# The Stacks Project

## Tag 08NG

### 7.22. Cocontinuous functors which have a left adjoint

It may happen that a cocontinuous functor $u$ has a left adjoint $w$.

Lemma 7.22.1. Let $\mathcal{C}$ and $\mathcal{D}$ be sites. Let $g : \mathop{\mathit{Sh}}\nolimits(\mathcal{C}) \to \mathop{\mathit{Sh}}\nolimits(\mathcal{D})$ be the morphism of topoi associated to a continuous and cocontinuous functor $u : \mathcal{C} \to \mathcal{D}$, see Lemmas 7.20.1 and 7.20.5.

1. If $w : \mathcal{D} \to \mathcal{C}$ is a left adjoint to $u$, then
1. $g_!\mathcal{F}$ is the sheaf associated to the presheaf $w^p\mathcal{F}$, and
2. $g_!$ is exact.
2. if $w$ is a continuous left adjoint, then $g_!$ has a left adjoint.
3. If $w$ is a cocontinuous left adjoint, then $g_! = h^{-1}$ and $g^{-1} = h_*$ where $h : \mathop{\mathit{Sh}}\nolimits(\mathcal{D}) \to \mathop{\mathit{Sh}}\nolimits(\mathcal{C})$ is the morphism of topoi associated to $w$.

Proof. Recall that $g_!\mathcal{F}$ is the sheafification of $u_p\mathcal{F}$. Hence (1)(a) follows from the fact that $u_p = w^p$ by Lemma 7.18.3.

To see (1)(b) note that $g_!$ commutes with all colimits as $g_!$ is a left adjoint (Categories, Lemma 4.24.5). Let $i \mapsto \mathcal{F}_i$ be a finite diagram in $\mathop{\mathit{Sh}}\nolimits(\mathcal{C})$. Then $\mathop{\mathrm{lim}}\nolimits \mathcal{F}_i$ is computed in the category of presheaves (Lemma 7.10.1). Since $w^p$ is a right adjoint (Lemma 7.5.4) we see that $w^p \mathop{\mathrm{lim}}\nolimits \mathcal{F}_i = \mathop{\mathrm{lim}}\nolimits w^p\mathcal{F}_i$. Since sheafification is exact (Lemma 7.10.14) we conclude by (1)(a).

Assume $w$ is continuous. Then $g_! = (w^p )^\# = w^s$ but sheafification isn't necessary and one has the left adjoint $w_s$, see Lemmas 7.13.2 and 7.13.3.

Assume $w$ is cocontinuous. The equality $g_! = h^{-1}$ follows from (1)(a) and the definitions. The equality $g^{-1} = h_*$ follows from the equality $g_! = h^{-1}$ and uniqueness of adjoint functor. Alternatively one can deduce it from Lemma 7.21.1. $\square$

The code snippet corresponding to this tag is a part of the file sites.tex and is located in lines 4398–4463 (see updates for more information).

\section{Cocontinuous functors which have a left adjoint}

\noindent
It may happen that a cocontinuous functor $u$ has a left adjoint $w$.

\begin{lemma}
Let $\mathcal{C}$ and $\mathcal{D}$ be sites. Let
$g : \Sh(\mathcal{C}) \to \Sh(\mathcal{D})$ be
the morphism of topoi associated to a continuous and cocontinuous functor
$u : \mathcal{C} \to \mathcal{D}$, see
Lemmas \ref{lemma-cocontinuous-morphism-topoi} and
\ref{lemma-when-shriek}.
\begin{enumerate}
\item If $w : \mathcal{D} \to \mathcal{C}$ is a left adjoint to $u$, then
\begin{enumerate}
\item $g_!\mathcal{F}$ is the sheaf associated to the presheaf
$w^p\mathcal{F}$, and
\item $g_!$ is exact.
\end{enumerate}
\item if $w$ is a continuous left adjoint, then $g_!$
\item If $w$ is a cocontinuous left adjoint, then $g_! = h^{-1}$ and
$g^{-1} = h_*$ where $h : \Sh(\mathcal{D}) \to \Sh(\mathcal{C})$ is
the morphism of topoi associated to $w$.
\end{enumerate}
\end{lemma}

\begin{proof}
Recall that $g_!\mathcal{F}$ is the sheafification of $u_p\mathcal{F}$.
Hence (1)(a) follows from the fact that $u_p = w^p$ by

\medskip\noindent
To see (1)(b) note that $g_!$ commutes with all colimits as $g_!$
Let $i \mapsto \mathcal{F}_i$ be a finite diagram in $\Sh(\mathcal{C})$.
Then $\lim \mathcal{F}_i$ is computed in the category of presheaves
(Lemma \ref{lemma-limit-sheaf}). Since $w^p$ is a right adjoint
we see that $w^p \lim \mathcal{F}_i = \lim w^p\mathcal{F}_i$. Since
sheafification is exact
(Lemma \ref{lemma-sheafification-exact})
we conclude by (1)(a).

\medskip\noindent
Assume $w$ is continuous. Then $g_! = (w^p\ )^\# = w^s$ but sheafification
isn't necessary and one has the left adjoint $w_s$, see

\medskip\noindent
Assume $w$ is cocontinuous. The equality $g_! = h^{-1}$ follows from (1)(a)
and the definitions. The equality $g^{-1} = h_*$ follows from the equality
$g_! = h^{-1}$ and uniqueness of adjoint functor. Alternatively one can deduce
it from Lemma \ref{lemma-have-functor-other-way}.
\end{proof}

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