Lemma 36.25.1. In the situation above, assume $a$ and $b$ are quasi-compact and quasi-separated and $X$ and $Y$ are tor independent over $S$. If $K$ is perfect, $K' \in D_\mathit{QCoh}(\mathcal{O}_ X)$, $M$ is perfect, and $M' \in D_\mathit{QCoh}(\mathcal{O}_ Y)$, then (36.25.0.1) is an isomorphism.

Proof. In this case we have $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, K') = K' \otimes ^\mathbf {L} K^\vee$, $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, M') = M' \otimes ^\mathbf {L} M^\vee$, and

$R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K \boxtimes M, K' \boxtimes M') = (K' \otimes ^\mathbf {L} K^\vee ) \boxtimes (M' \otimes ^\mathbf {L} M^\vee )$

See Cohomology, Lemma 20.47.5 and we also use that being perfect is preserved by pullback and by tensor products. Hence this case follows from Lemma 36.23.1. (We omit the verification that with these identifications we obtain the same map.) $\square$

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