The Stacks project

Proof. Assume $\mathcal{S}_1$ is a gerbe over $\mathcal{C}$. By Lemma 8.5.4 we see $\mathcal{S}_2$ is a stack in groupoids over $\mathcal{C}$. Let $F : \mathcal{S}_1 \to \mathcal{S}_2$, $G : \mathcal{S}_2 \to \mathcal{S}_1$ be equivalences of categories over $\mathcal{C}$. Given $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ we see that there exists a covering $\{ U_ i \to U\} $ such that $(\mathcal{S}_1)_{U_ i}$ is nonempty. Applying $F$ we see that $(\mathcal{S}_2)_{U_ i}$ is nonempty. Given $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $x, y \in \mathop{\mathrm{Ob}}\nolimits ((\mathcal{S}_2)_ U)$ there exists a covering $\{ U_ i \to U\} $ in $\mathcal{C}$ such that $G(x)|_{U_ i} \cong G(y)|_{U_ i}$ in $(\mathcal{S}_1)_{U_ i}$. By Categories, Lemma 4.35.9 this implies $x|_{U_ i} \cong y|_{U_ i}$ in $(\mathcal{S}_2)_{U_ i}$. $\square$

Proof. By Categories, Lemma 4.35.16 there exists a factorization $F = F' \circ a$ where $a : \mathcal{X} \to \mathcal{X}'$ is an equivalence of categories over $\mathcal{C}$ and $F'$ is fibred in groupoids. By Categories, Lemma 4.35.17 given any two such factorizations $F = F' \circ a = F'' \circ b$ we have that $\mathcal{X}'$ is equivalent to $\mathcal{X}''$ as categories over $\mathcal{Y}$. Hence Lemma 8.11.2 guarantees that the condition (1) is independent of the choice of the factorization. Moreover, this means that we may assume $\mathcal{X}' = \mathcal{X} \times _{F, \mathcal{Y}, \text{id}} \mathcal{Y}$ as in the proof of Categories, Lemma 4.35.16

Let us prove that (a) and (b) imply that $\mathcal{X}' \to \mathcal{Y}$ is a gerbe. First of all, by Lemma 8.10.5 we see that $\mathcal{X}' \to \mathcal{Y}$ is a stack in groupoids. Next, let $y$ be an object of $\mathcal{Y}$ lying over $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. By (a) we can find a covering $\{ U_ i \to U\} $ in $\mathcal{C}$ and objects $x_ i$ of $\mathcal{X}$ over $U_ i$ and isomorphisms $f_ i : F(x_ i) \to y|_{U_ i}$ in $\mathcal{Y}_{U_ i}$. Then $(U_ i, x_ i, y|_{U_ i}, f_ i)$ are objects of $\mathcal{X}'_{U_ i}$, i.e., the second condition of Definition 8.11.1 holds. Finally, let $(U, x, y, f)$ and $(U, x', y, f')$ be objects of $\mathcal{X}'$ lying over the same object $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{Y})$. Set $b = (f')^{-1} \circ f$. By condition (b) we can find a covering $\{ U_ i \to U\} $ and isomorphisms $a_ i : x|_{U_ i} \to x'|_{U_ i}$ in $\mathcal{X}_{U_ i}$ with $F(a_ i) = b|_{U_ i}$. Then

is a morphism in $\mathcal{X}'_{U_ i}$ as desired. This proves that (2) implies (1).

To prove that (1) implies (2) one reads the arguments in the preceding paragraph backwards. Details omitted. $\square$

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 $\mathcal{Y}' \times _\mathcal {Y} \mathcal{X} \to \mathcal{Y}'$.

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 $x_ i \in \mathcal{X}_{U_ i}$ with isomorphisms $\alpha _ i : G(y')|_{U_ i} \to F(x_ i)$. Then $(U_ i, y'|_{U_ i}, x_ i, \alpha _ i)$ is an object of $\mathcal{Y}' \times _\mathcal {Y} \mathcal{X}$ over $U_ i$ whose image in $\mathcal{Y}'$ is $y'|_{U_ i}$. Thus (2)(a) holds.

Let $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, let $x'_1, x'_2$ be objects of $\mathcal{Y}' \times _\mathcal {Y} \mathcal{X}$ over $U$, and let $b' : F'(x'_1) \to F'(x'_2)$ be a morphism in $\mathcal{Y}'_ U$. Write $x'_ i = (U, y'_ i, x_ i, \alpha _ i)$. Note that $F'(x'_ i) = x_ i$ and $G'(x'_ i) = y'_ i$. By assumption there exists a covering $\{ U_ i \to U\} $ in $\mathcal{C}$ and morphisms $a_ i : x_1|_{U_ i} \to x_2|_{U_ i}$ in $\mathcal{X}_{U_ i}$ with $F(a_ i) = G(b')|_{U_ i}$. Then $(b'|_{U_ i}, a_ i)$ is a morphism $x'_1|_{U_ i} \to x'_2|_{U_ i}$ as required in (2)(b). $\square$

Proof. Let us prove properties (2)(a) and (2)(b) of Lemma 8.11.3 for $\mathcal{X} \to \mathcal{Z}$.

Let $z$ be an object of $\mathcal{Z}$ lying over the object $U$ of $\mathcal{C}$. By assumption on $G$ there exists a covering $\{ U_ i \to U\} $ of $U$ and objects $y_ i \in \mathcal{Y}_{U_ i}$ such that $G(y_ i) \cong z|_{U_ i}$. By assumption on $F$ there exist coverings $\{ U_{ij} \to U_ i\} $ and objects $x_{ij} \in \mathcal{X}_{U_{ij}}$ such that $F(x_{ij}) \cong y_ i|_{U_{ij}}$. Then $\{ U_{ij} \to U\} $ is a covering of $\mathcal{C}$ and $(G \circ F)(x_{ij}) \cong z|_{U_{ij}}$. Thus (2)(a) holds.

Let $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, let $x_1, x_2$ be objects of $\mathcal{X}$ over $U$, and let $c : (G \circ F)(x_1) \to (G \circ F)(x_2)$ be a morphism in $\mathcal{Z}_ U$. By assumption on $G$ there exists a covering $\{ U_ i \to U\} $ of $U$ and morphisms $b_ i : F(x_1)|_{U_ i} \to F(x_2)|_{U_ i}$ in $\mathcal{Y}_{U_ i}$ such that $G(b_ i) = c|_{U_ i}$. By assumption on $F$ there exist coverings $\{ U_{ij} \to U_ i\} $ and morphisms $a_{ij} : x_1|_{U_{ij}} \to x_2|_{U_{ij}}$ in $\mathcal{X}_{U_{ij}}$ such that $F(a_{ij}) = b_ i|_{U_{ij}}$. Then $\{ U_{ij} \to U\} $ is a covering of $\mathcal{C}$ and $(G \circ F)(a_{ij}) = c|_{U_{ij}}$ as required in (2)(b). $\square$

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

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

Proof. Let $x, y$ be two objects of $\mathcal{S}$ with $U = p(x) = p(y)$.

If there is a morphism $\varphi : x \to y$ over $U$, then it is an isomorphism and then we indeed get an isomorphism $\mathit{Aut}(x) \to \mathit{Aut}(y)$ sending $\alpha $ to $\varphi \circ \alpha \circ \varphi ^{-1}$. Moreover, since we are assuming $\mathit{Aut}(x)$ is commutative, this isomorphism is independent of the choice of $\varphi $ by a simple computation: namely, if $\psi $ is a second such map, then

The upshot is a canonical isomorphism of sheaves $\mathit{Aut}(x) \to \mathit{Aut}(y)$. Furthermore, if there is a third object $z$ and a morphism $y \to z$ (and hence also a morphism $x \to z$), then the canonical isomorphisms $\mathit{Aut}(x) \to \mathit{Aut}(y)$, $\mathit{Aut}(y) \to \mathit{Aut}(z)$, and $\mathit{Aut}(x) \to \mathit{Aut}(z)$ are compatible in the sense that

commutes.

If there is no morphism from $x$ to $y$ over $U$, then we can choose a covering $\{ U_ i \to U\} $ such that there exist morphisms $x|_{U_ i} \to y|_{U_ i}$. This gives canonical isomorphisms

which agree over $U_ i \times _ U U_ j$ (by canonicity). By glueing of sheaves (Sites, Lemma 7.26.1) we get a unique isomorphism $\mathit{Aut}(x) \to \mathit{Aut}(y)$ whose restriction to any $U_ i$ is the canonical isomorphism of the previous paragraph. Similarly to the above these canonical isomorphisms satisfy a compatibility if we have a third object over $U$.

What if the fibre category of $\mathcal{S}$ over $U$ is empty? Well, in this case we can find a covering $\{ U_ i \to U\} $ and objects $x_ i$ of $\mathcal{S}$ over $U_ i$. Then we set $\mathcal{G}_ i = \mathit{Aut}(x_ i)$. By the above we obtain canonical isomorphisms

whose restrictions to $U_ i \times _ U U_ j \times _ U U_ k$ satisfy the cocycle condition explained in Sites, Section 7.26. By Sites, Lemma 7.26.4 we obtain a sheaf $\mathcal{G}$ over $U$ whose restriction to $U_ i$ gives $\mathcal{G}_ i$ in a manner compatible with the glueing maps $\varphi _{ij}$.

If $\mathcal{C}$ has a final object $U$, then this finishes the proof as we can take $\mathcal{G}$ equal to the sheaf we just constructed. In the general case we need to verify that the sheaves $\mathcal{G}$ constructed over varying $U$ are compatible in a canonical manner. This is omitted. $\square$