The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

8.11 Gerbes

Gerbes are a special kind of stacks in groupoids.

Definition 8.11.1. A gerbe over a site $\mathcal{C}$ is a category $p : \mathcal{S} \to \mathcal{C}$ over $\mathcal{C}$ such that

  1. $p : \mathcal{S} \to \mathcal{C}$ is a stack in groupoids over $\mathcal{C}$ (see Definition 8.5.1),

  2. for $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ there exists a covering $\{ U_ i \to U\} $ in $\mathcal{C}$ such that $\mathcal{S}_{U_ i}$ is nonempty, and

  3. for $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $x, y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U)$ there exists a covering $\{ U_ i \to U\} $ in $\mathcal{C}$ such that $x|_{U_ i} \cong y|_{U_ i}$ in $\mathcal{S}_{U_ i}$.

In other words, a gerbe is a stack in groupoids such that any two objects are locally isomorphic and such that objects exist locally.

Lemma 8.11.2. 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 gerbe over $\mathcal{C}$ if and only if $\mathcal{S}_2$ is a gerbe over $\mathcal{C}$.

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 4.34.8 this implies $x|_{U_ i} \cong y|_{U_ i}$ in $(\mathcal{S}_2)_{U_ i}$. $\square$

We want to generalize the definition of gerbes a bit. Namely, let $F : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of stacks in groupoids over a site $\mathcal{C}$. We want to say what it means for $\mathcal{X}$ to be a gerbe over $\mathcal{Y}$. By Section 8.10 the category $\mathcal{Y}$ inherits the structure of a site from $\mathcal{C}$. A naive guess is: Just require that $\mathcal{X} \to \mathcal{Y}$ is a gerbe in the sense above. Except the notion so obtained is not invariants under replacing $\mathcal{X}$ by an equivalent stack in groupoids over $\mathcal{C}$; this is even the case for the property of being fibred in groupoids over $\mathcal{Y}$. However, it turns out that we can replace $\mathcal{X}$ by an equivalent stack in groupoids over $\mathcal{Y}$ which is fibred in groupoids over $\mathcal{Y}$, and then the property of being a gerbe over $\mathcal{Y}$ is independent of this choice. Here is the precise formulation.

Lemma 8.11.3. Let $\mathcal{C}$ be a site. Let $p : \mathcal{X} \to \mathcal{C}$ and $q : \mathcal{Y} \to \mathcal{C}$ be stacks in groupoids. Let $F : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of categories over $\mathcal{C}$. The following are equivalent

  1. For some (equivalently any) 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, the map $F' : \mathcal{X}' \to \mathcal{Y}$ is a gerbe (with the topology on $\mathcal{Y}$ inherited from $\mathcal{C}$).

  2. The following two conditions are satisfied

    1. for $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{Y})$ lying over $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ there exists a covering $\{ U_ i \to U\} $ in $\mathcal{C}$ and objects $x_ i$ of $\mathcal{X}$ over $U_ i$ such that $F(x_ i) \cong y|_{U_ i}$ in $\mathcal{Y}_{U_ i}$, and

    2. for $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, $x, x' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_ U)$, and $b : F(x) \to F(x')$ in $\mathcal{Y}_ U$ there exists a covering $\{ U_ i \to U\} $ in $\mathcal{C}$ and morphisms $a_ i : x|_{U_ i} \to x'|_{U_ i}$ in $\mathcal{X}_{U_ i}$ with $F(a_ i) = b|_{U_ i}$.

Proof. By Categories, Lemma 4.34.15 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 4.34.16 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 4.34.15

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

\[ (a_ i, \text{id}) : (U, x, y, f)|_{U_ i} \to (U, x', y, f')|_{U_ i} \]

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. $\square$

Definition 8.11.4. Let $\mathcal{C}$ be a site. Let $\mathcal{X}$ and $\mathcal{Y}$ be stacks in groupoids over $\mathcal{C}$. Let $F : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of categories over $\mathcal{C}$. We say $\mathcal{X}$ is a gerbe over $\mathcal{Y}$ if the equivalent conditions of Lemma 8.11.3 are satisfied.

This definition does not conflict with Definition 8.11.1 when $\mathcal{Y} = \mathcal{C}$ because in this case we may take $\mathcal{X}' = \mathcal{X}$ in part (1) of Lemma 8.11.3. Note that conditions (2)(a) and (2)(b) of Lemma 8.11.3 are quite close in spirit to conditions (2) and (3) of Definition 8.11.1. Namely, (2)(a) says that the map of presheaves of isomorphism classes of objects becomes a surjection after sheafification. Moreover, (2)(b) says that

\[ \mathit{Isom}_\mathcal {X}(x, x') \longrightarrow \mathit{Isom}_\mathcal {Y}(F(x), F(x')) \]

is a surjection of sheaves on $\mathcal{C}/U$ for any $U$ and $x, x' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_ U)$.

Lemma 8.11.5. Let $\mathcal{C}$ be a site. Let

\[ \xymatrix{ \mathcal{X}' \ar[r]_{G'} \ar[d]_{F'} & \mathcal{X} \ar[d]^ F \\ \mathcal{Y}' \ar[r]^ G & \mathcal{Y} } \]

be a $2$-fibre product of stacks in groupoids over $\mathcal{C}$. If $\mathcal{X}$ is a gerbe over $\mathcal{Y}$, then $\mathcal{X}'$ is a gerbe over $\mathcal{Y}'$.

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 4.31.3. 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). $\square$

Lemma 8.11.6. Let $\mathcal{C}$ be a site. Let $F : \mathcal{X} \to \mathcal{Y}$ and $G : \mathcal{Y} \to \mathcal{Z}$ be $1$-morphisms of stacks in groupoids over $\mathcal{C}$. If $\mathcal{X}$ is a gerbe over $\mathcal{Y}$ and $\mathcal{Y}$ is a gerbe over $\mathcal{Z}$, then $\mathcal{X}$ is a gerbe over $\mathcal{Z}$.

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). $\square$

Lemma 8.11.7. Let $\mathcal{C}$ be a site. Let

\[ \xymatrix{ \mathcal{X}' \ar[r]_{G'} \ar[d]_{F'} & \mathcal{X} \ar[d]^ F \\ \mathcal{Y}' \ar[r]^ G & \mathcal{Y} } \]

be a $2$-cartesian diagram of stacks in groupoids over $\mathcal{C}$. If for every $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{Y}_ U)$ there exists a covering $\{ U_ i \to U\} $ such that $x|_{U_ i}$ is in the essential image of $G : \mathcal{Y}'_{U_ i} \to \mathcal{Y}_{U_ i}$ and $\mathcal{X}'$ is a gerbe over $\mathcal{Y}'$, then $\mathcal{X}$ is a gerbe over $\mathcal{Y}$.

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 4.31.3. 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

\[ x'_{1i} = (U_ i, y'_ i, x_1|_{U_ i}, \alpha _ i) \quad \text{and}\quad x'_{2i} = (U_ i, y'_ i, x_2|_{U_ i}, b|_{U_ i} \circ \alpha _ i) \]

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

Gerbes all of whose automorphism sheaves are abelian play an important role in algebraic geometry.

Lemma 8.11.8. Let $p : \mathcal{S} \to \mathcal{C}$ be a gerbe over a site $\mathcal{C}$. Assume that for all $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U)$ the sheaf of groups $\mathit{Aut}(x) = \mathit{Isom}(x, x)$ on $\mathcal{C}/U$ is abelian. Then there exist

  1. a sheaf $\mathcal{G}$ of abelian groups on $\mathcal{C}$,

  2. for every $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and every $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U)$ an isomorphism $\mathcal{G}|_ U \to \mathit{Aut}(x)$

such that for every $U$ and every morphism $\varphi : x \to y$ in $\mathcal{S}_ U$ the diagram

\[ \xymatrix{ \mathcal{G}|_ U \ar[d] \ar@{=}[rr] & & \mathcal{G}|_ U \ar[d] \\ \mathit{Aut}(x) \ar[rr]^{\alpha \mapsto \varphi \circ \alpha \circ \varphi ^{-1}} & & \mathit{Aut}(y) } \]

is commutative.

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

\[ \varphi \circ \alpha \circ \varphi ^{-1} = \psi \circ \psi ^{-1} \circ \varphi \circ \alpha \circ \varphi ^{-1} = \psi \circ \alpha \circ \psi ^{-1} \circ \varphi \circ \varphi ^{-1} = \psi \circ \alpha \circ \psi ^{-1} \]

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

\[ \xymatrix{ \mathit{Aut}(x) \ar[rd] \ar[rr] & & \mathit{Aut}(z) \\ & \mathit{Aut}(y) \ar[ru] } \]

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

\[ \mathit{Aut}(x)|_{U_ i} \longrightarrow \mathit{Aut}(y)|_{U_ i} \]

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

\[ \varphi _{ij} : \mathcal{G}_ i|_{U_ i \times _ U U_ j} \longrightarrow \mathcal{G}_ j|_{U_ i \times _ U U_ j} \]

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. $\square$


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