# Dual of quasi-isomorphism of chain complexes.

Let $$C_*$$, $$D_*$$ be chain complexes of modules over a ring $$R$$. Suppose that $$f\colon C_* \rightarrow D_*$$ is a quasi-isomorphism (i.e. an isomorphism in Homology).

I am wondering what conditions are needed (over either $$R,\ C_*$$ or $$D_*$$) so that the dual chain complexes $$\hom_R(C_*,R)$$ and $$\hom_R(D_*, R)$$ are quasi-isomorphic.

$$\newcommand{\Hom}{\mathrm{Hom}} \require{AMScd}$$ Suppose $$C$$ and $$D$$ are chain complexes of free $$R$$-modules, where $$R$$ is a PID. If $$f: C \to D$$ is a homology isomorphism, then $$f^*: \Hom(D,R) \to \Hom(C,R)$$ is as well. This follows from the naturality of the Universal Coefficient Theorem, as there is a commuting diagram of exact sequences $$\begin{CD} 0 @>>> Ext(H_{\ast-1}(D),R) @>>> H^\ast(\Hom(D,R)) @>>> \Hom(H_\ast(D),R) @>>>0 \\ @VVV @VVV @VVV @VVV @VVV\\ 0 @> >>Ext(H_{\ast-1}(C),R)@>>> H^\ast(\Hom(C,R)) @>>>\Hom(H_\ast(C),R) @>>>0\end{CD}$$ where the vertical maps are all induced by $$f: C \to D$$. As $$f$$ is a homology isomorphism, all maps except the middle are isomorphisms. Hence, the Five Lemma gives that the middle map is an isomorphism as well.
For more general $$R$$, there should be a spectral sequence instead of this short exact sequence giving the same result.