Quicker way to show that the restriction to a open subvariety is again proper? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T02:01:31Z http://mathoverflow.net/feeds/question/116703 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116703/quicker-way-to-show-that-the-restriction-to-a-open-subvariety-is-again-proper Quicker way to show that the restriction to a open subvariety is again proper? Joachim 2012-12-18T14:03:24Z 2012-12-18T14:03:24Z <p>Dear all,</p> <p>Let $f: X \rightarrow Y$ be a morphism of projective varieties over $\mathbb{C}$. Also let $V \subset Y$ be a nontrivial open subvariety and set $U:= f^{-1}(V)$.</p> <p>I would like to show that $f|_{U} : U \rightarrow V$ is again proper.</p> <p>I have found a proof, by exploiting $U = X \times_Y V$ to show that $f_U$ is universally closed. It then follows. My question however is the following.</p> <p>From Hartshorne II 4.8(f), we know that properness is local on the base, i.e. a moprphism is proper if and only if the target can be covered by opens $V_i$ such that $f^{-1}(V_i) \rightarrow V_i$ is again proper.</p> <p>Question: Does that imply that this holds for <em>any</em> open $V \subseteq Y$, hence proving what i want in a way quicker way?</p> <p>I am aware that that many properties of morphisms of schemes are defined using an open cover, and that under some circumstances this is equivalent to demanding it for all open covers. However i have never found the time to look into that properly, so apologies if this is very easy.</p> <p>Thanks a lot!</p>