Lemma 18.22.1. Let
f : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \longrightarrow (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}')
be a morphism of ringed topoi. Let \mathcal{G} be a sheaf on \mathcal{D}. Set \mathcal{F} = f^{-1}\mathcal{G}. Then there exists a commutative diagram of ringed topoi
\xymatrix{ (\mathop{\mathit{Sh}}\nolimits (\mathcal{C})/\mathcal{F}, \mathcal{O}_\mathcal {F}) \ar[rr]_{(j_\mathcal {F}, j_\mathcal {F}^\sharp )} \ar[d]_{(f', (f')^\sharp )} & & (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \ar[d]^{(f, f^\sharp )} \\ (\mathop{\mathit{Sh}}\nolimits (\mathcal{D})/\mathcal{G}, \mathcal{O}'_\mathcal {G}) \ar[rr]^{(j_\mathcal {G}, j_\mathcal {G}^\sharp )} & & (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}') }
We have f'_*j_\mathcal {F}^{-1} = j_\mathcal {G}^{-1}f_* and f'_*j_\mathcal {F}^* = j_\mathcal {G}^*f_*. Moreover, the morphism f' is characterized by the rule
(f')^{-1}(\mathcal{H} \xrightarrow {\varphi } \mathcal{G}) = (f^{-1}\mathcal{H} \xrightarrow {f^{-1}\varphi } \mathcal{F}).
Proof.
By Sites, Lemma 7.31.1 we have the diagram of underlying topoi, the equality f'_*j_\mathcal {F}^{-1} = j_\mathcal {G}^{-1}f_*, and the description of (f')^{-1}. To define (f')^\sharp we use the map
(f')^\sharp : \mathcal{O}'_\mathcal {G} = j_\mathcal {G}^{-1} \mathcal{O}' \xrightarrow {j_\mathcal {G}^{-1}f^\sharp } j_\mathcal {G}^{-1} f_*\mathcal{O} = f'_* j_\mathcal {F}^{-1}\mathcal{O} = f'_* \mathcal{O}_\mathcal {F}
or equivalently the map
(f')^\sharp : (f')^{-1}\mathcal{O}'_\mathcal {G} = (f')^{-1}j_\mathcal {G}^{-1} \mathcal{O}' = j_\mathcal {F}^{-1}f^{-1}\mathcal{O}' \xrightarrow {j_\mathcal {F}^{-1}f^\sharp } j_\mathcal {F}^{-1} \mathcal{O} = \mathcal{O}_\mathcal {F}.
We omit the verification that these two maps are indeed adjoint to each other. The second construction of (f')^\sharp shows that the diagram commutes in the 2-category of ringed topoi (as the maps j_\mathcal {F}^\sharp and j_\mathcal {G}^\sharp are identities). Finally, the equality f'_*j_\mathcal {F}^* = j_\mathcal {G}^*f_* follows from the equality f'_*j_\mathcal {F}^{-1} = j_\mathcal {G}^{-1}f_* and the fact that pullbacks of sheaves of modules and sheaves of sets agree, see Lemma 18.21.1.
\square
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