Isomorphism on Cohomology implies isomorphism on homology Say I am given a chain map $f:C \to D$ of complexes of (free if necessary) abelian groups. Assume that this map induces isomorphisms of cohomology with all coefficient rings. How do you prove that this map also induces isomorphisms on homology?
 A: First, let us note that a map induces isomorphisms on cohomology with coefficients in all rings iff it induces isomorphisms on cohomology with coefficients in all abelian groups.  Indeed, if $G$ is any abelian group, consider the ring $R=\mathbb{Z}\oplus G$, with multiplication defined by $(m,g)\cdot(n,h)=(mn,mh+ng)$.  Then there is an isomorphism $H^*(C,R)\cong H^*(C,\mathbb{Z})\oplus H^*(C,G)$ which is natural in $C$, so a map induces isomorphisms on cohomology with coefficients in $R$ iff it induces isomorphisms on cohomology with coefficients in $\mathbb{Z}$ and with coefficients in $G$.
Now by taking the mapping cone, it suffices to show that if a chain complex has vanishing cohomology with coefficients in all abelian groups, then it is exact.  That is, we want to show that if $X\stackrel{f}\to Y\stackrel{g}\to Z$ are maps of abelian groups such that $gf=0$ and $\operatorname{Hom}(Z,G)\stackrel{g^*}{\to}\operatorname{Hom}(Y,G)\stackrel{f^*}{\to}\operatorname{Hom}(X,G)$ is exact for all $G$, then $\ker(g)=\operatorname{im}(f)$.  To prove this, let $G=\operatorname{coker}(f)$ and let $p:Y\to G$ be the canonical map.  Then $p\in\operatorname{Hom}(Y,G)$, and $f^*(p)=0$.  By hypothesis, this means $p$ is in the image of $g^*$, so there is some $q:Z\to G$ such that $p=qg$.  But this implies $\ker(g)\subseteq\ker(p)$.  Since $\ker(p)=\operatorname{im}(f)$, we are done.
In fact, if you assume your chain complexes consist of free abelian groups, it suffices to consider coefficients in $\mathbb{Z}$.  Indeed, in that case the universal coefficient theorem tells you that the cohomology of a chain complex $E$ with coefficients in $\mathbb{Z}$ vanishes iff $\operatorname{Hom}(H_n(E),\mathbb{Z})=0$ and $\operatorname{Ext}(H_n(E),\mathbb{Z})=0$ for all $n\in\mathbb{Z}$.  But this implies $H_n(E)=0$; see my answer to this question, for instance.
