This is certainly an easy question... Why does a map of spaces $f:X\rightarrow Y$ which induces an isomorphism in cohomology $f^*:H^*(Y)\rightarrow H^*(X)$ induces an isomorphism in homology $f_*:H_*(X)\rightarrow H_*(Y)$, if the spaces are locally finite ? I know about the universal coefficient theorem, but I don't know how to apply it. For me $H$ can be any ordinary (co)homology.
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2What does locally finite mean here? – Qiaochu Yuan Jan 09 '15 at 06:58
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Yes I mean homology groups are finitely generated – mph Jan 09 '15 at 15:29
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Ah, I see. I would say something like "locally finite homology" instead (although I'm not a huge fan of that term either). "Locally finite space" sounds like it could mean a space such that each point has a finite open neighborhood; that would be consistent with the use of the terms "locally finite graph," "locally metric space," and "locally finite poset," for example. – Qiaochu Yuan Jan 09 '15 at 16:32
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I agree, I don't like this name either, but it seems very common in the litterature! – mph Jan 09 '15 at 19:47
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In this answer I assume that what you mean by "locally finite space" means "homology groups are finitely generated" (sometimes people call a graded module "locally finite" when it is are finitely generated in each degree), if that's not the case please clarify. Since you talk about the universal coefficient theorem I assume the base ring is a PID. I'm reasonably confident that using the universal coefficient spectral sequence leads to the same answer if the ring is arbitrary, maybe if we restrict to a domain...
We can replace $f$ up to homotopy by a cofibration, so that we can assume that $f$ is an inclusion $X \subset Y$. By looking at the long exact sequence, $f^*$ being an isomorphism in cohomology implies that $H^*(Y,X)$ vanishes. It is a general fact that the relative (co)homology is the same as the reduced (co)homology of the space with a cone attached to the subspace: $$H^*(Y,X) \cong \tilde{H}^*(Y \cup CX), \qquad H_*(Y,X) \cong \tilde{H}_*(Y \cup CX),$$ where $CX = X \times [0,1] / X \times 1$ and the attaching map $X \times 0 \to Y$ is the inclusion $X \subset Y$.
By this question (it uses the structure theorem for f.g. abelian groups, but it also works in general over a PID simply by using the structure theorem for f.g. modules over a PID) $\tilde{H}^*(Y \cup CX) = 0$ implies that $\tilde{H}_*(Y \cup CX) = 0$, which in turn implies that $H_*(Y,X) = 0$. Taking the relative homology long exact sequence, you can finally conclude that $f_*$ is an isomorphism in homology.
Najib Idrissi
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