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

$Lf^*(Rf_*K \otimes _ A^\mathbf {L} Rf_*M) = Lf^*Rf_*K \otimes _{\mathcal{O}_ X}^\mathbf {L} Lf^*Rf_*M \xrightarrow {\epsilon _ K \otimes \epsilon _ M} K \otimes _{\mathcal{O}_ X}^\mathbf {L} M$

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 map1

$A[-i - j] = A[-i] \otimes _ A^\mathbf {L} A[-j] \xrightarrow {\xi \otimes \eta } R\Gamma (X, K) \otimes _ A^\mathbf {L} R\Gamma (X, M)$

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

$(\epsilon _ K \otimes \epsilon _ M) \circ Lf^*(\xi \otimes \eta ) = (\epsilon _ K \circ Lf^*\xi ) \otimes (\epsilon _ M \circ Lf^*\eta )$

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. $\square$

 There is a sign hidden here, namely, the equality is defined by the composition
$A[-i - j] \to (A \otimes _ A^\mathbf {L} A)[-i - j] \to A[-i] \otimes _ A^\mathbf {L} A[-j]$
where in the second step we use the identification of More on Algebra, Item (7) which uses a sign in principle. Except, in this case the sign is $+1$ by our convention and even if it wasn't $+1$ it wouldn't matter since we used the same sign in the identification $\mathcal{O}_ X[-i - j] = \mathcal{O}_ X[-i] \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{O}_ X[-j]$.

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