Lemma 21.33.1. In the situation above the following diagram commutes
\[ \xymatrix{ f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {D}}^\mathbf {L} f_*\mathcal{M}^\bullet \ar[r] \ar[d] & Rf_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {D}}^\mathbf {L} Rf_*\mathcal{M}^\bullet \ar[d]^{\text{Remark 0B6C}} \\ \text{Tot}( f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {D}} f_*\mathcal{M}^\bullet ) \ar[d]_{\text{naive cup product}} & Rf_*(\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {C}}^\mathbf {L} \mathcal{M}^\bullet ) \ar[d] \\ f_*\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {C}} \mathcal{M}^\bullet ) \ar[r] & Rf_*\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {C}} \mathcal{M}^\bullet ) } \]
Proof.
By the construction in Remark 21.19.7 we see that going around the diagram clockwise the map
\[ f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {D}}^\mathbf {L} f_*\mathcal{M}^\bullet \longrightarrow Rf_*\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {C}} \mathcal{M}^\bullet ) \]
is adjoint to the map
\begin{align*} Lf^*(f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {D}}^\mathbf {L} f_*\mathcal{M}^\bullet ) & = Lf^*f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {D}}^\mathbf {L} Lf^*f_*\mathcal{M}^\bullet \\ & \to Lf^*Rf_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {D}}^\mathbf {L} Lf^*Rf_*\mathcal{M}^\bullet \\ & \to \mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {D}}^\mathbf {L} \mathcal{M}^\bullet \\ & \to \text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {C}} \mathcal{M}^\bullet ) \end{align*}
By Lemma 21.19.6 this is also equal to
\begin{align*} Lf^*(f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {D}}^\mathbf {L} f_*\mathcal{M}^\bullet ) & = Lf^*f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {D}}^\mathbf {L} Lf^*f_*\mathcal{M}^\bullet \\ & \to f^*f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {D}}^\mathbf {L} f^*f_*\mathcal{M}^\bullet \\ & \to \mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {D}}^\mathbf {L} \mathcal{M}^\bullet \\ & \to \text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {C}} \mathcal{M}^\bullet ) \end{align*}
Going around anti-clockwise we obtain the map adjoint to the map
\begin{align*} Lf^*(f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {D}}^\mathbf {L} f_*\mathcal{M}^\bullet ) & \to Lf^*\text{Tot}( f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {D}} f_*\mathcal{M}^\bullet ) \\ & \to Lf^*f_*\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {C}} \mathcal{M}^\bullet ) \\ & \to Lf^*Rf_*\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {C}} \mathcal{M}^\bullet ) \\ & \to \text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {C}} \mathcal{M}^\bullet ) \end{align*}
By Lemma 21.19.6 this is also equal to
\begin{align*} Lf^*(f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {D}}^\mathbf {L} f_*\mathcal{M}^\bullet ) & \to Lf^*\text{Tot}( f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {D}} f_*\mathcal{M}^\bullet ) \\ & \to Lf^*f_*\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {C}} \mathcal{M}^\bullet ) \\ & \to f^*f_*\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {C}} \mathcal{M}^\bullet ) \\ & \to \text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {C}} \mathcal{M}^\bullet ) \end{align*}
Now the proof is finished by a contemplation of the diagram
\[ \xymatrix{ Lf^*(f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {D}}^\mathbf {L} f_*\mathcal{M}^\bullet ) \ar[d] \ar[rr] & & Lf^*f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {C}}^\mathbf {L} Lf^*f_*\mathcal{M}^\bullet \ar[d] \\ Lf^*\text{Tot}( f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {D}} f_*\mathcal{M}^\bullet ) \ar[d]_{naive} \ar[r] & f^*\text{Tot}( f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {D}} f_*\mathcal{M}^\bullet ) \ar[ldd]^{naive} \ar[dd] & f^*f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {C}}^\mathbf {L} f^*f_*\mathcal{M}^\bullet \ar[dd] \ar[ldd] \\ Lf^*f_*\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {C}} \mathcal{M}^\bullet ) \ar[d] \\ f^*f_*\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {C}} \mathcal{M}^\bullet ) \ar[rd] & \text{Tot}(f^*f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {C}} f^*f_*\mathcal{M}^\bullet ) \ar[d] & \mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {C}}^\mathbf {L} \mathcal{M}^\bullet \ar[ld] \\ & \text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {C}} \mathcal{M}^\bullet ) } \]
All of the polygons in this diagram commute. The top one commutes by Lemma 21.18.8. The square with the two naive cup products commutes because $Lf^* \to f^*$ is functorial in the complex of modules. Similarly with the square involving the two maps $\mathcal{A}^\bullet \otimes ^\mathbf {L} \mathcal{B}^\bullet \to \text{Tot}(\mathcal{A}^\bullet \otimes \mathcal{B}^\bullet )$. Finally, the commutativity of the remaining square is true on the level of complexes and may be viewed as the definition of the naive cup product (by the adjointness of $f^*$ and $f_*$). The proof is finished because going around the diagram on the outside are the two maps given above.
$\square$
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