
## 8.6 Stacks in setoids

This is just a brief section saying that a stack in sets is the same thing as a sheaf of sets. Please consult Categories, Section 4.38 for notation.

Definition 8.6.1. Let $\mathcal{C}$ be a site.

1. A stack in setoids over $\mathcal{C}$ is a stack over $\mathcal{C}$ all of whose fibre categories are setoids.

2. A stack in sets, or a stack in discrete categories is a stack over $\mathcal{C}$ all of whose fibre categories are discrete.

From the discussion in Section 8.5 this is the same thing as a stack in groupoids whose fibre categories are setoids (resp. discrete). Moreover, it is also the same thing as a category fibred in setoids (resp. sets) which is a stack.

Lemma 8.6.2. Let $\mathcal{C}$ be a site. Under the equivalence

$\left\{ \begin{matrix} \text{the category of presheaves} \\ \text{of sets over }\mathcal{C} \end{matrix} \right\} \leftrightarrow \left\{ \begin{matrix} \text{the category of categories} \\ \text{fibred in sets over }\mathcal{C} \end{matrix} \right\}$

of Categories, Lemma 4.37.6 the stacks in sets correspond precisely to the sheaves.

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

Lemma 8.6.3. Let $\mathcal{C}$ be a site. Let $\mathcal{S}$ be a category fibred in setoids over $\mathcal{C}$. Then $\mathcal{S}$ is a stack in setoids if and only if the unique equivalent category $\mathcal{S}'$ fibred in sets (see Categories, Lemma 4.38.5) is a stack in sets. In other words, if and only if the presheaf

$U \longmapsto \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U)/\! \! \cong$

is a sheaf.

Proof. Omitted. $\square$

Lemma 8.6.4. Let $\mathcal{C}$ be a site. Let $\mathcal{S}_1$, $\mathcal{S}_2$ be categories over $\mathcal{C}$. Suppose that $\mathcal{S}_1$ and $\mathcal{S}_2$ are equivalent as categories over $\mathcal{C}$. Then $\mathcal{S}_1$ is a stack in setoids over $\mathcal{C}$ if and only if $\mathcal{S}_2$ is a stack in setoids over $\mathcal{C}$.

Proof. By Categories, Lemma 4.38.5 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. $\square$

The $2$-category of stacks in setoids over $\mathcal{C}$ is defined as follows.

Definition 8.6.5. Let $\mathcal{C}$ be a site. The $2$-category of stacks in setoids over $\mathcal{C}$ is the sub $2$-category of the $2$-category of stacks over $\mathcal{C}$ (see Definition 8.4.5) defined as follows:

1. Its objects will be stacks in setoids $p : \mathcal{S} \to \mathcal{C}$.

2. Its $1$-morphisms $(\mathcal{S}, p) \to (\mathcal{S}', p')$ will be functors $G : \mathcal{S} \to \mathcal{S}'$ such that $p' \circ G = p$. (Since every morphism is strongly cartesian every functor preserves them.)

3. Its $2$-morphisms $t : G \to H$ for $G, H : (\mathcal{S}, p) \to (\mathcal{S}', p')$ will be morphisms of functors such that $p'(t_ x) = \text{id}_{p(x)}$ for all $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S})$.

Note that any $2$-morphism is automatically an isomorphism, so that in fact the $2$-category of stacks in setoids over $\mathcal{C}$ is a (strict) $(2, 1)$-category.

Lemma 8.6.6. Let $\mathcal{C}$ be a site. The $2$-category of stacks in setoids over $\mathcal{C}$ has 2-fibre products, and they are described as in Categories, Lemma 4.31.3.

Lemma 8.6.7. Let $\mathcal{C}$ be a site. Let $\mathcal{S}, \mathcal{T}$ be stacks in groupoids over $\mathcal{C}$ and let $\mathcal{R}$ be a stack in setoids over $\mathcal{C}$. Let $f : \mathcal{T} \to \mathcal{S}$ and $g : \mathcal{R} \to \mathcal{S}$ be $1$-morphisms. If $f$ is faithful, then the $2$-fibre product

$\mathcal{T} \times _{f, \mathcal{S}, g} \mathcal{R}$

is a stack in setoids over $\mathcal{C}$.

Proof. Immediate from the explicit description of the $2$-fibre product in Categories, Lemma 4.31.3. $\square$

Lemma 8.6.8. Let $\mathcal{C}$ be a site. Let $\mathcal{S}$ be a stack in groupoids over $\mathcal{C}$ and let $\mathcal{S}_ i$, $i = 1, 2$ be stacks in setoids over $\mathcal{C}$. Let $f_ i : \mathcal{S}_ i \to \mathcal{S}$ be $1$-morphisms. Then the $2$-fibre product

$\mathcal{S}_1 \times _{f_1, \mathcal{S}, f_2} \mathcal{S}_2$

is a stack in setoids over $\mathcal{C}$.

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

Lemma 8.6.9. Let $\mathcal{C}$ be a site. Let

$\xymatrix{ \mathcal{T}_2 \ar[r] \ar[d]_{G'} & \mathcal{T}_1 \ar[d]^ G \\ \mathcal{S}_2 \ar[r]^ F & \mathcal{S}_1 }$

be a $2$-cartesian diagram of stacks in groupoids over $\mathcal{C}$. Assume

1. for every $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $x \in \mathop{\mathrm{Ob}}\nolimits ((\mathcal{S}_1)_ U)$ there exists a covering $\{ U_ i \to U\}$ such that $x|_{U_ i}$ is in the essential image of $F : (\mathcal{S}_2)_{U_ i} \to (\mathcal{S}_1)_{U_ i}$, and

2. $G'$ is faithful,

then $G$ is faithful.

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 4.31.3. By Categories, Lemma 4.34.8 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

$(1, \gamma ) : (U, x, y, f) \longrightarrow (U, x, y, f)$

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. $\square$

Lemma 8.6.10. Let $\mathcal{C}$ be a site. Let

$\xymatrix{ \mathcal{T}_2 \ar[r] \ar[d] & \mathcal{T}_1 \ar[d]^ G \\ \mathcal{S}_2 \ar[r]^ F & \mathcal{S}_1 }$

be a $2$-cartesian diagram of stacks in groupoids over $\mathcal{C}$. If

1. $F : \mathcal{S}_2 \to \mathcal{S}_1$ is fully faithful,

2. for every $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $x \in \mathop{\mathrm{Ob}}\nolimits ((\mathcal{S}_1)_ U)$ there exists a covering $\{ U_ i \to U\}$ such that $x|_{U_ i}$ is in the essential image of $F : (\mathcal{S}_2)_{U_ i} \to (\mathcal{S}_1)_{U_ i}$, and

3. $\mathcal{T}_2$ is a stack in setoids.

then $\mathcal{T}_1$ is a stack in setoids.

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 4.31.3. 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 4.31.3. Since by (3) the sheaf $\mathit{Aut}(y')$ is trivial we get the result of the lemma. $\square$

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. $\square$

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