The Stacks project

Proof. Omitted. Hint: Show that effectivity of descent corresponds exactly to the sheaf condition. $\square$

Proof. Omitted. $\square$

Proof. By Categories, Lemma 4.39.5 we see that a category $\mathcal{S}$ over $\mathcal{C}$ is fibred in setoids over $\mathcal{C}$ if and only if it is equivalent over $\mathcal{C}$ to a category fibred in sets. Hence we see that $\mathcal{S}_1$ is fibred in setoids over $\mathcal{C}$ if and only if $\mathcal{S}_2$ is fibred in setoids over $\mathcal{C}$. Hence now the lemma follows from Lemma 8.6.3. $\square$

Proof. This is clear from Categories, Lemmas 4.35.7 and 4.39.4 and Lemmas 8.5.2 and 8.4.6. $\square$

Proof. Immediate from the explicit description of the $2$-fibre product in Categories, Lemma 4.32.3. $\square$

Proof. This is a special case of Lemma 8.6.7 as $f_2$ is faithful. $\square$

Proof. We may assume that $\mathcal{T}_2$ is the category $\mathcal{S}_2 \times _{\mathcal{S}_1} \mathcal{T}_1$ described in Categories, Lemma 4.32.3. By Categories, Lemma 4.35.9 the faithfulness of $G, G'$ can be checked on fibre categories. Suppose that $y, y'$ are objects of $\mathcal{T}_1$ over the object $U$ of $\mathcal{C}$. Let $\alpha , \beta : y \to y'$ be morphisms of $(\mathcal{T}_1)_ U$ such that $G(\alpha ) = G(\beta )$. Our object is to show that $\alpha = \beta $. Considering instead $\gamma = \alpha ^{-1} \circ \beta $ we see that $G(\gamma ) = \text{id}_{G(y)}$ and we have to show that $\gamma = \text{id}_ y$. By assumption we can find a covering $\{ U_ i \to U\} $ such that $G(y)|_{U_ i}$ is in the essential image of $F :(\mathcal{S}_2)_{U_ i} \to (\mathcal{S}_1)_{U_ i}$. Since it suffices to show that $\gamma |_{U_ i} = \text{id}$ for each $i$, we may therefore assume that we have $f : F(x) \to G(y)$ for some object $x$ of $\mathcal{S}_2$ over $U$ and morphisms $f$ of $(\mathcal{S}_1)_ U$. In this case we get a morphism

in the fibre category of $\mathcal{S}_2 \times _{\mathcal{S}_1} \mathcal{T}_1$ over $U$ whose image under $G'$ in $\mathcal{S}_1$ is $\text{id}_ x$. As $G'$ is faithful we conclude that $\gamma = \text{id}_ y$ and we win. $\square$

Proof. We may assume that $\mathcal{T}_2$ is the category $\mathcal{S}_2 \times _{\mathcal{S}_1} \mathcal{T}_1$ described in Categories, Lemma 4.32.3. Pick $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $y \in \mathop{\mathrm{Ob}}\nolimits ((\mathcal{T}_1)_ U)$. We have to show that the sheaf $\mathit{Aut}(y)$ on $\mathcal{C}/U$ is trivial. To to this we may replace $U$ by the members of a covering of $U$. Hence by assumption (2) we may assume that there exists an object $x \in \mathop{\mathrm{Ob}}\nolimits ((\mathcal{S}_2)_ U)$ and an isomorphism $f : F(x) \to G(y)$. Then $y' = (U, x, y, f)$ is an object of $\mathcal{T}_2$ over $U$ which is mapped to $y$ under the projection $\mathcal{T}_2 \to \mathcal{T}_1$. Because $F$ is fully faithful by (1) the map $\mathit{Aut}(y') \to \mathit{Aut}(y)$ is surjective, use the explicit description of morphisms in $\mathcal{T}_2$ in Categories, Lemma 4.32.3. Since by (3) the sheaf $\mathit{Aut}(y')$ is trivial we get the result of the lemma. $\square$

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. $\square$