Definition 7.20.1. Let $\mathcal{C}$ and $\mathcal{D}$ be sites. Let $u : \mathcal{C} \to \mathcal{D}$ be a functor. The functor $u$ is called *cocontinuous* if for every $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and every covering $\{ V_ j \to u(U)\} _{j \in J}$ of $\mathcal{D}$ there exists a covering $\{ U_ i \to U\} _{i\in I}$ of $\mathcal{C}$ such that the family of maps $\{ u(U_ i) \to u(U)\} _{i \in I}$ refines the covering $\{ V_ j \to u(U)\} _{j \in J}$.

## 7.20 Cocontinuous functors

There is another way to construct morphisms of topoi. This involves using cocontinuous functors between sites defined as follows.

Note that $\{ u(U_ i) \to u(U)\} _{i \in I}$ is in general *not* a covering of the site $\mathcal{D}$.

Lemma 7.20.2. Let $\mathcal{C}$ and $\mathcal{D}$ be sites. Let $u : \mathcal{C} \to \mathcal{D}$ be cocontinuous. Let $\mathcal{F}$ be a sheaf on $\mathcal{C}$. Then ${}_ pu\mathcal{F}$ is a sheaf on $\mathcal{D}$, which we will denote ${}_ su\mathcal{F}$.

**Proof.**
Let $\{ V_ j \to V\} _{j \in J}$ be a covering of the site $\mathcal{D}$. We have to show that

is an equalizer diagram. Since ${}_ pu$ is right adjoint to $u^ p$ we have

Hence it suffices to show that

becomes a coequalizer diagram after sheafification. (Recall that a coproduct in the category of sheaves is the sheafification of the coproduct in the category of presheaves, see Lemma 7.10.13.)

We first show that the second arrow of (7.20.2.1) becomes surjective after sheafification. To do this we use Lemma 7.11.2. Thus it suffices to show a section $s$ of $u^ ph_ V$ over $U$ lifts to a section of $\coprod u^ p h_{V_ j}$ on the members of a covering of $U$. Note that $s$ is a morphism $s : u(U) \to V$. Then $\{ V_ j \times _{V, s} u(U) \to u(U)\} $ is a covering of $\mathcal{D}$. Hence, as $u$ is cocontinuous, there is a covering $\{ U_ i \to U\} $ such that $\{ u(U_ i) \to u(U)\} $ refines $\{ V_ j \times _{V, s} u(U) \to u(U)\} $. This means that each restriction $s|_{U_ i} : u(U_ i) \to V$ factors through a morphism $s_ i : u(U_ i) \to V_ j$ for some $j$, i.e., $s|_{U_ i}$ is in the image of $u^ ph_{V_ j}(U_ i) \to u^ ph_ V(U_ i)$ as desired.

Let $s, s' \in (\coprod u^ ph_{V_ j})^\# (U)$ map to the same element of $(u^ ph_ V)^\# (U)$. To finish the proof of the lemma we show that after replacing $U$ by the members of a covering that $s, s'$ are the image of the same section of $\coprod u^ p h_{V_ j \times _ V V_{j'}}$ by the two maps of (7.20.2.1). We may first replace $U$ by the members of a covering and assume that $s \in u^ ph_{V_ j}(U)$ and $s' \in u^ ph_{V_{j'}}(U)$. A second such replacement guarantees that $s$ and $s'$ have the same image in $u^ ph_ V(U)$ instead of in the sheafification. Hence $s : u(U) \to V_ j$ and $s' : u(U) \to V_{j'}$ are morphisms of $\mathcal{D}$ such that

is commutative. Thus we obtain $t = (s, s') : u(U) \to V_ j \times _ V V_{j'}$, i.e., a section $t \in u^ ph_{V_ j \times _ V V_{j'}}(U)$ which maps to $s, s'$ as desired. $\square$

Lemma 7.20.3. Let $\mathcal{C}$ and $\mathcal{D}$ be sites. Let $u : \mathcal{C} \to \mathcal{D}$ be cocontinuous. The functor $\mathop{\mathit{Sh}}\nolimits (\mathcal{D}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C})$, $\mathcal{G} \mapsto (u^ p\mathcal{G})^\# $ is a left adjoint to the functor ${}_ su$ introduced in Lemma 7.20.2 above. Moreover, it is exact.

**Proof.**
Let us prove the adjointness property as follows

Thus it is a left adjoint and hence right exact, see Categories, Lemma 4.24.6. We have seen that sheafification is left exact, see Lemma 7.10.14. Moreover, the inclusion $i : \mathop{\mathit{Sh}}\nolimits (\mathcal{D}) \to \textit{PSh}(\mathcal{D})$ is left exact by Lemma 7.10.1. Finally, the functor $u^ p$ is left exact because it is a right adjoint (namely to $u_ p$). Thus the functor is the composition ${}^\# \circ u^ p \circ i$ of left exact functors, hence left exact. $\square$

We finish this section with a technical lemma.

Lemma 7.20.4. In the situation of Lemma 7.20.3. For any presheaf $\mathcal{G}$ on $\mathcal{D}$ we have $(u^ p\mathcal{G})^\# = (u^ p(\mathcal{G}^\# ))^\# $.

**Proof.**
For any sheaf $\mathcal{F}$ on $\mathcal{C}$ we have

and the result follows from the Yoneda lemma. $\square$

Remark 7.20.5. Let $u : \mathcal{C} \to \mathcal{D}$ be a functor between categories. Given morphisms $g : u(U) \to V$ and $f : W \to V$ in $\mathcal{D}$ we can consider the functor

If this functor is representable, denote $U \times _{g, V, f} W$ the corresponding object of $\mathcal{C}$. Assume that $\mathcal{C}$ and $\mathcal{D}$ are sites. Consider the property $P$: for every covering $\{ f_ j : V_ j \to V\} $ of $\mathcal{D}$ and any morphism $g : u(U) \to V$ we have

$U \times _{g, V, f_ i} V_ i$ exists for all $i$, and

$\{ U \times _{g, V, f_ i} V_ i \to U\} $ is a covering of $\mathcal{C}$.

Please note the similarity with the definition of continuous functors. If $u$ has $P$ then $u$ is cocontinuous (details omitted). Many of the cocontinuous functors we will encounter satisfy $P$.

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