Lemma 99.11.14. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $x \in |\mathcal{X}|$ with image $y \in |\mathcal{Y}|$. Assume the residual gerbes $\mathcal{Z}_ x \subset \mathcal{X}$ and $\mathcal{Z}_ y \subset \mathcal{Y}$ of $x$ and $y$ exist and that there exists a morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ in the equivalence class of $x$ such that

\[ \mathop{\mathrm{Spec}}(k) \times _\mathcal {X} \mathop{\mathrm{Spec}}(k) \longrightarrow \mathop{\mathrm{Spec}}(k) \times _\mathcal {Y} \mathop{\mathrm{Spec}}(k) \]

is an isomorphism. Then $\mathcal{Z}_ x \to \mathcal{Z}_ y$ is an isomorphism.

**Proof.**
Let $k'/k$ be an extension of fields. Then

\[ \mathop{\mathrm{Spec}}(k') \times _\mathcal {X} \mathop{\mathrm{Spec}}(k') \longrightarrow \mathop{\mathrm{Spec}}(k') \times _\mathcal {Y} \mathop{\mathrm{Spec}}(k') \]

is the base change of the morphism in the lemma by the faithfully flat morphism $\mathop{\mathrm{Spec}}(k' \otimes k') \to \mathop{\mathrm{Spec}}(k \otimes k)$. Thus the property described in the lemma is independent of the choice of the morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ in the equivalence class of $x$. Thus we may assume that $\mathop{\mathrm{Spec}}(k) \to \mathcal{Z}_ x$ is surjective, flat, and locally of finite presentation. In this situation we have

\[ \mathcal{Z}_ x = [\mathop{\mathrm{Spec}}(k)/R] \]

with $R = \mathop{\mathrm{Spec}}(k) \times _\mathcal {X} \mathop{\mathrm{Spec}}(k)$. See proof of Lemma 99.11.5. Since also $R = \mathop{\mathrm{Spec}}(k) \times _\mathcal {Y} \mathop{\mathrm{Spec}}(k)$ we conclude that the morphism $\mathcal{Z}_ x \to \mathcal{Z}_ y$ of Lemma 99.11.13 is fully faithful by Algebraic Stacks, Lemma 93.16.1. We conclude for example by Lemma 99.11.12.
$\square$

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