Lemma 106.3.8. Let $(f, f') : (\mathcal{X} \subset \mathcal{X}') \to (\mathcal{Y} \subset \mathcal{Y}')$ be a morphism of thickenings of algebraic stacks. Let $\Delta : \mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ and $\Delta ' : \mathcal{X}' \to \mathcal{X}' \times _{\mathcal{Y}'} \mathcal{X}'$ be the corresponding diagonal morphisms. Then each property from the following list is satisfied by $\Delta $ if and only if it is satisfied by $\Delta '$: (a) representable by schemes, (b) affine, (c) surjective, (d) quasi-compact, (e) universally closed, (f) integral, (g) quasi-separated, (h) separated, (i) universally injective, (j) universally open, (k) locally quasi-finite, (l) finite, (m) unramified, (n) monomorphism, (o) immersion, (p) closed immersion, and (q) proper.
Proof. Observe that
is a morphism of thickenings (Lemma 106.3.7). Moreover $\Delta $ and $\Delta '$ are representable by algebraic spaces by Morphisms of Stacks, Lemma 101.3.3. Hence, via the discussion in Properties of Stacks, Section 100.3 the lemma follows for cases (a), (b), (c), (d), (e), (f), (g), (h), (i), and (j) by using More on Morphisms of Spaces, Lemma 76.10.1.
Lemma 106.3.7 tells us that $\mathcal{X} = (\mathcal{X} \times _\mathcal {Y} \mathcal{X}) \times _{(\mathcal{X}' \times _{\mathcal{Y}'} \mathcal{X}')} \mathcal{X}'$. Moreover, $\Delta $ and $\Delta '$ are locally of finite type by the aforementioned Morphisms of Stacks, Lemma 101.3.3. Hence the result for cases (k), (l), (m), (n), (o), (p), and (q) by using More on Morphisms of Spaces, Lemma 76.10.3. $\square$
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