Lemma 101.3.6. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks representable by algebraic spaces. Then the following are equivalent

1. $f$ is quasi-separated,

2. $\Delta _ f$ is quasi-compact, or

3. $\Delta _ f$ is of finite type.

Proof. The statements “$f$ is quasi-separated”, “$\Delta _ f$ is quasi-compact”, and “$\Delta _ f$ is of finite type” refer to the notions defined in Properties of Stacks, Section 100.3. Note that (2) and (3) are equivalent in view of the fact that $\Delta _ f$ is locally of finite type by Lemma 101.3.4 (and Algebraic Stacks, Lemma 94.10.9). Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Set $U = \mathcal{X} \times _\mathcal {Y} V$ which is an algebraic space by assumption, and the morphism $U \to \mathcal{X}$ is surjective and smooth. By Categories, Lemma 4.31.14 and Properties of Stacks, Lemma 100.3.3 we see that we have: $\Delta _ f$ is quasi-compact if and only if $\Delta _{U/V} : U \to U \times _ V U$ is quasi-compact. If (1) holds, then $U \to V$ is quasi-separated and we see that $\Delta _{U/V}$ is quasi-compact, i.e., (2) holds. Assume (2) holds. Let $T$ be a scheme, and $a : T \to \mathcal{Y}$ a morphism. Set $T' = \mathcal{X} \times _\mathcal {Y} T$. To prove (1) we have to show that the morphism of algebraic spaces $T' \to T$ is quasi-separated. Using Categories, Lemma 4.31.14 once more we see that $\Delta _{T'/T}$ is the base change of $\Delta _ f$. Hence our assumption (2) implies that $\Delta _{T'/T}$ is quasi-compact, hence $T' \to T$ is quasi-separated as desired. $\square$

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