# Isomorphism in cohomology is an isomorphism in homology

Let $f:X \to Y$ be a continuous map between topological spaces and $R$ some coefficients. From the universal coefficient theorem for homology we immediatly get, that if $H_*(f,\mathbb{Z})$ is an isomorphism then so is $H^*(f,\mathbb{Z})$ and $H^*(f,R)$.

Now for me, this implies three questions:

1) If $H_*(f,R)$ is an isomorphism, is $H^*(f,R)$ one, too?

2) If $H^*(f,\mathbb{Z})$ is an isomorphism, is $H_*(f,R)$ one, too?

3) If $H^*(f,R)$ is an isomorphism, is $H_*(f,R)$ one, too?

Obviously 3) implies 2) by setting $R=\mathbb{Z}$ and using the universal coefficient theorem for homology.

Feel free to add some prerequisites to $X,Y$ and $R$.

1) Yes, at least when $R$ is a ring. Indeed, $H_{*}(X, R)$ 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. Thus, the map is still a quasi-isomorphism after applying $Hom_{R}(-, R)$ which is one way to compute cohomology of a space.
For 2), 3) the following standard trick simplifies analysis. By replacing $Y$ by a mapping cyllinder in needed, we may assume that $f: X \rightarrow Y$ is an inclusion. Thus, for example to prove 2) and 3) it's enough to show that when $H^{*}(Y, X, \mathbb{Z}) = 0$ then also $H_{*}(Y, X, \mathbb{Z}) = 0$ (by the relevant long exact sequences). This is true under the added assumption that homology of $(Y, Z)$ is finitely generated, for which it is enough that both $Y, X$ have finitely generated homology.
Indeed, any non-trivial infinite cyclic summand in $H_{n}(Y, X, \mathbb{Z})$ would appear in $H^{n}(Y, Z, \mathbb{Z})$ (as its subquotient is $Hom(H_{n}(Y, Z, \mathbb{Z}), \mathbb{Z})$ by universal coefficient). On the other hand, any finite cylic summand would appear as $0 \neq Ext^{1}(\mathbb{Z}_{k}, \mathbb{Z}) \subseteq Ext^{1}(H_{n}(Y, X, \mathbb{Z}), \mathbb{Z}) \subseteq H^{n+1}(Y, X, \mathbb{Z})$ again by universal coefficient.
[Observe that above I have used universal coefficient for the relative co(homology) of $(Y, X)$. This is possible as the theorem is in fact a statement of homological algebra about bounded below free $\mathbb{Z}$-complexes (of which the relative singular complex of $(Y, Z)$ is an example.)]