Lemma 20.31.1. This construction gives the cup product.

**Proof.**
With $f : (X, \mathcal{O}_ X) \to (pt, A)$ as above we have $Rf_*(-) = R\Gamma (X, -)$ and our map $\mu $ is adjoint to the map

where $\epsilon $ is the counit of the adjunction between $Lf^*$ and $Rf_*$. If we think of $\xi $ and $\eta $ as maps $\xi : A[-i] \to R\Gamma (X, K)$ and $\eta : A[-j] \to R\Gamma (X, M)$, then the tensor $\xi \otimes \eta $ corresponds to the map^{1}

By definition the cup product $\xi \cup \eta $ is the map $A[-i - j] \to R\Gamma (X, K \otimes _{\mathcal{O}_ X}^\mathbf {L} M)$ which is adjoint to

However, it is easy to see that $\epsilon _ K \circ Lf^*\xi = \tilde\xi $ and $\epsilon _ M \circ Lf^*\eta = \tilde\eta $. We conclude that $\widetilde{\xi \cup \eta } = \tilde\xi \otimes \tilde\eta $ which means we have the desired agreement. $\square$

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