Lemma 24.9.1. In the situation above we have
\mathop{\mathrm{Hom}}\nolimits _{\textit{Mod}^{gr}(\mathcal{B})}( \mathcal{N}, f_*\mathcal{M}) = \mathop{\mathrm{Hom}}\nolimits _{\textit{Mod}^{gr}(\mathcal{A})}( f^*\mathcal{N}, \mathcal{M})
Lemma 24.9.1. In the situation above we have
Proof. Omitted. Hints: First prove that f^{-1} and f_* are adjoint as functors between \textit{Mod}(\mathcal{B}) and \textit{Mod}(f^{-1}\mathcal{B}) using the adjunction between f^{-1} and f_* on sheaves of abelian groups. Next, use the adjunction between base change and restriction given in Section 24.8. \square
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