Lemma 18.26.2. Let f : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}_\mathcal {C}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}_\mathcal {D}) be a morphism of ringed topoi. Let \mathcal{F}, \mathcal{G} be \mathcal{O}_\mathcal {D}-modules. Then f^*(\mathcal{F} \otimes _{\mathcal{O}_\mathcal {D}} \mathcal{G}) = f^*\mathcal{F} \otimes _{\mathcal{O}_\mathcal {C}} f^*\mathcal{G} functorially in \mathcal{F}, \mathcal{G}.
Proof. For a sheaf \mathcal{H} of \mathcal{O}_\mathcal {C} modules we have
\begin{align*} \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_\mathcal {C}}( f^*(\mathcal{F} \otimes _\mathcal {O} \mathcal{G}), \mathcal{H}) & = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_\mathcal {D}}( \mathcal{F} \otimes _\mathcal {O} \mathcal{G}, f_*\mathcal{H}) \\ & = \text{Bilin}_{\mathcal{O}_\mathcal {D}}( \mathcal{F} \times \mathcal{G}, f_*\mathcal{H}) \\ & = \text{Bilin}_{f^{-1}\mathcal{O}_\mathcal {D}}( f^{-1}\mathcal{F} \times f^{-1}\mathcal{G}, \mathcal{H}) \\ & = \mathop{\mathrm{Hom}}\nolimits _{f^{-1}\mathcal{O}_\mathcal {D}}( f^{-1}\mathcal{F} \otimes _{f^{-1}\mathcal{O}_\mathcal {D}} f^{-1}\mathcal{G}, \mathcal{H}) \\ & = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_\mathcal {C}}( f^*\mathcal{F} \otimes _{f^*\mathcal{O}_\mathcal {D}} f^*\mathcal{G}, \mathcal{H}) \end{align*}
The interesting “=” in this sequence of equalities is the third equality. It follows from the definition and adjointness of f_* and f^{-1} (as discussed in previous sections) in a straightforward manner. \square
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