# Isomorphism of homology groups implies isomorphism of cohomology groups

I have tried to read Hatcher's Algebraic Topology book. He said that one can prove the following corollary using the naturality of Universal Coefficient Theorem and Five-Lemma:

Corollary 3.4. If a chain map between chain complexes of free abelian groups induces an isomorphism on homology groups, then it induces an isomorphism on cohomology groups with any coefficient group $$G$$.

I cannot realize the role of Five-Lemma to prove the corollary. Is it something obvious?

Any help will be appreciated.

• You have a map of short exact sequences which is an isomorphism on the outer two terms. Well, it's actually an isomorphism between the outer 4 terms of a 5-term exact sequences (the 0's are the outermost terms). So the 5 lemma is what tells you that the middle vertical map is an isomorphism. – user98602 Dec 28 '18 at 16:33
• If I don't understand you wrong, you say that there are isomorphisms between Hom, Ext and outer 0-terms, so by Five-Lemma, we get an isomorphism between cohomology groups? – user625442 Dec 28 '18 at 16:39
• Yes. The universal coefficient theorem says the relevant short exact sequences are natural for chain maps of free abelian groups, and the assumption that your chain maps induce an isomorphism on homology imply they induce an isomorphism on the Hom and Ext terms. – user98602 Dec 28 '18 at 16:41
• @MikeMiller Perhaps you should give an official answer to clear the question from the unanswered queue. – Paul Frost Dec 28 '18 at 23:03
• @PaulF Normally I only comment when I feel unwilling to give an answer for whatever reason, but that is not the case here. I will write one up in an hour or so. Thanks for the reminder. – user98602 Dec 28 '18 at 23:11

Here is the situation. $$\require{AMScd}$$ The statement that the universal coefficient theorem is natural for maps $$f: C \to D$$ of chain complexes of free abelian groups means that there is a map of short exact sequences

$$\begin{CD}0 @>>> \text{Ext}^1_{\Bbb Z}(H_{i-1}(C), G) @>>> H^i(C;G) @>>> \text{Hom}(H_i(C), G) @>>> 0\\ @. @AAA @AAA @AAA \\ 0 @>>> \text{Ext}^1_{\Bbb Z}(H_{i-1}(D), G) @>>> H^i(D;G) @>>> \text{Hom}(H_i(D), G) @>>> 0 \end{CD}$$

Note that the arrows go the other way from $$f: C \to D$$; all three of $$\text{Ext}(-,G)$$, $$\text{Hom}(-, G)$$, and cohomology with $$G$$ coefficients are contravariant.

Because $$\text{Ext}$$ and $$\text{Hom}$$ are functorial in their inputs, they take isomorphisms to isomorphisms; in particular, because $$f_*: H_k(C) \to H_k(D)$$ is an isomorphism for any $$k$$, we see that the outer two vertical maps are isomorphisms.

Whenever you have a big diagram like this and some of those vertical arrows start being isomorphisms, you should think about the 5 lemma. In this case, though it looks like we only have 3 vertical arrows, don't forget about the implicit maps $$0 \to 0$$ on the outside (the trivial map!). This means that we have a map of 5-term exact sequences, which are isomorphisms on the "outer 4" vertical maps. Then the 5 lemma gets us that the middle vertical map is an isomorphism; this is what you wanted.