Lemma 66.8.3. Let $S$ be a scheme. If $f : X \to Y$ is a quasi-compact morphism of algebraic spaces over $S$, then the underlying map $|f| : |X| \to |Y|$ of topological space is quasi-compact.

**Proof.**
Let $V \subset |Y|$ be quasi-compact open. By Properties of Spaces, Lemma 65.4.8 there is an open subspace $Y' \subset Y$ with $V = |Y'|$. Then $Y'$ is a quasi-compact algebraic space by Properties of Spaces, Lemma 65.5.2 and hence $X' = Y' \times _ Y X$ is a quasi-compact algebraic space by Definition 66.8.2. On the other hand, $X' \subset X$ is an open subspace (Spaces, Lemma 64.12.3) and $|X'| = |f|^{-1}(|X'|) = |f|^{-1}(V)$ by Properties of Spaces, Lemma 65.4.3. We conclude using Properties of Spaces, Lemma 65.5.2 again that $|X'|$ is a quasi-compact open of $|X|$ as desired.
$\square$

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