Lemma 18.40.3. Assume given a commutative diagram

$\xymatrix{ (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}_{\mathcal{C}'}) \ar[r]_{(g', (g')^\sharp )} \ar[d]_{(f', (f')^\sharp )} & (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}_\mathcal {C}) \ar[d]^{(f, f^\sharp )} \\ (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}'), \mathcal{O}_{\mathcal{D}'}) \ar[r]^{(g, g^\sharp )} & (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}_\mathcal {D}) }$

of ringed topoi. Assume

1. $f$, $f'$, $g$, and $g'$ correspond to cocontinuous functors $u$, $u'$, $v$, and $v'$ as in Sites, Lemma 7.21.1,

2. $v \circ u' = u \circ v'$,

3. $v$ and $v'$ are continuous as well as cocontinuous,

4. for any object $V'$ of $\mathcal{D}'$ the functor ${}^{u'}_{V'}\mathcal{I} \to {}^{\ \ \ u}_{v(V')}\mathcal{I}$ given by $v$ is cofinal, and

5. $g^{-1}\mathcal{O}_{\mathcal{D}} = \mathcal{O}_{\mathcal{D}'}$ and $(g')^{-1}\mathcal{O}_{\mathcal{C}} = \mathcal{O}_{\mathcal{C}'}$.

Then we have $f'_* \circ (g')^* = g^* \circ f_*$ and $g'_! \circ (f')^{-1} = f^{-1} \circ g_!$ on modules.

Proof. We have $(g')^*\mathcal{F} = (g')^{-1}\mathcal{F}$ and $g^*\mathcal{G} = g^{-1}\mathcal{G}$ because of condition (5). Thus the first equality follows immediately from the corresponding equality in Sites, Lemma 7.28.6. Since the left adjoint functors $g_!$ and $g'_!$ to $g^*$ and $(g')^*$ exist by Lemma 18.40.1 we see that the second equality follows by uniqueness of adjoint functors. $\square$

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