Lemma 66.40.7. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $f : X \to Y$ be a morphism of algebraic spaces over $B$. If $X$ is universally closed over $B$ and $f$ is surjective then $Y$ is universally closed over $B$. In particular, if also $Y$ is separated and of finite type over $B$, then $Y$ is proper over $B$.

**Proof.**
Assume $X$ is universally closed and $f$ surjective. Denote $p : X \to B$, $q : Y \to B$ the structure morphisms. Let $B' \to B$ be a morphism of algebraic spaces over $S$. The base change $f' : X_{B'} \to Y_{B'}$ is surjective (Lemma 66.5.5), and the base change $p' : X_{B'} \to B'$ is closed. If $T \subset Y_{B'}$ is closed, then $(f')^{-1}(T) \subset X_{B'}$ is closed, hence $p'((f')^{-1}(T)) = q'(T)$ is closed. So $q'$ is closed.
$\square$

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