Show that $\hom_R(f, M)$ is a quasi-isomorphism if $f:P \to P'$ is a quasi-isomorphism of $K$-projectives complexes

Question

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.

Answer 1

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.