$f : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \longrightarrow (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}')$

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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