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.
 A: 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.
