Lemma 99.3.5. 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 separated,

2. $\Delta _ f$ is a closed immersion,

3. $\Delta _ f$ is proper, or

4. $\Delta _ f$ is universally closed.

Proof. The statements “$f$ is separated”, “$\Delta _ f$ is a closed immersion”, “$\Delta _ f$ is universally closed”, and “$\Delta _ f$ is proper” refer to the notions defined in Properties of Stacks, Section 98.3. 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 98.3.3 we see that for any property $P$ (as in that lemma) we have: $\Delta _ f$ has $P$ if and only if $\Delta _{U/V} : U \to U \times _ V U$ has $P$. Hence the equivalence of (2), (3) and (4) follows from Morphisms of Spaces, Lemma 65.40.9 applied to $U \to V$. Moreover, if (1) holds, then $U \to V$ is separated and we see that $\Delta _{U/V}$ is a closed immersion, i.e., (2) holds. Finally, 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 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 a closed immersion, hence $T' \to T$ is separated as desired. $\square$

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