most recent 30 from http://mathoverflow.net 2013-05-20T22:15:17Z http://mathoverflow.net/feeds/question/17678 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/17678/example-of-restriction-of-a-finite-morphism-which-is-not-finite Paul Yuryev 2010-03-10T00:51:23Z 2010-03-10T06:18:13Z

<p>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?</p> <p>PS. Sorry the <a href="http://mathoverflow.net/questions/17658/example-of-inclusion-which-is-not-a-finite-morphism" rel="nofollow">original version</a> of this question was hilariously stupid.</p>

http://mathoverflow.net/questions/17678/example-of-restriction-of-a-finite-morphism-which-is-not-finite/17701#17701 Anton Geraschenko 2010-03-10T06:18:13Z 2010-03-10T06:18:13Z

<p>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).</p>