Lemma 20.31.1. The diagram relating the cup product and the naive cup product commutes. The diagram relating the cup product with the cup product (20.25.3.2) on Čech complexes commutes.

Proof. Consider the diagram relating the cup product and the naive cup product. Recall that $f : (X, \mathcal{O}_ X) \to (pt, A)$ is the projection to a point. By construction of $\mu$ the composition clockwise

$\Gamma (X, \mathcal{K}^\bullet ) \otimes _ A^\mathbf {L} \Gamma (X, \mathcal{M}^\bullet ) \to R\Gamma (X, \text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}^\bullet ))$

$a : Lf^*\left(\Gamma (X, \mathcal{K}^\bullet ) \otimes _ A^\mathbf {L} \Gamma (X, \mathcal{M}^\bullet )\right) \to \text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}^\bullet )$

The lower horizontal arrow is adjoint to the map

$b : Lf^*\Gamma (X, \mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}^\bullet ) \to \text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}^\bullet )$

Hence it suffices to see that $a$ is the composition of $b$ with $Lf^*$ applied to the composition of $\Gamma (X, \mathcal{K}^\bullet ) \otimes _ A^\mathbf {L} \Gamma (X, \mathcal{M}^\bullet ) \to \Gamma (X, \mathcal{K}^\bullet ) \otimes _ A \Gamma (X, \mathcal{M}^\bullet )$ with $\mu '$. We omit the details.

Consider the diagram relating the cup product with the cup product (20.25.3.2) on Čech complexes. Choose quasi-isomorphisms of complexes $a : \mathcal{K}^\bullet \to \mathcal{K}_1^\bullet$ and $b : \mathcal{M}^\bullet \to \mathcal{M}_1^\bullet$ as in Lemma 20.30.2. Since the maps $a$ and $b$ on stalks are homotopy equivalences we see that the induced map

$\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}^\bullet ) \to \text{Tot}(\mathcal{K}_1^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}_1^\bullet )$

is a homotopy equivalence on stalks too (More on Algebra, Lemma 15.57.1) and hence a quasi-isomorphism. Thus the targets

$R\Gamma (X, \text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}^\bullet )) = R\Gamma (X, \text{Tot}(\mathcal{K}_1^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}_1^\bullet ))$

of the two diagrams are the same in $D(A)$. It follows that it suffices to prove the diagram commutes for $\mathcal{K}$ and $\mathcal{M}$ replaced by $\mathcal{K}_1$ and $\mathcal{M}_1$. This reduces us to the case discussed in the next paragraph.

Assume $\mathcal{K}^\bullet$ and $\mathcal{M}^\bullet$ are bounded below complexes of flasque $\mathcal{O}_ X$-modules and consider the diagram relating the cup product with the cup product (20.25.3.2) on Čech complexes. Then we can consider the commutative diagram

$\xymatrix{ \Gamma (X, \mathcal{K}^\bullet ) \otimes _ A^\mathbf {L} \Gamma (X, \mathcal{M}^\bullet ) \ar[d] \ar[r] & \text{Tot}(\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{K}^\bullet )) \otimes _ A^\mathbf {L} \text{Tot}(\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{M}^\bullet )) \ar[d] \\ \text{Tot}(\Gamma (X, \mathcal{K}^\bullet ) \otimes _ A \Gamma (X, \mathcal{M}^\bullet )) \ar[d] \ar[r] & \text{Tot}( \text{Tot}(\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{K}^\bullet )) \otimes _ A \text{Tot}(\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{M}^\bullet ))) \ar[d]^{(07MB)} \\ \Gamma (X, \text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}^\bullet )) \ar[r] & \text{Tot}( \check{\mathcal{C}}^\bullet ({\mathcal U}, \text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}^\bullet ) )) }$

In this diagram the horizontal arrows are isomorphisms in $D(A)$ because for a bounded below complex of flasque modules such as $\mathcal{K}^\bullet$ we have

$\Gamma (X, \mathcal{K}^\bullet ) = \text{Tot}(\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{K}^\bullet )) = R\Gamma (X, \mathcal{K}^\bullet )$

in $D(A)$. This follows from Lemma 20.12.3, Derived Categories, Lemma 13.16.7, and Lemma 20.25.2. Hence the commutativity of the diagram involving (20.25.3.2) follows from the already commutativity proven in the first paragraph of the proof. $\square$

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