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Math 6802 Algebraic Topology II Math 6802 Algebraic Topology II Nathan Broaddus Ohio State University February 11, 2015 Homotopy inverse of the cross product I We have the cross product chain map − × − : Cp(X ) ⊗ Cq(Y ) ! Cp+q(X × Y ) I and the Alexander-Whitney chain map A : Cn(X × Y ) ! (C(X ) ⊗ C(Y ))n I Thus we have chain maps A ◦ (− × −): Cp(X ) ⊗ Cq(Y ) ! (C(X ) ⊗ C(Y ))p+q I and (− × −) ◦ A : Cn(X × Y ) ! Cn(X × Y ) The acyclic models theorem Definition 1 (Category with models) A category with models is a category C with specified set of objects M ⊂ C called the models of C . Definition 2 (Free functor with respect to the models) Given a category C with models M a functor T : C ! Ab is free with respect to M if for each M 2 M there is eM 2 T (M) such that for all X 2 Ob(C ) T (f )(eM ) M 2 M ; f 2 Mor(M; X ) is a free basis for T (X ). Definition 3 (Free functor with respect to the models) Given a category C with models M = [nMn a functor T : C ! Chain is free with respect to M = [nMn if each Tn is free with respect to Mn Theorem 4 (Acyclic models theorem) I C a category with models M ⊂ C I S; T : C ! Chain functors with Sn = Tn = 0 for n < 0 I T is free with respect to the models M = [nMn I Hi (S(M)) = 0 for i > 0 for all models M 2 M (M is S-acyclic) I There is a natural transformation Φ: H0(T ) ! H0(S) Then there is a natural transformation ' : T ! S which induces Φ unique up to chain homotopy. Notation for product spaces For a product space X × Y we have continuous projection maps π0 = πX : X × Y ! X and π1 = πY : 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)) Observe that for any u : W ! X × Y we have u = (πX u) × (πY u) Category of ordered pairs of spaces Let Top × Top be the category 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. Proof of Eilenberg-Zilber. I For now assume Acyclic Models Theorem. I We have two functors S; T : Top × Top ! Chain where S(X ; Y ) = C(X ) ⊗ C(Y ) T (X ; Y ) = C(X × Y ) I 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) = (f πX σ) × (gπY σ) I Can you show T (f ; g) is a chain map? Proof of Eilenberg-Zilber (continued). I 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) I Can you show S(f ; g) is a chain map? Proof of Eilenberg-Zilber (continued). n n I Set Mn = (∆ ; ∆ ) I Choose models for Top × Top the singleton sets Mn = fMng n n I Choose en 2 Tn(Mn) = Cn(∆ × ∆ ) to be the n-chain 1 × 1 : ∆n ! ∆n × ∆n This is the diagonal map for ∆n. Note that π0en = π1en = 1∆n Proof of Eilenberg-Zilber (continued). I Claim that the functor Tn is free with models Mn I In other words, we must show that T (X ; Y ) = C(X × Y ) has free basis n n B = Tn(f ; g)(en) (f ; g) : (∆ ; ∆ ) ! (X ; Y ) I We claim that B is exactly the set of singular n-simplices of X × Y . n I Given a singular simplex σ : ∆ ! X × Y Tn(π0σ; π1σ)(en) = (π0σπ0en) × (π1σπ1en) = (π0σ1∆n ) × (π1σ1∆n ) = (π0σ) × (π1σ) = σ I Hence B contains the singular n-simplices of X × Y . n I Tn(f ; g)(en) = (f π0en) × (gπ1en) = f × g is a map from ∆ to X × Y so the set of singular simplices of X × Y contains B. 0 0 I Finally if (f ; g) 6= (f ; g ) then 0 0 0 0 Tn(f ; g)(en) = f × g 6= f × g = Tn(f ; g )(en) Proof of Eilenberg-Zilber (continued). I Claim that each Mn is S-acyclic I In other words, we must show that for k > 0 we have Hk (S(Mn)) = 0 I n n Hk (S(Mn)) = Hk (S(∆ ; ∆ )) n n = Hk (C(∆ ) ⊗ C(∆ )) M n n =∼ Hi (C(∆ )) ⊗ Hk−i (C(∆ )) i M n n ⊕ Tor(Hi (C(∆ )); Hk−i−1(C(∆ ))) i M n n =∼ Hi (∆ ) ⊗ Hk−i (∆ ) i M n n ⊕ Tor(Hi (∆ ); Hk−i−1(∆ )) i 0; k 6= 0 = Z; k = 0 Proof of Eilenberg-Zilber (continued). I Claim there is a natural isomorphism Φ : H0(T ) ! H0(S) I In other words for each (X ; Y ) we have an isomorphism Φ(X ;Y ) : H0(T (X ; Y )) ! H0(S(X ; Y )) such that for any morphism (f ; g):(X ; Y ) ! (Z; W ) we get commutative Φ(X ;Y ) H0(T (X ; Y )) / H0(S(X ; Y )) H0T (f ;g) H0S(f ;g) Φ(Z;W ) H0(T (Z; W )) / H0(S(Z; W )) I Set Φ(X ;Y ) = H0 ◦ A : H0(C(X × Y )) ! H0(C(X ) ⊗ C(Y )) where A : C(X × Y ) ! C(X ) ⊗ C(Y ) is the Alexander-Whitney chain map. Proof of Eilenberg-Zilber (continued). I First we address naturality of Φ = H0 ◦ A I We will actually show that A is natural and since H0 : Chain ! Ab is natural we get naturality of H0 ◦ A. I That is for any morphism (f ; g):(X ; Y ) ! (Z; W ) we get commutative C(X × Y ) A / C(X ) ⊗ C(Y ) (f ×g)∗ f∗⊗g∗ C(Z × W ) A / C(Z) ⊗ C(W ) Proof of Eilenberg-Zilber (continued). n I Let (f ; g):(X ; Y ) ! (Z; W ) be a morphism and σ : ∆ ! X × Y be a singular simplex. A(f × g)∗σ = A(f πX σ × gπY σ) Pn = i=0 πZ (f πX σ×gπY σ)◦fi ⊗πW (f πX σ×gπY σ)◦bn−i n X = f πX σ ◦ fi ⊗ gπY σ ◦ bn−i i=0 n X (f∗ ⊗ g∗)Aσ = (f∗ ⊗ g∗) πX σ ◦ fi ⊗ πY σ ◦ bn−i i=0 n X = f πX σ ◦ fi ⊗ gπY σ ◦ bn−i i=0 I Thus A : Top × Top ! Chain is natural. I Hence Φ = H0 ◦ A is natural. Proof of Eilenberg-Zilber (continued). I Next we claim that H0 ◦ A : H0(C(X × Y )) ! H0(C(X ) ⊗ C(Y )) is an isomorphism. 0 I For p a point in the space P let σp : ∆ ! P be the 0-simplex with σp(∗) = p. I Given a singular 0-simplex with σ(x;y) of X × Y Aσ(x;y) = πX σ(x;y)f0 ⊗ πY σ(x;y)b0 = = σx f0 ⊗ σy b0 = σx ⊗ σy I Let [σ] denote the class of σ in H0 I Then we've shown H0 ◦ A[σ(x;y)] = [Aσ(x;y)] = [σx ⊗ σy ] I [σ(x;y)] = [σ(x0;y 0)] in H0(X × Y ) 0 0 I iff there are paths from x to x in X and y to y in Y I iff [σx ] = [σx0 ] in H0(X ) and [σy ] = [σy 0 ] in H0(Y ) Proof of Eilenberg-Zilber (continued). I H0(S(X ; Y )) = H0(C(X ) ⊗ C(Y )) ∼ M = Hi (X ) ⊗ H−i (Y ) i M ⊕ Tor(Hi (X ); H−i−1(Y )) i = H0(X ) ⊗ H0(Y ) I H0(T (X ; Y )) = H0(C(X × Y )) = H0(X × Y ) I Thus H0 ◦ A[σ(x;y)] = [σx ⊗ σy ] = [σx ] ⊗ [σy ] is surjective and injective. I This isomorphism amounts to observation that path components of X × Y is (setwise) the cartesian product of path components of X with path components of Y . Proof of Eilenberg-Zilber (continued). I We've fulfilled hypotheses of Acyclic Models Theorem so we get a natural chain map A0 : C(X × Y ) ! C(X ) ⊗ C(Y ) defined up to chain homotopy inducing isomorphism H0 ◦ A. 0 I But A is such a natural chain map so A ' A . Proof of Eilenberg-Zilber (continued). I Now we need a futher chain homotopy γ : S ! T ∼ inducing the reverse isomorphism H0(S) = H0(T ). I Here we need larger set of models n o S p q Mn = (∆ ; ∆ ) p + q = n S p q L p q I Let ep;q 2 Sn(∆ ; ∆ ) = i2Z Ci (∆ ) ⊗ Cn−i (∆ ) be S ep;q = 1∆p ⊗ 1∆q S I Sn is free with models Mn (check this) Proof of Eilenberg-Zilber (continued). p q S I Claim that each (∆ ; ∆ ) 2 Mn is T -acyclic I In other words, we must show that for k > 0 we have p q Hk (T (∆ ; ∆ )) = 0 I p q p q Hk (T (∆ ; ∆ )) = Hk (C(∆ × ∆ )) p q = Hk (∆ × ∆ ) ∼ = Hk ({∗}) 0; k 6= 0 = Z; k = 0 I Cross product − × − is natural and induces isomorphism ∼ H0(C(X ) ⊗ C(Y )) = H0(C(X × Y )) I Thus it is unique such chain map (up to chain homotopy). Proof of Eilenberg-Zilber (continued).
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