# The Stacks Project

## Tag 0DTI

Lemma 91.11.13. 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 91.11.5. 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 91.11.12 is fully faithful by Algebraic Stacks, Lemma 85.16.1. We conclude for example by Lemma 91.11.11. $\square$

The code snippet corresponding to this tag is a part of the file stacks-properties.tex and is located in lines 2874–2889 (see updates for more information).

\begin{lemma}
\label{lemma-residual-gerbe-isomorphic}
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 $\Spec(k) \to \mathcal{X}$
in the equivalence class of $x$ such that
$$\Spec(k) \times_\mathcal{X} \Spec(k) \longrightarrow \Spec(k) \times_\mathcal{Y} \Spec(k)$$
is an isomorphism. Then $\mathcal{Z}_x \to \mathcal{Z}_y$
is an isomorphism.
\end{lemma}

\begin{proof}
Let $k'/k$ be an extension of fields. Then
$$\Spec(k') \times_\mathcal{X} \Spec(k') \longrightarrow \Spec(k') \times_\mathcal{Y} \Spec(k')$$
is the base change of the morphism in the lemma by the
faithfully flat morphism $\Spec(k' \otimes k') \to \Spec(k \otimes k)$.
Thus the property described in the lemma is independent of the
choice of the morphism $\Spec(k) \to \mathcal{X}$ in the
equivalence class of $x$. Thus we may assume that
$\Spec(k) \to \mathcal{Z}_x$ is surjective, flat, and
locally of finite presentation. In this situation we have
$$\mathcal{Z}_x = [\Spec(k)/R]$$
with $R = \Spec(k) \times_\mathcal{X} \Spec(k)$. See
proof of Lemma \ref{lemma-improve-unique-point}.
Since also $R = \Spec(k) \times_\mathcal{Y} \Spec(k)$
we conclude that the morphism $\mathcal{Z}_x \to \mathcal{Z}_y$
of Lemma \ref{lemma-residual-gerbe-functorial}
is fully faithful by
Algebraic Stacks, Lemma \ref{algebraic-lemma-map-space-into-stack}.
We conclude for example by Lemma \ref{lemma-residual-gerbe-unique}.
\end{proof}

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