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

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.