Lemma 29.41.7. Suppose given a commutative diagram of schemes

$\xymatrix{ X \ar[rr] \ar[rd] & & Y \ar[ld] \\ & S & }$

with $Y$ separated over $S$.

1. If $X \to S$ is universally closed, then the morphism $X \to Y$ is universally closed.

2. If $X$ is proper over $S$, then the morphism $X \to Y$ is proper.

In particular, in both cases the image of $X$ in $Y$ is closed.

Proof. Assume that $X \to S$ is universally closed (resp. proper). We factor the morphism as $X \to X \times _ S Y \to Y$. The first morphism is a closed immersion, see Schemes, Lemma 26.21.10. Hence the first morphism is proper (Lemma 29.41.6). The projection $X \times _ S Y \to Y$ is the base change of a universally closed (resp. proper) morphism and hence universally closed (resp. proper), see Lemma 29.41.5. Thus $X \to Y$ is universally closed (resp. proper) as the composition of universally closed (resp. proper) morphisms (Lemma 29.41.4). $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).