Isomorphism in cohomology is an isomorphism in homology

Solution 1:

This is a great question! I know no references for the following facts and I am pretty sure one can make more general statements, but these are the ones that I have seen used in practice.

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.)]