Timeline for equivalence of Grothendieck-style versus Cech-style sheaf cohomology
Current License: CC BY-SA 4.0
21 events
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| Sep 15, 2021 at 17:03 | comment | added | Z. M | Supplementing the comment by @ACL, Petersen's note (arxiv.org/abs/2102.06927) explains the equivalence between singular and sheaf cohomology conceptually. | |
| Sep 4, 2019 at 21:48 | history | edited | Georges Elencwajg | CC BY-SA 4.0 |
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| Nov 2, 2018 at 14:44 | history | edited | Georges Elencwajg | CC BY-SA 4.0 |
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| Nov 2, 2018 at 14:43 | comment | added | Georges Elencwajg | Dear @Damien L., yes, for me too paracompact implies Hausdorff by definition. I thought this convention was universal but for clarity I have added Hausdorffness in point 4). | |
| Nov 2, 2018 at 3:56 | comment | added | Damien L | Dear Georges, ‘paracompact’ in French always means Hausdorff. In particular 4) is not true without the Hausdorff hypothesis. In fact, I don't know any useful theorem regarding non-Hausdorff paracompact spaces. | |
| Dec 4, 2016 at 21:37 | comment | added | Denis Nardin | To answer your (implicit) question, there is a definition of nonabelian $H^1(X;G)$ that does not use Čech cohomology: Just take the $\pi_1$ of the sheafification of the sheaf of pointed spaces $U\mapsto BG(U)$. I haven't checked the details but I think that the usual proof should imply that it coincides with the Čech $H^1$ for all spaces (that is, both compute the pointed set of torsors for $G$). | |
| Dec 4, 2016 at 20:36 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
Corrected spelling of "Wofsey"
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| Dec 4, 2016 at 20:32 | comment | added | Georges Elencwajg | Dear @ACL, thanks for the interesting reference. I find it a bit depressing that such an elementary result is technically so difficult to prove . The silver lining is that I have the pleasant fantasy that the author is a bright young student who doesn't yet know much advanced mathematics but is not afraid to tackle complicated situations with bare hands (not that I have the slightest idea whether sophisticated methods would be of any use here!) | |
| Dec 4, 2016 at 20:25 | comment | added | Eric Wofsey | @ACL: Thanks for mentioning that. I actually saw that a while ago (and wrote out an answer sketching the argument on the MSE question I linked to above), but forgot to mention it here. | |
| Dec 4, 2016 at 20:13 | comment | added | ACL | @EricWofsey There is a paper by Yehonatan Sella on arXiv, arxiv.org/abs/1602.06674v3, that deletes the paracompactness assumption from the comparison between singular and sheaf cohomology. (Moreover, semi-locally contractible seems to be enough.) I have not checked that paper though. | |
| May 22, 2016 at 19:40 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
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| May 22, 2016 at 16:53 | comment | added | Eric Wofsey | Wedhorn's proof seems to gloss over the exact same point where Ramanan implicitly needs to assume paracompactness, as pointed out in this question at MSE. Specifically, he says "one easily checks that for $U\subseteq X$ open one has $\mathscr{S}^n_R(U)=S^n(U,R)/S^n(U,R)_0$, but this seems to be not so easy... | |
| May 22, 2016 at 8:39 | comment | added | Georges Elencwajg | @Eric: I have added a reference at the end of my answer. | |
| May 22, 2016 at 8:38 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
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| May 22, 2016 at 4:31 | comment | added | Eric Wofsey | Do you have a specific reference for (1), without any paracompactness hypothesis? I could not find this result in Godement (though I may have missed it), and while it is asserted in Ramanan as Theorem 4.14 of Chapter 4, Ramanan's proof actually requires the space to be paracompact (notice that the proof of Proposition 4.12 relies on Proposition 1.14 of Chapter 1, which assumes paracompactness). | |
| Mar 30, 2014 at 20:57 | comment | added | Keenan Kidwell | Dear @Georges, I've seen a couple slightly (a priori) different definitions of locally contractible: the more natural one, in my opinion, that there is a base of open contractible sets, and then the (apparently weaker) one, that for each open $V\subseteq X$ and $x\in V$, there is an open $x\in U\subseteq V$ and a continuous map $F:U\times[0,1]\to V$ such that $F(u,0)=u$ and $F(u,1)=x$ for all $u\in U$ (so $U$ is "contractible to $x$ in $V$"). Which definition is being used in the comparison result you cite in (1)? | |
| Feb 26, 2013 at 11:19 | comment | added | Torsten Schoeneberg | Dear Georges, no need to worry for my soul, I'm a faithful Bourbakist. But the master has (justly) struck the dissenters with confusion here: mathoverflow.net/questions/4214/… and here: mathoverflow.net/questions/19312/… , so it is our task to care and help them. | |
| Feb 21, 2013 at 5:48 | comment | added | Georges Elencwajg | Dear Torsten: of course paracompact implies Hausdorff ! Beware Bourbaki's wrath if you think otherwise:-) | |
| Feb 1, 2013 at 14:15 | comment | added | Torsten Schoeneberg | Regarding 4) it should be mentioned that in Godement's book, a paracompact space is, by definition, Hausdorff (see II.3.2 there). | |
| Mar 15, 2011 at 7:00 | history | edited | Anton Geraschenko | CC BY-SA 2.5 |
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| Nov 5, 2009 at 8:40 | history | answered | Georges Elencwajg | CC BY-SA 2.5 |