# an equivalent statement of “morphisms of projective varieties are closed”

I am interested in seeing why the statement

(1) "If $Y$ is any variety and $Z$ a closed subset of $\mathbb{P}^n \times Y$, then the projection of $Z$ on $Y$ is closed."

implies the statement

(1) "A morphism of projective varieties takes closed sets to closed sets."

In particular, let $X,Y$ be projective varieties of $\mathbb{P}^m,\mathbb{P}^n$ respectively and let $f: X \rightarrow Y$ be a morphism. If the graph $G_f = \left\{ (x,f(x)), x \in X \right\} \subset \mathbb{P}^m \times Y$ was a closed subset of $\mathbb{P}^m \times Y$, then statement (1) would readily apply and $f(X)$ would be closed in $Y$ as desired. However, why need $G_f$ be closed in $\mathbb{P}^m \times Y$?

Reference: See Theorem 1 and above in Akhil Mathew's post here: http://amathew.wordpress.com/2010/10/23/a-projective-morphism-is-proper/#comments

PS: Thanks to Mariano for a correction.

• $X\times f(X)$ is not the graph of $f$. – Mariano Suárez-Álvarez Jun 30 '14 at 18:41
• Why the downvotes? – Manos Jun 30 '14 at 19:08

I'm not sure what you have available to you, but it might be helpful to note that there's a map $f \times \operatorname{id}\colon X \times Y \to Y \times Y$, under which the inverse image of the diagonal is the graph of $f$. So if $Y$ is separated then the graph is closed.
• @Manos I see. I'll try to think of another way to say it, but in algebraic geometry a morphism $f\colon X \to Y$ is separated if the associated diagonal morphism $X \to X \times_Y X$ which is the identity on each factor is a closed immersion. Happily, all projective varieties have this property. No worries about the votes -- it matters little to me. – Hoot Jun 30 '14 at 19:22
• So the graph of a projective variety $X$ under the morphism $f:X \rightarrow Y$ is always closed in the Zariski topology of $\mathbb{P}^n \times Y$? If yes, then this will satisfy me as an answer together with a reference (Hartshorne would do). – Manos Jun 30 '14 at 19:28