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How to prove that a chain complex is a projective object in $ {Ch} $ (chain complexes of $R$-modules) iff it is a split exact complex of projectives?

A chain complex of projectives means a chain complex $\mathbb P_{\bullet}$ in which each $P_n$ is projective.

Hint: To see that P must be split exact, consider the surjection $\operatorname{cone}(\operatorname{id})\to P[-1]$. To see that split exact complexes are projective objects, consider the special case $0 \to P_1 \cong P_0 \to 0 $

user26857
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user150245
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1 Answers1

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I will give a more detailed hint on how to show a projective object $P$ of the category of chain complexes must be a split exact complex of projectives. There is an exact sequence $$ 0 \to P \to \operatorname{cone}(\operatorname{id}) \to P[-1] \to 0. $$ $P$ is projective, so $P[-1]$ is too, so this short exact sequence splits. Thus $\operatorname{cone}(\operatorname{id}) \cong P \oplus P[-1]$. But the cone is always exact, so $P$ is exact too.

The surjection you mentioned is $(p_{n-1}, p_n)\mapsto -p_{n-1}$, so the splitting $\phi : P[-1] \to \operatorname{cone}(\operatorname{id})$ has the form $p_{m-1} \mapsto (-p_{m-1}, \theta_{m-1}(p_{m-1}))$ where $\theta: P[-1] \to P$. Now use the fact that $\phi$ is a map of chain complexes, and your knowledge of the differential on the cone, to see that $\theta$ is a chain contraction of the identity and therefore that $P$ is split exact. Remember that the differential on $P[-1]$ is minus that on $P$.

Finally you need to show each $P_n$ is projective. Since $P$ is split exact, $P_n=B_n\oplus B'_n$ and the differential is zero on $B_n$ and an isomorphism from $B'_n$ to $B_{n-1}$. It's enough to show each $B_r'$ is projective, so suppose there is a module map $f:B_r' \to M$, and consider a module surjection $N \to M$ and the resulting surjection of stalk complexes $$ (\cdots 0 \to N \to 0 \cdots ) \to (\cdots 0 \to M \to 0 \cdots) $$ with $M$ and $N$ in degree $r$. There is a map of complexes from $P$ to the second of these which is $f\oplus 0$ in degree $r$ and zero everywhere else. By projectivity of $P$, this factors through the surjection of stalk complexes above. You can use this to get that $B_r'$ is projective.

Aryaman Maithani
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Matthew Towers
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  • I know that this answer is quite old, but I have a problem with understanding the first line - why do I know that since $P$ is projective then this SES splits? – Igor Sikora Oct 23 '18 at 14:40
  • @IgorSikora If $P$ is projective then so is $P[-1]$, and surjections onto projective objects split by definition of projectivity. – Matthew Towers Oct 23 '18 at 16:20
  • All right, now I see ;) Actually, as my friend pointed out, it would be the same as working with shifted cone, and then we don't have to think why $P[-1]$ has to be projective. – Igor Sikora Oct 23 '18 at 16:41
  • @IgorSikora sure, that's a good way to do it. I'll edit to clarify that it's the projectivity of $P[-1]$ that is used. – Matthew Towers Oct 23 '18 at 17:01
  • I know this questions is ancient at this point but hopefully someone will see this. I am unclear why $cone(id) \cong P \oplus P[-1]$ implies $P$ is exact and I am also unclear about the definition of $\theta$. – RhythmInk Sep 02 '20 at 14:10
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    @RhythmInk: $\DeclareMathOperator{\cone}{cone}$ To answer your doubt about the $\cone$: Note that homology commutes with direct sums. Thus, we have $$0 = H(\cone(P)) = H(P) \oplus H(P[-1])$$ and hence, $H(P) = 0$. – Aryaman Maithani Sep 01 '21 at 22:41