Lemma 7.29.4. Let $\mathcal{C}$ be a site. Let $\mathcal{C}' \subset \mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ be a full subcategory (with a set of objects) such that

$h_ U^\# \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}')$ for all $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, and

$\mathcal{C}'$ is preserved under fibre products in $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})$.

Declare a covering of $\mathcal{C}'$ to be any family $\{ \mathcal{F}_ i \to \mathcal{F}\} _{i \in I}$ of maps such that $\coprod _{i \in I} \mathcal{F}_ i \to \mathcal{F}$ is a surjective map of sheaves. Then

$\mathcal{C}'$ is a site (after choosing a set of coverings, see Sets, Lemma 3.11.1),

representable presheaves on $\mathcal{C}'$ are sheaves (i.e., the topology on $\mathcal{C}'$ is subcanonical, see Definition 7.12.2),

the functor $v : \mathcal{C} \to \mathcal{C}'$, $U \mapsto h_ U^\# $ is a special cocontinuous functor, hence induces an equivalence $g : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C}')$,

for any $\mathcal{F} \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}')$ we have $g^{-1}h_\mathcal {F} = \mathcal{F}$, and

for any $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ we have $g_*h_ U^\# = h_{v(U)} = h_{h_ U^\# }$.

**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

\[ \xymatrix{ \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}( \coprod _{i \in I} \mathcal{F}_ i, \mathcal{G}) \ar@<1ex>[r] \ar@<-1ex>[r] & \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}( \coprod _{(i_0, i_1) \in I \times I} \mathcal{F}_{i_0} \times _\mathcal {F} \mathcal{F}_{i_1}, \mathcal{G}) } \]

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

\[ g^{-1}h_\mathcal {F}(U) = h_\mathcal {F}(v(U)) = \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(h_ U^\# , \mathcal{F}) = \mathcal{F}(U) \]

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

\begin{align*} g_*h_ U^\# (\mathcal{F}) & = \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C}')}(h_\mathcal {F}, g_*h_ U^\# ) \\ & = \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(g^{-1}h_\mathcal {F}, h_ U^\# ) \\ & = \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(\mathcal{F}, h_ U^\# ) \\ & = \mathop{\mathrm{Mor}}\nolimits _{\mathcal{C}'}(\mathcal{F}, h_ U^\# ) \end{align*}

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

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