Lemma 36.24.1. In the situation above the cup product (36.24.0.2) is an isomorphism in $D(A)$ if the following assumptions hold

1. $S = \mathop{\mathrm{Spec}}(A)$ is affine,

2. $X$ and $Y$ are quasi-compact with affine diagonal,

3. $\mathcal{F}^\bullet$ is bounded,

4. $\mathcal{G}^\bullet$ is bounded below,

5. $\mathcal{F}^ n$ is $S$-flat, and

6. $\mathcal{G}^ m$ is $S$-flat.

Proof. We will use the notation $\mathcal{A}_{X/S}$ and $\mathcal{A}_{Y/S}$ introduced in Morphisms, Remark 29.33.3. Suppose that we have maps of complexes

$\mathcal{F}_1^\bullet \to \mathcal{F}_2^\bullet \to \mathcal{F}_3^\bullet \to \mathcal{F}_1^\bullet [1]$

in the category $\mathcal{A}_{X/S}$. Then by the functoriality of the cup product we obtain a commutative diagram

$\xymatrix{ R\Gamma (X, \mathcal{F}_1^\bullet ) \otimes _ A^\mathbf {L} R\Gamma (Y, \mathcal{G}^\bullet ) \ar[r] \ar[d] & R\Gamma (X \times _ S Y, \text{Tot}(\mathcal{F}_1^\bullet \boxtimes \mathcal{G}^\bullet )) \ar[d] \\ R\Gamma (X, \mathcal{F}_2^\bullet ) \otimes _ A^\mathbf {L} R\Gamma (Y, \mathcal{G}^\bullet ) \ar[r] \ar[d] & R\Gamma (X \times _ S Y, \text{Tot}(\mathcal{F}_2^\bullet \boxtimes \mathcal{G}^\bullet )) \ar[d] \\ R\Gamma (X, \mathcal{F}_3^\bullet ) \otimes _ A^\mathbf {L} R\Gamma (Y, \mathcal{G}^\bullet ) \ar[r] \ar[d] & R\Gamma (X \times _ S Y, \text{Tot}(\mathcal{F}_3^\bullet \boxtimes \mathcal{G}^\bullet )) \ar[d] \\ R\Gamma (X, \mathcal{F}_1^\bullet [1]) \otimes _ A^\mathbf {L} R\Gamma (Y, \mathcal{G}^\bullet ) \ar[r] & R\Gamma (X \times _ S Y, \text{Tot}(\mathcal{F}_1^\bullet [1] \boxtimes \mathcal{G}^\bullet )) }$

If the original maps form a distinguished triangle in the homotopy category of $\mathcal{A}_{X/S}$, then the columns of this diagram form distinguished triangles in $D(A)$.

In the situation of the lemma, suppose that $\mathcal{F}^ n = 0$ for $n < i$. Then we may consider the termwise split short exact sequence of complexes

$0 \to \sigma _{\geq i + 1}\mathcal{F}^\bullet \to \mathcal{F}^\bullet \to \mathcal{F}^ i[-i] \to 0$

where the truncation is as in Homology, Section 12.15. This produces the distinguished triangle

$\sigma _{\geq i + 1}\mathcal{F}^\bullet \to \mathcal{F}^\bullet \to \mathcal{F}^ i[-i] \to (\sigma _{\geq i + 1}\mathcal{F}^\bullet )[1]$

in the homotopy category of $\mathcal{A}_{X/S}$ where the final arrow is given by the boundary map $\mathcal{F}^ i \to \mathcal{F}^{i + 1}$. It follows from the discussion above that it suffices to prove the lemma for $\mathcal{F}^ i[-i]$ and $\sigma _{\geq i + 1}\mathcal{F}^\bullet$. Since $\sigma _{\geq i + 1}\mathcal{F}^\bullet$ has fewer nonzero terms, by induction, if we can prove the lemma if $\mathcal{F}^\bullet$ is nonzero only in single degree, then the lemma follows. Thus we may assume $\mathcal{F}^\bullet$ is nonzero only in one degree.

Assume $\mathcal{F}^\bullet$ is the complex which has an $S$-flat quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}$ sitting in degree $0$ and is zero in other degrees. Observe that $R\Gamma (X, \mathcal{F})$ has finite tor dimension by Lemma 36.23.2 for example. Say it has tor amplitude in $[i, j]$. Pick $N \gg 0$ and consider the distinguished triangle

$\sigma _{\geq N + 1}\mathcal{G}^\bullet \to \mathcal{G}^\bullet \to \sigma _{\leq N}\mathcal{G}^\bullet \to (\sigma _{\geq N + 1}\mathcal{G}^\bullet )[1]$

in the homotopy category of $\mathcal{A}_{Y/S}$. Now observe that both

$R\Gamma (X, \mathcal{F}) \otimes _ A^\mathbf {L} R\Gamma (Y, \sigma _{\geq N + 1}\mathcal{G}^\bullet ) \quad \text{and}\quad R\Gamma (X \times _ S Y, \text{Tot}(\mathcal{F} \boxtimes \sigma _{\geq N + 1}\mathcal{G}^\bullet ))$

have vanishing cohomology in degrees $\leq N + i$. Thus, using the arguments given above, if we want to prove our statement in a given degree, then we may assume $\mathcal{G}^\bullet$ is bounded. Repeating the arguments above one more time we may also assume $\mathcal{G}^\bullet$ is nonzero only in one degree. This case is handled by Lemma 36.23.4. $\square$

Comment #4650 by on

First, of all, we should add a variant of this lemma where $X$ and $Y$ are flat over $S$ in which case this lemma is a lot easier to prove. Secondly, in the formula $R\Gamma (X, \sigma _{\geq n}\mathcal{F}^\bullet ) \otimes _ A^\mathbf {L} R\Gamma (Y, \mathcal{G}^\bullet )$ there is a typo and $n$ should be $N$. Thirdly, the statement that this lives in degrees $\geq N + b$ needs to be justified by proving a bound on the tor dimension of $R\Gamma (Y, \mathcal{G}^\bullet )$ which follows from the results in the next paragraph.

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