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 $S$.

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

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).
$\square$

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