The Stacks project

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 $\mathcal{X} \to \mathcal{Y}$. $\square$

Comments (3)

Comment #1214 by JuanPablo on

I do not understand this step:

is a morphism in , but then it says that is a morphism in to lift it locally to .

I think this lemma is false as stated, take for example 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 is just the 2-category of groupoids.

The conditions for a functor to be a gerbe becomes: a) is essentially surjective and b) is full. And the condition over here becomes: is essentially surjective.

Under the hypothesis of the lemma is essentially surjective. But we may take such that fullness of is trivial, the discrete category with objects . And if is essentially surjective, so is , but is not necessarily full.

If is a gerbe then (a morphism isomorphic to) in the proof may be lifted locally to morphisms in 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 is discrete, then the condition that identity morphisms in lift still imposes a condition on .

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.

Post a comment

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 toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 06P4. Beware of the difference between the letter 'O' and the digit '0'.