Definition 101.28.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. We say $\mathcal{X}$ is a *gerbe over* $\mathcal{Y}$ if $\mathcal{X}$ is a gerbe over $\mathcal{Y}$ as stacks in groupoids over $(\mathit{Sch}/S)_{fppf}$, see Stacks, Definition 8.11.4. We say an algebraic stack $\mathcal{X}$ is a *gerbe* if there exists a morphism $\mathcal{X} \to X$ where $X$ is an algebraic space which turns $\mathcal{X}$ into a gerbe over $X$.

## 101.28 Gerbes

An important type of algebraic stack are the stacks of the form $[B/G]$ where $B$ is an algebraic space and $G$ is a flat and locally finitely presented group algebraic space over $B$ (acting trivially on $B$), see Criteria for Representability, Lemma 97.18.3. It turns out that an algebraic stack is a gerbe when it locally in the fppf topology is of this form, see Lemma 101.28.7. In this section we briefly discuss this notion and the corresponding relative notion.

The condition that $\mathcal{X}$ be a gerbe over $\mathcal{Y}$ is defined purely in terms of the topology and category theory underlying the given algebraic stacks; but as we will see later this condition has geometric consequences. For example it implies that $\mathcal{X} \to \mathcal{Y}$ is surjective, flat, and locally of finite presentation, see Lemma 101.28.8. The absolute notion is trickier to parse, because it may not be at first clear that $X$ is well determined. Actually, it is.

Lemma 101.28.2. Let $\mathcal{X}$ be an algebraic stack. If $\mathcal{X}$ is a gerbe, then the sheafification of the presheaf

is an algebraic space and $\mathcal{X}$ is a gerbe over it.

**Proof.**
(In this proof the abuse of language introduced in Section 101.2 really pays off.) Choose a morphism $\pi : \mathcal{X} \to X$ where $X$ is an algebraic space which turns $\mathcal{X}$ into a gerbe over $X$. It suffices to prove that $X$ is the sheafification of the presheaf $\mathcal{F}$ displayed in the lemma. It is clear that there is a map $c : \mathcal{F} \to X$. We will use Stacks, Lemma 8.11.3 properties (2)(a) and (2)(b) to see that the map $c^\# : \mathcal{F}^\# \to X$ is surjective and injective, hence an isomorphism, see Sites, Lemma 7.11.2. Surjective: Let $T$ be a scheme and let $f : T \to X$. By property (2)(a) there exists an fppf covering $\{ h_ i : T_ i \to T\} $ and morphisms $x_ i : T_ i \to \mathcal{X}$ such that $f \circ h_ i$ corresponds to $\pi \circ x_ i$. Hence we see that $f|_{T_ i}$ is in the image of $c$. Injective: Let $T$ be a scheme and let $x, y : T \to \mathcal{X}$ be morphisms such that $c \circ x = c \circ y$. By (2)(b) we can find a covering $\{ T_ i \to T\} $ and morphisms $x|_{T_ i} \to y|_{T_ i}$ in the fibre category $\mathcal{X}_{T_ i}$. Hence the restrictions $x|_{T_ i}, y|_{T_ i}$ are equal in $\mathcal{F}(T_ i)$. This proves that $x, y$ give the same section of $\mathcal{F}^\# $ over $T$ as desired.
$\square$

Lemma 101.28.3. Let

be a fibre product of algebraic stacks. If $\mathcal{X}$ is a gerbe over $\mathcal{Y}$, then $\mathcal{X}'$ is a gerbe over $\mathcal{Y}'$.

**Proof.**
Immediate from the definitions and Stacks, Lemma 8.11.5.
$\square$

Lemma 101.28.4. Let $\mathcal{X} \to \mathcal{Y}$ and $\mathcal{Y} \to \mathcal{Z}$ be morphisms of algebraic stacks. If $\mathcal{X}$ is a gerbe over $\mathcal{Y}$ and $\mathcal{Y}$ is a gerbe over $\mathcal{Z}$, then $\mathcal{X}$ is a gerbe over $\mathcal{Z}$.

**Proof.**
Immediate from Stacks, Lemma 8.11.6.
$\square$

Lemma 101.28.5. Let

be a fibre product of algebraic stacks. If $\mathcal{Y}' \to \mathcal{Y}$ is surjective, flat, and locally of finite presentation and $\mathcal{X}'$ is a gerbe over $\mathcal{Y}'$, then $\mathcal{X}$ is a gerbe over $\mathcal{Y}$.

**Proof.**
Follows immediately from Lemma 101.27.13 and Stacks, Lemma 8.11.7.
$\square$

Lemma 101.28.6. Let $\pi : \mathcal{X} \to U$ be a morphism from an algebraic stack to an algebraic space and let $x : U \to \mathcal{X}$ be a section of $\pi $. Set $G = \mathit{Isom}_\mathcal {X}(x, x)$, see Definition 101.5.3. If $\mathcal{X}$ is a gerbe over $U$, then

there is a canonical equivalence of stacks in groupoids

\[ x_{can} : [U/G] \longrightarrow \mathcal{X}. \]where $[U/G]$ is the quotient stack for the trivial action of $G$ on $U$,

$G \to U$ is flat and locally of finite presentation, and

$U \to \mathcal{X}$ is surjective, flat, and locally of finite presentation.

**Proof.**
Set $R = U \times _{x, \mathcal{X}, x} U$. The morphism $R \to U \times U$ factors through the diagonal $\Delta _ U : U \to U \times U$ as it factors through $U \times _ U U = U$. Hence $R = G$ because

for the fourth equality use Categories, Lemma 4.31.12. Let $t, s : R \to U$ be the projections. The composition law $c : R \times _{s, U, t} R \to R$ constructed on $R$ in Algebraic Stacks, Lemma 94.16.1 agrees with the group law on $G$ (proof omitted). Thus Algebraic Stacks, Lemma 94.16.1 shows we obtain a canonical fully faithful $1$-morphism

of stacks in groupoids over $(\mathit{Sch}/S)_{fppf}$. To see that it is an equivalence it suffices to show that it is essentially surjective. To do this it suffices to show that any object of $\mathcal{X}$ over a scheme $T$ comes fppf locally from $x$ via a morphism $T \to U$, see Stacks, Lemma 8.4.8. However, this follows the condition that $\pi $ turns $\mathcal{X}$ into a gerbe over $U$, see property (2)(a) of Stacks, Lemma 8.11.3.

By Criteria for Representability, Lemma 97.18.3 we conclude that $G \to U$ is flat and locally of finite presentation. Finally, $U \to \mathcal{X}$ is surjective, flat, and locally of finite presentation by Criteria for Representability, Lemma 97.17.1. $\square$

Lemma 101.28.7. Let $\pi : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The following are equivalent

$\mathcal{X}$ is a gerbe over $\mathcal{Y}$, and

there exists an algebraic space $U$, a group algebraic space $G$ flat and locally of finite presentation over $U$, and a surjective, flat, and locally finitely presented morphism $U \to \mathcal{Y}$ such that $\mathcal{X} \times _\mathcal {Y} U \cong [U/G]$ over $U$.

**Proof.**
Assume (2). By Lemma 101.28.5 to prove (1) it suffices to show that $[U/G]$ is a gerbe over $U$. This is immediate from Groupoids in Spaces, Lemma 78.27.2.

Assume (1). Any base change of $\pi $ is a gerbe, see Lemma 101.28.3. As a first step we choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Thus we may assume that $\pi : \mathcal{X} \to V$ is a gerbe over a scheme. This means that there exists an fppf covering $\{ V_ i \to V\} $ such that the fibre category $\mathcal{X}_{V_ i}$ is nonempty, see Stacks, Lemma 8.11.3 (2)(a). Note that $U = \coprod V_ i \to V$ is surjective, flat, and locally of finite presentation. Hence we may replace $V$ by $U$ and assume that $\pi : \mathcal{X} \to U$ is a gerbe over a scheme $U$ and that there exists an object $x$ of $\mathcal{X}$ over $U$. By Lemma 101.28.6 we see that $\mathcal{X} = [U/G]$ over $U$ for some flat and locally finitely presented group algebraic space $G$ over $U$. $\square$

Lemma 101.28.8. Let $\pi : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. If $\mathcal{X}$ is a gerbe over $\mathcal{Y}$, then $\pi $ is surjective, flat, and locally of finite presentation.

**Proof.**
By Properties of Stacks, Lemma 100.5.4 and Lemmas 101.25.4 and 101.27.11 it suffices to prove to the lemma after replacing $\pi $ by a base change with a surjective, flat, locally finitely presented morphism $\mathcal{Y}' \to \mathcal{Y}$. By Lemma 101.28.7 we may assume $\mathcal{Y} = U$ is an algebraic space and $\mathcal{X} = [U/G]$ over $U$. Then $U \to [U/G]$ is surjective, flat, and locally of finite presentation, see Lemma 101.28.6. This implies that $\pi $ is surjective, flat, and locally of finite presentation by Properties of Stacks, Lemma 100.5.5 and Lemmas 101.25.5 and 101.27.12.
$\square$

Proposition 101.28.9. Let $\mathcal{X}$ be an algebraic stack. The following are equivalent

$\mathcal{X}$ is a gerbe, and

$\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is flat and locally of finite presentation.

**Proof.**
Assume (1). Choose a morphism $\mathcal{X} \to X$ into an algebraic space $X$ which turns $\mathcal{X}$ into a gerbe over $X$. Let $X' \to X$ be a surjective, flat, locally finitely presented morphism and set $\mathcal{X}' = X' \times _ X \mathcal{X}$. Note that $\mathcal{X}'$ is a gerbe over $X'$ by Lemma 101.28.3. Then both squares in

are fibre product squares, see Lemma 101.5.5. Hence to prove $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is flat and locally of finite presentation it suffices to do so after such a base change by Lemmas 101.25.4 and 101.27.11. Thus we can apply Lemma 101.28.7 to assume that $\mathcal{X} = [U/G]$. By Lemma 101.28.6 we see $G$ is flat and locally of finite presentation over $U$ and that $x : U \to [U/G]$ is surjective, flat, and locally of finite presentation. Moreover, the pullback of $\mathcal{I}_\mathcal {X}$ by $x$ is $G$ and we conclude that (2) holds by descent again, i.e., by Lemmas 101.25.4 and 101.27.11.

Conversely, assume (2). Choose a smooth presentation $\mathcal{X} = [U/R]$, see Algebraic Stacks, Section 94.16. Denote $G \to U$ the stabilizer group algebraic space of the groupoid $(U, R, s, t, c, e, i)$, see Groupoids in Spaces, Definition 78.16.2. By Lemma 101.5.7 we see that $G \to U$ is flat and locally of finite presentation as a base change of $\mathcal{I}_\mathcal {X} \to \mathcal{X}$, see Lemmas 101.25.3 and 101.27.3. Consider the following action

of $G$ on $R$. This action is free on $T$-valued points for any scheme $T$ as $R$ is a groupoid. Hence $R' = R/G$ is an algebraic space and the quotient morphism $\pi : R \to R'$ is surjective, flat, and locally of finite presentation by Bootstrap, Lemma 80.11.7. The projections $s, t : R \to U$ are $G$-invariant, hence we obtain morphisms $s' , t' : R' \to U$ such that $s = s' \circ \pi $ and $t = t' \circ \pi $. Since $s, t : R \to U$ are flat and locally of finite presentation we conclude that $s', t'$ are flat and locally of finite presentation, see Morphisms of Spaces, Lemmas 67.31.5 and Descent on Spaces, Lemma 74.16.1. Consider the morphism

We claim this is a monomorphism. Namely, suppose that $T$ is a scheme and that $a, b : T \to R'$ are morphisms which have the same image in $U \times U$. By definition of the quotient $R' = R/G$ there exists an fppf covering $\{ h_ j : T_ j \to T\} $ such that $a \circ h_ j = \pi \circ a_ j$ and $b \circ h_ j = \pi \circ b_ j$ for some morphisms $a_ j, b_ j : T_ j \to R$. Since $a_ j, b_ j$ have the same image in $U \times U$ we see that $g_ j = c(a_ j, i(b_ j))$ is a $T_ j$-valued point of $G$ such that $c(g_ j, b_ j) = a_ j$. In other words, $a_ j$ and $b_ j$ have the same image in $R'$ and the claim is proved. Since $j : R \to U \times U$ is a pre-equivalence relation (see Groupoids in Spaces, Lemma 78.11.2) and $R \to R'$ is surjective (as a map of sheaves) we see that $j' : R' \to U \times U$ is an equivalence relation. Hence Bootstrap, Theorem 80.10.1 shows that $X = U/R'$ is an algebraic space. Finally, we claim that the morphism

turns $\mathcal{X}$ into a gerbe over $X$. This follows from Groupoids in Spaces, Lemma 78.27.1 as $R \to R'$ is surjective, flat, and locally of finite presentation (if needed use Bootstrap, Lemma 80.4.6 to see this implies the required hypothesis). $\square$

Lemma 101.28.10. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks which makes $\mathcal{X}$ a gerbe over $\mathcal{Y}$. Then

$\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X}$ is flat and locally of finite presentation,

$\mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ is surjective, flat, and locally of finite presentation,

given algebraic spaces $T_ i$, $i = 1, 2$ and morphisms $x_ i : T_ i \to \mathcal{X}$, with $y_ i = f \circ x_ i$ the morphism

\[ T_1 \times _{x_1, \mathcal{X}, x_2} T_2 \longrightarrow T_1 \times _{y_1, \mathcal{Y}, y_2} T_2 \]is surjective, flat, and locally of finite presentation,

given an algebraic space $T$ and morphisms $x_ i : T \to \mathcal{X}$, $i = 1, 2$, with $y_ i = f \circ x_ i$ the morphism

\[ \mathit{Isom}_\mathcal {X}(x_1, x_2) \longrightarrow \mathit{Isom}_\mathcal {Y}(y_1, y_2) \]is surjective, flat, and locally of finite presentation.

**Proof.**
Proof of (1). Choose a scheme $Y$ and a surjective smooth morphism $Y \to \mathcal{Y}$. Set $\mathcal{X}' = \mathcal{X} \times _\mathcal {Y} Y$. By Lemma 101.5.5 we obtain cartesian squares

By Lemmas 101.25.4 and 101.27.11 it suffices to prove that $\mathcal{I}_{\mathcal{X}'} \to \mathcal{X}'$ is flat and locally of finite presentation. This follows from Proposition 101.28.9 (because $\mathcal{X}'$ is a gerbe over $Y$ by Lemma 101.28.3).

Proof of (2). With notation as above, note that we may assume that $\mathcal{X}' = [Y/G]$ for some group algebraic space $G$ flat and locally of finite presentation over $Y$, see Lemma 101.28.7. The base change of the morphism $\Delta : \mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ over $\mathcal{Y}$ by the morphism $Y \to \mathcal{Y}$ is the morphism $\Delta ' : \mathcal{X}' \to \mathcal{X}' \times _ Y \mathcal{X}'$. Hence it suffices to show that $\Delta '$ is surjective, flat, and locally of finite presentation (see Lemmas 101.25.4 and 101.27.11). In other words, we have to show that

is surjective, flat, and locally of finite presentation. This is true because the base change by the surjective, flat, locally finitely presented morphism $Y \to [Y/G \times _ Y G]$ is the morphism $G \to Y$.

Proof of (3). Observe that the diagram

is cartesian. Hence (3) follows from (2).

Proof of (4). This is true because

hence the morphism in (4) is a base change of the morphism in (3). $\square$

Proposition 101.28.11. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The following are equivalent

$\mathcal{X}$ is a gerbe over $\mathcal{Y}$, and

$f : \mathcal{X} \to \mathcal{Y}$ and $\Delta : \mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ are surjective, flat, and locally of finite presentation.

**Proof.**
The implication (1) $\Rightarrow $ (2) follows from Lemmas 101.28.8 and 101.28.10.

Assume (2). It suffices to prove (1) for the base change of $f$ by a surjective, flat, and locally finitely presented morphism $\mathcal{Y}' \to \mathcal{Y}$, see Lemma 101.28.5 (note that the base change of the diagonal of $f$ is the diagonal of the base change). Thus we may assume $\mathcal{Y}$ is a scheme $Y$. In this case $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is a base change of $\Delta $ and we conclude that $\mathcal{X}$ is a gerbe by Proposition 101.28.9. We still have to show that $\mathcal{X}$ is a gerbe over $Y$. Let $\mathcal{X} \to X$ be the morphism of Lemma 101.28.2 turning $\mathcal{X}$ into a gerbe over the algebraic space $X$ classifying isomorphism classes of objects of $\mathcal{X}$. It is clear that $f : \mathcal{X} \to Y$ factors as $\mathcal{X} \to X \to Y$. Since $f$ is surjective, flat, and locally of finite presentation, we conclude that $X \to Y$ is surjective as a map of fppf sheaves (for example use Lemma 101.27.13). On the other hand, $X \to Y$ is injective too: for any scheme $T$ and any two $T$-valued points $x_1, x_2$ of $X$ which map to the same point of $Y$, we can first fppf locally on $T$ lift $x_1, x_2$ to objects $\xi _1, \xi _2$ of $\mathcal{X}$ over $T$ and second deduce that $\xi _1$ and $\xi _2$ are fppf locally isomorphic by our assumption that $\Delta : \mathcal{X} \to \mathcal{X} \times _ Y \mathcal{X}$ is surjective, flat, and locally of finite presentation. Whence $x_1 = x_2$ by construction of $X$. Thus $X = Y$ and the proof is complete. $\square$

At this point we have developed enough machinery to prove that residual gerbes (when they exist) are gerbes.

Lemma 101.28.12. Let $\mathcal{Z}$ be a reduced, locally Noetherian algebraic stack such that $|\mathcal{Z}|$ is a singleton. Then $\mathcal{Z}$ is a gerbe over a reduced, locally Noetherian algebraic space $Z$ with $|Z|$ a singleton.

**Proof.**
By Properties of Stacks, Lemma 100.11.3 there exists a surjective, flat, locally finitely presented morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{Z}$ where $k$ is a field. Then $\mathcal{I}_ Z \times _\mathcal {Z} \mathop{\mathrm{Spec}}(k) \to \mathop{\mathrm{Spec}}(k)$ is representable by algebraic spaces and locally of finite type (as a base change of $\mathcal{I}_\mathcal {Z} \to \mathcal{Z}$, see Lemmas 101.5.1 and 101.17.3). Therefore it is locally of finite presentation, see Morphisms of Spaces, Lemma 67.28.7. Of course it is also flat as $k$ is a field. Hence we may apply Lemmas 101.25.4 and 101.27.11 to see that $\mathcal{I}_\mathcal {Z} \to \mathcal{Z}$ is flat and locally of finite presentation. We conclude that $\mathcal{Z}$ is a gerbe by Proposition 101.28.9. Let $\pi : \mathcal{Z} \to Z$ be a morphism to an algebraic space such that $\mathcal{Z}$ is a gerbe over $Z$. Then $\pi $ is surjective, flat, and locally of finite presentation by Lemma 101.28.8. Hence $\mathop{\mathrm{Spec}}(k) \to Z$ is surjective, flat, and locally of finite presentation as a composition, see Properties of Stacks, Lemma 100.5.2 and Lemmas 101.25.2 and 101.27.2. Hence by Properties of Stacks, Lemma 100.11.3 we see that $|Z|$ is a singleton and that $Z$ is locally Noetherian and reduced.
$\square$

Lemma 101.28.13. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. If $\mathcal{X}$ is a gerbe over $\mathcal{Y}$ then $f$ is a universal homeomorphism.

**Proof.**
By Lemma 101.28.3 the assumption on $f$ is preserved under base change. Hence it suffices to show that the map $|\mathcal{X}| \to |\mathcal{Y}|$ is a homeomorphism of topological spaces. Let $k$ be a field and let $y$ be an object of $\mathcal{Y}$ over $\mathop{\mathrm{Spec}}(k)$. By Stacks, Lemma 8.11.3 property (2)(a) there exists an fppf covering $\{ T_ i \to \mathop{\mathrm{Spec}}(k)\} $ and objects $x_ i$ of $\mathcal{X}$ over $T_ i$ with $f(x_ i) \cong y|_{T_ i}$. Choose an $i$ such that $T_ i \not= \emptyset $. Choose a morphism $\mathop{\mathrm{Spec}}(K) \to T_ i$ for some field $K$. Then $k \subset K$ and $x_ i|_ K$ is an object of $\mathcal{X}$ lying over $y|_ K$. Thus we see that $|\mathcal{Y}| \to |\mathcal{X}|$. is surjective. The map $|\mathcal{Y}| \to |\mathcal{X}|$ is also injective. Namely, if $x, x'$ are objects of $\mathcal{X}$ over $\mathop{\mathrm{Spec}}(k)$ whose images $f(x), f(x')$ become isomorphic (over an extension) in $\mathcal{Y}$, then Stacks, Lemma 8.11.3 property (2)(b) guarantees the existence of an extension of $k$ over which $x$ and $x'$ become isomorphic (details omitted). Hence $|\mathcal{X}| \to |\mathcal{Y}|$ is continuous and bijective and it suffices to show that it is also open. This follows from Lemmas 101.28.8 and 101.27.15.
$\square$

Lemma 101.28.14. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks such that $\mathcal{X}$ is a gerbe over $\mathcal{Y}$. If $\Delta _\mathcal {X}$ is quasi-compact, so is $\Delta _\mathcal {Y}$.

**Proof.**
Consider the diagram

By Proposition 101.28.11 we find that the arrow on the top left is surjective. Since the composition of the top horizontal arrows is quasi-compact, we conclude that the top right arrow is quasi-compact by Lemma 101.7.6. The square is cartesian and the right vertical arrow is surjective, flat, and locally of finite presentation. Thus we conclude by Lemma 101.27.16. $\square$

The following lemma tells us that residual gerbes exist for all points on any algebraic stack which is a gerbe.

Lemma 101.28.15. Let $\mathcal{X}$ be an algebraic stack. If $\mathcal{X}$ is a gerbe then for every $x \in |\mathcal{X}|$ the residual gerbe of $\mathcal{X}$ at $x$ exists.

**Proof.**
Let $\pi : \mathcal{X} \to X$ be a morphism from $\mathcal{X}$ into an algebraic space $X$ which turns $\mathcal{X}$ into a gerbe over $X$. Let $Z_ x \to X$ be the residual space of $X$ at $x$, see Decent Spaces, Definition 68.13.6. Let $\mathcal{Z} = \mathcal{X} \times _ X Z_ x$. By Lemma 101.28.3 the algebraic stack $\mathcal{Z}$ is a gerbe over $Z_ x$. Hence $|\mathcal{Z}| = |Z_ x|$ (Lemma 101.28.13) is a singleton. Since $\mathcal{Z} \to Z_ x$ is locally of finite presentation as a base change of $\pi $ (see Lemmas 101.28.8 and 101.27.3) we see that $\mathcal{Z}$ is locally Noetherian, see Lemma 101.17.5. Thus the residual gerbe $\mathcal{Z}_ x$ of $\mathcal{X}$ at $x$ exists and is equal to $\mathcal{Z}_ x = \mathcal{Z}_{red}$ the reduction of the algebraic stack $\mathcal{Z}$. Namely, we have seen above that $|\mathcal{Z}_{red}|$ is a singleton mapping to $x \in |\mathcal{X}|$, it is reduced by construction, and it is locally Noetherian (as the reduction of a locally Noetherian algebraic stack is locally Noetherian, details omitted).
$\square$

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