Lemma 62.11.5. Let $f : X \to Y$ be a separated finite type morphism of quasi-compact and quasi-separated schemes. Let $\Lambda$ be a torsion ring. For every $K \in D(Y_{\acute{e}tale}, \Lambda )$ and $L \in D(X_{\acute{e}tale}, \Lambda )$ the map (62.11.4.1)

$Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\Lambda (L, Rf^!K) \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\Lambda (Rf_!L, K)$

is an isomorphism.

Proof. To prove the lemma we have to show that for any $M \in D(Y_{\acute{e}tale}, \Lambda )$ the map (62.11.4.1) induces an bijection

$\mathop{\mathrm{Hom}}\nolimits _ Y(M, Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\Lambda (L, Rf^!K)) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ Y(M, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\Lambda (Rf_!L, K))$

To see this we use the following string of equalities

\begin{align*} \mathop{\mathrm{Hom}}\nolimits _ Y(M, Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\Lambda (L, Rf^!K)) & = \mathop{\mathrm{Hom}}\nolimits _ X(f^{-1}M, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\Lambda (L, Rf^!K)) \\ & = \mathop{\mathrm{Hom}}\nolimits _ X(f^{-1}M \otimes _\Lambda ^\mathbf {L} L, Rf^!K) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(Rf_!(f^{-1}M \otimes _\Lambda ^\mathbf {L} L), K) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(M \otimes _\Lambda ^\mathbf {L} Rf_!L, K) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(M, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\Lambda (Rf_!L, K)) \end{align*}

The first equality holds by Cohomology on Sites, Lemma 21.19.1. The second equality by Cohomology on Sites, Lemma 21.35.2. The third equality by construction of $Rf^!$. The fourth equality by Lemma 62.10.7 (this is the important step). The fifth by Cohomology on Sites, Lemma 21.35.2. $\square$

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