Lemma 36.23.4. In the situation above the map (36.23.3.1) is an isomorphism if $S$ is affine, $\mathcal{F}$ and $\mathcal{G}$ are $S$-flat and quasi-coherent and $X$ and $Y$ are quasi-compact with affine diagonal.
Proof. We strongly urge the reader to read the proof of Varieties, Lemma 33.29.1 first. Choose finite affine open coverings $\mathcal{U} : X = \bigcup _{i \in I} U_ i$ and $\mathcal{V} : Y = \bigcup _{j \in J} V_ j$. This determines an affine open covering $\mathcal{W} : X \times _ S Y = \bigcup _{(i, j) \in I \times J} U_ i \times _ S V_ j$. Note that $\mathcal{W}$ is a refinement of $\text{pr}_1^{-1}\mathcal{U}$ and of $\text{pr}_2^{-1}\mathcal{V}$. Thus by the discussion in Cohomology, Section 20.25 we obtain maps
well defined up to homotopy and compatible with pullback maps on cohomology. In Cohomology, Equation (20.25.3.2) we have constructed a map of complexes
which is compatible with the cup product on cohomology by Cohomology, Lemma 20.31.4. Combining the above we obtain a map of complexes
We claim this is the map in the statement of the lemma, i.e., the source and target of this arrow are the same as the source and target of (36.23.3.1). Namely, by Cohomology of Schemes, Lemma 30.2.2 and Cohomology, Lemma 20.25.2 the canonical maps
and
are isomorphisms. On the other hand, the complex $\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F})$ is K-flat by Lemma 36.23.3 and we conclude that $\text{Tot}( \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F}) \otimes _ A \check{\mathcal{C}}^\bullet (\mathcal{V}, \mathcal{G}))$ represents the derived tensor product $R\Gamma (X, \mathcal{F}) \otimes _ A^\mathbf {L} R\Gamma (Y, \mathcal{G})$ as claimed.
We still have to show that (36.23.4.1) is a quasi-isomorphism. We will do this using dimension shifting. Set $d(\mathcal{F}) = \max \{ d \mid H^ d(X, \mathcal{F}) \not= 0\} $. Assume $d(\mathcal{F}) > 0$. Set $U = \coprod \nolimits _{i \in I} U_ i$. This is an affine scheme as $I$ is finite. Denote $j : U \to X$ the morphism which is the inclusion $U_ i \to X$ on each $U_ i$. Since the diagonal of $X$ is affine, the morphism $j$ is affine, see Morphisms, Lemma 29.11.11. It follows that $\mathcal{F}' = j_*j^*\mathcal{F}$ is $S$-flat, see Morphisms, Lemma 29.25.4. It also follows that $d(\mathcal{F}') = 0$ by combining Cohomology of Schemes, Lemmas 30.2.4 and 30.2.2. For all $x \in X$ we have $\mathcal{F}_ x \to \mathcal{F}'_ x$ is the inclusion of a direct summand: if $x \in U_ i$, then $\mathcal{F}' \to (U_ i \to X)_*\mathcal{F}|_{U_ i}$ gives a splitting. We conclude that $\mathcal{F} \to \mathcal{F}'$ is injective and $\mathcal{F}'' = \mathcal{F}'/\mathcal{F}$ is $S$-flat as well. The short exact sequence $0 \to \mathcal{F} \to \mathcal{F}' \to \mathcal{F}'' \to 0$ of flat quasi-coherent $\mathcal{O}_ X$-modules produces a short exact sequence of complexes
and a short exact sequence of complexes
Moreover, the maps (36.23.4.1) between these are compatible with these short exact sequences. Hence it suffices to prove (36.23.4.1) is an isomorphism for $\mathcal{F}'$ and $\mathcal{F}''$. Finally, we have $d(\mathcal{F}'') < d(\mathcal{F})$. In this way we reduce to the case $d(\mathcal{F}) = 0$.
Arguing in the same fashion for $\mathcal{G}$ we find that we may assume that both $\mathcal{F}$ and $\mathcal{G}$ have nonzero cohomology only in degree $0$. Observe that this means that $\Gamma (X, \mathcal{F})$ is quasi-isomorphic to the $K$-flat complex $\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F})$ of $A$-modules sitting in degrees $\geq 0$. It follows that $\Gamma (X, \mathcal{F})$ is a flat $A$-module (because we can compute higher Tor's against this module by tensoring with the Cech complex). Let $V \subset Y$ be an affine open. Consider the affine open covering $\mathcal{U}_ V : X \times _ S V = \bigcup _{i \in I} U_ i \times _ S V$. It is immediate that
(equality of complexes). By the flatness of $\mathcal{G}(V)$ over $A$ we see that $\Gamma (X, \mathcal{F}) \otimes _ A \mathcal{G}(V) \to \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F}) \otimes _ A \mathcal{G}(V)$ is a quasi-isomorphism. Since the sheafification of $V \mapsto \check{\mathcal{C}}^\bullet (\mathcal{U}_ V, p^*\mathcal{F} \otimes _{\mathcal{O}_{X \times Y}} q^*\mathcal{G})$ represents $Rq_*(p^*\mathcal{F} \otimes _{\mathcal{O}_{X \times Y}} q^*\mathcal{G})$ by Cohomology of Schemes, Lemma 30.7.1 we conclude that
on $Y$ where the notation on the right hand side indicates the module
Using the Leray spectral sequence for $q$ we find
Using Lemma 36.22.1 for the morphism $b : Y \to S = \mathop{\mathrm{Spec}}(A)$ and using that $\Gamma (X, \mathcal{F})$ is $A$-flat we conclude that $H^ n(X \times _ S Y, p^*\mathcal{F} \otimes _{\mathcal{O}_{X \times Y}} q^*\mathcal{G})$ is zero for $n > 0$ and isomorphic to $H^0(X, \mathcal{F}) \otimes _ A H^0(Y, \mathcal{G})$ for $n = 0$. Of course, here we also use that $\mathcal{G}$ only has cohomology in degree $0$. This finishes the proof (except that we should check that the isomorphism is indeed given by cup product in degree $0$; we omit the verification). $\square$
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