The Stacks project

Lemma 8.6.11. Let $\mathcal{C}$ be a site. Let $F : \mathcal{S} \to \mathcal{T}$ be a $1$-morphism of categories fibred in groupoids over $\mathcal{C}$. Assume that

  1. $\mathcal{T}$ is a stack in groupoids over $\mathcal{C}$,

  2. for every $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ the functor $\mathcal{S}_ U \to \mathcal{T}_ U$ of fibre categories is faithful,

  3. for each $U$ and each $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{T}_ U)$ the presheaf

    \[ (h : V \to U) \longmapsto \{ (x, f) \mid x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ V), f : F(x) \to f^*y\text{ over }V\} /\cong \]

    is a sheaf on $\mathcal{C}/U$.

Then $\mathcal{S}$ is a stack in groupoids over $\mathcal{C}$.

Proof. We have to prove descent for morphisms and descent for objects.

Descent for morphisms. Let $\{ U_ i \to U\} $ be a covering of $\mathcal{C}$. Let $x, x'$ be objects of $\mathcal{S}$ over $U$. For each $i$ let $\alpha _ i : x|_{U_ i} \to x'|_{U_ i}$ be a morphism over $U_ i$ such that $\alpha _ i$ and $\alpha _ j$ restrict to the same morphism $x|_{U_ i \times _ U U_ j} \to x'|_{U_ i \times _ U U_ j}$. Because $\mathcal{T}$ is a stack in groupoids, there is a morphism $\beta : F(x) \to F(x')$ over $U$ whose restriction to $U_ i$ is $F(\alpha _ i)$. Then we can think of $\xi = (x, \beta )$ and $\xi ' = (x', \text{id}_{F(x')})$ as sections of the presheaf associated to $y = F(x')$ over $U$ in assumption (3). On the other hand, the restrictions of $\xi $ and $\xi '$ to $U_ i$ are $(x|_{U_ i}, F(\alpha _ i))$ and $(x'|_{U_ i}, \text{id}_{F(x'|_{U_ i})})$. These are isomorphic to each other by the morphism $\alpha _ i$. Thus $\xi $ and $\xi '$ are isomorphic by assumption (3). This means there is a morphism $\alpha : x \to x'$ over $U$ with $F(\alpha ) = \beta $. Since $F$ is faithful on fibre categories we obtain $\alpha |_{U_ i} = \alpha _ i$.

Descent of objects. Let $\{ U_ i \to U\} $ be a covering of $\mathcal{C}$. Let $(x_ i, \varphi _{ij})$ be a descent datum for $\mathcal{S}$ with respect to the given covering. Because $\mathcal{T}$ is a stack in groupoids, there is an object $y$ in $\mathcal{T}_ U$ and isomorphisms $\beta _ i : F(x_ i) \to y|_{U_ i}$ such that $F(\varphi _{ij}) = \beta _ j|_{U_ i \times _ U U_ j} \circ (\beta _ i|_{U_ i \times _ U U_ j})^{-1}$. Then $(x_ i, \beta _ i)$ are sections of the presheaf associated to $y$ over $U$ defined in assumption (3). Moreover, $\varphi _{ij}$ defines an isomorphism from the pair $(x_ i, \beta _ i)|_{U_ i \times _ U U_ j}$ to the pair $(x_ j, \beta _ j)|_{U_ i \times _ U U_ j}$. Hence by assumption (3) there exists a pair $(x, \beta )$ over $U$ whose restriction to $U_ i$ is isomorphic to $(x_ i, \beta _ i)$. This means there are morphisms $\alpha _ i : x_ i \to x|_{U_ i}$ with $\beta _ i = \beta |_{U_ i} \circ F(\alpha _ i)$. Since $F$ is faithful on fibre categories a calculation shows that $\varphi _{ij} = \alpha _ j|_{U_ i \times _ U U_ j} \circ (\alpha _ i|_{U_ i \times _ U U_ j})^{-1}$. This finishes the proof. $\square$


Comments (0)


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 0CKJ. Beware of the difference between the letter 'O' and the digit '0'.