most recent 30 from http://mathoverflow.net 2013-05-20T17:54:31Z http://mathoverflow.net/feeds/user/4738 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18720/morphism-of-schemes-that-is-closed-at-topological-space-level saurav 2010-03-19T08:53:24Z 2010-03-19T13:28:28Z

<p>Suppose $X\stackrel f\to Y$ be a morphism of finite type $k$-schemes, where $k$ is a field; for the time being let me say that $k$ is algebraically closed.</p> <p>Then one knows that $f$ takes $k$-valued points to $k$-valued points. Now suppose the scheme morphism is a closed map i.e. takes closed subsets to closed subsets. Take the restriction of $f$ to closed points i.e. $k$-valued points. We get a map of topological spaces $\tilde f:X_0\to Y_0$ where $X_0$, $Y_0$ are the subsets of closed points. We have topology on $X$ and $Y$ since they are schemes, so we get induced topology on $X_0,Y_0$ also. The question is, whether the restriction map $\tilde f$ is still a closed map.</p> <p>I think it is easy for $X$ and $Y$ affine. But especially when $X$ is not affine, I have no idea.</p> <p>Best regards, Saurav</p>

http://mathoverflow.net/questions/18720/morphism-of-schemes-that-is-closed-at-topological-space-level/18725#18725 saurav 2010-03-19T13:18:28Z 2010-03-19T13:18:28Z

But $f(E_c)[=f(E\cap X_c)]$ may not be equal to $f(E)_c[=f(E)\cap Y_c]$. To say that $f(E_c)$ is closed in $Y_c$, we probably need this equality.