Lemma 6.24.4. Let $f : X \to Y$ be a continuous map of topological spaces. Let $\mathcal{O}$ be a presheaf of rings on $X$. Let $\mathcal{F}$ be a presheaf of $\mathcal{O}$-modules. Let $\mathcal{G}$ be a presheaf of $f_*\mathcal{O}$-modules. Then
\[ \mathop{\mathrm{Mor}}\nolimits _{\textit{PMod}(\mathcal{O})}( \mathcal{O} \otimes _{p, f_ pf_*\mathcal{O}} f_ p\mathcal{G}, \mathcal{F}) = \mathop{\mathrm{Mor}}\nolimits _{\textit{PMod}(f_*\mathcal{O})}(\mathcal{G}, f_*\mathcal{F}). \]
Here we use Lemmas 6.24.2 and 6.24.1, and we use the map $c_\mathcal {O} : f_ pf_*\mathcal{O} \to \mathcal{O}$ in the definition of the tensor product.
Proof.
This follows from the equalities
\begin{eqnarray*} \mathop{\mathrm{Mor}}\nolimits _{\textit{PMod}(\mathcal{O})}( \mathcal{O} \otimes _{p, f_ pf_*\mathcal{O}} f_ p\mathcal{G}, \mathcal{F}) & = & \mathop{\mathrm{Mor}}\nolimits _{\textit{PMod}(f_ pf_*\mathcal{O})}( f_ p\mathcal{G}, \mathcal{F}_{f_ pf_*\mathcal{O}}) \\ & = & \mathop{\mathrm{Mor}}\nolimits _{\textit{PMod}(f_*\mathcal{O})}(\mathcal{G}, f_*(\mathcal{F}_{f_ pf_*\mathcal{O}})) \\ & = & \mathop{\mathrm{Mor}}\nolimits _{\textit{PMod}(f_*\mathcal{O})}(\mathcal{G}, f_*\mathcal{F}). \end{eqnarray*}
The first equality is Lemma 6.6.2. The second equality is Lemma 6.24.3. The third equality is given by the equality $f_*(\mathcal{F}_{f_ pf_*\mathcal{O}}) = f_*\mathcal{F}$ of abelian sheaves which is $f_*\mathcal{O}$-linear. Namely, $\text{id}_{f_*\mathcal{O}}$ corresponds to $c_\mathcal {O}$ under the adjunction described in the proof of Lemma 6.21.3 and thus $\text{id}_{f_*\mathcal{O}} = f_*c_\mathcal {O} \circ i_{f_*\mathcal{O}}$.
$\square$
Comments (0)
There are also: