Math 6802 Algebraic Topology II

Transcription

1 Math 6802 Algebraic Topology II Nathan Broaddus Ohio State University February 9, 2015

2 Theorem 1 (The Künneth Formula) If X and Y are spaces then there is 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 This sequence splits (unnaturally). In particular H n (X Y ) = i H i (X ) H n i (Y ) Tor(H i (X ), H n i 1 (Y ))

3 Proof of the Künneth Formula follows from two results: Theorem 2 (The Künneth Formula for Chain Complexes) For free abelian chain complexes C and D there is a natural short exact sequence 0 i H i (C) H n i (D) H n (C D) i Tor(H i (C), H n i 1 (D)) 0 Theorem 3 (Eilenberg-Zilber Theorem) For spaces X and Y there is a chain homotopy equivalence C(X ) C(Y ) C(X Y )

4 The cross product for singular simplicial chains To prove the Eilenberg-Zilber Theorem we need a chain map from C(X ) C(Y ) to C(X Y ) which induces the desired chain homotopy. This chain map will be called the cross product : C p (X ) C q (Y ) C p+q (X Y ) Given a singular p-simplex σ : p X and q-simplex η : q Y We want (p + q)-chain in X Y. Key point is to triangulate p q.

5 Definition 4 (Poset) A poset is a pair (S, ) where S is a set and is a binary relation on S satisfying: 1. Reflexivity (a a) 2. Antisymmetry (a b and b a implies a = b) 3. Transitivity (a b and b c implies a c) Two elements of a poset s, t S are comparable if s t or t s. A subset L of a poset S is linearly ordered if the elements of L are pairwise comparable. Definition 5 (Geometric realization of a poset) If S is a poset then the geometric realization of S is the simplicial complex S with 1. 0-skeleton S 0 = S 2. And an n-simplex joining the vertices s 0, s 1,, s n S for each finite, linearly ordered subset {s 0, s 1,, s n } S.

6 Note that the n-simplex n is the geometric realization of the linearly ordered poset {x 0 < x 1 < < x n }. We turn to the definition of the cross product : C p (X ) C q (Y ) C p+q (X Y ) Recall our triangulation of n I = n 1 from the proof of Homotopy Axiom of homology: Let S = {s0 < s 1 < < s n} and T = {t 0 < t 1}. Then n = S and 1 = T Vertex set of n 1 is S T We triangulated n 1 with maximal simplices of form {(s 0, t 0), (s 1, t 0),, (s i, t 0), (s i, t 1), (s n, t 1)} Notice that this is exactly the geometric realization of the poset S T with the product order (s, t) (u, v) if s u and t v.

7 Let X and Y spaces with singular simplices σ : p X and η : q Y Let S = {s 0 < s 1 < < s p } so that p = S. Let T = {t 0 < t 1 < < t q } so that q = T. We may identify each maximal linearly ordered subset L = {l 0 = (s 0, t 0 ) < < l p+q = (s p, t q )} S T with a path in the p q grid moving from (0, 0) to (p, q) monotonically upwards and rightwards. Define sgn L = ( 1) b L where b L is number of boxes in p q grid below path for L. We may view L as a subset of the affine space S T = p q

8 Definition 6 (Cross product of for singular simplicial chains) The cross product is the chain map : C(X ) C(Y ) C(X Y ) where σ η = (sgn L)σ η [l0,,l p+q] L = {l 0 < < l p+q} S T maximal linearly ordered Note that we are defining the symbol on the LHS and writing σ η on the RHS to denote the product of the continuous maps σ and η. Lemma 7 The cross product is a chain map.

9 Proof of Lemma 7. ( )(σ η) = (σ η) = M = {m 0 < < m p+q} maximal linearly ordered subset of S T (sgn M) ( ) σ η [m0,,m p+q] For a linearly ordered set L = {l 0 < < l k } S T let θ L = σ η [l0,,l k ] be the singular simplex. After applying we get a formal linear combination of maps of the form θ L where L = {l 0 < < l p+q 1 } S T is one term short of being maximal.

10 Proof of Lemma 7(continued). There are two possibilities for such a linearly ordered L S T : 1. L is a subset of precisely two maximal linearly ordered sets M, M S T. 2. L is a subset of precisely one maximal linearly ordered set M S T. If L is of the first type then sgn M = sgn M and θ L terms of (sgn M) θ M and (sgn M ) θ M cancel. Hence θ L does not appear in ( )(σ η). If L is of second type then L = M {m i } for some unique maximal linearly ordered M = {m 0 < < m p+q }. Set k L = i. Thus we see that ( )(σ η) = L = {l 0 < < l p+q 1 }! maximal linear M with L M S T (sgn M)( 1) k L θ L

11 Proof of Lemma 7 (continued). Now apply the cross product and boundary map in the reverse order. ( ) (σ η) = ( )( σ η + ( 1) p σ η) = ( σ) η + ( 1) p σ ( η) = (sgn L)( 1) i θ L L = {l 0 < < l p+q 1 } maximal linearly ordered subset of (S {s i }) T + L = {l 0 < < l p+q 1 } maximal linearly ordered subset of S (T {t i }) ( 1) p (sgn L)( 1) i θ L

12 Proof of Lemma 7 (continued). The maximal linearly ordered sets M (S {s i }) T or M S (T {t i }) in the two sums above are exactly the linearly ordered subsets of S T which extend uniquely to a maximal linearly ordered subset of S T. Thus nonzero terms agree. It remains to show that signs agree.

13 The Alexander-Whitney map We wish to show that the cross product : C(X ) C(Y ) C(X Y ) induces a chain homotopy equivalence. It will be convenient to define its chain homotopy inverse the Alexander-Whitney map A : C(X Y ) C(X ) C(Y ) Let S = {s 0 < < s n } be the linearly ordered set and let n = S. Define the front p-face map of the simplex n to be the inclusion where p = {s 0,, s p } S. f p : p n Define the back q-face map of the simplex n to be the inclusion b q : q n where q = {s n q,, s n } S.

14 Definition 8 (The Alexander-Whitney map) Let σ : n X Y be a singular n-simplex of X Y. We define the Alexander-Whitney map on σ with the formula: A(σ) = A : C(X Y ) C(X ) C(Y ) n (π X σ f i ) (π Y σ b n i ). i=0 Lemma 9 The Alexander-Whitney map is a chain map.

15 Proof of Lemma 9. n Aσ = (σ f i ) (σ b n i ). = = = i=0 n [ (σ f i ) (σ b n i ) ] i=0 n ( (σ f i )) (σ b n i ) + ( 1) i (σ f i ) ( (σ b n i )) ] i=0 ( n i ( 1) j (σ f i [s0,,ŝ j,,s i ])) (σ b n i ) i=0 + j=0 ) n ( 1) i ( 1) j i (σ f i ) (σ b n i [si,,ŝ j,,s n]) j=i

16 Proof of Lemma 9 (continued). ( n i = ( 1) j (σ f i [s0,,ŝ j,,s i ])) (σ b n i ) i=0 + j=0 ) n ( 1) j (σ f i ) (σ b n i [si,,ŝ j,,s n]) j=i = ( 1) j (σ f i [s0,,ŝ j,,s i ])) (σ b n i ) 0 j<i n + ( 1) j (σ f i ) (σ b n i [si,,ŝ j,,s n]) 0 i<j n

17 Proof of Lemma 9 (continued). n A σ = A ( 1) j σ [s0,,ŝ j,,s n] j=0 = ( 1) j (σ f i [s0,,ŝ j,,s i ])) (σ b n i ) 0 j<i n + ( 1) j (σ f i ) (σ b n i [si,,ŝ j,,s n]) 0 i<j n