morphism of schemes that is closed at topological space level 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.
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.
I think it is easy for $X$ and $Y$ affine. But especially when $X$ is not affine, I have no idea.
Best regards,
Saurav
 A: I think you can just reduce to the case of  of $X,Y$ affine by covering $Y$ by open affines.
Alternatively, the sets $X_{cl}, Y_{cl}$ of closed (i.e., $k$-valued) points in $X,Y$ map quasi-homeomorphically onto $X,Y$ (cf. EGA 0.3).  Thus a constructible set $E \subset X$  is closed iff $E \cap X_{cl}$ is closed, and similarly for $Y$.  Thus by Chevalley's theorem and this fact, it follows that $\tilde{f}$ is closed.  In detail, given a closed $E_c \subset X_{cl}$, there is a unique constructible $E \subset X$ with $E \cap X_{cl} = E_c$ (by quasi-homeomorphism).  Then $f(E)$ is closed in $Y$, so $f(E_c)$ must be closed in $Y_{cl}$.
A: Dear saurav, yes the restriction  is closed and you don't  have at all to assume $k$ algebraically closed. 
Reminder: For a scheme $S$ of finite type over $k$, the set $T_0$ of closed points of $T$ is very dense in $T$, i.e. dense in every closed subset of $T$ .
Here then is the statement you need :
Let $f:X\to Y$ be a closed morphism between schemes of finite type over a field $k$. Then the restriction $f_0:X_0\to Y_0$  to the subspaces of respective closed points is closed.
Proof:
1) We may assume $X$ and $Y$ reduced. We have to show that, for $F$ closed in $X$, the subset  $f(F\cap X_0)$ is closed in $Y_0$. By endowing $F$ and $f(F)$ with their reduced  scheme structure we can assume $F=X, f(F)=Y$ and we have reduced (!) the problem to showing that $Y_0=f(X_0)$ : call this "closed surjectivity". 
2) Here is why "closed surjectivity" is  true. Take any closed point $y_0\in Y_0$. It has a residual field $\kappa (y_0)$ which is a finite extension of $k$.The fibre $Z=f^{-1}(y_0)$ is a closed non-empty subscheme of $X$ (recall that after our reduction  $f$ is assumed surjective!).
But $Z$ is also a $\kappa (y_0)$-scheme of finite type. Hence it has a closed point $x_0$, since those closed points are even very dense in $Z$. Since $Z$ is closed in $X$, $x_0$ is closed also in $X$ i.e. $x_0\in X_0$. So we have shown "closed surjectivity" and everything is proved.
(Everything I used is contained in EGA I)
