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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