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7.20 Cocontinuous functors

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

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 $\{ V_ j \to u(U)\} _{j \in J}$.

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

\[ \xymatrix{ & \ \phantom{}_ pu\mathcal{F}(V) \ar[r] & \prod {}_ pu\mathcal{F}(V_ j) \ar@<1ex>[r] \ar@<-1ex>[r] & \prod {}_ pu\mathcal{F}(V_ j \times _ V V_{j'}) & } \]

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

\[ {}_ pu\mathcal{F}(V) = \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{D})}(h_ V, {}_ pu\mathcal{F}) = \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})}(u^ ph_ V, \mathcal{F}) = \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}((u^ ph_ V)^\# , \mathcal{F}) \]

Hence it suffices to show that
\begin{equation} \label{sites-equation-coequalizer} \xymatrix{ \coprod u^ p h_{V_ j \times _ V V_{j'}} \ar@<1ex>[r] \ar@<-1ex>[r] & \coprod u^ p h_{V_ j} \ar[r] & u^ p h_ V } \end{equation}

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 ( 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 ( 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

\[ \xymatrix{ u(U) \ar[r]_{s'} \ar[d]_ s & V_{j'} \ar[d] \\ V_ j \ar[r] & V } \]

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

\begin{eqnarray*} \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})} ((u^ p\mathcal{G})^\# , \mathcal{F}) & = & \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})} (u^ p\mathcal{G}, \mathcal{F}) \\ & = & \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{D})} (\mathcal{G}, {}_ pu\mathcal{F}) \\ & = & \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{D})} (\mathcal{G}, {}_ su\mathcal{F}). \end{eqnarray*}

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

\begin{eqnarray*} \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}((u^ p(\mathcal{G}^\# ))^\# , \mathcal{F}) & = & \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{D})}(\mathcal{G}^\# , {}_ su\mathcal{F}) \\ & = & \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{D})}(\mathcal{G}^\# , {}_ pu\mathcal{F}) \\ & = & \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{D})}(\mathcal{G}, {}_ pu\mathcal{F}) \\ & = & \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})}(u^ p\mathcal{G}, \mathcal{F}) \\ & = & \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}((u^ p\mathcal{G})^\# , \mathcal{F}) \end{eqnarray*}

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

\[ \mathcal{C}^{opp} \longrightarrow \textit{Sets},\quad T \longmapsto \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(T, U) \times _{\mathop{\mathrm{Mor}}\nolimits _\mathcal {D}(u(T), V)} \mathop{\mathrm{Mor}}\nolimits _\mathcal {D}(u(T), W) \]

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

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

  2. $\{ 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 $P$.

Comments (5)

Comment #544 by Niels Borne on

In proof of Lemma 7.19.2 : parse error at or near "{}_pu\mathcal{F}(V) \ar[r] & "

Comment #554 by on

Tried to fix this, but I guess xyjax doesn't know \empty and cannot handle {}. Sometimes what helps is putting in spaces...

Comment #2032 by Dragos Fratila on

extra word in Remark 7.19.5 "we can define consider the functor"

Comment #2103 by on

It's a dirty hack, but XyJax does know the \phantom command. So replacing {}_pu\mathcal{F}(V) by \phantom{}_pu\mathcal{F}(V) does the trick (tested locally).

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