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Every closed immersion is a finite morphism. Therefore, restriction of a finite morphism to a closed subset is always a finite morphism itself. Can you give an example of quasi-projective varieties $X\subset Y$, $Z$ and a finite morphism $f:Y\to Z$ such that restriction $f:X\to f(X)$ is not finite? Same with Y -- projective?

PS. Sorry the original version of this question was hilariously stupid.

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You can always get the counterexample from your old question by setting Y=Z, setting f to be the identity map, and X an open subset of Y that isn't closed. –  Dinakar Muthiah Mar 10 '10 at 0:56
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I think you're being a bit hard on yourself using the term "hilariously stupid" here. –  S. Carnahan Mar 10 '10 at 2:20
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Thanks Dinakar -- I edited the question to reveal what I meant exactly. –  Paul Yuryev Mar 10 '10 at 4:37
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up vote 1 down vote accepted

Almost the same counterexample works. Take any non-closed (so non-finite) open immersion $U\hookrightarrow Z$. Then the trivial double cover $Z\sqcup Z\to Z$ is finite, but the restriction to $U\sqcup Z\to Z$ is not (but is still surjective).

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Thanks Anton! Will I get banned, if I ask for example where Y is irreducible? –  Paul Yuryev Mar 10 '10 at 15:09
    
I think the following example works in that case. Take $Y=\mathbb P^1$, $Z$ the nodal cubic, and $Y\to Z$ the normalization map (which is finite). Then take $X\subseteq Y$ to be the complement of one of the points lying over the node. Then $X\to Z$ is surjective but not finite. –  Anton Geraschenko Mar 10 '10 at 16:16
    
Oh, great! This is the example I was looking for. Even simpler: $V(y^2 - x) \to \mathbf{A}^1$ and now take X as a complement of $(1,1)$ in $Y$. So this shows there are (at least) two mechanisms how morphism can fail to be finite: if the preimage runs away to infinity'', or if one of the preimages disappears without merging''. –  Paul Yuryev Mar 10 '10 at 22:40
    
BTW, notice that if f(X) is open and X saturated wrt X then $f: X\to f(X)$ is finite. –  Paul Yuryev Mar 10 '10 at 22:41
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