The image of a quasi-compact algebraic space under a surjective morphism is quasi-compact.

Lemma 65.8.6. Let $S$ be a scheme.

1. If $X \to Y$ is a surjective morphism of algebraic spaces over $S$, and $X$ is quasi-compact then $Y$ is quasi-compact.

2. If

$\xymatrix{ X \ar[rr]_ f \ar[rd]_ p & & Y \ar[dl]^ q \\ & Z }$

is a commutative diagram of morphisms of algebraic spaces over $S$ and $f$ is surjective and $p$ is quasi-compact, then $q$ is quasi-compact.

Proof. Assume $X$ is quasi-compact and $X \to Y$ is surjective. By Definition 65.5.2 the map $|X| \to |Y|$ is surjective, hence we see $Y$ is quasi-compact by Properties of Spaces, Lemma 64.5.2 and the topological fact that the image of a quasi-compact space under a continuous map is quasi-compact, see Topology, Lemma 5.12.7. Let $f, p, q$ be as in (2). Let $T \to Z$ be a morphism whose source is a quasi-compact algebraic space. By assumption $T \times _ Z X$ is quasi-compact. By Lemma 65.5.5 the morphism $T \times _ Z X \to T \times _ Z Y$ is surjective. Hence by part (1) we see $T \times _ Z Y$ is quasi-compact too. Thus $q$ is quasi-compact. $\square$

## Comments (3)

Comment #1370 by Herman Rohrbach on

Suggested slogan: Surjective morphisms of algebraic spaces preserve quasi-compactness over the base.

Comment #5024 by Leo on

Should be "commutative diagram of spaces over Z" and "let Z be a scheme."

Comment #5260 by on

@#5024: No! Please see https://www.math.columbia.edu/~dejong/wordpress/?p=327

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