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Apr 7, 2016 at 10:45 vote accept Saal Hardali
Apr 7, 2016 at 9:13 comment added Zhen Lin I don't deal with analytic spaces, but this kind of thing is considered in my thesis. Preservation of finite limits is not necessary if you are willing to give up having a geometric morphism.
Apr 7, 2016 at 9:06 answer added Simon Henry timeline score: 8
Apr 7, 2016 at 8:47 comment added Saal Hardali @SimonHenry Well that's super easy. So basically since $f$ sends zariski covers to analytic covers and preserves finite limits it defines the geometric morphism above. In particular now $f^*$ being a left adjoint preserves colimits so schemes are sent to analytic spaces. That's it? seems to good to be true...
Apr 7, 2016 at 8:22 comment added Simon Henry Oh wait ! I just realized you started with all finitely presented algebras, without any restriction. So your category does have all finite limits. So the flatness condition is equivalent to the fact those limit are preserved by the functor that attach the analytic space, which is easier to check than flatness.
Apr 7, 2016 at 8:18 comment added Simon Henry The existence of the geometric morphism follow from general non-sense: if your functor $f$ send covering to covering and is flat, then it defines a geometric morphism (see ncatlab.org/nlab/show/morphism+of+sites ). The flatness is not completely obvious as the categories involved don't have all finite limit, but I'm convinced it is not very hard to check. I am not sure how is defined the analytification functor in this framework so I cannot answer the question more, but maybe it has some kind of universal properties equivalent to the fact that it is indeed $f^*$ ?
Apr 7, 2016 at 7:44 history edited Saal Hardali CC BY-SA 3.0
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Apr 7, 2016 at 7:39 history asked Saal Hardali CC BY-SA 3.0