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Proposition Let $P_{\ast }$ be a chain complex taking values in an abelian category $\operatorname{\mathcal{A}}$. Assume that $P_{\ast }$ is acyclic, concentrated in degrees $\geq 0$, and that each $P_{n}$ is a projective object of $\operatorname{\mathcal{A}}$. Then $P_{\ast }$ is a projective object of the category $\operatorname{Ch}(\operatorname{\mathcal{A}})$. In other words, every epimorphism of chain complexes $f: M_{\ast } \twoheadrightarrow P_{\ast }$ admits a section.

Proof. Our assumption that $P_{\ast }$ is acyclic guarantees that for every integer $n \geq 0$, we have a short exact sequence

\[ 0 \rightarrow \mathrm{Z}_{n}(P) \rightarrow P_{n} \xrightarrow {\partial } \mathrm{Z}_{n-1}(P) \rightarrow 0. \]

It follows by induction on $n$ that each of these exact sequences splits and that each $\mathrm{Z}_ n(P)$ is also a projective object of $\operatorname{\mathcal{A}}$. We can therefore choose a direct sum decomposition $P_{n} \simeq \mathrm{Z}_ n(P) \oplus Q_ n$, where the differential on $P_{\ast }$ restricts to isomorphisms $\partial : Q_{n} \simeq \mathrm{Z}_{n-1}(P)$. Since each $Q_{n}$ is projective and $f$ is an epimorphism in each degree, we can choose maps $u_{n}: Q_{n} \rightarrow M_ n$ for which the composition $f_ n \circ u_ n$ equal to the identity on $Q_{n}$. The maps $u_ n$ then extend uniquely to a map of chain complexes $s = \{ s_ n \} _{n \in \operatorname{\mathbf{Z}}}$, characterized by the requirement that each composition

\[ Q_{n+1} \oplus Q_{n} \xrightarrow { \partial \oplus \operatorname{id}} \mathrm{Z}_{n}(P) \oplus Q_ n = P_ n \xrightarrow {s_ n} M_ n \]

is the sum of the maps $\partial u_{n+1}$ and $u_ n$. $\square$