Timeline for If an abelian category $\mathcal{A}$ has enough injectives then so does $\mathrm{Ch}^{\geq 0}(\mathcal{A})$

12 events

when toggle format	what		by	license	comment
S Jul 31, 2014 at 9:31	history	suggested	R.P.		added algebraic topology tag
Jul 31, 2014 at 9:23	review	Suggested edits
S Jul 31, 2014 at 9:31
Jul 31, 2014 at 8:48	comment	added	Qiaochu Yuan		What do you need the Dold-Kan correspondence for? The category of chain complexes is already a presheaf category.
Jul 30, 2014 at 15:04	answer	added	Jeremy Rickard		timeline score: 6
Jul 30, 2014 at 14:54	comment	added	abx		@Jeremy Rickard: Right! Strange that Grothendieck doesn't put it that way.
Jul 30, 2014 at 14:53	comment	added	John Wiltshire-Gordon		By the Dold-Kan correspondence, the category of chain complexes is a presheaf category.
Jul 30, 2014 at 14:48	history	edited	Jeremy Rickard	CC BY-SA 3.0	grammar in title
Jul 30, 2014 at 14:07	comment	added	Jeremy Rickard		@abx I think that follows from the other conditions. By induction on $i$, $\operatorname{ker}(d^i)$ is injective and so $0\to\operatorname{ker}(d^i)\to A^{i+1}\to\operatorname{ker}(d^{i+1})\to0$ splits.
Jul 30, 2014 at 13:30	comment	added	abx		I think you need more, namely that $\mathrm{Ker}\, d^i$ is a direct factor of $A^i$ for each $i\geq 0$. See A. Grothendieck, Sur quelques points d'algèbre homologique, p. 146.
Jul 30, 2014 at 13:15	comment	added	Zhen Lin		The title question is related to this one, but the body question seems to be something different.
Jul 30, 2014 at 13:11	review	First posts
Jul 30, 2014 at 15:14
Jul 30, 2014 at 13:10	history	asked	W.Z.	CC BY-SA 3.0