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This is exercise 2.2.2 in Weibel's AIHA. We already know that a chain complex $P_{\bullet}$ is projective in $Ch(\mathcal{A})$ iff it is a split exact complex of projectives. Here's my proof, but it looks too easy. I'm not sure about that. Could anyone please help me check it?

Given any chain complex $M_{\bullet}$ in $Ch({\mathcal{A}})$. For each $M_n \in \mathcal{A}$, we have an epimorphism $f_n: P_n \rightarrow M_n$ where $P_n$ is projective. Now define $P_{\bullet}$ with the $n$-th object $P_{n} \oplus P_{n+1}$ and $d_n: (a,b)\mapsto (d(a),a-d(b))$ and define $s_n: (a,b)\mapsto (b,0)$. Then we have $sd+ds=id$, which means the complex $P_{\bullet}$ is split exact, hence projective in $Ch(\mathcal{A})$. Naturally, the map from $P_{n} \oplus P_{n+1}$ to $M_n$ just maps $(a,b)$ to $f(a)$. Hence we get $P_{\bullet}\rightarrow M_{\bullet}\rightarrow 0$.

ZYX
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  • Those differentials will not make the projections $P_n\rightarrow M_n$ into a map of chain compelxes. – Thorgott Mar 28 '22 at 16:12
  • Well, there’s only one natural map $P_n \rightarrow M_n$ and only one natural map $P_{n+1} \rightarrow M_n$, that doesn’t give you much choice… – Aphelli Mar 28 '22 at 16:19
  • This answer gives a construction of split chain complex of projectives, while it’s not exact. Could anyone complete the proof? Thanks a lot. https://math.stackexchange.com/a/4240068/791697 – ZYX Mar 28 '22 at 16:25
  • @Thorgott Hi, I modified the answer. Could you check it for me again? Thanks a lot. – ZYX Mar 29 '22 at 01:03
  • This is not clear to me. What are $d(a),d(b)$ supposed to mean? You've not provided us with a map $d$ on the $P_n$. – Thorgott Mar 29 '22 at 10:38
  • @Thorgott Since $P_n$ is projective and $P_{n-1} \rightarrow M_{n-1}$ is epi, there exists $d_n: P_n \rightarrow P_{n-1}$ s.t. $fd^P=d^Mf$. For simplicity, I omit the index. – ZYX Mar 29 '22 at 10:44
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    Ok, that would be very important to clarify if you want your solution to be readable. But yes, I agree this works then. – Thorgott Mar 29 '22 at 11:01

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