## 21.32 Cup product

Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $K, M$ be objects of $D(\mathcal{O})$. Set $A = \Gamma (\mathcal{C}, \mathcal{O})$. The (global) cup product in this setting is a map

$R\Gamma (\mathcal{C}, K) \otimes _ A^\mathbf {L} R\Gamma (\mathcal{C}, M) \longrightarrow R\Gamma (\mathcal{C}, K \otimes _\mathcal {O}^\mathbf {L} M)$

in $D(A)$. We define it as the relative cup product for the morphism of ringed topoi $(\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (pt), A)$ as in Remark 21.19.7.

Let us formulate and prove a natural compatibility of the relative cup product. Namely, suppose that we have a morphism $f : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}_\mathcal {C}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}_\mathcal {D})$ of ringed topoi. Let $\mathcal{K}^\bullet$ and $\mathcal{M}^\bullet$ be complexes of $\mathcal{O}_\mathcal {C}$-modules. There is a naive cup product

$\text{Tot}( f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {D}} f_*\mathcal{M}^\bullet ) \longrightarrow f_*\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {C}} \mathcal{M}^\bullet )$

We claim that this is related to the relative cup product.

Lemma 21.32.1. 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

$f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {D}}^\mathbf {L} f_*\mathcal{M}^\bullet \longrightarrow Rf_*\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_\mathcal {C}} \mathcal{M}^\bullet )$

\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 definiton 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. $\square$

Lemma 21.32.2. Let $f : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}')$ be a morphism of ringed topoi. The relative cup product of Remark 21.19.7 is associative in the sense that the diagram

$\xymatrix{ Rf_*K \otimes _{\mathcal{O}'}^\mathbf {L} Rf_*L \otimes _{\mathcal{O}'}^\mathbf {L} Rf_*M \ar[r] \ar[d] & Rf_*(K \otimes _\mathcal {O}^\mathbf {L} L) \otimes _{\mathcal{O}'}^\mathbf {L} Rf_*M \ar[d] \\ Rf_*K \otimes _{\mathcal{O}'}^\mathbf {L} Rf_*(L \otimes _\mathcal {O}^\mathbf {L} M) \ar[r] & Rf_*(K \otimes _\mathcal {O}^\mathbf {L} L \otimes _\mathcal {O}^\mathbf {L} M) }$

is commutative in $D(\mathcal{O}')$ for all $K, L, M$ in $D(\mathcal{O})$.

Proof. Going around either side we obtain the map adjoint to the obvious map

\begin{align*} Lf^*(Rf_*K \otimes _{\mathcal{O}'}^\mathbf {L} Rf_*L \otimes _{\mathcal{O}'}^\mathbf {L} Rf_*M) & = Lf^*(Rf_*K) \otimes _\mathcal {O}^\mathbf {L} Lf^*(Rf_*L) \otimes _\mathcal {O}^\mathbf {L} Lf^*(Rf_*M) \\ & \to K \otimes _\mathcal {O}^\mathbf {L} L \otimes _\mathcal {O}^\mathbf {L} M \end{align*}

in $D(\mathcal{O})$. $\square$

Lemma 21.32.3. Let $f : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}')$ be a morphism of ringed topoi. The relative cup product of Remark 21.19.7 is commutative in the sense that the diagram

$\xymatrix{ Rf_*K \otimes _{\mathcal{O}'}^\mathbf {L} Rf_*L \ar[r] \ar[d]_\psi & Rf_*(K \otimes _\mathcal {O}^\mathbf {L} L) \ar[d]^{Rf_*\psi } \\ Rf_*L \otimes _{\mathcal{O}'}^\mathbf {L} Rf_*K \ar[r] & Rf_*(L \otimes _\mathcal {O}^\mathbf {L} K) }$

is commutative in $D(\mathcal{O}')$ for all $K, L$ in $D(\mathcal{O})$. Here $\psi$ is the commutativity constraint on the derived category (Lemma 21.46.5).

Proof. Omitted. $\square$

Lemma 21.32.4. Let $f : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}')$ and $f' : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}') \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}''), \mathcal{O}'')$ be morphisms of ringed topoi. The relative cup product of Remark 21.19.7 is compatible with compositions in the sense that the diagram

$\xymatrix{ R(f' \circ f)_*K \otimes _{\mathcal{O}''}^\mathbf {L} R(f' \circ f)_*L \ar@{=}[rr] \ar[d] & & Rf'_*Rf_*K \otimes _{\mathcal{O}''}^\mathbf {L} Rf'_*Rf_*L \ar[d] \\ R(f' \circ f)_*(K \otimes _\mathcal {O}^\mathbf {L} L) \ar@{=}[r] & Rf'_*Rf_*(K \otimes _\mathcal {O}^\mathbf {L} L) & Rf'_*(Rf_*K \otimes _{\mathcal{O}'}^\mathbf {L} Rf_*L) \ar[l] }$

is commutative in $D(\mathcal{O}'')$ for all $K, L$ in $D(\mathcal{O})$.

Proof. This is true because going around the diagram either way we obtain the map adjoint to the map

\begin{align*} & L(f' \circ f)^*\left(R(f' \circ f)_*K \otimes _{\mathcal{O}''}^\mathbf {L} R(f' \circ f)_*L\right) \\ & = L(f' \circ f)^*R(f' \circ f)_*K \otimes _\mathcal {O}^\mathbf {L} L(f' \circ f)^*R(f' \circ f)_*L) \\ & \to K \otimes _\mathcal {O}^\mathbf {L} L \end{align*}

in $D(\mathcal{O})$. To see this one uses that the composition of the counits like so

$L(f' \circ f)^*R(f' \circ f)_* = Lf^* L(f')^* Rf'_* Rf_* \to Lf^* Rf_* \to \text{id}$

is the counit for $L(f' \circ f)^*$ and $R(f' \circ f)_*$. See Categories, Lemma 4.24.9. $\square$

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