Lemma 29.41.9. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of schemes over $S$. If $X$ is universally closed over $S$ and $f$ is surjective then $Y$ is universally closed over $S$. In particular, if also $Y$ is separated and locally of finite type over $S$, then $Y$ is proper over $S$.

**Proof.**
Assume $X$ is universally closed and $f$ surjective. Denote $p : X \to S$, $q : Y \to S$ the structure morphisms. Let $S' \to S$ be a morphism of schemes. The base change $f' : X_{S'} \to Y_{S'}$ is surjective (Lemma 29.9.4), and the base change $p' : X_{S'} \to S'$ is closed. If $T \subset Y_{S'}$ is closed, then $(f')^{-1}(T) \subset X_{S'}$ is closed, hence $p'((f')^{-1}(T)) = q'(T)$ is closed. So $q'$ is closed. This proves the first statement. Thus $Y \to S$ is quasi-compact by Lemma 29.41.8 and hence $Y \to S$ is proper by definition if in addition $Y \to S$ is locally of finite type and separated.
$\square$

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