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.
36.25 Künneth formula for Ext
Consider a cartesian diagram of schemes
\[ \xymatrix{ & X \times _ S Y \ar[ld]^ p \ar[rd]_ q \ar[dd]^ f \\ X \ar[rd]_ a & & Y \ar[ld]^ b \\ & S } \]
For $K \in D(\mathcal{O}_ X)$ and $M \in D(\mathcal{O}_ Y)$ in this section let us define
\[ K \boxtimes M = Lp^*K \otimes _{\mathcal{O}_{X \times _ S Y}}^\mathbf {L} Lq^*M \]
We claim there is a canonical map
36.25.0.1
\begin{equation} \label{perfect-equation-kunneth-ext} Ra_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, K') \otimes _{\mathcal{O}_ S}^\mathbf {L} Rb_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, M') \longrightarrow Rf_*(R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K \boxtimes M, K' \boxtimes M')) \end{equation}
for $K, K' \in D(\mathcal{O}_ X)$ and $M, M' \in D(\mathcal{O}_ Y)$. Namely, we can take the map adjoint to the map
\[ \begin{matrix} Lf^*\left(Ra_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, K') \otimes _{\mathcal{O}_ S}^\mathbf {L} Rb_* R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, M')\right) =
\\ Lf^* Ra_* R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, K') \otimes _{\mathcal{O}_{X \times _ S Y}}^\mathbf {L} Lf^* Rb_* R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, M') =
\\ Lp^* La^* Ra_* R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, K') \otimes _{\mathcal{O}_{X \times _ S Y}}^\mathbf {L} Lq^* Lb^* Rb_* R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, M') \to
\\ Lp^* R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, K') \otimes _{\mathcal{O}_{X \times _ S Y}}^\mathbf {L} Lq^* R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, M') \to
\\ R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (Lp^*K, Lp^*K') \otimes _{\mathcal{O}_{X \times _ S Y}}^\mathbf {L} R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (Lq^*M, Lq^*M') \to
\\ R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K \boxtimes M, K' \boxtimes M')
\end{matrix} \]
Here the first equality is compatibility of pullbacks with tensor products, Cohomology, Lemma 20.27.3. The second equality is $f = a \circ p = b \circ q$ and composition of pullbacks, Cohomology, Lemma 20.27.2. The first arrow is given by the adjunction maps $La^* Ra_* \to \text{id}$ and $Lb^* Rb_* \to \text{id}$ because pushforward and pullback are adjoint, Cohomology, Lemma 20.28.1. The second arrow is given by Cohomology, Remark 20.42.13. The third and final arrow is Cohomology, Remark 20.42.10. A simple special case of this is the following result.
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 20.50.5 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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