Constructing projective resolution of a chain complex I am trying to construct the projective resolution in the category of chain complexes of
$\dots \to 0 \to M \to 0 \to \dots$
It seems like it should be possible to do this in terms of the projective resolution of $M$ but I am completely stuck.
I know a projective chain complex is split exact and formed by projectives, so if we think of the resolution as a half plane double complex, the column with $M$ must be a projective resolution of $M$.
I was trying to use the trick of $0 \to P \to P \to 0$ is a projective complex whenever $P$ is projective, but if I put that on top of our complex we don't necessarily get exactness.

In this case, you are in the category of bounded above complexes, where a $\textit{projective resolution}$ of a complex (in this case $\bar{M}:\cdots\rightarrow 0\rightarrow M\rightarrow0\rightarrow\cdots$) means a bounded-above complex of projectives $P$ with a quasi-isomorphism $P\rightarrow \bar{M}$. So, if you take the usual projective resolution of $M$ as a module, $$\cdots\rightarrow P^{-n}\rightarrow P^{-n+1}\rightarrow\cdots\rightarrow P^{-1}\rightarrow P^{0}\rightarrow M\rightarrow0\rightarrow\cdots$$
we can construct the projective resolution of $\bar{M}$ as follows
$\require{AMScd}$
\begin{CD}
\cdots @>>>P^{-1} @>>>P^{0} @>>> 0  @>>>\cdots\\
@V{f^{-2}}VV @V{f^{-1}}VV @V{f^{0}}VV @V{f^{1}}VV @V{f^{1}}VV\\
\cdots@>>>0@>>>M @>>> 0 @>>> \cdots
\end{CD}
where the arrow $f:\bar{P}\rightarrow \bar{M}$ is obviously a quasi-isomorphism.
In the homotopic category $K(\mathscr{A})$ (where $\mathscr{A}$ is an abelian category such as the category of modules over a ring ) you can generalize this and talk about $K$-projective resolutions, complexes $X$ in $K(\mathscr{A})$ which verify that $Hom(X,Z)=0\ ,\ \forall Z\in\mathscr{Z}=\lbrace Z\in K(\mathscr{A})\ \text{such that}\ H^{n}(Z)=0\ \forall \ n\in\ \mathbb{N} \rbrace $.
The good thing is that if $P$ is a bounded above complex of projectives, then is $K$-projective.
