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Feb 14, 2011 at 12:33 vote accept evgeniamerkulova
Feb 13, 2011 at 21:34 answer added Qing Liu timeline score: 7
Feb 13, 2011 at 21:31 comment added Ariyan Javanpeykar Ok so you're not (really) asking about affine morphisms because U doesn't need to be the inverse image of V. Again apologies. Thank you Qing Liu.
Feb 13, 2011 at 21:29 comment added Ariyan Javanpeykar Oops. Sorry. I guess I should have looked better.
Feb 13, 2011 at 21:23 comment added Ariyan Javanpeykar To see that you are asking about "affine morphisms" you could look at chap. II, exercise 5.17 of Hartshorne's Algebraic geometry. (Not completely sure about the exercise number because I don't have access to the book right now.)
Feb 13, 2011 at 21:23 comment added Qing Liu @Ariyan: if $Y$ is one point, any any non-empty affine open subset of $X$ maps surjectively to $Y$. So $f$ always satisfies the required property in this case.
Feb 13, 2011 at 21:12 comment added Ariyan Javanpeykar Take Y=Spec k, with k a field. Let f:X---> Y be a morphism. (It is automatically surjective.) Then f satisfies your condition if and only if X is affine. What you are asking for is the "relativization" of this. That is, your morphism f will satisfy your condition if it is "affine". Let me emphasize here that the morphism needs to be affine (and not the schemes necessarily.) Finite morphisms are affine. When they are surjective they are called (branched) covers. It is not true that every surjective affine morphism is finite. Consider for example the projection of A^n to A^1.
Feb 13, 2011 at 19:52 history edited evgeniamerkulova CC BY-SA 2.5
I ask if question become true with more hypotheses
Feb 13, 2011 at 19:16 answer added Kevin Ventullo timeline score: 6
Feb 13, 2011 at 18:25 history asked evgeniamerkulova CC BY-SA 2.5