Timeline for Unbounded acyclic resolutions
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Dec 15, 2022 at 9:26 | comment | added | Jeremy Rickard | @NikolasKuhn That proof assumes exact countable products, which isn't the case for a general Grothendieck category. | |
| Dec 15, 2022 at 8:19 | comment | added | Nikolas Kuhn | It seems like this is shown in the proof of existence of K-injective resolutions [Stacks, Tag 090Y]. | |
| Dec 14, 2022 at 23:00 | history | edited | R. van Dobben de Bruyn | CC BY-SA 4.0 |
Added extra hypothesis suggested by Rickard's answer
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| Dec 14, 2022 at 22:22 | comment | added | R. van Dobben de Bruyn | @NikolasKuhn oh do I? I am aware of results in AB4* categories [Bökstedt–Neeman, Rmk. 2.3], but that doesn't apply here. There are results in the Stacks project (essentially due to Spaltenstein) under some technical hypotheses [Tag 0D62 and its corollaries], which apply in the other answer I gave. But I am wondering if I can remove these conditions when the $Z^i$ and $B^i$ are totally $F$-acyclic. | |
| Dec 14, 2022 at 19:25 | comment | added | Nikolas Kuhn | Maybe I'm misunderstanding something... Do you not know that the map to the homotopy limit of the truncations is an isomorphism in the setting that you're interested in (of a Grothendieck abelian category)? (I guess I'm suggesting that you already solved the question) | |
| Dec 14, 2022 at 18:10 | answer | added | Jeremy Rickard | timeline score: 6 | |
| Dec 14, 2022 at 13:29 | comment | added | R. van Dobben de Bruyn | @D.-C.Cisinski what I wrote is actually the condition that I mean, but you're right that the classical notion of $F$-acyclicity means $RF(T) = F(T)$. If $C^\bullet$ is a bounded below complex of $F$-acyclic objects in that sense, you don't get $RF(C^\bullet) = 0$ but $RF(C^\bullet) = F(C^\bullet)$. So I should have chosen a different name (totally $F$-acyclic?). I have rewritten the question with this improved terminology. But your version also sounds like a sensible question, and maybe more analogous to the classical story of $F$-acyclic resolutions. My question is a baby case of that. | |
| Dec 14, 2022 at 13:18 | history | edited | R. van Dobben de Bruyn | CC BY-SA 4.0 |
Changed the terminology
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| Dec 14, 2022 at 12:01 | comment | added | D.-C. Cisinski | Your notion of $F$-acyclicity is a little too strong it seems: since $F$ is left exact, we have $H^0(RF(T))=T$ for any $T$ in $\mathcal A$. Your definition of acyclicity seems to say that the complex $C$ is actually zero in each degree! We should define acyclicity of $T$ by saying $H^i(RF(T))=0$ for $i\neq 0$. Then we might want to know if $F(C)\simeq RF(C)$ whenever $Z^i(C)$ and $B^i(C)$ are $F$-acyclic for all $i$. | |
| Dec 14, 2022 at 11:20 | history | edited | R. van Dobben de Bruyn | CC BY-SA 4.0 |
Fixed another typoe
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| Dec 14, 2022 at 11:15 | history | edited | R. van Dobben de Bruyn | CC BY-SA 4.0 |
Made more precise where the spectral sequence argument fails
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| Dec 14, 2022 at 11:04 | history | edited | R. van Dobben de Bruyn | CC BY-SA 4.0 |
Fixed some typos
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| Dec 13, 2022 at 21:20 | history | edited | R. van Dobben de Bruyn | CC BY-SA 4.0 |
Fixed a typo and streamlined the question.
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| Dec 13, 2022 at 21:14 | comment | added | R. van Dobben de Bruyn | My question is close to part (2) of this question, but made more precise. The other question never received a very satisfying answer. | |
| Dec 13, 2022 at 21:00 | history | edited | R. van Dobben de Bruyn | CC BY-SA 4.0 |
Added a remark on why the hypotheses are what they are.
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| Dec 13, 2022 at 17:22 | history | asked | R. van Dobben de Bruyn | CC BY-SA 4.0 |