H n (X ) relate? Theorem 110 ( and sing. simp. homology agree) If X is a CW complex then H CW n (X ) = H n (X ) Advantages of H CW 1 If X is finite then C CW (X ) is finitely generated 2 boundary maps in C CW (X ) can be easily understood.
theory The proof of Theorem 110 will use the following properties of homology for CW complexes. Lemma 111 ( of CW complexes) Let X is a CW complex. Let Γ n be the set of n-cells of X 1 H k (X n, X n 1 ) = { Z[Γ n ] k = n 0, k n 2 If k > n H k (X n ) = 0 3 If k < n and i : X n X is the inclusion map then H k (X n ) = i H k (X )
theory Proof of Lemma 111 Claim 1. We have a commuting quotient maps of good pairs ( α Γ n Dn, α Γ S n 1) r (X n, X n 1 ) q (X n /X n 1, { }) n By Theorem 95 we get that p p : H n ( α Γ n D n, α Γ n S n 1 ) H n (X n /X n 1 ) and q : H n (X n, X n 1 ) H n (X n /X n 1 ) are isomorphisms. So we get an isomorphism r : H n ( α Γ n D n, α Γ n S n 1 ) H n (X n, X n 1 )
theory Proof of Lemma 111 Claim 1 (continued). Applying the additivity axiom ( H n (X n, X n 1 ) = H n D n, ) S n 1 α Γ n α Γ n = H n (D n, S n 1 ) α Γ n = α Γ n Hn (S n ) = Z α Γ n = Z[Γ n ]