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Mar 31, 2011 at 17:08 answer added Arnav Tripathy timeline score: 5
Mar 31, 2011 at 15:52 comment added user5117 @Sandor: Right, that's what I meant (but expressed badly). In his answer he clarifies that this is indeed what is meant.
Mar 31, 2011 at 14:49 comment added Sándor Kovács @Francesco: ivor probably means that there exists an open set $V\subset Y$ such that $f$ is regular (i.e., defined) on $f^{-1}(V)$.
Mar 31, 2011 at 13:06 history edited user5117
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Mar 31, 2011 at 13:03 comment added user5117 Dear Francesco, I think the point of the question (which was not immediately clear to me) is that f should have some proper fibres.
Mar 31, 2011 at 12:58 answer added ivor timeline score: 2
Mar 31, 2011 at 12:50 comment added Francesco Polizzi If $f\colon X \to Y$ is dominant, then there are always non-empty, open subsets $U \subset X$ and $V \subset Y$ such that the restriction $f \colon U \to V$ is a regular map. Why are you requiring $Y$ non-uniruled?
Mar 31, 2011 at 12:04 comment added user5117 Dear Ivor: it might be useful to give some motivation for the question. For instance, do you have a reason to think it is true?
Mar 31, 2011 at 9:00 history asked ivor CC BY-SA 2.5