Lemma 6.24.8. Let $f : X \to Y$ be a continuous map of topological spaces. Let $\mathcal{O}$ be a sheaf of rings on $X$. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}$-modules. Let $\mathcal{G}$ be a sheaf of $f_*\mathcal{O}$-modules. Then

$\mathop{\mathrm{Mor}}\nolimits _{\textit{Mod}(\mathcal{O})}( \mathcal{O} \otimes _{f^{-1}f_*\mathcal{O}} f^{-1}\mathcal{G}, \mathcal{F}) = \mathop{\mathrm{Mor}}\nolimits _{\textit{Mod}(f_*\mathcal{O})}(\mathcal{G}, f_*\mathcal{F}).$

Here we use Lemmas 6.24.6 and 6.24.5, and we use the canonical map $f^{-1}f_*\mathcal{O} \to \mathcal{O}$ in the definition of the tensor product.

Proof. This follows from the equalities

\begin{eqnarray*} \mathop{\mathrm{Mor}}\nolimits _{\textit{Mod}(\mathcal{O})}( \mathcal{O} \otimes _{f^{-1}f_*\mathcal{O}} f^{-1}\mathcal{G}, \mathcal{F}) & = & \mathop{\mathrm{Mor}}\nolimits _{\textit{Mod}(f^{-1}f_*\mathcal{O})}( f^{-1}\mathcal{G}, \mathcal{F}_{f^{-1}f_*\mathcal{O}}) \\ & = & \mathop{\mathrm{Mor}}\nolimits _{\textit{Mod}(f_*\mathcal{O})}(\mathcal{G}, f_*\mathcal{F}). \end{eqnarray*}

which are a combination of Lemma 6.20.2 and 6.24.7. $\square$

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