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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.

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$\newcommand{\Hom}{\mathrm{Hom}} \require{AMScd}$ 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.

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