Induced isomorphism of homology implies isomorphism with coefficients in any group?

If $\alpha\colon C \to C'$ is a map of chain complexes (of free abelian groups) that induces an isomorphism on homology $a_{*} \colon H_n(C) \simeq H_n(C')$, then I know that $\alpha$ induces an isomorphism on cohomology $a^{*} \colon H^n(C;G) \simeq H^n(C';G)$ with coefficients in any abelian group $G$. But is it also true that $\alpha$ induces an isomorphism on homology $a_{*} \colon H_n(C;G) \simeq H_n(C';G)$ with coefficients in any abelian group $G$ or (more likely) is there some counterexample?

• Does The Universal Coefficient Theorem cover what you want? – zibadawa timmy Nov 12 '13 at 19:25
• But how would you construct such a map? You are changing your coefficients. In the cohomology case you just use the $\text{Hom}(-, G)$ functor. – M.B. Nov 12 '13 at 19:29
• Are you assuming that your map of chain complexes is also a chain map? Then, indeed, the UCT does the job (since it defines a natural exact sequence). – Moishe Kohan Nov 12 '13 at 20:56
• math.toronto.edu/~mat1300/oldnotes/homology-coefficients.pdf is Proposition 9.11.3 the thing you're looking for? – Dan Rust Nov 12 '13 at 21:51
• @M.B.: $- \otimes G$ is also a functor. – Najib Idrissi Nov 12 '13 at 22:09

Here is an answer assuming that $\alpha$ is a chain map; without this assumption I do not see how you can talk about induced maps of integer homology groups. Then the answer to your question is positive and follows from the Universal coefficient Theorem (UCT). Indeed, since $\alpha$ is a chain map, it induces maps $\alpha_{i*}: H_i(C; G)\to H_i(C'; G)$ for arbitrary abelian groups $G$. Since the statement of the UCT provides a natural short exact sequence, we get isomorphisms $$\beta_i: H_i(C)\otimes G \to H_i(C')\otimes G$$ $$\gamma_i: Tor(H_{i-1}(C), G)\to Tor(H_{i-1}(C'), G)$$ which (together with $\alpha_{i*}$) make a commutative diagram with UCT sequences for $C, C'$ on the top and on the bottom. (Sorry, I am not very good at drawing diagrams in the version of TeX used at MSE.) Now, the fact that $\alpha_{i*}$ is an isomorphism follows from the diagram chasing in this commutative diagram. For instance, to see injectivity, note that an element $x$ of the kernel has to project to zero in $Tor(H_{i-1}(C), G)$. Hence, it belongs to $H_i(C; {\mathbb Z})\otimes G$, but then injectivity of $\beta_i$ implies that $x=0$.