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Let $X$(resp. $Y$) be a scheme of finite type over a field $k$. Let $f\colon X \rightarrow Y$ be a closed morphism. Let $X_0$(resp. $Y_0$) be the set of closed points of $X$(resp. $Y$). Then $f$ induces a map $f_0\colon X_0 \rightarrow Y_0$(right?). We consider $X_0$(resp. $Y_0$) as a subspace of $X$(resp. $Y$). Is $f_0$ a closed map?

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1 Answer 1

Let $W$ be closed in $X_0$. Then $W = C\cap X_0$ for some closed $C$ of $X$. The image of $W$ via $f_0$ is $f_0(W) = f_0(C\cap X_0) = f(C) \cap Y_0$. This is closed in $Y_0$ by the fact that $f(C)$ is closed in $Y$.

Note that closed points of a finite type $k$-scheme correspond to algebraic points. This point of view makes it easy to see that $f:X\to Y$ induces a map $f_0:X_0\to Y_0$.

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The equality $f_0(C\cap X_0)=f(C)\cap Y_0$ is true but needs a proof. –  user18119 Nov 26 '12 at 21:51
A proof of the above equality can be found at math.stackexchange.com/questions/245317 –  user18119 Nov 27 '12 at 13:03
Thank you QiL! I only just read your comment. –  Harry Nov 27 '12 at 20:55
@Harry If you add a proof of the above equality, I will accept your answer. –  Makoto Kato Nov 28 '12 at 18:43

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