Lemma 100.51.1. Let $\pi : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $x \in |\mathcal{X}|$ with image $y \in |\mathcal{Y}|$. Assume the residual gerbe $\mathcal{Z}_ y \subset \mathcal{Y}$ of $\mathcal{Y}$ at $y$ exists and that $\mathcal{X}$ is a gerbe over $\mathcal{Y}$. Then $\mathcal{Z}_ x = \mathcal{Z}_ y \times _\mathcal {Y} \mathcal{X}$ is the residual gerbe of $\mathcal{X}$ at $x$.

**Proof.**
The morphism $\mathcal{Z}_ x \to \mathcal{X}$ is a monomorphism as the base change of the monomorphism $\mathcal{Z}_ y \to \mathcal{Y}$. The morphism $\pi $ is a univeral homeomorphism by Lemma 100.28.13 and hence $|\mathcal{Z}_ x| = \{ x\} $. Finally, the morphism $\mathcal{Z}_ x \to \mathcal{Z}_ y$ is smooth as a base change of the smooth morphism $\pi $, see Lemma 100.33.8. Hence as $\mathcal{Z}_ y$ is reduced and locally Noetherian, so is $\mathcal{Z}_ x$ (details omitted). Thus $\mathcal{Z}_ x$ is the residual gerbe of $\mathcal{X}$ at $x$ by Properties of Stacks, Definition 99.11.8.
$\square$

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