Lemma 106.3.2. Let $i : \mathcal{X} \to \mathcal{X}'$ be a morphism of algebraic stacks. The following are equivalent
$i$ is a thickening of algebraic stacks (abuse of language as above), and
$i$ is representable by algebraic spaces and is a thickening in the sense of Properties of Stacks, Section 100.3.
In this case $i$ is a closed immersion and a universal homeomorphism.
Proof. By More on Morphisms of Spaces, Lemmas 76.9.10 and 76.9.8 the property $P$ that a morphism of algebraic spaces is a (first order) thickening is fpqc local on the base and stable under base change. Thus the discussion in Properties of Stacks, Section 100.3 indeed applies. Having said this the equivalence of (1) and (2) follows from the fact that $P = P_1 + P_2$ where $P_1$ is the property of being a closed immersion and $P_2$ is the property of being surjective. (Strictly speaking, the reader should also consult More on Morphisms of Spaces, Definition 76.9.1, Properties of Stacks, Definition 100.9.1 and the discussion following, Morphisms of Spaces, Lemma 67.5.1, Properties of Stacks, Section 100.5 to see that all the concepts all match up.) The final assertion is clear from the foregoing. $\square$
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