Timeline for Homeomorphic endomorphism of schemes inducing equivalence of sheaves
Current License: CC BY-SA 4.0
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| Jun 16 at 23:43 | comment | added | R. van Dobben de Bruyn | The Zariski sheaves only depend on the underlying topological space of a scheme, so it is indeed true that a homeomorphism induces an equivalence of topoi. The reason universal homeomorphisms induce equivalences on étale topoi is roughly that they moreover induce purely inseparable extensions on residue fields [Tags 01S3 and 01S4], so the 'Galois direction is unchanged' as well. | |
| Jun 16 at 14:18 | history | edited | user267839 | CC BY-SA 4.0 |
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| Jun 16 at 14:12 | history | edited | user267839 | CC BY-SA 4.0 |
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| Jun 16 at 13:41 | history | edited | user267839 | CC BY-SA 4.0 |
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| Jun 16 at 13:34 | history | edited | LSpice | CC BY-SA 4.0 |
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| Jun 16 at 13:21 | comment | added | user267839 | sheaves or only Zariski sheaves? So short does for $F$ the discrepance between beeing universally homeomorphism vs just homeomorphism amount to discrepance if such $F$ induces equiv between étale sheaves or only Zariski sheaves? Or do I missing here the point? | |
| Jun 16 at 13:18 | comment | added | user267839 | Another point concerning generally the equivalence of the sheaf topoi above: The quoted statement from SGA5 requires in order to have such equivalence of étale topoi induced by adjoints $F_*\!:\mathbf{Sh}(X_{\mathrm{ét}})\leftrightarrows\mathbf{Sh}(X_{\mathrm{ét}}):\!F^*$ that $F$ is universally homeo. Can this latter assumption be weakened to just homeomorphism ( ie not neccess universal) if we want only to have equivalence of topoi of Zariski sheaves instead? Is this "universality" assumption presisely the "pivotal" point between if we want that $F$ induce equiv of topoi of étale | |
| Jun 16 at 12:48 | comment | added | user267839 | @PiotrAchinger: on the affiness of $F$ under assumption that $| F|$ is homeo: is there a say "standard argument" why preimage of an affine open $U \subset X$ under such $F$ should be still affine? The only one I know sor far is via Serre's cohomological characterisation of affineness via vanishing of $H^j(U, A), j>0$ for quasicoh $A$, but it appears to me to be a bit "around the corner". Is there any "conceptional" way to see the affine of such preimage of affine subscheme? | |
| Jun 16 at 12:13 | comment | added | user267839 | @PiotrAchinger: I see the problem, so it's much Frobenius specific issue then I thought, due to neccessarity of this equivariant structure as additional data which in general not given under the assumptions I above posed but Frobenius carries it intrinsically. | |
| Jun 16 at 12:09 | comment | added | Piotr Achinger | Even worse, $F$ does not a priori act on the cohomology of some sheaf $\mathcal{F}$ (unless e.g. $\mathcal{F}$ is constant). Rather, it induces a map $F^* \colon H^*(X, \mathcal{F}) \to H^*(X, F^* \mathcal{F})$ between two different groups. To compare it with the identity, you need an "equivariant structure" i.e. some choice of isomorphism $F^*\mathcal{F}\to \mathcal{F}$. In case of the Frobenius, such a canonical structure is described in the paragraph preceding the result in SGA5 you cite. P.S. $F$ is affine since the preimage of an affine open $U$ is $U$, which is affine. | |
| Jun 16 at 12:09 | comment | added | user267839 | @PiotrAchinger: let assume we want $F$ to be identity on top space as you suggested. But why it is then trivial from Zariski topology? Is $F$ then neccessarily affine? | |
| Jun 16 at 12:04 | comment | added | user267839 | @PiotrAchinger: yes that's a good catch. Initially I indeed $F$ to be any morphism which induces a homeom on underlying topology, but your example shows that overlooked that the claim in 2. paragraph is plainly wrong. Yes thanks, the assumption with $F$ identity on underlying top space would had been more resonable as this is exactly "modeled" on the case with absolute Frobenius. Then everything clear. | |
| Jun 16 at 11:51 | history | edited | user267839 | CC BY-SA 4.0 |
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| Jun 16 at 11:50 | comment | added | Piotr Achinger | I disagree with the statement in the second paragraph: consider the case when $F$ is an isomorphism. Certainly not every automorphism induces the identity on cohomology! Did you mean to assume that $F$ is the identity on the underlying space? But then, the case of the Zariski topology is trivial. | |
| Jun 16 at 11:48 | history | edited | user267839 | CC BY-SA 4.0 |
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| Jun 16 at 11:46 | history | edited | YCor | CC BY-SA 4.0 |
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| Jun 16 at 11:39 | history | edited | user267839 | CC BY-SA 4.0 |
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| Jun 16 at 11:24 | history | asked | user267839 | CC BY-SA 4.0 |