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 $R$.


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.)]


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.