# Exercise 2.2.1 in Charles A. Weibel's book An Introduction To Homological Algebra

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.
• 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