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 $Y$.)