7.13 Continuous functors
Definition 7.13.1. Let $\mathcal{C}$ and $\mathcal{D}$ be sites. A functor $u : \mathcal{C} \to \mathcal{D}$ is called continuous if for every $\{ V_ i \to V\} _{i\in I} \in \text{Cov}(\mathcal{C})$ we have the following
$\{ u(V_ i) \to u(V)\} _{i\in I}$ is in $\text{Cov}(\mathcal{D})$, and
for any morphism $T \to V$ in $\mathcal{C}$ the morphism $u(T \times _ V V_ i) \to u(T) \times _{u(V)} u(V_ i)$ is an isomorphism.
Recall that given a functor $u$ as above, and a presheaf of sets $\mathcal{F}$ on $\mathcal{D}$ we have defined $u^ p\mathcal{F}$ to be simply the presheaf $\mathcal{F} \circ u$, in other words
\[ u^ p\mathcal{F} (V) = \mathcal{F}(u(V)) \]
for every object $V$ of $\mathcal{C}$.
Lemma 7.13.2. Let $\mathcal{C}$ and $\mathcal{D}$ be sites. Let $u : \mathcal{C} \to \mathcal{D}$ be a continuous functor. If $\mathcal{F}$ is a sheaf on $\mathcal{D}$ then $u^ p\mathcal{F}$ is a sheaf as well.
Proof.
Let $\{ V_ i \to V\} $ be a covering. By assumption $\{ u(V_ i) \to u(V)\} $ is a covering in $\mathcal{D}$ and $u(V_ i \times _ V V_ j) = u(V_ i)\times _{u(V)}u(V_ j)$. Hence the sheaf condition for $u^ p\mathcal{F}$ and the covering $\{ V_ i \to V\} $ is precisely the same as the sheaf condition for $\mathcal{F}$ and the covering $\{ u(V_ i) \to u(V)\} $.
$\square$
In order to avoid confusion we sometimes denote
\[ u^ s : \mathop{\mathit{Sh}}\nolimits (\mathcal{D}) \longrightarrow \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \]
the functor $u^ p$ restricted to the subcategory of sheaves of sets. Recall that $u^ p$ has a left adjoint $u_ p : \textit{PSh}(\mathcal{C}) \to \textit{PSh}(\mathcal{D})$, see Section 7.5.
Lemma 7.13.3. In the situation of Lemma 7.13.2. The functor $u_ s : \mathcal{G} \mapsto (u_ p \mathcal{G})^\# $ is a left adjoint to $u^ s$.
Proof.
Follows directly from Lemma 7.5.4 and Proposition 7.10.12.
$\square$
Here is a technical lemma.
Lemma 7.13.4. In the situation of Lemma 7.13.2. For any presheaf $\mathcal{G}$ on $\mathcal{C}$ we have $(u_ p\mathcal{G})^\# = (u_ p(\mathcal{G}^\# ))^\# $.
Proof.
For any sheaf $\mathcal{F}$ on $\mathcal{D}$ we have
\begin{eqnarray*} \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{D})}(u_ s(\mathcal{G}^\# ), \mathcal{F}) & = & \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(\mathcal{G}^\# , u^ s\mathcal{F}) \\ & = & \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})}(\mathcal{G}^\# , u^ p\mathcal{F}) \\ & = & \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})}(\mathcal{G}, u^ p\mathcal{F}) \\ & = & \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{D})}(u_ p\mathcal{G}, \mathcal{F}) \\ & = & \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{D})}((u_ p\mathcal{G})^\# , \mathcal{F}) \end{eqnarray*}
and the result follows from the Yoneda lemma.
$\square$
Lemma 7.13.5. Let $u : \mathcal{C} \to \mathcal{D}$ be a continuous functor between sites. For any object $U$ of $\mathcal{C}$ we have $u_ sh_ U^\# = h_{u(U)}^\# $.
Proof.
Follows from Lemmas 7.5.6 and 7.13.4.
$\square$
We are going to see that Lemmas 7.13.2 and 7.13.3 hold in case $u$ is quasi-continuous as well.
We first remark that the morphisms $u(V_ i) \to u(V)$ are representable, since they are isomorphic to representable morphisms (by the first condition). In particular, the family $u(\mathcal{V}) = \{ u(V_ i) \to u(V)\} _{i\in I}$ gives rise to a zeroth Čech cohomology group $H^0(u(\mathcal{V}), \mathcal{F})$ for any presheaf $\mathcal{F}$ on $\mathcal{D}$. Let $\mathcal{U} = \{ U_ j \to u(V)\} _{j \in J}$ be an element of $\text{Cov}(\mathcal{D})$ tautologically equivalent to $\{ u(V_ i) \to u(V)\} _{i \in I}$. Note that $u(\mathcal{V})$ is a refinement of $\mathcal{U}$ and vice versa. Hence by Remark 7.10.7 we see that $H^0(u(\mathcal{V}), \mathcal{F}) = H^0(\mathcal{U}, \mathcal{F})$. In particular, if $\mathcal{F}$ is a sheaf, then $\mathcal{F}(u(V)) = H^0(u(\mathcal{V}), \mathcal{F})$ because of the sheaf property expressed in terms of zeroth Čech cohomology groups. We conclude that $u^ p\mathcal{F}$ is a sheaf if $\mathcal{F}$ is a sheaf, since $H^0(\mathcal{V}, u^ p\mathcal{F}) = H^0(u(\mathcal{V}), \mathcal{F})$ which we just observed is equal to $\mathcal{F}(u(V)) = u^ p\mathcal{F}(V)$. Thus Lemma 7.13.2 holds. Lemma 7.13.3 follows immediately.
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Comment #3520 by Laurent Moret-Bailly on
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