Lemma 99.11.12. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $x \in |\mathcal{X}|$ with image $y \in |\mathcal{Y}|$. If the residual gerbes $\mathcal{Z}_ x \subset \mathcal{X}$ and $\mathcal{Z}_ y \subset \mathcal{Y}$ of $x$ and $y$ exist, then $f$ induces a commutative diagram

**Proof.**
Choose a field $k$ and a surjective, flat, locally finitely presented morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{Z}_ x$. The morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{Y}$ factors through $\mathcal{Z}_ y$ by Lemma 99.11.10. Thus $\mathcal{Z}_ x \times _\mathcal {Y} \mathcal{Z}_ y$ is a nonempty substack of $\mathcal{Z}_ x$ hence equal to $\mathcal{Z}_ x$ by Lemma 99.11.4.
$\square$

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## Comments (2)

Comment #2647 by Daniel Dore on

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