Exercise 2.2.2. from Weibel's Homological Algebra. If $A$ has enough projectives, then so does $\operatorname{Ch}(A)$. $\newcommand{\A}{\mathcal{A}}\newcommand{\Ch}{\mathbf{Ch}}$The following is an exercise from Weibel's An Introduction to Homological Algebra.

Show that if $\A$ has enough projectives, then so does the category $\Ch(\A)$ of chain complexes over $\A$.

The hint given is to use the previous exercise, where one characterises projective objects in $\Ch(\A)$. Namely, we saw that $P_{\bullet}$ is projective in $\Ch(\A)$ iff it is a split exact complex of projectives.

My attempt.
Let $M_{\bullet}$ be an arbitrary chain complex in $\Ch(\A)$. Then, since $\A$ has enough projectives, we can find epics $$P_{n} \to M_{n}$$ for all $n$. Moreover, using projectivity and the differentials of $M_{\bullet}$, we get maps $$d_{n} : P_{n} \to P_{n - 1}$$
such that $d^2= 0$ and they fit together to give a chain map $$P_{\bullet} \to M_{\bullet}.$$
The issue now is showing that the $P_{\bullet}$ is projective. (The above map being epic is clear since it is epic in each degree.)
In fact, I am quite sure that the construction above will actually not work in general. For example, if each $M_{n}$ was projective to begin with, then we could've picked each $P_{n}$ to be $M_{n}$ and the maps to be the identity maps. But an arbitrary complex of projectives need not be split exact.
This makes me think that one needs to do a different construction, but I don't see one.

Proceed as follows:

*

*Components. Pick epic maps $Q_n \longrightarrow H_n$ and $P_n'' \longrightarrow B_n$ from projectives that then, by the proof of the Horseshoe lemma fit into a SEC $ 0\longrightarrow P_n'' \longrightarrow  P_n' \longrightarrow Q_n \longrightarrow  0$ covering $0 \longrightarrow B_n \longrightarrow  Z_n \longrightarrow  H_n \longrightarrow  0$.
This gives surjections $P_n' \to Z_n$.


*More components. There now exists another SEC $ 0\longrightarrow P_n' \longrightarrow  P_n \longrightarrow P_{n-1}'' \longrightarrow  0$
covering the SEC$0 \longrightarrow Z_n \longrightarrow  C_n \longrightarrow  B_{n-1} \longrightarrow  0$.


*Differential. Note that $P_n = P_n'\oplus P_{n-1}''$ maps into $P_{n-1} = P_{n-1}'\oplus P_{n-2}''$ via the matrix
$$d = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$$
where $1$ is the inclusion $P''\subseteq P'$ in (1), and that $d^2=0$. This makes $P$ into a complex $(P,d)$ of projectives.


*Split. To see it is split, note that $P_n' = P_n''\oplus Q_n$ (another copy of $P'')$ and $P_{n+1} = (P_{n+1}''\oplus Q_{n+1})\oplus P_n''$ so you can project and include into this copy. In short:
$$d(x,y,z) = (z,0,0), \quad s(x,y,z) = (0,0,x)$$
and then it is clear that $dsd=d$, so $(P,d)$ is split.
