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Let $F: X \to X$ to be an endomorphism of scheme $X$, which is additionally assumed to induce an universal homeomorphism on the underlying topological space $| X|$. Then it is known that this induces an selfequivalence of the etale topos $\mathbf{Sh}(X_{\text{ét}})$ in the sense that adjoint functors $F_*\!:\mathbf{Sh}(X_{\text{ét}})\leftrightarrows\mathbf{Sh}(X_{\text{ét}}):\!F^*$ are even inverse.

Now it is known that this has as consequence that induced $F^*$ acts trivially ($=$ literally as identity, and not as, say, a certain isom) on the étale cohomology groups $H_{\text{ét}}^i(X, \mathcal{F})$ (SGA 5, Exp. XV, §2, Prop. 2(c); the idea is also elaborated for case $F$ beeing the absolute Frobenius by Remy van Dobben de Bruyn in comments below the answer here ; see also p. 89 of these notes on étale cohomology by Brian Conrad).

#EDIT: As Piotr Achinger elaboreted in the comments the initial question below is badly posed, as the assumptions on $F:X \to X$ above in that generality are even not sufficient to assure that $F$ even acts on the cohomology groups; one needs additionally to impose a kind of equivariance condition for each involved sheaf as additional datum to make such action welldefined, which the absolute Frobenius - which inspires this question - intrinsically has, but not every endomorphism $F$ as above assumed.

So the initially posed question & explanations below can be ignored; (I leave it undeleted in brackets for simplifying to keep better track the hostory where the original motivation came from even as remarked the original question flaws)

The modified QUESTION I would like to pose instead with same assumptions on $F$ as above is how "far" is the difference for $F$ as above to induce equivalence between étale sheaves and just Zariski sheaves in dependence on if we require $F$ to be a "universal homeomorphism" or "just a homeomorphism"?

Namely, the quoted statement above from SGA5 states that if $F$ is universally homeomorphism, then this induces equivalence on étale topos. Now if we pose weaker assumption $F$ to be only homeomorphism, does it still induce an equivalence of sheaves but maybe on a coarser site than étale, eg Zariski, or maybe something finer intermediate?
Can it be made precise to "which amount" of sheaf theoretic/topos theoretic information the assumption $F$ to be universally homeomorphism instead of only beeing homeomorphism is "finer" in the sense of reflecting between sheaves on which site $F$ induces an equivalence?

Expectation/intuitive (...probably wrong?) picture: $F$ universally homeo vs vs homeo induces equivalence of sheaves on étale site vs Zariski (or maybe something finer between étale and Zariski)?

[FROM HERE THE ORIGINALLY posed flawed question & considerations (can be ignored): What can be said about the induced action by $F$ on Zariski cohomology $H^i_\text{zar}(X, -)$ instead of étale if we weaken the assumption on $F$ from universal homeomorphism to only homeomorphism?
As étale topos is finer than Zariski topos, under above "strong" assumptions on $F$ ($=$ $F$ be universal homeom) the equivalence by $F^*$ above for étale topos implies equiv on Zariski topoi, so once $F$ is universal homeom, identical argument shows that $F$ acts also on Zariski cohomology trivially.

But the question is, can the assumption on $F$ be weakened from universal homeomorphism with the "price" we are ready to pay be that we haven't any more a statement as above for triviality of action by $F$ on étale cohomology, but maybe passing to coarser topology we can say something interesting about induced action by $F$ on Zariski cohomology? (E.g., is it trivial, or at least controllable in appropriate sense?)

Note, that for any morphism $G: X \to Y$ between sheaves which is a homeomorphism (not necessarily universal homeo(!)), that $G_*$ still gives rise to equivalence of cats $\mathbf{Sh}(X_{\text{zar}})\to \mathbf{Sh}(Y_{\text{zar}})$ (a priori so far I know this holds only for the Zariski site if I'm not confusing something; see somewhere in Mac Lane and Moerdijk's "Sheaves in Logic").
(…btw is this something "Zariski site specific" or does it hold also for reasonable refinements? Invoking the quoted result above about étale site, it seems without assuming additionally universality.)

Especially such equivalence implies that just $G$ being a homeomorphism alone (especially a pure "topological" property, although one should still keep in mind that $G$ has to be a schematic morphism; otherwise refined constructions could steer into serious troubles…) suffice to assure that already the cohomology groups $H^i(X, \mathcal{F})$ and $H^i(Y, G_* \mathcal{F})$ are somehow isomorphic.

But coming back to case $Y=X$ and $G=F$: The question is which information about $F$ we need to assure that $F$ not only acts as an isomorphism of the cohomology groups, but even as the identity?

Above on étale site it suffice to require that $F$ is universal homeo, and not just homeo. But if we restrict to coarser Zariski site, how far can the assumption on $F$ to be universal homeom to be weakened to still have that $F$ acts as identity on cohomology groups?

So the key issue in my concern is how the difference between the assumptions on $F$ to be universal homeomorphism" vs "just homoemorphism" affects the statement about the action by $F$ on Zariski cohomology groups.]

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    $\begingroup$ 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. $\endgroup$ Commented Jun 16 at 11:50
  • $\begingroup$ @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. $\endgroup$
    – user267839
    Commented Jun 16 at 12:04
  • $\begingroup$ @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? $\endgroup$
    – user267839
    Commented Jun 16 at 12:09
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    $\begingroup$ 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. $\endgroup$ Commented Jun 16 at 12:09
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    $\begingroup$ 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. $\endgroup$ Commented Jun 16 at 23:43

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