TITLE: Show that $\hom_R(f, M)$ is a quasi-isomorphism if $f:P \to P'$ is a quasi-isomorphism of $K$-projectives complexes
QUESTION [4 upvotes]: For any complexes of $R$-modules, $P$ and $M$, $\hom_{\mathcal{C}(R)}(P,M)$ is the complexe defined by,
$$\forall n \in \mathbb{Z}\ \ \ \hom_{\mathcal{C}(R)}(P,M)_n = \prod_{i \in \mathbb{Z}} \hom_R(P_i, M_{i+n})$$
We say that a complexe $P$ is $\pi$-projective or K-projective, if for all quasi-isomorphisme $f: M \to M'$
$$\hom_{\mathcal{C}(R)}(P,f): \hom_{\mathcal{C}(R)}(P,M') \to \hom_{\mathcal{C}(R)}(P,M)$$
is a quasi-isomorphisme too. Equivalently we says that the complexe of $R$-modules $P$ is $\pi$-projective, if for all exacte complex $M$, $\hom_{\mathcal{C}(R)}(P,M)$ is also exacte.
My question is: Suppose $f:P \to P'$ is a quasi-isomorphism of $\pi$-projective complexes of $R$-modules. I want to show that for any complexe of $R$-module M, $\hom_{\mathcal{C}(R)}(f,M)$, which is at degrée $n$:
\begin{align}
\hom_{\mathcal{C}(R)}(f, M)_n :\hom_{\mathcal{C}(R)}(P', M)_n &\longmapsto \hom_{\mathcal{C}(R)}(P, M)_n\\
(\alpha_i : P'_i \to M_{i+n})_{i \in \mathbb{Z}} &\longmapsto (\alpha_i \circ f_i )_{i \in \mathbb{Z}}
\end{align}
is a quasi-isomorphism.

REPLY [6 votes]: What you call $\pi$-projective is what Spaltenstein calls $K$-projective (and this is the only term I've heard). Spaltenstein shows in his original paper that a quasi-isomorphism between $K$-projective complexes is a chain homotopy equivalence, which implies the result you want.
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