Let $f: X \rightarrow Y$ be a morphism of schemes of finite type over an algebraically closed field $k$. This means $X$, $Y$ are schemes of finite type over $k$ and $f$ is a $k$ -morphism. Let $M, N$ be the set of closed points in $X, Y$ respectively. Then it is clear that $f(M) \subset N$ and for any closed subset $C$ of $X$ we have $f(C) \cap N = f(C\cap M)$. So if $f$ is a closed map then the restriction $f|_M : M \rightarrow N$ is so. Is the converse true? We know that $f(C)$ is constructible subset of $Y$ for any closed set $C$ in $X$. Can I claim that $f(C)$ is closed in $Y$ if $f(C) \cap N$ is closed in $N$? I can see this if $f(C)$ is locally closed but not in general i.e. when $f(C)$ is union of finite number of locally closed sets.
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I found that this question has already been asked here http://math.stackexchange.com/questions/245317/is-a-morphism-between-schemes-of-finite-type-over-a-field-closed-if-it-induces-a?rq=1. I therefore close the question. – A.G Jun 18 '13 at 02:36