# Morphic Image of a Complete Variety

A variety $X$ is called separated if $\Delta_X = \{ (x,x) \mid x \in X \} \subset X \times X$ is closed in $X \times X$.

A variety $X$ is called complete if it is separated and for any other variety $Y$ and a close subset $Z \subset X \times Y$ we have that $\pi_Y(Z)$ is closed (here $\pi_Y : X \times Y \to Y$ is the canonical projection).

I want to prove:

If $X'$ is separated, $f : X \to X'$ is a morphism and $X$ is complete variety then $f(X) \subset X'$ is complete.

I know that a closed subvariety of a complete variety is complete. I also know that since $X'$ is separated then the graph $\Gamma_f = \{ (x, f(x) ) \mid x \in X \}$ is closed in $X \times X'$.

Have you shown that $f(X)$ is closed in $X'$? If not, do this first. Now you know that $f(X)$ is a variety (being closed in a variety). (Without this, the question doesn't quite make sense; we need to knowthat $f(X)$ is a variety to talk about it being complete.)

Next, show that if $X \to Z$ is a surjection of varieties and $X$ is complete, so is $Z$. (Just work directly with the definition. This is basically a simple exercise in topology, using that the preimage of closed under a continuous map is closed, and using that if $X \to Z$ is surjective, so is $X \times Y \to Z \times Y$ for any $Y$.)

• I was typing my answer while you posted yours. Didn't mean to double post! – Dori Bejleri Dec 19 '13 at 0:12

We expect this to be true since proper is the algebraic geometry version of compact and we know continuous images of compact sets are compact. In fact over $\mathbb{C}$, we could view this as the proof since $X/\mathbb{C}$ is proper if and only if it is compact in the analytic topology.

In general, suppose the base is $S$ (you could pick your favorite field but since you didn't specify I figured I'd do it in general). Then we will show $Z = f(X)$ is proper over $S$ by first showing $f(X)$ is separated over $S$ and that the morphism $f(X) \to S$ is universally closed.

For separated, by assumption $X'$ is separated over $S$. Consider the diagram

$$\require{AMScd} \begin{CD} Z @>>> Z \times_S Z \\ @VVV @VVV \\ {X'} @>>> {X' \times_S X'} \end{CD}$$

You can check this is a fiber product diagram. The bottom morphism is a closed immersion so its pullback is also a closed immersion and $Z$ is separated as required.

Next, for universally closed, first note that $f: X \to X'$ is proper. This is a general fact about proper morphisms that if we have $f:X \to X'$ is a morphism and $X$ is proper and $X'$ is separated then $f$ is proper. Thus it is closed.

Now we want to show that the structure morphism $Z \to S$ is universally closed. We know $X \to S$ is universally closed and $f$ is closed. $f$ is surjective onto $Z$ by definition so a subset of $Z$ is closed if and only if it is the image of a closed subset of $X$. Then for any $S$-scheme $U$, we have

$$X \times_S U \to Z \times_S U \to U$$

where the first morphism is the pullback of $f$ and the composition is the pullback of $X \to S$. By the discussion above, the first morphism is closed and surjective and the composition is closed so we can deduce the second morphism is closed. Thus $Z \to S$ is universally closed so $Z$ is proper over $S$.