• A Variety $X$ is complete iff for all other varieties $Y$, the projection map $X \times Y \to Y$ is closed.
• Intuitively, given equations of the form $f(x, y) = 0$ cutting a closed subset of $X \times Y$, one can always 'eliminate' the dependency on $y$ to get a new equation $g(y) = 0$.
• Alternatively, we can write down the image as $\{ y | \exists x, f(x, y) = 0 \}$.
• So we are told that we can quantifier eliminate to get a different equation $\{ y | g(y) = 0 \}$(mild lie, we need a family of equations $g_i$ and whatnot, but still, that's the idea.

§ Proof strategy

• We will show that $\mathbb P^n$ is complete, which will imply that a projective variety $X$ is complete, by including it