
## 7.18 Colimits of sites

We need an analogue of Lemma 7.17.5 in the case that the site is the limit of an inverse system of sites. For simplicity we only explain the construction in case the index sets of coverings are finite.

Situation 7.18.1. Here we are given

1. a cofiltered index category $\mathcal{I}$,

2. for $i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$ a site $\mathcal{C}_ i$ such that every covering in $\mathcal{C}_ i$ has a finite index set,

3. for a morphism $a : i \to j$ in $\mathcal{I}$ a morphism of sites $f_ a : \mathcal{C}_ i \to \mathcal{C}_ j$ given by a continuous functor $u_ a : \mathcal{C}_ j \to \mathcal{C}_ i$,

such that $f_ a \circ f_ b = f_ c$ whenever $c = a \circ b$ in $\mathcal{I}$.

Lemma 7.18.2. In Situation 7.18.1 we can construct a site $(\mathcal{C}, \text{Cov}(\mathcal{C}))$ as follows

1. as a category $\mathcal{C} = \mathop{\mathrm{colim}}\nolimits \mathcal{C}_ i$, and

2. $\text{Cov}(\mathcal{C})$ is the union of the images of $\text{Cov}(\mathcal{C}_ i)$ by $u_ i : \mathcal{C}_ i \to \mathcal{C}$.

Proof. Our definition of composition of morphisms of sites implies that $u_ b \circ u_ a = u_ c$ whenever $c = a \circ b$ in $\mathcal{I}$. The formula $\mathcal{C} = \mathop{\mathrm{colim}}\nolimits \mathcal{C}_ i$ means that $\mathop{\mathrm{Ob}}\nolimits (\mathcal{C}) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_ i)$ and $\text{Arrows}(\mathcal{C}) = \mathop{\mathrm{colim}}\nolimits \text{Arrows}(\mathcal{C}_ i)$. Then source, target, and composition are inherited from the source, target, and composition on $\text{Arrows}(\mathcal{C}_ i)$. In this way we obtain a category. Denote $u_ i : \mathcal{C}_ i \to \mathcal{C}$ the obvious functor. Remark that given any finite diagram in $\mathcal{C}$ there exists an $i$ such that this diagram is the image of a diagram in $\mathcal{C}_ i$.

Let $\{ U^ t \to U\}$ be a covering of $\mathcal{C}$. We first prove that if $V \to U$ is a morphism of $\mathcal{C}$, then $U^ t \times _ U V$ exists. By our remark above and our definition of coverings, we can find an $i$, a covering $\{ U_ i^ t \to U_ i\}$ of $\mathcal{C}_ i$ and a morphism $V_ i \to U_ i$ whose image by $u_ i$ is the given data. We claim that $U^ t \times _ U V$ is the image of $U^ t_ i \times _{U_ i} V_ i$ by $u_ i$. Namely, for every $a : j \to i$ in $\mathcal{I}$ the functor $u_ a$ is continuous, hence $u_ a(U^ t_ i \times _{U_ i} V_ i) = u_ a(U^ t_ i) \times _{u_ a(U_ i)} u_ a(V_ i)$. In particular we can replace $i$ by $j$, if we so desire. Thus, if $W$ is another object of $\mathcal{C}$, then we may assume $W = u_ i(W_ i)$ and we see that

\begin{align*} & \mathop{Mor}\nolimits _\mathcal {C}(W, u_ i(U^ t_ i \times _{U_ i} V_ i)) \\ & = \mathop{\mathrm{colim}}\nolimits _{a : j \to i} \mathop{Mor}\nolimits _{\mathcal{C}_ j}(u_ a(W_ i), u_ a(U^ t_ i \times _{U_ i} V_ i)) \\ & = \mathop{\mathrm{colim}}\nolimits _{a : j \to i} \mathop{Mor}\nolimits _{\mathcal{C}_ j}(u_ a(W_ i), u_ a(U^ t_ i)) \times _{\mathop{Mor}\nolimits _{\mathcal{C}_ j}(u_ a(W_ i), u_ a(U_ i))} \mathop{Mor}\nolimits _{\mathcal{C}_ j}(u_ a(W_ i), u_ a(V_ i)) \\ & = \mathop{Mor}\nolimits _\mathcal {C}(W, U^ t) \times _{\mathop{Mor}\nolimits _\mathcal {C}(W, U)} \mathop{Mor}\nolimits _\mathcal {C}(W, V) \end{align*}

as filtered colimits commute with finite limits (Categories, Lemma 4.19.2). It also follows that $\{ U^ t \times _ U V \to V\}$ is a covering in $\mathcal{C}$. In this way we see that axiom (3) of Definition 7.6.2 holds.

To verify axiom (2) of Definition 7.6.2 let $\{ U^ t \to U\} _{t \in T}$ be a covering of $\mathcal{C}$ and for each $t$ let $\{ U^{ts} \to U^ t\}$ be a covering of $\mathcal{C}$. Then we can find an $i$ and a covering $\{ U^ t_ i \to U_ i\} _{t \in T}$ of $\mathcal{C}_ i$ whose image by $u_ i$ is $\{ U^ t \to U\}$. Since $T$ is finite we may choose an $a : j \to i$ in $\mathcal{I}$ and coverings $\{ U^{ts}_ j \to u_ a(U^ t_ i)\}$ of $\mathcal{C}_ j$ whose image by $u_ j$ gives $\{ U^{ts} \to U^ t\}$. Then we conclude that $\{ U^{ts} \to U\}$ is a covering of $\mathcal{C}$ by an application of axiom (2) to the site $\mathcal{C}_ j$.

We omit the proof of axiom (1) of Definition 7.6.2. $\square$

Lemma 7.18.3. In Situation 7.18.1 let $u_ i : \mathcal{C}_ i \to \mathcal{C}$ be as constructed in Lemma 7.18.2. Then $u_ i$ defines a morphism of sites $f_ i : \mathcal{C} \to \mathcal{C}_ i$. For $U_ i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_ i)$ and sheaf $\mathcal{F}$ on $\mathcal{C}_ i$ we have

7.18.3.1
$$\label{sites-equation-compute-pullback-to-limit} f_ i^{-1}\mathcal{F}(u_ i(U_ i)) = \mathop{\mathrm{colim}}\nolimits _{a : j \to i} f_ a^{-1}\mathcal{F}(u_ a(U_ i))$$

Proof. It is immediate from the arguments in the proof of Lemma 7.18.2 that the functors $u_ i$ are continuous. To finish the proof we have to show that $f_ i^{-1} := u_{i, s}$ is an exact functor $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_ i) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C})$. In fact it suffices to show that $f_ i^{-1}$ is left exact, because it is right exact as a left adjoint (Categories, Lemma 4.24.6). We first prove (7.18.3.1) and then we deduce exactness.

For an arbitrary object $V$ of $\mathcal{C}$ we can pick a $a : j \to i$ and an object $V_ j \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ with $V = u_ j(V_ j)$. Then we can set

$\mathcal{G}(V) = \mathop{\mathrm{colim}}\nolimits _{b : k \to j} f_{a \circ b}^{-1}\mathcal{F}(u_ b(V_ j))$

The value $\mathcal{G}(V)$ of the colimit is independent of the choice of $b : j \to i$ and of the object $V_ j$ with $u_ j(V_ j) = V$; we omit the verification. Moreover, if $\alpha : V \to V'$ is a morphism of $\mathcal{C}$, then we can choose $b : j \to i$ and a morphism $\alpha _ j : V_ j \to V'_ j$ with $u_ j(\alpha _ j) = \alpha$. This induces a map $\mathcal{G}(V') \to \mathcal{G}(V)$ by using the restrictions along the morphisms $u_ b(\alpha _ j) : u_ b(V_ j) \to u_ b(V'_ j)$. A check shows that $\mathcal{G}$ is a presheaf (omitted). In fact, $\mathcal{G}$ satisfies the sheaf condition. Namely, any covering $\mathcal{U} = \{ U^ t \to U\}$ in $\mathcal{C}$ comes from a finite level. Say $\mathcal{U}_ j = \{ U^ t_ j \to U_ j\}$ is mapped to $\mathcal{U}$ by $u_ j$ for some $a : j \to i$ in $\mathcal{I}$. Then we have

$H^0(\mathcal{U}, \mathcal{G}) = \mathop{\mathrm{colim}}\nolimits _{b : k \to j} H^0(u_ b(\mathcal{U}_ j), f_{b \circ a}^{-1}\mathcal{F}) = \mathop{\mathrm{colim}}\nolimits _{b : k \to j} f_{b \circ a}^{-1}\mathcal{F}(u_ b(U_ j)) = \mathcal{G}(U)$

as desired. The first equality holds because filtered colimits commute with finite limits (Categories, Lemma 4.19.2). By construction $\mathcal{G}(U)$ is given by the right hand side of (7.18.3.1). Hence (7.18.3.1) is true if we can show that $\mathcal{G}$ is equal to $f_ i^{-1}\mathcal{F}$.

In this paragraph we check that $\mathcal{G}$ is canonically isomorphic to $f_ i^{-1}\mathcal{F}$. We strongly encourage the reader to skip this paragraph. To check this we have to show there is a bijection $\mathop{Mor}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(\mathcal{G}, \mathcal{H}) = \mathop{Mor}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_ i)}(\mathcal{F}, f_{i, *}\mathcal{H})$ functorial in the sheaf $\mathcal{H}$ on $\mathcal{C}$ where $f_{i, *} = u_ i^ p$. A map $\mathcal{G} \to \mathcal{H}$ is the same thing as a compatible system of maps

$\varphi _{a, b, V_ j} : f_{a \circ b}^{-1}\mathcal{F}(u_ b(V_ j)) \longrightarrow \mathcal{H}(u_ j(V_ j))$

for all $a : j \to i$, $b : k \to j$ and $V_ j \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_ j)$. The compatibilities force the maps $\varphi _{a, b, V_ j}$ to be equal to $\varphi _{a \circ b, \text{id}, u_ b(V_ j)}$. Given $a : j \to i$, the family of maps $\varphi _{a, \text{id}, V_ j}$ corresponds to a map of sheaves $\varphi _ a : f_ a^{-1}\mathcal{F} \to f_{j, *}\mathcal{H}$. The compatibilities between the $\varphi _{a, \text{id}, u_ a(V_ i)}$ and the $\varphi _{\text{id}, \text{id}, V_ i}$ implies that $\varphi _ a$ is the adjoint of the map $\varphi _{id}$ via

$\mathop{Mor}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_ j)}(f_ a^{-1}\mathcal{F}, f_{j, *}\mathcal{H}) = \mathop{Mor}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_ i)}(\mathcal{F}, f_{a, *}f_{j, *}\mathcal{H}) = \mathop{Mor}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_ i)}(\mathcal{F}, f_{i, *}\mathcal{H})$

Thus finally we see that the whole system of maps $\varphi _{a, b, V_ j}$ is determined by the map $\varphi _{id} : \mathcal{F} \to f_{i, *}\mathcal{H}$. Conversely, given such a map $\psi : \mathcal{F} \to f_{i, *}\mathcal{H}$ we can read the argument just given backwards to construct the family of maps $\varphi _{a, b, V_ j}$. This finishes the proof that $\mathcal{G} = f_ i^{-1}\mathcal{F}$.

Assume (7.18.3.1) holds. Then the functor $\mathcal{F} \mapsto f_ i^{-1}\mathcal{F}(U)$ commutes with finite limits because finite limits of sheaves are computed in the category of presheaves (Lemma 7.10.1), the functors $f_ a^{-1}$ commutes with finite limits, and filtered colimits commute with finite limits. To see that $\mathcal{F} \mapsto f_ i^{-1}\mathcal{F}(V)$ commutes with finite limits for a general object $V$ of $\mathcal{C}$, we can use the same argument using the formula for $f_ i^{-1}\mathcal{F}(V) = \mathcal{G}(V)$ given above. Thus $f_ i^{-1}$ is left exact and the proof of the lemma is complete. $\square$

Lemma 7.18.4. In Situation 7.18.1 assume given

1. a sheaf $\mathcal{F}_ i$ on $\mathcal{C}_ i$ for all $i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$,

2. for $a : j \to i$ a map $\varphi _ a : f_ a^{-1}\mathcal{F}_ i \to \mathcal{F}_ j$ of sheaves on $\mathcal{C}_ j$

such that $\varphi _ c = \varphi _ b \circ f_ b^{-1}\varphi _ a$ whenever $c = a \circ b$. Set $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits f_ i^{-1}\mathcal{F}_ i$ on the site $\mathcal{C}$ of Lemma 7.18.2. Let $i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$ and $X_ i \in \text{Ob}(\mathcal{C}_ i)$. Then

$\mathop{\mathrm{colim}}\nolimits _{a : j \to i} \mathcal{F}_ j(u_ a(X_ i)) = \mathcal{F}(u_ i(X_ i))$

Proof. A formal argument shows that

$\mathop{\mathrm{colim}}\nolimits _{a : j \to i} \mathcal{F}_ i(u_ a(X_ i)) = \mathop{\mathrm{colim}}\nolimits _{a : j \to i} \mathop{\mathrm{colim}}\nolimits _{b : k \to j} f_ b^{-1}\mathcal{F}_ j(u_{a \circ b}(X_ i))$

By (7.18.3.1) we see that the inner colimit is equal to $f_ j^{-1}\mathcal{F}_ j(u_ i(X_ i))$ hence we conclude by Lemma 7.17.5. $\square$

Lemma 7.18.5. In Situation 7.18.1 assume we have a sheaf $\mathcal{F}$ on $\mathcal{C}$. Then

$\mathcal{F} = \mathop{\mathrm{colim}}\nolimits f_ i^{-1}f_{i, *}\mathcal{F}$

where the transition maps are $f_ j^{-1}\varphi _ a$ for $a : j \to i$ where $\varphi _ a : f_ a^{-1}f_{i, *}\mathcal{F} \to f_{j, *}\mathcal{F}$ is a canonical map satisfying a cocycle condition as in Lemma 7.18.4.

Proof. For the morphism

$\varphi _ a : f_ a^{-1}f_{i, *}\mathcal{F} \to f_{j, *}\mathcal{F}$

we choose the adjoint to the identity map

$f_{i, *}\mathcal{F} \to f_{a, *}f_{j, *}\mathcal{F}$

Hence $\varphi _ a$ is the counit for the adjunction given by $(f_ a^{-1}, f_{a, *})$. We must prove that for all $a : j \to i$ and $b : k \to i$ with composition $c = a \circ b$ we have $\varphi _ c = \varphi _ b \circ f_ b^{-1}\varphi _ a$. This follows from Categories, Lemma 4.24.8. Lastly, we must prove that the map given by adjunction

$\mathop{\mathrm{colim}}\nolimits _{i \in I} f_ i^{-1}f_{i, *}\mathcal{F} \longrightarrow \mathcal{F}$

is an isomorphism. For an object $U$ of $\mathcal{C}$ we need to show the map

$(\mathop{\mathrm{colim}}\nolimits _{i \in I} f_ i^{-1}\mathcal{F}_ i)(U) \to \mathcal{F}(U)$

is bijective. Choose an $i$ and an object $U_ i$ of $\mathcal{C}_ i$ with $u_ i(U_ i) = U$. Then the left hand side is equal to

$(\mathop{\mathrm{colim}}\nolimits _{i \in I} f_ i^{-1}\mathcal{F}_ i)(U) = \mathop{\mathrm{colim}}\nolimits _{a : j \to i} f_{j, *}\mathcal{F}(u_ a(U_ i))$

by Lemma 7.18.4. Since $u_ j(u_ a(U_ i)) = U$ we have $f_{j, *}\mathcal{F}(u_ a(U_ i)) = \mathcal{F}(U)$ for all $a : j \to i$ by definition. Hence the value of the colimit is $\mathcal{F}(U)$ and the proof is complete. $\square$

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