The Stacks project

Lemma 18.12.3. Let $\mathcal{C}$, $\mathcal{D}$ be sites. Let $f : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{D})$ be a morphism of topoi. Let $\mathcal{O}$ be a sheaf of rings on $\mathcal{D}$. Let $\mathcal{G}$ be a sheaf of $\mathcal{O}$-modules. Let $\mathcal{F}$ be a sheaf of $f^{-1}\mathcal{O}$-modules. Then

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

Here we use Lemmas 18.12.2 and 18.12.1, and we think of $f_*\mathcal{F}$ as an $\mathcal{O}$-module by restriction via $\mathcal{O} \to f_*f^{-1}\mathcal{O}$.

Proof. First we note that we have

\[ \mathop{\mathrm{Mor}}\nolimits _{\textit{Ab}(\mathcal{C})}(f^{-1}\mathcal{G}, \mathcal{F}) = \mathop{\mathrm{Mor}}\nolimits _{\textit{Ab}(\mathcal{D})}(\mathcal{G}, f_*\mathcal{F}). \]

by Sites, Proposition 7.44.3. Suppose that $\alpha : f^{-1}\mathcal{G} \to \mathcal{F}$ and $\beta : \mathcal{G} \to f_*\mathcal{F}$ are morphisms of abelian sheaves which correspond via the formula above. We have to show that $\alpha $ is $f^{-1}\mathcal{O}$-linear if and only if $\beta $ is $\mathcal{O}$-linear. For example, suppose $\alpha $ is $f^{-1}\mathcal{O}$-linear, then clearly $f_*\alpha $ is $f_*f^{-1}\mathcal{O}$-linear, and hence (as restriction is a functor) is $\mathcal{O}$-linear. Hence it suffices to prove that the adjunction map $\mathcal{G} \to f_*f^{-1}\mathcal{G}$ is $\mathcal{O}$-linear. Using that both $f_*$ and $f^{-1}$ commute with products (on sheaves of sets) this comes down to showing that

\[ \xymatrix{ \mathcal{O} \times \mathcal{G} \ar[r] \ar[d] & f_*f^{-1}(\mathcal{O} \times \mathcal{G}) \ar[d] \\ \mathcal{G} \ar[r] & f_*f^{-1}\mathcal{G} } \]

is commutative. This holds because the adjunction mapping $\text{id}_{\mathop{\mathit{Sh}}\nolimits (\mathcal{D})} \to f_*f^{-1}$ is a transformation of functors. We omit the proof of the implication $\beta $ linear $\Rightarrow $ $\alpha $ linear. $\square$

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