Timeline for Can one characterize those sheaves which have Hausdorff etale spaces?
Current License: CC BY-SA 3.0
9 events
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| Jul 19, 2012 at 9:23 | comment | added | David Carchedi | @Flippo: No worries whatsoever :). | |
| Jul 19, 2012 at 2:06 | comment | added | Filippo Alberto Edoardo | @ David: Ok, I am sorry, I read your answer better. | |
| Jul 18, 2012 at 12:13 | comment | added | David Carchedi | @Filippo: That is why I am asking though. Sometimes a sheaf is Hausdorff, even when the underlying space isn't. I'd like to understand precisely when this happens. | |
| Jul 18, 2012 at 10:57 | history | edited | David Carchedi | CC BY-SA 3.0 |
added 1053 characters in body; deleted 6 characters in body
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| Jul 18, 2012 at 3:14 | comment | added | Filippo Alberto Edoardo | Actually, as Serre observes in FAC, chapter 1, section 1 it might happen that $E(F)$ is not separated although $X$ is. | |
| Jul 17, 2012 at 23:14 | comment | added | David Carchedi | Thanks Mike. I'll have a look. I wrote this up rather hastily because my collaborator is visiting me. I'll have a think and revise this answer. | |
| Jul 17, 2012 at 20:21 | comment | added | Mike Shulman | And actually, doesn't your condition basically amount to saying that the diagonal $E(F) \to E(F) \times_X E(F)$ is closed, i.e. that the geometric morphism $Sh(X)/F \to Sh(X)$ is separated? Which, given that $Sh(X)$ is a separated topos, is equivalent to $Sh(X)/F$ being separated by B3.2.25. | |
| Jul 17, 2012 at 20:19 | comment | added | Mike Shulman | This can't possibly be right unless $X$ is Hausdorff, since the terminal sheaf always satisfies your condition, but its etale space is just $X$. I think the flaw in your proof is that $W$ might contain $x$ or $y$, so that $f(\tilde{U})$ and $g(\tilde{V})$ need not contain $\tilde{x}$ and $\tilde{y}$ respectively. | |
| Jul 17, 2012 at 17:05 | history | answered | David Carchedi | CC BY-SA 3.0 |