For any complexes of $R$-modules, $P$ and $M$, $\hom_R(P,M)$ is the complexe defined by,For any complexes of $R$-modules, $P$ and $M$, $\hom_{\mathcal{C}(R)}(P,M)$ is the complexe defined by, $$\forall n \in \mathbb{Z}\ \ \ \hom_R(P,M)_n = \prod_{i \in \mathbb{Z}} \hom_R(P_i, M_{i+n})$$$$\forall n \in \mathbb{Z}\ \ \ \hom_{\mathcal{C}(R)}(P,M)_n = \prod_{i \in \mathbb{Z}} \hom_R(P_i, M_{i+n})$$
We say that a complexe $P$ is $\pi$-projective or K-projective, if for all quasi-isomorphisme $f: M \to M'$We say that a complexe $P$ is $\pi$-projective or K-projective, if for all quasi-isomorphisme $f: M \to M'$ $$\hom_R(P,f): \hom_R(P,M') \to \hom_R(P,M)$$$$\hom_{\mathcal{C}(R)}(P,f): \hom_{\mathcal{C}(R)}(P,M') \to \hom_{\mathcal{C}(R)}(P,M)$$ is a quasi-isomorphisme to.
Equivalently we says that the complexe of $R$-modules $P$ is $\pi$-projective, if for all exacte complex $M$, $\hom_R(P,M)$ is also exacte.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.
Now supposeMy 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_R(f,M)$$\hom_{\mathcal{C}(R)}(f,M)$, which is at degrée $n$: \begin{align} \hom_R(f, M)_n :\hom_R(P', M)_n &\longmapsto \hom_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}\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.