Lemma 66.40.6. Let $S$ be a scheme. Consider a commutative diagram of algebraic spaces

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

over $S$.

If $X \to B$ is universally closed and $Y \to B$ is separated, then the morphism $X \to Y$ is universally closed. In particular, the image of $|X|$ in $|Y|$ is closed.

If $X \to B$ is proper and $Y \to B$ is separated, then the morphism $X \to Y$ is proper.

**Proof.**
Assume $X \to B$ is universally closed and $Y \to B$ is separated. We factor the morphism as $X \to X \times _ B Y \to Y$. The first morphism is a closed immersion, see Lemma 66.4.6 hence universally closed. The projection $X \times _ B Y \to Y$ is the base change of a universally closed morphism and hence universally closed, see Lemma 66.9.3. Thus $X \to Y$ is universally closed as the composition of universally closed morphisms, see Lemma 66.9.4. This proves (1). To deduce (2) combine (1) with Lemmas 66.4.10, 66.8.9, and 66.23.6.
$\square$

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