Timeline for What conditions on an Abelian category allow members of a direct sum to be determined entirely by their components?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Dec 22, 2023 at 17:04 | comment | added | FShrike | @DenisT Yes, I’m aware! I try to be very vigilant when I read any book and I’ve noted several errors or nontrivialities in generalising results for non-module categories | |
| Dec 22, 2023 at 16:31 | comment | added | Denis T | Also I'd like to note/warn that there are several erroneous claims in Weibel's "Homological algebra", mostly ones related to generalisations of statements from module categories to arbitrary abelian ones; for example, it's claimed that the category of torsion abelian groups is not complete (it actually is, but forgetful functor to all abelian groups does not commute with products). | |
| Dec 22, 2023 at 14:18 | vote | accept | FShrike | ||
| Dec 22, 2023 at 13:45 | answer | added | Denis T | timeline score: 3 | |
| Dec 21, 2023 at 17:38 | history | edited | FShrike | CC BY-SA 4.0 |
added 386 characters in body
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| Dec 21, 2023 at 17:38 | comment | added | Maxime Ramzi | @DenisT of course, this being a mono is just a reformulation of the claim. I was giving an example of a common situation where this happens, namely exact filtered colimits. It might be equivalent but I'm not sure. FShrike : yes, exactly ! :) | |
| Dec 21, 2023 at 17:36 | comment | added | FShrike | @MaximeRamzi Aha! So under your axiom on $\mathscr{A}$, I can assemble these monos to get a monomorphism $\bigoplus_i X_i\to\prod_i X_i$ and then use the obviousness of the claim for $\prod_iX_i$ to conclude. Very nice, thanks. | |
| Dec 21, 2023 at 16:46 | comment | added | Denis T | @MaximeRamzi I think it's enough to require canonical transformation $\coprod \to \prod$ to be mono, but the "proof" I have in mind uses injective cogenerator, which obviously implies exactness of coproducts. | |
| Dec 21, 2023 at 16:38 | comment | added | Maxime Ramzi | Cokernels always commute with filtered colimits, the point of exactness here is to guarantee that filtered colimits of monomorphisms are monomorphisms. But for each finite subset $F$, the map from the finite direct sum to the infinite product is a (split) monomorphism ! | |
| Dec 21, 2023 at 15:07 | comment | added | FShrike | @MaximeRamzi Ok, so following Daniel's point I realise $\bigoplus_i X_i$ is the filtered colimit of the finite sums=finite products $\prod_{i\in F}X_i$ where $F\subset J$ is finite; I know my claim about members holds for products simply by the universal property of products; if I wlog consider the case $\pi_j a=0\forall j$ I might want to factor $a$ through the objects of this filtered colimit and use exactness to say something about the cokernel of $a$ being some colimit of cokernels for the finite products but ! $a$ need not factor through, unless $Y$ is compact. Would you mind elaborating? | |
| Dec 21, 2023 at 14:27 | comment | added | Daniel Bruegmann | W.r.t the last question: Every colimit is a filtered colimit of finite colimits. In particular, direct sums are filtered colimits of finite direct sums. | |
| Dec 21, 2023 at 14:21 | comment | added | Maxime Ramzi | You're right that this is not true in general. It is true at least when filtered colimits in $A$ are exact, and I suspect the converse might hold, but I don't have a proof | |
| Dec 21, 2023 at 11:16 | history | asked | FShrike | CC BY-SA 4.0 |