The Stacks project

7.23 Cocontinuous functors which have a left adjoint

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

Lemma 7.23.1. 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 7.21.1 and 7.21.5.

  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 7.19.3.

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 7.22.1. $\square$

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