Surjective implies local affine surjective? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T04:48:40Z http://mathoverflow.net/feeds/question/55333 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/55333/surjective-implies-local-affine-surjective Surjective implies local affine surjective? evgeniamerkulova 2011-02-13T18:25:35Z 2011-02-13T21:46:26Z <p>Take scheme morphism $f: X\to Y$ and suppose $f$ surjective. If $y \in Y$ can one find affine open $V \subset Y$ containing $y$ and affine open $U \subset X$ such $f(U) = V$ ? Thank you.</p> <p>Later: Very good answer of Kevin shows it is not true. Is there hypothese which make it true ? For example $X$ irreducible and/or $f$ faithfuly flat ?</p> http://mathoverflow.net/questions/55333/surjective-implies-local-affine-surjective/55337#55337 Answer by Kevin Ventullo for Surjective implies local affine surjective? Kevin Ventullo 2011-02-13T19:16:35Z 2011-02-13T19:16:35Z <p>Consider the disjoint union Spec$(\mathbb{Q})\coprod_{p}$Spec$(\mathbb{F}_p)$ with its canonical map to Spec$(\mathbb{Z})$. This is bijective on points, but the preimage of any open in Spec$(\mathbb{Z})$ won't even be compact, much less affine.</p> http://mathoverflow.net/questions/55333/surjective-implies-local-affine-surjective/55346#55346 Answer by Qing Liu for Surjective implies local affine surjective? Qing Liu 2011-02-13T21:34:40Z 2011-02-13T21:46:26Z <p>If $f$ is open (e.g. $f$ finite type and flat over noetherian $Y$), then your condition is trivially satisfied: let $V'$ be any affine open neighborhood of $y$ and let $U'$ be an affine open subset of $X$ such that $y\in f(U')\subseteq V'$. Take a principal open subset $V'_h$ such that $y\in V'_h\subseteq f(U')$, then $V:=V'_h$ and $U:=U'_h$ are what you want. </p> <p>A counterexample with $X$ irreducible and $f$ projective : consider $Y$ the affine plan, $y$ the origin and $f : X\to Y$ the blowing-up of $y$. For any affine open subset $V$ containing $y$, $f^{-1}(V) \to V$ is the blowing-up of $y$. If $f(U)=V$, then the complement of $U$ in $f^{-1}(V)$ is finite because $f$ is an isomorphism out of $y$. By Zariski's extension theorem, $O_X(U)=O_X(f^{-1}(V))=O_Y(V)$. This is impossible as $U$ is affine, because $U$ would be the image of a section $V\to f^{-1}(V)$, hence closed in (and then equal to) $f^{-1}(V)$.</p>