# Prove that if $\mathcal{A}$ has enough projectives, then so does $Ch(\mathcal{A})$

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

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