Timeline for Is every "nice" abelian category with enough projectives an additive presheaf category?
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24 events
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| Feb 1, 2018 at 3:44 | comment | added | user40276 | I'm not sure if you are aware about it win.ua.ac.be/~wlowen/pdf%20papers/JPAA-GabrielPopescu.pdf . | |
| Jan 18, 2018 at 2:42 | history | edited | Tim Campion | CC BY-SA 3.0 |
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| Jan 9, 2018 at 20:13 | answer | added | Tim Campion | timeline score: 0 | |
| Jan 9, 2018 at 9:09 | comment | added | მამუკა ჯიბლაძე | @QiaochuYuan I had in mind "set-theoretic" presheaves - just plain functors from a plain category to the category of abelian groups | |
| Jan 9, 2018 at 8:30 | comment | added | Qiaochu Yuan | მამუკა ჯიბლაძე: Yes, that's what I'm talking about too; the category of modules over $\mathbb{F}_2^{\mathbb{N}}$ is the category of (linear) presheaves, valued in abelian groups, over the one-object linear category with endomorphism ring $\mathbb{F}_2^{\mathbb{N}}$... | |
| Jan 9, 2018 at 7:54 | comment | added | მამუკა ჯიბლაძე | @QiaochuYuan Sorry I should be more clear. What I said applies to presheaves with values in abelian groups (more generally, in modules over rings with atomic Pierce spectra). However I must confess I am confused myself by the last comment of the OP. | |
| Jan 9, 2018 at 6:41 | comment | added | Qiaochu Yuan | @მამუკა ჯიბლაძე: I'm also confused by your claim. Take the category of modules over $\mathbb{F}_2^{\mathbb{N}}$. I believe every projective is decomposable, but this is still a module category. | |
| Jan 8, 2018 at 18:24 | comment | added | Tim Campion | @მამუკაჯიბლაძე Why do the indecomposable projectives generate in an additive presheaf category? For example, suppose I start with a representable $X$ (which is projective), and decompose it into successively smaller retracts. If this process terminates after finitely many steps, then I can reconstruct $X$ via a direct sum. But if it takes infinitely many steps, I don't see how to get $X$ as a sum of indecomposables. | |
| Jan 8, 2018 at 17:29 | comment | added | მამუკა ჯიბლაძე | I would say this is an obstruction "in a different direction". In any presheaf category, the family of indecomposable projectives generates, in particular, every projective is a coproduct of indecomposables. In $\mathsf{Ab}^{\mathbb N}$, these are $(0,...,0,0,\mathbb Z,0,0,...)$. An example of enough projectives without any indecomposables is given, I believe, by sheaves over an atomless complete Boolean algebra. | |
| Jan 8, 2018 at 17:28 | comment | added | Tim Campion | Er -- no, under my guessed definition for "enough indecomposable projectieves", $\mathsf{Ab}^\mathbb{N}$ does have enough of them. So I must be wrong. | |
| Jan 8, 2018 at 17:21 | history | edited | Tim Campion | CC BY-SA 3.0 |
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| Jan 8, 2018 at 17:16 | comment | added | Tim Campion | @მამუკაჯიბლაძე What does "enough indecomposable projectives" mean? For instance, even $\mathsf{Ab}$ doesn't have the property that every object admits an epimorphism from an indecomposable projective. Does it mean that there is a set $I$ of indecomposable projectives such that every object admits an epimorphism from a coproduct of objects from $I$? If so, I suppose Qiaochu's example of $\mathsf{Ab}^\mathbb{N}$ is an example with enough projectives but not enough indecomposable projectives. | |
| Jan 8, 2018 at 17:09 | comment | added | მამუკა ჯიბლაძე | You need enough indecomposable projectives. I believe there are examples with every projective nontrivially decomposable into a coproduct. | |
| Jan 8, 2018 at 17:07 | history | edited | Tim Campion | CC BY-SA 3.0 |
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| Jan 8, 2018 at 17:02 | history | edited | Tim Campion | CC BY-SA 3.0 |
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| Jan 8, 2018 at 16:56 | vote | accept | Tim Campion | ||
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| Jan 7, 2018 at 19:49 | answer | added | Qiaochu Yuan | timeline score: 15 | |
| Jan 7, 2018 at 17:07 | history | edited | Tim Campion | CC BY-SA 3.0 |
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| Jan 7, 2018 at 16:59 | history | edited | Tim Campion | CC BY-SA 3.0 |
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| Jan 7, 2018 at 16:52 | history | edited | Tim Campion | CC BY-SA 3.0 |
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| Jan 7, 2018 at 16:23 | history | edited | Tim Campion | CC BY-SA 3.0 |
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| Jan 7, 2018 at 15:49 | comment | added | Tim Campion | Yes. And I'm aware that any cocomplete abelian category with a compact projective generator is a module category. Am I missing something? | |
| Jan 7, 2018 at 15:36 | comment | added | abx | Are you aware of Mitchell's embedding theorem? | |
| Jan 7, 2018 at 15:14 | history | asked | Tim Campion | CC BY-SA 3.0 |