The Groove of Math

or in other words, the Devil is in the Details

Random Exercise #1 “Isomorphism in cohomology induces Isomorphism in Homology”

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I’ve decided to start collecting here some random exercises arose during discussions with my colleagues (usually during the weekly pizza-time) and I found interesting.

Let f\colon X \to Y be a continuous map between topological spaces, and R some coefficients. Let Z be the integers.

 

  1. If H_*(f;R) is an isomorphism, is H^*(f;R) one, too?
  2. If H^*(f,Z) is an isomorphism, is H_*(f; R) one, too (for every R)?

Assume that R is a ring. H_*(X;R) can be defined as homology of C_*(X;R), the chain complex which in degree n is the free R-module generated by singular n-simplices of X. This is a bounded below projective chain complex and thus if the map induced by f is a quasi-isomorphism, it must be already a homotopy equivalence. Abstract no sense: in the standard model category structure for \text{Chain}_R whose weak equiv are quasi-isomorphisms (See https://www.math.rochester.edu/people/faculty/doug/otherpapers/dwyer-spalinski.pdf Thm 7.2) C_*(X;R) and C_*(Y;R) are both fibrant and cofibrat objects, since the inclusion 0 \hookrightarrow C_*(X;R) is a monomorphism whose cokernel is degree-wise projective and the map C_*(Y;R)\to 0 is a degreewise epimorphism. Therefore by Whitehead theorem for Model Cats, a weak-equiv between fibrant-cofibrant objects is already an homotopy equivalence. Thus the map is still a quasi-isomorphism after applying \hom_R(-,R) which is one way to compute cohomology of a space.

For the second point, by replacing Y with the mapping cylinder of f, we can always assume that f is the inclusion. Thus it is enough to prove that when H^*(Y,X;R)=0 then also H_*(Y,X;R)=0. This is true under the added assumption that the homology of (Y,X) is finitely generated, for which it is enough that both Y,X have finitely generated homology. In fact using this UCT for homology (See Davis & Kirk’s Lecture Notes in Algebraic Topology page 48 Theorem 2.36) we have

0 \to \text{Ext}^1_Z(H^{q+1}(Y,X),R)\to H_q(Y,X;R) \to \hom(H^q(Y,X),R)\to 0

Therefore if both ends are zero, the middle one is zero as well, proving that H_*(Y,X;R)=0.

Actually this last version of the UCT is the reason why I decided to write a little article about it, it’s a version which is always forgotten, and it’s a pity since it turns out it is very powerful.

 

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Author: RicPed

Young mathematician :)

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