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 $


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.

  • $\begingroup$ 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? $\endgroup$ – Igor Sikora Oct 23 '18 at 14:40
  • $\begingroup$ @IgorSikora If $P$ is projective then so is $P[-1]$, and surjections onto projective objects split by definition of projectivity. $\endgroup$ – Matthew Towers Oct 23 '18 at 16:20
  • $\begingroup$ 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. $\endgroup$ – Igor Sikora Oct 23 '18 at 16:41
  • $\begingroup$ @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. $\endgroup$ – Matthew Towers Oct 23 '18 at 17:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.