Suppose that $p \colon P \to X$ is a projective bundle. Is $p$ an open map ?
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1$\begingroup$ At least over $\mathbb{C}$, any flat map is open, see mathoverflow.net/questions/41158/… $\endgroup$– Francesco PolizziCommented Feb 19, 2013 at 12:20
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$\begingroup$ thanks.. i saw the references, just to be sure, does it also prove in the zariski topology (instead of complex topology) for the map P-->X over the field of complex numbers. $\endgroup$– johnCommented Feb 19, 2013 at 12:37
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2 Answers
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For any $X,Y$ the projection $X \times Y \to Y$ is an open map. Since being open is a local property you know this holds also for any fiber bundle and in particular any projective bundle.
answered Feb 19, 2013 at 12:28
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Isn't any fiber bundle $p: E \to B$ projection an open map? If $e \in E$ belongs to an open $U$, then there is a trivializing neighborhood $V_e$ of $p(e)$ so that $p^{-1}(V_e) \to V_e$ is isomorphic to a projection map $V_e \times F \to V_e$, which is certainly open. Hence $p(U \cap p^{-1}(V_e))$ is open. Since direct images preserve unions, we see that
$$p(U) = p(\bigcup_{e \in U} U \cap p^{-1}(V_e)) = \bigcup_{e \in U} p(U \cap p^{-1}(V_e))$$
is open.
answered Feb 19, 2013 at 12:32