*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.



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


