# Induced isomorphism of cohomology group implies induced isomorphism of cohomology with any coefficient

Let $f,g:C_* \rightarrow D_*$ be maps of nonnegative chains of free abelian group. Suppose that $f^*=g^*: H^*(C,\mathbb{Z}) \rightarrow H^*(D,\mathbb{Z})$. Prove that $f$ induces isomorphism of cohomology group with coefficient in any abelian group, i.e. $f^*:H^*(C,G)\rightarrow H^*(D,G)$ is isomorphism.

Is it true that it induces the same isomorphisms $f^*=g^*:H^*(C,G)\rightarrow H^*(D,G)$ for any abelian group $G$?

I thought that first statement implies from Universal Coefficient Theorem, but I made stupid mistake... Could anyone help me how to prove this statement? Maybe I can use UCT, but in more subtle way?

I have also problem to decide if for two the same induced isomorphism of cohomology with $\mathbb{Z}$-coefficients have to induce the same isomorphism in any coefficient. I think that it isn't true (and probably it is the case that in UTC this sequence don't splits naturally), but I can't give a counterexample or prove.

• I think you have two different statements confused here. It's certainly not true that $f^*$ must be an isomorphism (nothing in your assumptions prevents having $f=g=0$). – Eric Wofsey Feb 17 '18 at 22:26
• math.stackexchange.com/questions/1782321/… answers what might be your intended question. – Eric Wofsey Feb 17 '18 at 22:28