Lemma 100.11.14. 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 100.11.5. 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 100.11.13 is fully faithful by Algebraic Stacks, Lemma 94.16.1. We conclude for example by Lemma 100.11.12.
\square
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