Lemma 36.23.1. In the situation above, if $a$ and $b$ are quasi-compact and quasi-separated and $X$ and $Y$ are tor-independent over $S$, then (36.23.0.1) is an isomorphism for $K \in D_\mathit{QCoh}(\mathcal{O}_ X)$ and $M \in D_\mathit{QCoh}(\mathcal{O}_ Y)$. If in addition $S = \mathop{\mathrm{Spec}}(A)$ is affine, then the map (36.23.0.2) is an isomorphism.

**First proof.**
This follows from the following sequence of isomorphisms

The first equality holds because $f = a \circ p$. The second equality by Lemma 36.22.1. The third equality by Lemma 36.22.5. The fourth equality by Lemma 36.22.1. We omit the verification that the composition of these isomorphisms is the same as the map (36.23.0.1). If $S$ is affine, then the source and target of the arrow (36.23.0.2) are the result of applying $R\Gamma (S, -)$ to the source and target of (36.23.0.1) and we obtain the final statement; details omitted. $\square$

**Second proof.**
The construction of the arrow (36.23.0.1) is compatible with restricting to open subschemes of $S$ as is immediate from the construction of the relative cup product. Thus it suffices to prove that (36.23.0.1) is an isomorphism when $S$ is affine.

Assume $S = \mathop{\mathrm{Spec}}(A)$ is affine. By Leray we have $R\Gamma (S, Rf_*K) = R\Gamma (X, K)$ and similarly for the other cases. By Cohomology, Lemma 20.31.7 the map (36.23.0.1) induces the map (36.23.0.2) on taking $R\Gamma (S, -)$. On the other hand, by Lemmas 36.4.1 and 36.3.9 the source and target of the map (36.23.0.1) are in $D_\mathit{QCoh}(\mathcal{O}_ S)$. Thus, by Lemma 36.3.5, it suffices to prove that (36.23.0.2) is an isomorphism.

Assume $S = \mathop{\mathrm{Spec}}(A)$ and $X = \mathop{\mathrm{Spec}}(B)$ and $Y = \mathop{\mathrm{Spec}}(C)$ are all affine. We will use Lemma 36.3.5 without further mention. In this case we can choose a K-flat complex $K^\bullet $ of $B$-modules whose terms are flat such that $K$ is represented by $\widetilde{K}^\bullet $. Similarly, we can choose a K-flat complex $M^\bullet $ of $C$-modules whose terms are flat such that $M$ is represented by $\widetilde{M}^\bullet $. See More on Algebra, Lemma 15.59.10. Then $\widetilde{K}^\bullet $ is a K-flat complex of $\mathcal{O}_ X$-modules and similarly for $\widetilde{M}^\bullet $, see Lemma 36.3.6. Thus $La^*K$ is represented by

and similarly for $Lb^*M$. This in turn is a K-flat complex of $\mathcal{O}_{X \times _ S Y}$-modules by the lemma cited above and More on Algebra, Lemma 15.59.3. Thus we finally see that the complex of $\mathcal{O}_{X \times _ S Y}$-modules associated to

represents $La^*K \otimes _{\mathcal{O}_{X \times _ S Y}}^\mathbf {L} Lb^*M$ in the derived category of $X \times _ S Y$. Taking global sections we obtain $\text{Tot}(K^\bullet \otimes _ A M^\bullet )$ which of course is also the complex representing $R\Gamma (X, K) \otimes _ A^\mathbf {L} R\Gamma (Y, M)$. The fact that the isomorphism is given by cup product follows from the relationship between the genuine cup product and the naive one in Cohomology, Section 20.31 (and in particular Cohomology, Lemma 20.31.3 and the discussion following it).

Assume $S = \mathop{\mathrm{Spec}}(A)$ and $Y$ are affine. We will use the induction principle of Cohomology of Schemes, Lemma 30.4.1 to prove the statement. To do this we only have to show: if $X = U \cup V$ is an open covering with $U$ and $V$ quasi-compact and if the map (36.23.0.2)

for $U$ and $Y$ over $S$, the map (36.23.0.2)

for $V$ and $Y$ over $S$, and the map (36.23.0.2)

for $U \cap V$ and $Y$ over $S$ are isomorphisms, then so is the map (36.23.0.2) for $X$ and $Y$ over $S$. However, by Cohomology, Lemma 20.33.7 these maps fit into a map of distinguished triangles with (36.23.0.2) the final leg and hence we conclude by Derived Categories, Lemma 13.4.3.

Assume $S = \mathop{\mathrm{Spec}}(A)$ is affine. To finish the proof we can use the induction principle of Cohomology of Schemes, Lemma 30.4.1 on $Y$. Namely, by the above we already know that our map is an isomorphism when $Y$ is affine. The rest of the argument is exactly the same as in the previous paragraph but with the roles of $X$ and $Y$ switched. $\square$

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