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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