The Stacks project

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. $\square$