Lemma 18.40.4. Consider 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 and suppose we have functors

$\xymatrix{ \mathcal{C}' \ar[r]_{v'} & \mathcal{C} \\ \mathcal{D}' \ar[r]^ v \ar[u]^{u'} & \mathcal{D} \ar[u]_ u }$

such that (with notation as in Sites, Sections 7.14 and 7.21) we have

1. $u$ and $u'$ are continuous and give rise to the morphisms $f$ and $f'$,

2. $v$ and $v'$ are cocontinuous giving rise to the morphisms $g$ and $g'$,

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

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

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

Then $f'_* \circ (g')^* = g^* \circ f_*$ 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 7.28.7. 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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