## Tag `06ML`

## 90.11. Residual gerbes

In the Stacks project we would like to define the

residual gerbeof an algebraic stack $\mathcal{X}$ at a point $x \in |\mathcal{X}|$ to be a monomorphism of algebraic stacks $m_x : \mathcal{Z}_x \to \mathcal{X}$ where $\mathcal{Z}_x$ is a reduced algebraic stack having a unique point which is mapped by $m_x$ to $x$. It turns out that there are many issues with this notion; existence is not clear in general and neither is uniqueness. We resolve the uniqueness issue by imposing a slightly stronger condition on the algebraic stacks $\mathcal{Z}_x$. We discuss this in more detail by working through a few simple lemmas regarding reduced algebraic stacks having a unique point.Lemma 90.11.1. Let $\mathcal{Z}$ be an algebraic stack. Let $k$ be a field and let $\mathop{\rm Spec}(k) \to \mathcal{Z}$ be surjective and flat. Then any morphism $\mathop{\rm Spec}(k') \to \mathcal{Z}$ where $k'$ is a field is surjective and flat.

Proof.Consider the fibre square $$ \xymatrix{ T \ar[d] \ar[r] & \mathop{\rm Spec}(k) \ar[d] \\ \mathop{\rm Spec}(k') \ar[r] & \mathcal{Z} } $$ Note that $T \to \mathop{\rm Spec}(k')$ is flat and surjective hence $T$ is not empty. On the other hand $T \to \mathop{\rm Spec}(k)$ is flat as $k$ is a field. Hence $T \to \mathcal{Z}$ is flat and surjective. It follows from Morphisms of Spaces, Lemma 58.30.5 (via the discussion in Section 90.3) that $\mathop{\rm Spec}(k') \to \mathcal{Z}$ is flat. It is clear that it is surjective as by assumption $|\mathcal{Z}|$ is a singleton. $\square$Lemma 90.11.2. Let $\mathcal{Z}$ be an algebraic stack. The following are equivalent

- $\mathcal{Z}$ is reduced and $|\mathcal{Z}|$ is a singleton,
- there exists a surjective flat morphism $\mathop{\rm Spec}(k) \to \mathcal{Z}$ where $k$ is a field, and
- there exists a locally of finite type, surjective, flat morphism $\mathop{\rm Spec}(k) \to \mathcal{Z}$ where $k$ is a field.

Proof.Assume (1). Let $W$ be a scheme and let $W \to \mathcal{Z}$ be a surjective smooth morphism. Then $W$ is a reduced scheme. Let $\eta \in W$ be a generic point of an irreducible component of $W$. Since $W$ is reduced we have $\mathcal{O}_{W, \eta} = \kappa(\eta)$. It follows that the canonical morphism $\eta = \mathop{\rm Spec}(\kappa(\eta)) \to W$ is flat. We see that the composition $\eta \to \mathcal{Z}$ is flat (see Morphisms of Spaces, Lemma 58.29.3). It is also surjective as $|\mathcal{Z}|$ is a singleton. In other words (2) holds.Assume (2). Let $W$ be a scheme and let $W \to \mathcal{Z}$ be a surjective smooth morphism. Choose a field $k$ and a surjective flat morphism $\mathop{\rm Spec}(k) \to \mathcal{Z}$. Then $W \times_\mathcal{Z} \mathop{\rm Spec}(k)$ is an algebraic space smooth over $k$, hence regular (see Spaces over Fields, Lemma 63.12.1) and in particular reduced. Since $W \times_\mathcal{Z} \mathop{\rm Spec}(k) \to W$ is surjective and flat we conclude that $W$ is reduced (Descent on Spaces, Lemma 65.8.2). In other words (1) holds.

It is clear that (3) implies (2). Finally, assume (2). Pick a nonempty affine scheme $W$ and a smooth morphism $W \to \mathcal{Z}$. Pick a closed point $w \in W$ and set $k = \kappa(w)$. The composition $$ \mathop{\rm Spec}(k) \xrightarrow{w} W \longrightarrow \mathcal{Z} $$ is locally of finite type by Morphisms of Spaces, Lemmas 58.23.2 and 58.36.6. It is also flat and surjective by Lemma 90.11.1. Hence (3) holds. $\square$

The following lemma singles out a slightly better class of singleton algebraic stacks than the preceding lemma.

Lemma 90.11.3. Let $\mathcal{Z}$ be an algebraic stack. The following are equivalent

- $\mathcal{Z}$ is reduced, locally Noetherian, and $|\mathcal{Z}|$ is a singleton, and
- there exists a locally finitely presented, surjective, flat morphism $\mathop{\rm Spec}(k) \to \mathcal{Z}$ where $k$ is a field.

Proof.Assume (2) holds. By Lemma 90.11.2 we see that $\mathcal{Z}$ is reduced and $|\mathcal{Z}|$ is a singleton. Let $W$ be a scheme and let $W \to \mathcal{Z}$ be a surjective smooth morphism. Choose a field $k$ and a locally finitely presented, surjective, flat morphism $\mathop{\rm Spec}(k) \to \mathcal{Z}$. Then $W \times_\mathcal{Z} \mathop{\rm Spec}(k)$ is an algebraic space smooth over $k$, hence locally Noetherian (see Morphisms of Spaces, Lemma 58.23.5). Since $W \times_\mathcal{Z} \mathop{\rm Spec}(k) \to W$ is flat, surjective, and locally of finite presentation, we see that $\{W \times_\mathcal{Z} \mathop{\rm Spec}(k) \to W\}$ is an fppf covering and we conclude that $W$ is locally Noetherian (Descent on Spaces, Lemma 65.8.3). In other words (1) holds.Assume (1). Pick a nonempty affine scheme $W$ and a smooth morphism $W \to \mathcal{Z}$. Pick a closed point $w \in W$ and set $k = \kappa(w)$. Because $W$ is locally Noetherian the morphism $w : \mathop{\rm Spec}(k) \to W$ is of finite presentation, see Morphisms, Lemma 28.20.7. Hence the composition $$ \mathop{\rm Spec}(k) \xrightarrow{w} W \longrightarrow \mathcal{Z} $$ is locally of finite presentation by Morphisms of Spaces, Lemmas 58.28.2 and 58.36.5. It is also flat and surjective by Lemma 90.11.1. Hence (2) holds. $\square$

Lemma 90.11.4. Let $\mathcal{Z}' \to \mathcal{Z}$ be a monomorphism of algebraic stacks. Assume there exists a field $k$ and a locally finitely presented, surjective, flat morphism $\mathop{\rm Spec}(k) \to \mathcal{Z}$. Then either $\mathcal{Z}'$ is empty or $\mathcal{Z}' \to \mathcal{Z}$ is an equivalence.

Proof.We may assume that $\mathcal{Z}'$ is nonempty. In this case the fibre product $T = \mathcal{Z}' \times_\mathcal{Z} \mathop{\rm Spec}(k)$ is nonempty, see Lemma 90.4.3. Now $T$ is an algebraic space and the projection $T \to \mathop{\rm Spec}(k)$ is a monomorphism. Hence $T = \mathop{\rm Spec}(k)$, see Morphisms of Spaces, Lemma 58.10.8. We conclude that $\mathop{\rm Spec}(k) \to \mathcal{Z}$ factors through $\mathcal{Z}'$. Suppose the morphism $z : \mathop{\rm Spec}(k) \to \mathcal{Z}$ is given by the object $\xi$ over $\mathop{\rm Spec}(k)$. We have just seen that $\xi$ is isomorphic to an object $\xi'$ of $\mathcal{Z}'$ over $\mathop{\rm Spec}(k)$. Since $z$ is is surjective, flat, and locally of finite presentation we see that every object of $\mathcal{Z}$ over any scheme is fppf locally isomorphic to a pullback of $\xi$, hence also to a pullback of $\xi'$. By descent of objects for stacks in groupoids this implies that $\mathcal{Z}' \to \mathcal{Z}$ is essentially surjective (as well as fully faithful, see Lemma 90.8.4). Hence we win. $\square$Lemma 90.11.5. Let $\mathcal{Z}$ be an algebraic stack. Assume $\mathcal{Z}$ satisfies the equivalent conditions of Lemma 90.11.2. Then there exists a unique strictly full subcategory $\mathcal{Z}' \subset \mathcal{Z}$ such that $\mathcal{Z}'$ is an algebraic stack which satisfies the equivalent conditions of Lemma 90.11.3. The inclusion morphism $\mathcal{Z}' \to \mathcal{Z}$ is a monomorphism of algebraic stacks.

Proof.The last part is immediate from the first part and Lemma 90.8.4. Pick a field $k$ and a morphism $\mathop{\rm Spec}(k) \to \mathcal{Z}$ which is surjective, flat, and locally of finite type. Set $U = \mathop{\rm Spec}(k)$ and $R = U \times_\mathcal{Z} U$. The projections $s, t : R \to U$ are locally of finite type. Since $U$ is the spectrum of a field, it follows that $s, t$ are flat and locally of finite presentation (by Morphisms of Spaces, Lemma 58.28.7). We see that $\mathcal{Z}' = [U/R]$ is an algebraic stack by Criteria for Representability, Theorem 87.17.2. By Algebraic Stacks, Lemma 84.16.1 we obtain a canonical morphism $$ f : \mathcal{Z}' \longrightarrow \mathcal{Z} $$ which is fully faithful. Hence this morphism is representable by algebraic spaces, see Algebraic Stacks, Lemma 84.15.2 and a monomorphism, see Lemma 90.8.4. By Criteria for Representability, Lemma 87.17.1 the morphism $U \to \mathcal{Z}'$ is surjective, flat, and locally of finite presentation. Hence $\mathcal{Z}'$ is an algebraic stack which satisfies the equivalent conditions of Lemma 90.11.3. By Algebraic Stacks, Lemma 84.12.4 we may replace $\mathcal{Z}'$ by its essential image in $\mathcal{Z}$. Hence we have proved all the assertions of the lemma except for the uniqueness of $\mathcal{Z}' \subset \mathcal{Z}$. Suppose that $\mathcal{Z}'' \subset \mathcal{Z}$ is a second such algebraic stack. Then the projections $$ \mathcal{Z}' \longleftarrow \mathcal{Z}' \times_\mathcal{Z} \mathcal{Z}'' \longrightarrow \mathcal{Z}'' $$ are monomorphisms. The algebraic stack in the middle is nonempty by Lemma 90.4.3. Hence the two projections are isomorphisms by Lemma 90.11.4 and we win. $\square$Example 90.11.6. Here is an example where the morphism constructed in Lemma 90.11.5 isn't an isomorphism. This example shows that imposing that residual gerbes are locally Noetherian is necessary in Definition 90.11.8. In fact, the example is even an algebraic space! Let $\text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ be the absolute Galois group of $\mathbf{Q}$ with the pro-finite topology. Let $$ U = \mathop{\rm Spec}(\overline{\mathbf{Q}}) \times_{\mathop{\rm Spec}(\mathbf{Q})} \mathop{\rm Spec}(\overline{\mathbf{Q}}) = \text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q}) \times \mathop{\rm Spec}(\overline{\mathbf{Q}}) $$ (we omit a precise explanation of the meaning of the last equal sign). Let $G$ denote the absolute Galois group $\text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ with the discrete topology viewed as a constant group scheme over $\mathop{\rm Spec}(\overline{\mathbf{Q}})$, see Groupoids, Example 38.5.6. Then $G$ acts freely and transitively on $U$. Let $X = U/G$, see Spaces, Definition 56.14.4. Then $X$ is a non-noetherian reduced algebraic space with exactly one point. Furthermore, $X$ has a (locally) finite type point: $$ x : \mathop{\rm Spec}(\overline{\mathbf{Q}}) \longrightarrow U \longrightarrow X $$ Indeed, every point of $U$ is actually closed! As $X$ is an algebraic space over $\overline{\mathbf{Q}}$ it follows that $x$ is a monomorphism. So $x$ is the morphism constructed in Lemma 90.11.5 but $x$ is not an isomorphism. In fact $\mathop{\rm Spec}(\overline{\mathbf{Q}}) \to X$ is the residual gerbe of $X$ at $x$.

It will turn out later that under mild assumptions on the algebraic stack $\mathcal{X}$ the equivalent conditions of the following lemma are satisfied for every point $x \in |\mathcal{X}|$ (see Morphisms of Stacks, Section 91.30).

Lemma 90.11.7. Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$ be a point. The following are equivalent

- there exists an algebraic stack $\mathcal{Z}$ and a monomorphism $\mathcal{Z} \to \mathcal{X}$ such that $|\mathcal{Z}|$ is a singleton and such that the image of $|\mathcal{Z}|$ in $|\mathcal{X}|$ is $x$,
- there exists a reduced algebraic stack $\mathcal{Z}$ and a monomorphism $\mathcal{Z} \to \mathcal{X}$ such that $|\mathcal{Z}|$ is a singleton and such that the image of $|\mathcal{Z}|$ in $|\mathcal{X}|$ is $x$,
- there exists an algebraic stack $\mathcal{Z}$, a monomorphism $f : \mathcal{Z} \to \mathcal{X}$, and a surjective flat morphism $z : \mathop{\rm Spec}(k) \to \mathcal{Z}$ where $k$ is a field such that $x = f(z)$.
Moreover, if these conditions hold, then there exists a unique strictly full subcategory $\mathcal{Z}_x \subset \mathcal{X}$ such that $\mathcal{Z}_x$ is a reduced, locally Noetherian algebraic stack and $|\mathcal{Z}_x|$ is a singleton which maps to $x$ via the map $|\mathcal{Z}_x| \to |\mathcal{X}|$.

Proof.If $\mathcal{Z} \to \mathcal{X}$ is as in (1), then $\mathcal{Z}_{red} \to \mathcal{X}$ is as in (2). (See Section 90.10 for the notion of the reduction of an algebraic stack.) Hence (1) implies (2). It is immediate that (2) implies (1). The equivalence of (2) and (3) is immediate from Lemma 90.11.2.At this point we've seen the equivalence of (1) – (3). Pick a monomorphism $f : \mathcal{Z} \to \mathcal{X}$ as in (2). Note that this implies that $f$ is fully faithful, see Lemma 90.8.4. Denote $\mathcal{Z}' \subset \mathcal{X}$ the essential image of the functor $f$. Then $f : \mathcal{Z} \to \mathcal{Z}'$ is an equivalence and hence $\mathcal{Z}'$ is an algebraic stack, see Algebraic Stacks, Lemma 84.12.4. Apply Lemma 90.11.5 to get a strictly full subcategory $\mathcal{Z}_x \subset \mathcal{Z}'$ as in the statement of the lemma. This proves all the statements of the lemma except for uniqueness.

In order to prove the uniqueness suppose that $\mathcal{Z}_x \subset \mathcal{X}$ and $\mathcal{Z}'_x \subset \mathcal{X}$ are two strictly full subcategories as in the statement of the lemma. Then the projections $$ \mathcal{Z}'_x \longleftarrow \mathcal{Z}'_x \times_\mathcal{X} \mathcal{Z}_x \longrightarrow \mathcal{Z}_x $$ are monomorphisms. The algebraic stack in the middle is nonempty by Lemma 90.4.3. Hence the two projections are isomorphisms by Lemma 90.11.4 and we win. $\square$

Having explained the above we can now make the following definition.

Definition 90.11.8. Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$.

- We say the
residual gerbe of $\mathcal{X}$ at $x$ existsif the equivalent conditions (1), (2), and (3) of Lemma 90.11.7 hold.- If the residual gerbe of $\mathcal{X}$ at $x$ exists, then the
residual gerbe of $\mathcal{X}$ at $x$^{1}is the strictly full subcategory $\mathcal{Z}_x \subset \mathcal{X}$ constructed in Lemma 90.11.7.

In particular we know that $\mathcal{Z}_x$ (if it exists) is a locally Noetherian, reduced algebraic stack and that there exists a field and a surjective, flat, locally finitely presented morphism $$ \mathop{\rm Spec}(k) \longrightarrow \mathcal{Z}_x. $$ We will see in Morphisms of Stacks, Lemma 91.27.12 that $\mathcal{Z}_x$ is a gerbe. Existence of residual gerbes is discussed in Morphisms of Stacks, Section 91.30. It turns out that $\mathcal{Z}_x$ is a regular algebraic stack as follows from the following lemma.

Lemma 90.11.9. A reduced, locally Noetherian algebraic stack $\mathcal{Z}$ such that $|\mathcal{Z}|$ is a singleton is regular.

Proof.Let $W \to \mathcal{Z}$ be a surjective smooth morphism where $W$ is a scheme. Let $k$ be a field and let $\mathop{\rm Spec}(k) \to \mathcal{Z}$ be surjective, flat, and locally of finite presentation (see Lemma 90.11.3). The algebraic space $T = W \times_\mathcal{Z} \mathop{\rm Spec}(k)$ is smooth over $k$ in particular regular, see Spaces over Fields, Lemma 63.12.1. Since $T \to W$ is locally of finite presentation, flat, and surjective it follows that $W$ is regular, see Descent on Spaces, Lemma 65.8.4. By definition this means that $\mathcal{Z}$ is regular. $\square$Lemma 90.11.10. Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$. Assume that the residual gerbe $\mathcal{Z}_x$ of $\mathcal{X}$ exists. Let $f : \mathop{\rm Spec}(K) \to \mathcal{X}$ be a morphism where $K$ is a field in the equivalence class of $x$. Then $f$ factors through the inclusion morphism $\mathcal{Z}_x \to \mathcal{X}$.

Proof.Choose a field $k$ and a surjective flat locally finite presentation morphism $\mathop{\rm Spec}(k) \to \mathcal{Z}_x$. Set $T = \mathop{\rm Spec}(K) \times_\mathcal{X} \mathcal{Z}_x$. By Lemma 90.4.3 we see that $T$ is nonempty. As $\mathcal{Z}_x \to \mathcal{X}$ is a monomorphism we see that $T \to \mathop{\rm Spec}(K)$ is a monomorphism. Hence by Morphisms of Spaces, Lemma 58.10.8 we see that $T = \mathop{\rm Spec}(K)$ which proves the lemma. $\square$Lemma 90.11.11. Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$. Let $\mathcal{Z}$ be an algebraic stack satisfying the equivalent conditions of Lemma 90.11.3 and let $\mathcal{Z} \to \mathcal{X}$ be a monomorphism such that the image of $|\mathcal{Z}| \to |\mathcal{X}|$ is $x$. Then the residual gerbe $\mathcal{Z}_x$ of $\mathcal{X}$ at $x$ exists and $\mathcal{Z} \to \mathcal{X}$ factors as $\mathcal{Z} \to \mathcal{Z}_x \to \mathcal{X}$ where the first arrow is an equivalence.

Proof.Let $\mathcal{Z}_x \subset \mathcal{X}$ be the full subcategory corresponding to the essential image of the functor $\mathcal{Z} \to \mathcal{X}$. Then $\mathcal{Z} \to \mathcal{Z}_x$ is an equivalence, hence $\mathcal{Z}_x$ is an algebraic stack, see Algebraic Stacks, Lemma 84.12.4. Since $\mathcal{Z}_x$ inherits all the properties of $\mathcal{Z}$ from this equivalence it is clear from the uniqueness in Lemma 90.11.7 that $\mathcal{Z}_x$ is the residual gerbe of $\mathcal{X}$ at $x$. $\square$Lemma 90.11.12. 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 $\mathop{\rm Spec}(k) \to \mathcal{Z}_x$. The morphism $\mathop{\rm Spec}(k) \to \mathcal{Y}$ factors through $\mathcal{Z}_y$ by Lemma 90.11.10. 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 90.11.4. $\square$Lemma 90.11.13. 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 $\mathop{\rm Spec}(k) \to \mathcal{X}$ in the equivalence class of $x$ such that $$ \mathop{\rm Spec}(k) \times_\mathcal{X} \mathop{\rm Spec}(k) \longrightarrow \mathop{\rm Spec}(k) \times_\mathcal{Y} \mathop{\rm Spec}(k) $$ is an isomorphism. Then $\mathcal{Z}_x \to \mathcal{Z}_y$ is an isomorphism.

Proof.Let $k'/k$ be an extension of fields. Then $$ \mathop{\rm Spec}(k') \times_\mathcal{X} \mathop{\rm Spec}(k') \longrightarrow \mathop{\rm Spec}(k') \times_\mathcal{Y} \mathop{\rm Spec}(k') $$ is the base change of the morphism in the lemma by the faithfully flat morphism $\mathop{\rm Spec}(k' \otimes k') \to \mathop{\rm Spec}(k \otimes k)$. Thus the property described in the lemma is independent of the choice of the morphism $\mathop{\rm Spec}(k) \to \mathcal{X}$ in the equivalence class of $x$. Thus we may assume that $\mathop{\rm Spec}(k) \to \mathcal{Z}_x$ is surjective, flat, and locally of finite presentation. In this situation we have $$ \mathcal{Z}_x = [\mathop{\rm Spec}(k)/R] $$ with $R = \mathop{\rm Spec}(k) \times_\mathcal{X} \mathop{\rm Spec}(k)$. See proof of Lemma 90.11.5. Since also $R = \mathop{\rm Spec}(k) \times_\mathcal{Y} \mathop{\rm Spec}(k)$ we conclude that the morphism $\mathcal{Z}_x \to \mathcal{Z}_y$ of Lemma 90.11.12 is fully faithful by Algebraic Stacks, Lemma 84.16.1. We conclude for example by Lemma 90.11.11. $\square$

- This clashes with [LM-B] in spirit, but not in fact. Namely, in Chapter 11 they associate to any point on any quasi-separated algebraic stack a gerbe (not necessarily algebraic) which they call the residual gerbe. We will see in Morphisms of Stacks, Lemma 91.30.1 that on a quasi-separated algebraic stack every point has a residual gerbe in our sense which is then equivalent to theirs. For more information on this topic see [rydh_etale_devissage, Appendix B]. ↑

The code snippet corresponding to this tag is a part of the file `stacks-properties.tex` and is located in lines 2371–2924 (see updates for more information).

```
\section{Residual gerbes}
\label{section-residual-gerbe}
\noindent
In the Stacks project we would like to define the {\it residual gerbe}
of an algebraic stack $\mathcal{X}$ at a point $x \in |\mathcal{X}|$
to be a monomorphism of algebraic stacks
$m_x : \mathcal{Z}_x \to \mathcal{X}$ where $\mathcal{Z}_x$ is a reduced
algebraic stack having a unique point which is mapped by $m_x$ to $x$.
It turns out that there are many issues with this notion; existence is not
clear in general and neither is uniqueness.
We resolve the uniqueness issue by imposing a slightly stronger
condition on the algebraic stacks $\mathcal{Z}_x$.
We discuss this in more detail by working through a few simple lemmas
regarding reduced algebraic stacks having a unique point.
\begin{lemma}
\label{lemma-flat-cover-by-field}
Let $\mathcal{Z}$ be an algebraic stack. Let $k$ be a field and let
$\Spec(k) \to \mathcal{Z}$ be surjective and flat. Then any
morphism $\Spec(k') \to \mathcal{Z}$ where $k'$ is a field is
surjective and flat.
\end{lemma}
\begin{proof}
Consider the fibre square
$$
\xymatrix{
T \ar[d] \ar[r] & \Spec(k) \ar[d] \\
\Spec(k') \ar[r] & \mathcal{Z}
}
$$
Note that $T \to \Spec(k')$ is flat and surjective hence $T$
is not empty. On the other hand $T \to \Spec(k)$ is flat as
$k$ is a field. Hence $T \to \mathcal{Z}$ is flat and surjective.
It follows from
Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-flat-permanence}
(via the discussion in
Section \ref{section-properties-morphisms})
that $\Spec(k') \to \mathcal{Z}$ is flat. It is clear that it
is surjective as by assumption $|\mathcal{Z}|$ is a singleton.
\end{proof}
\begin{lemma}
\label{lemma-unique-point}
Let $\mathcal{Z}$ be an algebraic stack. The following are equivalent
\begin{enumerate}
\item $\mathcal{Z}$ is reduced and $|\mathcal{Z}|$ is a singleton,
\item there exists a surjective flat morphism $\Spec(k) \to \mathcal{Z}$
where $k$ is a field, and
\item there exists a locally of finite type, surjective, flat morphism
$\Spec(k) \to \mathcal{Z}$ where $k$ is a field.
\end{enumerate}
\end{lemma}
\begin{proof}
Assume (1). Let $W$ be a scheme and
let $W \to \mathcal{Z}$ be a surjective smooth morphism. Then $W$ is
a reduced scheme. Let $\eta \in W$ be a generic point of an irreducible
component of $W$. Since $W$ is reduced we have
$\mathcal{O}_{W, \eta} = \kappa(\eta)$. It follows that the canonical
morphism $\eta = \Spec(\kappa(\eta)) \to W$ is flat. We see that the
composition $\eta \to \mathcal{Z}$ is flat (see
Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-composition-flat}).
It is also surjective as $|\mathcal{Z}|$ is a singleton. In other words
(2) holds.
\medskip\noindent
Assume (2). Let $W$ be a scheme and
let $W \to \mathcal{Z}$ be a surjective smooth morphism. Choose a field
$k$ and a surjective flat morphism $\Spec(k) \to \mathcal{Z}$.
Then $W \times_\mathcal{Z} \Spec(k)$ is an algebraic space smooth
over $k$, hence regular (see
Spaces over Fields, Lemma \ref{spaces-over-fields-lemma-smooth-regular})
and in particular reduced. Since $W \times_\mathcal{Z} \Spec(k) \to W$
is surjective and flat we conclude that $W$ is reduced
(Descent on Spaces, Lemma \ref{spaces-descent-lemma-descend-reduced}).
In other words (1) holds.
\medskip\noindent
It is clear that (3) implies (2). Finally, assume (2). Pick a nonempty
affine scheme $W$ and a smooth morphism $W \to \mathcal{Z}$. Pick a closed
point $w \in W$ and set $k = \kappa(w)$. The composition
$$
\Spec(k) \xrightarrow{w} W \longrightarrow \mathcal{Z}
$$
is locally of finite type by
Morphisms of Spaces, Lemmas
\ref{spaces-morphisms-lemma-composition-finite-type} and
\ref{spaces-morphisms-lemma-smooth-locally-finite-type}.
It is also flat and surjective by
Lemma \ref{lemma-flat-cover-by-field}.
Hence (3) holds.
\end{proof}
\noindent
The following lemma singles out a slightly better class of singleton
algebraic stacks than the preceding lemma.
\begin{lemma}
\label{lemma-unique-point-better}
Let $\mathcal{Z}$ be an algebraic stack. The following are equivalent
\begin{enumerate}
\item $\mathcal{Z}$ is reduced, locally Noetherian, and $|\mathcal{Z}|$
is a singleton, and
\item there exists a locally finitely presented, surjective, flat morphism
$\Spec(k) \to \mathcal{Z}$ where $k$ is a field.
\end{enumerate}
\end{lemma}
\begin{proof}
Assume (2) holds. By
Lemma \ref{lemma-unique-point}
we see that $\mathcal{Z}$ is reduced and $|\mathcal{Z}|$ is a singleton.
Let $W$ be a scheme and let $W \to \mathcal{Z}$ be a surjective smooth
morphism. Choose a field $k$ and a locally finitely presented, surjective,
flat morphism $\Spec(k) \to \mathcal{Z}$.
Then $W \times_\mathcal{Z} \Spec(k)$ is an algebraic space
smooth over $k$, hence locally Noetherian (see
Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-locally-finite-type-locally-noetherian}).
Since $W \times_\mathcal{Z} \Spec(k) \to W$
is flat, surjective, and locally of finite presentation, we see
that $\{W \times_\mathcal{Z} \Spec(k) \to W\}$ is an fppf covering
and we conclude that $W$ is locally Noetherian
(Descent on Spaces, Lemma
\ref{spaces-descent-lemma-descend-locally-Noetherian}).
In other words (1) holds.
\medskip\noindent
Assume (1). Pick a nonempty affine scheme $W$ and a smooth morphism
$W \to \mathcal{Z}$. Pick a closed point $w \in W$ and set
$k = \kappa(w)$. Because $W$ is locally Noetherian the morphism
$w : \Spec(k) \to W$ is of finite presentation, see
Morphisms, Lemma \ref{morphisms-lemma-closed-immersion-finite-presentation}.
Hence the composition
$$
\Spec(k) \xrightarrow{w} W \longrightarrow \mathcal{Z}
$$
is locally of finite presentation by
Morphisms of Spaces, Lemmas
\ref{spaces-morphisms-lemma-composition-finite-presentation} and
\ref{spaces-morphisms-lemma-smooth-locally-finite-presentation}.
It is also flat and surjective by
Lemma \ref{lemma-flat-cover-by-field}.
Hence (2) holds.
\end{proof}
\begin{lemma}
\label{lemma-monomorphism-into-point}
Let $\mathcal{Z}' \to \mathcal{Z}$ be a monomorphism of algebraic stacks.
Assume there exists a field $k$ and a locally finitely presented, surjective,
flat morphism $\Spec(k) \to \mathcal{Z}$. Then either $\mathcal{Z}'$
is empty or $\mathcal{Z}' \to \mathcal{Z}$ is an equivalence.
\end{lemma}
\begin{proof}
We may assume that $\mathcal{Z}'$ is nonempty. In this case the
fibre product $T = \mathcal{Z}' \times_\mathcal{Z} \Spec(k)$
is nonempty, see
Lemma \ref{lemma-points-cartesian}.
Now $T$ is an algebraic space and the projection $T \to \Spec(k)$
is a monomorphism. Hence $T = \Spec(k)$, see
Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-monomorphism-toward-field}.
We conclude that $\Spec(k) \to \mathcal{Z}$ factors through
$\mathcal{Z}'$. Suppose the morphism $z : \Spec(k) \to \mathcal{Z}$
is given by the object $\xi$ over $\Spec(k)$. We have just seen that
$\xi$ is isomorphic to an object $\xi'$ of $\mathcal{Z}'$ over
$\Spec(k)$. Since $z$ is
is surjective, flat, and locally of finite presentation we see that
every object of $\mathcal{Z}$ over any scheme is fppf locally isomorphic
to a pullback of $\xi$, hence also to a pullback of $\xi'$. By descent of
objects for stacks in groupoids this implies that
$\mathcal{Z}' \to \mathcal{Z}$ is essentially surjective (as well as
fully faithful, see
Lemma \ref{lemma-monomorphism}).
Hence we win.
\end{proof}
\begin{lemma}
\label{lemma-improve-unique-point}
Let $\mathcal{Z}$ be an algebraic stack. Assume $\mathcal{Z}$ satisfies
the equivalent conditions of
Lemma \ref{lemma-unique-point}.
Then there exists a unique strictly full subcategory
$\mathcal{Z}' \subset \mathcal{Z}$ such that
$\mathcal{Z}'$ is an algebraic stack which satisfies the equivalent
conditions of
Lemma \ref{lemma-unique-point-better}.
The inclusion morphism $\mathcal{Z}' \to \mathcal{Z}$ is a monomorphism
of algebraic stacks.
\end{lemma}
\begin{proof}
The last part is immediate from the first part and
Lemma \ref{lemma-monomorphism}.
Pick a field $k$ and a morphism $\Spec(k) \to \mathcal{Z}$
which is surjective, flat, and locally of finite type.
Set $U = \Spec(k)$ and $R = U \times_\mathcal{Z} U$.
The projections $s, t : R \to U$ are locally of finite type.
Since $U$ is the spectrum of a field, it follows that
$s, t$ are flat and locally of finite presentation (by
Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-noetherian-finite-type-finite-presentation}).
We see that $\mathcal{Z}' = [U/R]$ is an algebraic stack by
Criteria for Representability,
Theorem \ref{criteria-theorem-flat-groupoid-gives-algebraic-stack}.
By
Algebraic Stacks, Lemma \ref{algebraic-lemma-map-space-into-stack}
we obtain a canonical morphism
$$
f : \mathcal{Z}' \longrightarrow \mathcal{Z}
$$
which is fully faithful. Hence this morphism is representable by
algebraic spaces, see
Algebraic Stacks, Lemma
\ref{algebraic-lemma-characterize-representable-by-algebraic-spaces}
and a monomorphism, see
Lemma \ref{lemma-monomorphism}.
By
Criteria for Representability,
Lemma \ref{criteria-lemma-flat-quotient-flat-presentation}
the morphism $U \to \mathcal{Z}'$ is surjective, flat, and locally of finite
presentation. Hence $\mathcal{Z}'$ is an algebraic stack which satisfies
the equivalent conditions of
Lemma \ref{lemma-unique-point-better}.
By
Algebraic Stacks, Lemma \ref{algebraic-lemma-equivalent}
we may replace $\mathcal{Z}'$ by its essential image in $\mathcal{Z}$.
Hence we have proved all the assertions of the lemma except for the
uniqueness of $\mathcal{Z}' \subset \mathcal{Z}$. Suppose that
$\mathcal{Z}'' \subset \mathcal{Z}$ is a second such algebraic stack.
Then the projections
$$
\mathcal{Z}'
\longleftarrow
\mathcal{Z}' \times_\mathcal{Z} \mathcal{Z}''
\longrightarrow
\mathcal{Z}''
$$
are monomorphisms. The algebraic stack in the middle is nonempty by
Lemma \ref{lemma-points-cartesian}.
Hence the two projections are isomorphisms by
Lemma \ref{lemma-monomorphism-into-point}
and we win.
\end{proof}
\begin{example}
\label{example-distinct}
Here is an example where the morphism constructed in
Lemma \ref{lemma-improve-unique-point}
isn't an isomorphism. This example shows that imposing
that residual gerbes are locally Noetherian is necessary in
Definition \ref{definition-residual-gerbe}.
In fact, the example is even an algebraic space!
Let $\text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ be the absolute
Galois group of $\mathbf{Q}$ with the pro-finite topology.
Let
$$
U =
\Spec(\overline{\mathbf{Q}})
\times_{\Spec(\mathbf{Q})}
\Spec(\overline{\mathbf{Q}}) =
\text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q}) \times
\Spec(\overline{\mathbf{Q}})
$$
(we omit a precise explanation of the meaning of the last equal sign).
Let $G$ denote the absolute Galois group
$\text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ with the discrete
topology viewed as a constant group scheme over
$\Spec(\overline{\mathbf{Q}})$, see
Groupoids, Example \ref{groupoids-example-constant-group}.
Then $G$ acts freely and transitively on $U$.
Let $X = U/G$, see
Spaces, Definition \ref{spaces-definition-quotient}.
Then $X$ is a non-noetherian reduced algebraic space with exactly one point.
Furthermore, $X$ has a (locally) finite type point:
$$
x :
\Spec(\overline{\mathbf{Q}}) \longrightarrow
U \longrightarrow X
$$
Indeed, every point of $U$ is actually closed!
As $X$ is an algebraic space over $\overline{\mathbf{Q}}$
it follows that $x$ is a monomorphism. So $x$ is the morphism
constructed in
Lemma \ref{lemma-improve-unique-point}
but $x$ is not an isomorphism. In fact
$\Spec(\overline{\mathbf{Q}}) \to X$ is the residual gerbe of $X$ at $x$.
\end{example}
\noindent
It will turn out later that under mild assumptions on the algebraic stack
$\mathcal{X}$ the equivalent conditions of the following lemma are satisfied
for every point $x \in |\mathcal{X}|$ (see Morphisms of Stacks, Section
\ref{stacks-morphisms-section-existence-residual-gerbes}).
\begin{lemma}
\label{lemma-residual-gerbe}
Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$ be a point.
The following are equivalent
\begin{enumerate}
\item there exists an algebraic stack $\mathcal{Z}$ and a monomorphism
$\mathcal{Z} \to \mathcal{X}$ such that $|\mathcal{Z}|$ is a singleton
and such that the image of $|\mathcal{Z}|$ in $|\mathcal{X}|$ is $x$,
\item there exists a reduced algebraic stack $\mathcal{Z}$ and a monomorphism
$\mathcal{Z} \to \mathcal{X}$ such that $|\mathcal{Z}|$ is a singleton
and such that the image of $|\mathcal{Z}|$ in $|\mathcal{X}|$ is $x$,
\item there exists an algebraic stack $\mathcal{Z}$, a monomorphism
$f : \mathcal{Z} \to \mathcal{X}$, and a surjective flat morphism
$z : \Spec(k) \to \mathcal{Z}$ where $k$ is a field such that
$x = f(z)$.
\end{enumerate}
Moreover, if these conditions hold, then there exists a unique
strictly full subcategory $\mathcal{Z}_x \subset \mathcal{X}$
such that $\mathcal{Z}_x$ is a reduced, locally Noetherian algebraic
stack and $|\mathcal{Z}_x|$ is a singleton which maps to $x$
via the map $|\mathcal{Z}_x| \to |\mathcal{X}|$.
\end{lemma}
\begin{proof}
If $\mathcal{Z} \to \mathcal{X}$ is as in (1), then
$\mathcal{Z}_{red} \to \mathcal{X}$ is as in (2). (See
Section \ref{section-reduced}
for the notion of the reduction of an algebraic stack.)
Hence (1) implies (2).
It is immediate that (2) implies (1).
The equivalence of (2) and (3) is immediate from
Lemma \ref{lemma-unique-point}.
\medskip\noindent
At this point we've seen the equivalence of (1) -- (3).
Pick a monomorphism $f : \mathcal{Z} \to \mathcal{X}$ as in (2).
Note that this implies that $f$ is fully faithful, see
Lemma \ref{lemma-monomorphism}.
Denote $\mathcal{Z}' \subset \mathcal{X}$ the essential image of the functor
$f$. Then $f : \mathcal{Z} \to \mathcal{Z}'$ is an equivalence and hence
$\mathcal{Z}'$ is an algebraic stack, see
Algebraic Stacks, Lemma \ref{algebraic-lemma-equivalent}.
Apply
Lemma \ref{lemma-improve-unique-point}
to get a strictly full subcategory $\mathcal{Z}_x \subset \mathcal{Z}'$
as in the statement of the lemma.
This proves all the statements of the lemma except for uniqueness.
\medskip\noindent
In order to prove the uniqueness suppose that
$\mathcal{Z}_x \subset \mathcal{X}$
and
$\mathcal{Z}'_x \subset \mathcal{X}$
are two strictly full subcategories as in the statement of the lemma.
Then the projections
$$
\mathcal{Z}'_x
\longleftarrow
\mathcal{Z}'_x \times_\mathcal{X} \mathcal{Z}_x
\longrightarrow
\mathcal{Z}_x
$$
are monomorphisms. The algebraic stack in the middle is nonempty by
Lemma \ref{lemma-points-cartesian}.
Hence the two projections are isomorphisms by
Lemma \ref{lemma-monomorphism-into-point}
and we win.
\end{proof}
\noindent
Having explained the above we can now make the following definition.
\begin{definition}
\label{definition-residual-gerbe}
Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$.
\begin{enumerate}
\item We say the {\it residual gerbe of $\mathcal{X}$ at $x$ exists}
if the equivalent conditions (1), (2), and (3) of
Lemma \ref{lemma-residual-gerbe}
hold.
\item If the residual gerbe of $\mathcal{X}$ at $x$ exists, then the
{\it residual gerbe of $\mathcal{X}$ at $x$}\footnote{This clashes with
\cite{LM-B} in spirit, but not in fact. Namely, in Chapter 11 they associate
to any point on any quasi-separated algebraic stack a gerbe (not necessarily
algebraic) which they call the residual gerbe. We will see in
Morphisms of Stacks, Lemma
\ref{stacks-morphisms-lemma-every-point-residual-gerbe}
that on a quasi-separated algebraic stack every point
has a residual gerbe in our sense which is then equivalent to theirs. For
more information on this topic see
\cite[Appendix B]{rydh_etale_devissage}.}
is the strictly full
subcategory $\mathcal{Z}_x \subset \mathcal{X}$ constructed in
Lemma \ref{lemma-residual-gerbe}.
\end{enumerate}
\end{definition}
\noindent
In particular we know that $\mathcal{Z}_x$ (if it exists) is a
locally Noetherian, reduced algebraic stack and that there exists a
field and a surjective, flat, locally finitely presented morphism
$$
\Spec(k) \longrightarrow \mathcal{Z}_x.
$$
We will see in
Morphisms of Stacks, Lemma
\ref{stacks-morphisms-lemma-noetherian-singleton-stack-gerbe}
that $\mathcal{Z}_x$ is a gerbe. Existence of residual gerbes is
discussed in Morphisms of Stacks, Section
\ref{stacks-morphisms-section-existence-residual-gerbes}.
It turns out that $\mathcal{Z}_x$
is a regular algebraic stack as follows from the following lemma.
\begin{lemma}
\label{lemma-residual-gerbe-regular}
A reduced, locally Noetherian algebraic stack $\mathcal{Z}$ such that
$|\mathcal{Z}|$ is a singleton is regular.
\end{lemma}
\begin{proof}
Let $W \to \mathcal{Z}$ be a surjective smooth morphism where $W$ is a scheme.
Let $k$ be a field and let $\Spec(k) \to \mathcal{Z}$ be surjective,
flat, and locally of finite presentation (see
Lemma \ref{lemma-unique-point-better}).
The algebraic space $T = W \times_\mathcal{Z} \Spec(k)$ is
smooth over $k$ in particular regular, see
Spaces over Fields, Lemma \ref{spaces-over-fields-lemma-smooth-regular}.
Since $T \to W$ is locally of finite presentation, flat, and surjective it
follows that $W$ is regular, see
Descent on Spaces, Lemma \ref{spaces-descent-lemma-descend-regular}.
By definition this means that $\mathcal{Z}$ is regular.
\end{proof}
\begin{lemma}
\label{lemma-residual-gerbe-points}
Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$.
Assume that the residual gerbe $\mathcal{Z}_x$ of $\mathcal{X}$ exists.
Let $f : \Spec(K) \to \mathcal{X}$ be a morphism where $K$ is a field
in the equivalence class of $x$. Then $f$ factors through the inclusion
morphism $\mathcal{Z}_x \to \mathcal{X}$.
\end{lemma}
\begin{proof}
Choose a field $k$ and a surjective flat locally finite presentation
morphism $\Spec(k) \to \mathcal{Z}_x$. Set
$T = \Spec(K) \times_\mathcal{X} \mathcal{Z}_x$. By
Lemma \ref{lemma-points-cartesian}
we see that $T$ is nonempty. As $\mathcal{Z}_x \to \mathcal{X}$
is a monomorphism we see that $T \to \Spec(K)$ is a monomorphism.
Hence by
Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-monomorphism-toward-field}
we see that $T = \Spec(K)$ which proves the lemma.
\end{proof}
\begin{lemma}
\label{lemma-residual-gerbe-unique}
Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$.
Let $\mathcal{Z}$ be an algebraic stack satisfying the equivalent conditions of
Lemma \ref{lemma-unique-point-better}
and let $\mathcal{Z} \to \mathcal{X}$ be a monomorphism such that the image
of $|\mathcal{Z}| \to |\mathcal{X}|$ is $x$. Then the residual gerbe
$\mathcal{Z}_x$ of $\mathcal{X}$ at $x$ exists and
$\mathcal{Z} \to \mathcal{X}$ factors as
$\mathcal{Z} \to \mathcal{Z}_x \to \mathcal{X}$ where the first arrow
is an equivalence.
\end{lemma}
\begin{proof}
Let $\mathcal{Z}_x \subset \mathcal{X}$ be the full subcategory corresponding
to the essential image of the functor $\mathcal{Z} \to \mathcal{X}$.
Then $\mathcal{Z} \to \mathcal{Z}_x$ is an equivalence, hence
$\mathcal{Z}_x$ is an algebraic stack, see
Algebraic Stacks, Lemma \ref{algebraic-lemma-equivalent}.
Since $\mathcal{Z}_x$ inherits all the properties of $\mathcal{Z}$ from
this equivalence it is clear from the uniqueness in
Lemma \ref{lemma-residual-gerbe}
that $\mathcal{Z}_x$ is the residual gerbe of $\mathcal{X}$ at $x$.
\end{proof}
\begin{lemma}
\label{lemma-residual-gerbe-functorial}
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
$$
\xymatrix{
\mathcal{X} \ar[d]_f & \mathcal{Z}_x \ar[l] \ar[d] \\
\mathcal{Y} & \mathcal{Z}_y \ar[l]
}
$$
\end{lemma}
\begin{proof}
Choose a field $k$ and a surjective, flat, locally finitely presented
morphism $\Spec(k) \to \mathcal{Z}_x$. The morphism
$\Spec(k) \to \mathcal{Y}$ factors through $\mathcal{Z}_y$ by
Lemma \ref{lemma-residual-gerbe-points}.
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 \ref{lemma-monomorphism-into-point}.
\end{proof}
\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}
```

## Comments (2)

## Add a comment on tag `06ML`

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).

All contributions are licensed under the GNU Free Documentation License.