The Stacks project

Lemma 63.9.8. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $f : X \to Y$ be a surjective universally closed morphism of algebraic spaces over $B$.

  1. If $X$ is quasi-separated, then $Y$ is quasi-separated.

  2. If $X$ is separated, then $Y$ is separated.

  3. If $X$ is quasi-separated over $B$, then $Y$ is quasi-separated over $B$.

  4. If $X$ is separated over $B$, then $Y$ is separated over $B$.

Proof. Parts (1) and (2) are a consequence of (3) and (4) for $S = B = \mathop{\mathrm{Spec}}(\mathbf{Z})$ (see Properties of Spaces, Definition 62.3.1). Consider the commutative diagram

\[ \xymatrix{ X \ar[d] \ar[rr]_{\Delta _{X/B}} & & X \times _ B X \ar[d] \\ Y \ar[rr]^{\Delta _{Y/B}} & & Y \times _ B Y } \]

The left vertical arrow is surjective (i.e., universally surjective). The right vertical arrow is universally closed as a composition of the universally closed morphisms $X \times _ B X \to X \times _ B Y \to Y \times _ B Y$. Hence it is also quasi-compact, see Lemma 63.9.7.

Assume $X$ is quasi-separated over $B$, i.e., $\Delta _{X/B}$ is quasi-compact. Then if $Z$ is quasi-compact and $Z \to Y \times _ B Y$ is a morphism, then $Z \times _{Y \times _ B Y} X \to Z \times _{Y \times _ B Y} Y$ is surjective and $Z \times _{Y \times _ B Y} X$ is quasi-compact by our remarks above. We conclude that $\Delta _{Y/B}$ is quasi-compact, i.e., $Y$ is quasi-separated over $B$.

Assume $X$ is separated over $B$, i.e., $\Delta _{X/B}$ is a closed immersion. Then if $Z$ is affine, and $Z \to Y \times _ B Y$ is a morphism, then $Z \times _{Y \times _ B Y} X \to Z \times _{Y \times _ B Y} Y$ is surjective and $Z \times _{Y \times _ B Y} X \to Z$ is universally closed by our remarks above. We conclude that $\Delta _{Y/B}$ is universally closed. It follows that $\Delta _{Y/B}$ is representable, locally of finite type, a monomorphism (see Lemma 63.4.1) and universally closed, hence a closed immersion, see √Čtale Morphisms, Lemma 40.7.2 (and also the abstract principle Spaces, Lemma 61.5.8). Thus $Y$ is separated over $B$. $\square$


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