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This is Exercise 2.2.2. from Weibel's An Introduction to Homological Algebra.

Suppose $A$ is an abelian category, if $A$ has enough projectives, then so does the category $Ch(A)$ of chain complex over $A$.

Weibel's hint is to use the fact: a chain complex $P$ ($P$ means each $P_n$ is a projective object in $A$) iff it's a split exact complex of projectives. And I have proven this. (For reference: one can find the two directions of the proof here and here.)

Let $B$ be a chain complex over $A$. Then for each $B_n$ we have a $P_n$ such that $P_n\rightarrow B_n$ is surjective.

I don't know how to get a differential $d_n: P_n\rightarrow P_{n-1}$ s.t. complex $P$ is split exact. I tried use the definition of $P_n$ to get a map from $P_n$ to $P_{n-1}$. But I don't know what to do next.

Aryaman Maithani
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Cille
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2 Answers2

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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. 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. $fd^{P}=d^{M}f$. For simplicity, I omit the index.

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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Let $M_{\bullet}$ be some chain complex. Choose epic maps $f_n:P_n\to M_n$ from projective objects $P_n$. Define $P_{\bullet}$ with the $n$th object $P_n\oplus P_{n+1}$ and define $d^P(a,b)=(0,a).$ Then $P_{\bullet}$ is split exact. Define an epic chain map $f:P_{\bullet}\to M_{\bullet}$ by $f(a,b)=f_na+d^Mf_{n+1}b$. It's a chain map because $$ d^Mf(a,b)=d^M(f_na+d^Mf_{n+1}b)=d^Mf_na=f(0,a)=fd^P(a,b). $$

subrosar
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