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

$\xymatrix{ \mathcal{X} \ar[d]_ f & \mathcal{Z}_ x \ar[l] \ar[d] \\ \mathcal{Y} & \mathcal{Z}_ y \ar[l] }$

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. $\square$

Comment #2647 by Daniel Dore on

Shouldn't the application of Lemma 90.11.12 be to show the morphism $\mathrm{Spec}(k) \rightarrow \mathcal{Y}$ factors through $\mathcal{Z}_y$, not through $\mathcal{Z}_x$?

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