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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 $P_{\bullet}\rightarrow M_{\bullet}\rightarrow 0$.

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  • $\begingroup$ Those differentials will not make the projections $P_n\rightarrow M_n$ into a map of chain compelxes. $\endgroup$
    – Thorgott
    Mar 28 at 16:12
  • $\begingroup$ 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… $\endgroup$
    – Aphelli
    Mar 28 at 16:19
  • $\begingroup$ This answer gives a construction of split chain complex of projectives, while it’s not exact. Could anyone complete the proof? Thanks a lot. math.stackexchange.com/a/4240068/791697 $\endgroup$
    – ZYX
    Mar 28 at 16:25
  • $\begingroup$ @Thorgott Hi, I modified the answer. Could you check it for me again? Thanks a lot. $\endgroup$
    – ZYX
    Mar 29 at 1:03
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    $\begingroup$ Ok, that would be very important to clarify if you want your solution to be readable. But yes, I agree this works then. $\endgroup$
    – Thorgott
    Mar 29 at 11:01

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