Lemma 8.11.7. Let $\mathcal{C}$ be a site. Let

$\xymatrix{ \mathcal{X}' \ar[r]_{G'} \ar[d]_{F'} & \mathcal{X} \ar[d]^ F \\ \mathcal{Y}' \ar[r]^ G & \mathcal{Y} }$

be a $2$-cartesian diagram of stacks in groupoids over $\mathcal{C}$. If for every $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{Y}_ U)$ there exists a covering $\{ U_ i \to U\}$ such that $x|_{U_ i}$ is in the essential image of $G : \mathcal{Y}'_{U_ i} \to \mathcal{Y}_{U_ i}$ and $\mathcal{X}'$ is a gerbe over $\mathcal{Y}'$, then $\mathcal{X}$ is a gerbe over $\mathcal{Y}$.

Proof. By the uniqueness property of a $2$-fibre product may assume that $\mathcal{X}' = \mathcal{Y}' \times _\mathcal {Y} \mathcal{X}$ as in Categories, Lemma 4.32.3. Let us prove properties (2)(a) and (2)(b) of Lemma 8.11.3 for $\mathcal{X} \to \mathcal{Y}$.

Let $y$ be an object of $\mathcal{Y}$ lying over the object $U$ of $\mathcal{C}$. By assumption there exists a covering $\{ U_ i \to U\}$ of $U$ and objects $y'_ i \in \mathcal{Y}'_{U_ i}$ with $G(y'_ i) \cong y|_{U_ i}$. By (2)(a) for $\mathcal{X}' \to \mathcal{Y}'$ there exist coverings $\{ U_{ij} \to U_ i\}$ and objects $x'_{ij}$ of $\mathcal{X}'$ over $U_{ij}$ with $F'(x'_{ij})$ isomorphic to the restriction of $y'_ i$ to $U_{ij}$. Then $\{ U_{ij} \to U\}$ is a covering of $\mathcal{C}$ and $G'(x'_{ij})$ are objects of $\mathcal{X}$ over $U_{ij}$ whose images in $\mathcal{Y}$ are isomorphic to the restrictions $y|_{U_{ij}}$. This proves (2)(a) for $\mathcal{X} \to \mathcal{Y}$.

Let $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, let $x_1, x_2$ be objects of $\mathcal{X}$ over $U$, and let $b : F(x_1) \to F(x_2)$ be a morphism in $\mathcal{Y}_ U$. By assumption we may choose a covering $\{ U_ i \to U\}$ and objects $y'_ i$ of $\mathcal{Y}'$ over $U_ i$ such that there exist isomorphisms $\alpha _ i : G(y'_ i) \to F(x_1)|_{U_ i}$. Then we get objects

$x'_{1i} = (U_ i, y'_ i, x_1|_{U_ i}, \alpha _ i) \quad \text{and}\quad x'_{2i} = (U_ i, y'_ i, x_2|_{U_ i}, b|_{U_ i} \circ \alpha _ i)$

of $\mathcal{X}'$ over $U_ i$. The identity morphism on $y'_ i$ is a morphism $F'(x'_{1i}) \to F'(x'_{2i})$. By (2)(b) for $\mathcal{X}' \to \mathcal{Y}'$ there exist coverings $\{ U_{ij} \to U_ i\}$ and morphisms $a'_{ij} : x'_{1i}|_{U_{ij}} \to x'_{2i}|_{U_{ij}}$ such that $F'(a'_{ij}) = \text{id}_{y'_ i}|_{U_{ij}}$. Unwinding the definition of morphisms in $\mathcal{Y}' \times _\mathcal {Y} \mathcal{X}$ we see that $G'(a'_{ij}) : x_1|_{U_{ij}} \to x_2|_{U_{ij}}$ are the morphisms we're looking for, i.e., (2)(b) holds for $\mathcal{X} \to \mathcal{Y}$. $\square$

Comment #1214 by JuanPablo on

I do not understand this step:

$b:F(x_1)\rightarrow F(x_2)$ is a morphism in $\mathcal{Y}$, but then it says that $b|_{U_i}$ is a morphism in $\mathcal{Y}'$ to lift it locally to $\mathcal{X}'$.

I think this lemma is false as stated, take for example $\mathcal{C}$ to be a category with one object and only the identity morphism for the object with chaotic topology. Then the 2-category of stacks in groupoids over $\mathcal{C}$ is just the 2-category of groupoids.

The conditions for a functor $F:\mathcal{X}\rightarrow \mathcal{Y}$ to be a gerbe becomes: a) $F$ is essentially surjective and b) $F$ is full. And the condition over $G$ here becomes: $G$ is essentially surjective.

Under the hypothesis of the lemma $F$ is essentially surjective. But we may take $\mathcal{Y}'$ such that fullness of $F'$ is trivial, the discrete category with objects $\text{Ob}\mathcal{Y}$. And if $F$ is essentially surjective, so is $F'$, but $F$ is not necessarily full.

If $G$ is a gerbe then (a morphism isomorphic to) $b|_{U_i}$ in the proof may be lifted locally to morphisms in $\mathcal{Y}'$ and the rest of the proof follows.

Comment #1244 by on

Hi! Thanks very much for pointing out this mistake in the proof. But, I think the lemma is correct anyway. Please see here for the fix (hopefully correct).

Your counterexample won't work for the following reason: If $\mathcal{Y}'$ is discrete, then the condition that identity morphisms in $\mathcal{Y}'$ lift still imposes a condition on $F$.

Comment #1253 by JuanPablo on

Ah yes, you are right. Sorry for the bad example.

I had somehow convinced myself that a functor with discrete codomain is full, which is not true.

I think the proof is ok now.

There are also:

• 2 comment(s) on Section 8.11: Gerbes

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