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Apr 16, 2020 at 23:44 answer added R. van Dobben de Bruyn timeline score: 6
Apr 16, 2020 at 16:44 comment added Harry Gindi @TimCampion Yeah, I added it in an edit.
Apr 16, 2020 at 16:30 answer added Pavel Čoupek timeline score: 3
Apr 16, 2020 at 16:10 comment added Tim Campion @HarryGindi Ah, thanks! I see that 4.8 is the equivalence of etale sites that you mentioned -- 4.4.1 is an affirmative answer to my Question 1 (in the current form). I think I initially missed your link -- maybe it was added as an edit to your comment?
Apr 16, 2020 at 16:02 comment added Harry Gindi @TimCampion This is proved as theorem 4.8 in Barwick's notes that I linked. It's also somewhere in SGA or EGA.
Apr 16, 2020 at 16:02 comment added Tim Campion @Anonymous Oh wow -- I was feeling pessimistic about this! I'm having difficulty convincing myself of either implication. I'd be grateful if you could elaborate (perhaps this would be easier in the form of an answer rather than a comment).
Apr 16, 2020 at 15:59 history edited Tim Campion CC BY-SA 4.0
refined question in light of insightful comments
Apr 16, 2020 at 15:51 comment added Anonymous Then etale maps are strongly formally etale (by the topological invariance of the etale site) and conversely under finite presentation assumption (by the infinitesimal lifting criterion). Not sure about the last question.
Apr 16, 2020 at 15:39 comment added Tim Campion @Anonymous Thanks, point well taken.So the answer to my main question as stated is no. I wonder if anything changes if one substitutes "universal homeomorphism" as Harry suggests...
Apr 16, 2020 at 15:16 comment added Anonymous Being "weakly nilpotent" is a really weak condition: any field extension gives such a map. Corresponding, being "strongly formally etale" is a ridiculously strong condition. For example, given a field $k$, a $k$-algebra domain $A$ is strongly formally etale over $k$ only when $A=k$.
Apr 16, 2020 at 15:02 comment added Harry Gindi A universal homeomorphism induces an equivalence of étale topoi. Barwick has some notes about this whole question: maths.ed.ac.uk/~cbarwick/papers/trig1.pdf
Apr 16, 2020 at 15:02 comment added Tim Campion @HarryGindi Thanks! What does the theorem of Grothendieck say?
Apr 16, 2020 at 15:00 comment added Harry Gindi I'm not the downvoter, but the notion of 'weakly nilpotent' that you have is too weak. You need the condition to be pullback stable. As soon as you do that, you get the notion of universal homeomorphism, and I think this becomes related to a theorem of Grothendieck and also to some open problems in anabelian geometry.
Apr 16, 2020 at 14:59 comment added Tim Campion @downvoter I'd welcome any criticism you may have.
Apr 16, 2020 at 14:39 history asked Tim Campion CC BY-SA 4.0