The Stacks project

Proof. Assume $u$ satisfies properties (1) – (5). We will show that the adjunction mappings

are isomorphisms.

Note that Lemma 7.21.5 applies and we have $g^{-1}\mathcal{G}(U) = \mathcal{G}(u(U))$ for any sheaf $\mathcal{G}$ on $\mathcal{D}$. Next, let $\mathcal{F}$ be a sheaf on $\mathcal{C}$, and let $V$ be an object of $\mathcal{D}$. By definition we have $g_*\mathcal{F}(V) = \mathop{\mathrm{lim}}\nolimits _{u(U) \to V} \mathcal{F}(U)$. Hence

where the morphisms $\psi : u(U') \to u(U)$ need not be of the form $u(\alpha )$. The category of such pairs $(U', \psi )$ has a final object, namely $(U, \text{id})$, which gives rise to the map from the limit into $\mathcal{F}(U)$. Let $(s_{(U', \psi )})$ be an element of the limit. We want to show that $s_{(U', \psi )}$ is uniquely determined by the value $s_{(U, \text{id})} \in \mathcal{F}(U)$. By property (4) given any $(U', \psi )$ there exists a covering $\{ U'_ i \to U'\} $ such that the compositions $u(U'_ i) \to u(U') \to u(U)$ are of the form $u(c_ i)$ for some $c_ i : U'_ i \to U$ in $\mathcal{C}$. Hence

Since $\mathcal{F}$ is a sheaf it follows that indeed $s_{(U', \psi )}$ is determined by $s_{(U, \text{id})}$. This proves uniqueness. For existence, assume given any $s \in \mathcal{F}(U)$, $\psi : u(U') \to u(U)$, $\{ f_ i : U_ i' \to U'\} $ and $c_ i : U_ i' \to U$ such that $\psi \circ u(f_ i) = u(c_ i)$ as above. We claim there exists a (unique) element $s_{(U', \psi )} \in \mathcal{F}(U')$ such that

Namely, a priori it is not clear the elements $c_ i^*(s)|_{U_ i' \times _{U'} U_ j'}$ and $c_ j^*(s)|_{U_ i' \times _{U'} U_ j'}$ agree, since the diagram

need not commute. But condition (3) of the lemma guarantees that there exist coverings $\{ f_{ijk} : U'_{ijk} \to U_ i' \times _{U'} U_ j'\} _{k \in K_{ij}}$ such that $c_ i \circ \text{pr}_1 \circ f_{ijk} = c_ j \circ \text{pr}_2 \circ f_{ijk}$. Hence

Hence $c_ i^*(s)|_{U_ i' \times _{U'} U_ j'} = c_ j^*(s)|_{U_ i' \times _{U'} U_ j'}$ by the sheaf condition for $\mathcal{F}$ and hence the existence of $s_{(U', \psi )}$ also by the sheaf condition for $\mathcal{F}$. The uniqueness guarantees that the collection $(s_{(U', \psi )})$ so obtained is an element of the limit with $s_{(U, \psi )} = s$. This proves that $g^{-1}g_*\mathcal{F} \to \mathcal{F}$ is an isomorphism.

Let $\mathcal{G}$ be a sheaf on $\mathcal{D}$. Let $V$ be an object of $\mathcal{D}$. Then we see that

By the preceding paragraph we see that the value of the sheaf $g_*g^{-1}\mathcal{G}$ on an object $V$ of the form $V = u(U)$ is equal to $\mathcal{G}(u(U))$. (Formally, this holds because we have $g^{-1}g_*g^{-1} \cong g^{-1}$, and the description of $g^{-1}$ given at the beginning of the proof; informally just by comparing limits here and above.) Hence the adjunction mapping $\mathcal{G} \to g_*g^{-1}\mathcal{G}$ has the property that it is a bijection on sections over any object of the form $u(U)$. Since by axiom (5) there exists a covering of $V$ by objects of the form $u(U)$ we see easily that the adjunction map is an isomorphism. $\square$

Proof. We have seen the existence and commutativity of the diagrams in Lemma 7.28.4. We have to check hypotheses (1) – (5) of Lemma 7.29.1 for the induced functor $u : \mathcal{C}/U \to \mathcal{D}/u(U)$. This is completely mechanical.

Property (1). This is Lemma 7.28.4.

Property (2). Let $\{ U_ i'/U \to U'/U\} _{i \in I}$ be a covering of $U'/U$ in $\mathcal{C}/U$. Because $u$ is continuous we see that $\{ u(U_ i')/u(U) \to u(U')/u(U)\} _{i \in I}$ is a covering of $u(U')/u(U)$ in $\mathcal{D}/u(U)$. Hence (2) holds for $u : \mathcal{C}/U \to \mathcal{D}/u(U)$.

Property (3). Let $a, b : U''/U \to U'/U$ in $\mathcal{C}/U$ be morphisms such that $u(a) = u(b)$ in $\mathcal{D}/u(U)$. Because $u$ satisfies (3) we see there exists a covering $\{ f_ i : U''_ i \to U''\} $ in $\mathcal{C}$ such that $a \circ f_ i = b \circ f_ i$. This gives a covering $\{ f_ i : U''_ i/U \to U''/U\} $ in $\mathcal{C}/U$ such that $a \circ f_ i = b \circ f_ i$. Hence (3) holds for $u : \mathcal{C}/U \to \mathcal{D}/u(U)$.

Property (4). Let $U''/U, U'/U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}/U)$ and a morphism $c : u(U'')/u(U) \to u(U')/u(U)$ in $\mathcal{D}/u(U)$ be given. Because $u$ satisfies property (4) there exists a covering $\{ f_ i : U_ i'' \to U''\} $ in $\mathcal{C}$ and morphisms $c_ i : U_ i'' \to U'$ such that $u(c_ i) = c \circ u(f_ i)$. We think of $U_ i''$ as an object over $U$ via the composition $U_ i'' \to U'' \to U$. It may not be true that $c_ i$ is a morphism over $U$! But since $u(c_ i)$ is a morphism over $u(U)$ we may apply property (3) for $u$ and find coverings $\{ f_{ik} : U''_{ik} \to U''_ i\} $ such that $c_{ik} = c_ i \circ f_{ik} : U''_{ik} \to U'$ are morphisms over $U$. Hence $\{ f_ i \circ f_{ik} : U''_{ik}/U \to U''/U\} $ is a covering in $\mathcal{C}/U$ such that $u(c_{ik}) = c \circ u(f_{ik})$. Hence (4) holds for $u : \mathcal{C}/U \to \mathcal{D}/u(U)$.

Property (5). Let $h : V \to u(U)$ be an object of $\mathcal{D}/u(U)$. Because $u$ satisfies property (5) there exists a covering $\{ c_ i : u(U_ i) \to V\} $ in $\mathcal{D}$. By property (4) we can find coverings $\{ f_{ij} : U_{ij} \to U_ i\} $ and morphisms $c_{ij} : U_{ij} \to U$ such that $u(c_{ij}) = h \circ c_ i \circ u(f_{ij})$. Hence $\{ u(U_{ij})/u(U) \to V/u(U)\} $ is a covering in $\mathcal{D}/u(U)$ of the desired shape and we conclude that (5) holds for $u : \mathcal{C}/U \to \mathcal{D}/u(U)$. $\square$

Proof. Warning: Some of the statements above may look be a bit confusing at first; this is because objects of $\mathcal{C}'$ can also be viewed as sheaves on $\mathcal{C}$! We omit the proof that the coverings of $\mathcal{C}'$ as described in the lemma satisfy the conditions of Definition 7.6.2.

Suppose that $\{ \mathcal{F}_ i \to \mathcal{F}\} $ is a surjective family of morphisms of sheaves. Let $\mathcal{G}$ be another sheaf. Part (2) of the lemma says that the equalizer of

is $\mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(\mathcal{F}, \mathcal{G}).$ This is clear (for example use Lemma 7.11.3).

To prove (3) we have to check conditions (1) – (5) of Lemma 7.29.1. The fact that $v$ is cocontinuous is equivalent to the description of surjective maps of sheaves in Lemma 7.11.2. The functor $v$ is continuous because $U \mapsto h_ U^\# $ commutes with fibre products, and transforms coverings into coverings (see Lemma 7.10.14, and Lemma 7.12.4). Properties (3), (4) of Lemma 7.29.1 are statements about morphisms $f : h_{U'}^\# \to h_ U^\# $. Such a morphism is the same thing as an element of $h_ U^\# (U')$. Hence (3) and (4) are immediate from the construction of the sheafification. Property (5) of Lemma 7.29.1 is Lemma 7.12.5. Denote $g : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C}')$ the equivalence of topoi associated with $v$ by Lemma 7.29.1.

Let $\mathcal{F}$ be as in part (4) of the lemma. For any $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ we have

The first equality by Lemma 7.21.5. Thus part (4) holds.

Let $\mathcal{F} \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}')$. Let $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. Then

as desired (where the third equality was shown above). $\square$

Proof. Consider the full subcategory $\mathcal{C}_1 \subset \mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ consisting of all $h_ U^\# $ for all $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, the given sheaves $\mathcal{F}_ i$ and the final sheaf $*$ (see Example 7.10.2). We are going to inductively define full subcategories

Namely, given $\mathcal{C}_ n$ let $\mathcal{C}_{n + 1}$ be the full subcategory consisting of all fibre products and subsheaves of objects of $\mathcal{C}_ n$. (Note that $\mathcal{C}_{n + 1}$ has a set of objects.) Set $\mathcal{C}' = \bigcup _{n \geq 1} \mathcal{C}_ n$. A covering in $\mathcal{C}'$ is any family $\{ \mathcal{G}_ j \to \mathcal{G}\} _{j \in J}$ of morphisms of objects of $\mathcal{C}'$ such that $\coprod \mathcal{G}_ j \to \mathcal{G}$ is surjective as a map of sheaves on $\mathcal{C}$. The functor $v : \mathcal{C} \to \mathcal{C'}$ is given by $U \mapsto h_ U^\# $. Apply Lemma 7.29.4. $\square$

Proof. Consider the full subcategory $\mathcal{C}_1 \subset \mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ consisting of all $h_ U^\# $ and all $f^{-1}h_ V^\# $ for all $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and all $V \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$. Let $\mathcal{C}_{n + 1}$ be a full subcategory consisting of all fibre products of objects of $\mathcal{C}_ n$. Set $\mathcal{C}' = \bigcup _{n \geq 1} \mathcal{C}_ n$. A covering in $\mathcal{C}'$ is any family $\{ \mathcal{F}_ i \to \mathcal{F}\} _{i \in I}$ such that $\coprod _{i \in I} \mathcal{F}_ i \to \mathcal{F}$ is surjective as a map of sheaves on $\mathcal{C}$. The functor $v : \mathcal{C} \to \mathcal{C'}$ is given by $U \mapsto h_ U^\# $. The functor $u : \mathcal{D} \to \mathcal{C'}$ is given by $V \mapsto f^{-1}h_ V^\# $.

Part (1) follows from Lemma 7.29.4.

Proof of (2) and (3) of the lemma. The functor $u$ commutes with fibre products as both $V \mapsto h_ V^\# $ and $f^{-1}$ do. Moreover, since $f^{-1}$ is exact and commutes with arbitrary colimits we see that it transforms a covering into a surjective family of morphisms of sheaves. Hence $u$ is continuous. To see that it defines a morphism of sites we still have to see that $u_ s$ is exact. In order to do this we will show that $g^{-1} \circ u_ s = f^{-1}$. Namely, then since $g^{-1}$ is an equivalence and $f^{-1}$ is exact we will conclude. Because $g^{-1}$ is adjoint to $g_*$, and $u_ s$ is adjoint to $u^ s$, and $f^{-1}$ is adjoint to $f_*$ it also suffices to prove that $u^ s \circ g_* = f_*$. Let $\mathcal{F}$ be a sheaf on $\mathcal{C}$ and let $V$ be an object of $\mathcal{D}$. Then

The first equality because $u^ s = u^ p$. The second equality is the Yoneda lemma. The third equality by adjointness of $g^{-1}$ and $g_*$. The fourth equality is by Lemma 7.29.4 (4). The fifth equality by adjointness of $f^{-1}$ and $f_*$. The sixth equality by the Yoneda lemma. Hence $u^ s g_*\mathcal{F} = f_*\mathcal{F}$ and this finishes the proof of the lemma. $\square$