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Apr 13, 2017 at 12:58 history edited CommunityBot
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Apr 6, 2017 at 0:57 answer added Artur Jackson timeline score: 1
Mar 4, 2017 at 10:51 comment added Ingo Blechschmidt Exactly. Yes, it is a scheme since $f$ is a local homeomorphism: For any point of $\mathrm{Ét}(\mathcal{E})$ there is an open neighbourhood $U$ which is homeomorphic to some open subset $V$ of $X$, and $(f^{-1} \mathcal{O}_X)|_U$ corresponds to $(\mathcal{O}_X)|_V$ under this homeomorphism.
Mar 4, 2017 at 10:44 comment added მამუკა ჯიბლაძე For the latter you view $\operatorname{Ét}(\mathcal{E})$ as a ringed space with $f^{-1}\mathcal O_X$? Is it a scheme?
Mar 4, 2017 at 10:37 comment added Ingo Blechschmidt Yes, and afterwards we can equip $f_! f^{-1} \mathcal{O}_X$ with the structure of an $\mathcal{O}_X$-module. But we also directly use the versions of these functors for modules instead of abelian groups.
Mar 4, 2017 at 10:29 comment added მამუკა ჯიბლაძე Concerning $f_! f^{-1} \mathcal{O}_X$, where $f : \operatorname{Ét}(\mathcal{E}) \to X$ is the projection of the étalé space associated to $\mathcal{E}$: do you mean the functors $f_!$ and $f^{-1}$ acting on sheaves of abelian groups?
Mar 4, 2017 at 9:47 history asked Ingo Blechschmidt CC BY-SA 3.0