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edited Mar 20 2011 at 21:39

Karl Schwede
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Can't we base-change $f : X \to Y$ with $Z$ and obtain: $g : f^{-1}(Z) \to Z$? This also is a universal homeomorphism by construction, right? So now we have a universal homeomorphism to a regular scheme, but a regular scheme is weakly normal, see A. Andreotti and E. Bombieri, ``Sugli omeomorfismi delle varietà algebriche''.

Therefore $g$ is an isomorphism at least as long as $f^{-1}(Z)$ is reduced and the map $g$ is birational.

EDIT: My argument that $f^{-1}(Z)$ was reduced was junk. I shouldn't have tried to do math while on the run. But as long as $f^{-1}(Z)$ is reduced and $g$ is birationalreduced, then I think things are ok.

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edited Mar 20 2011 at 19:41

Karl Schwede
11.3k●1●14●35

Therefore $g$ is an isomorphism at least as long as $f^{-1}(Z)$ is reduced and the map $g$ is birational.

EDIT: My argument that $f^{-1}(Z)$ was reduced was junk. I shouldn't have tried to do math while on the run. But as long as $f^{-1}(Z)$ is and $g$ is birational reduced, then I think things are ok.

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edited Mar 20 2011 at 19:03

Karl Schwede
11.3k●1●14●35

Therefore $g$ is an isomorphism at least as long as $f^{-1}(Z)$ is reduced. But

EDIT: My argument that $f^{-1}(Z)$ has to be reduced, because if it isn't was reduced this would imply inseparability of $f$ at some (possibly generic) point of $Z$, which contradicts the universal homeomorphism of $f$.was junk. I shouldn't have tried to do math while on the runfor a couple hours. But as long as $f^{-1}(Z)$ is reduced, but did then I misunderstand the question?think things are ok.

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occurred Mar 20 2011 at 18:57

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answered Mar 20 2011 at 12:54

Karl Schwede
11.3k●1●14●35