## 20.31 Cup product

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

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

in $D(A)$. We define it as the relative cup product for the morphism of ringed spaces $f : (X, \mathcal{O}_ X) \to (pt, A)$ as in Remark 20.28.7 via $D(pt, A) = D(A)$. This map in particular defines pairings

$\cup : H^ i(X, K) \times H^ j(X, M) \longrightarrow H^{i + j}(X, K \otimes _{\mathcal{O}_ X}^\mathbf {L} M)$

Namely, given $\xi \in H^ i(X, K) = H^ i(R\Gamma (X, K))$ and $\eta \in H^ j(X, M) = H^ j(R\Gamma (X, M))$ we can first “tensor” them to get an element $\xi \otimes \eta$ in $H^{i + j}(R\Gamma (X, K) \otimes _ A^\mathbf {L} R\Gamma (X, M))$, see More on Algebra, Section 15.61. Then we can apply $\mu$ to get the desired element $\xi \cup \eta = \mu (\xi \otimes \eta )$ of $H^{i + j}(X, K \otimes _{\mathcal{O}_ X}^\mathbf {L} M)$.

Here is another way to think of the cup product of $\xi$ and $\eta$. Namely, we can write

$R\Gamma (X, K) = R\mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{O}_ X, K) \quad \text{and}\quad R\Gamma (X, M) = R\mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{O}_ X, M)$

because $\mathop{\mathrm{Hom}}\nolimits (\mathcal{O}_ X, -) = \Gamma (X, -)$. Thus $\xi$ and $\eta$ are the “same” thing as maps

$\tilde\xi : \mathcal{O}_ X[-i] \to K \quad \text{and}\quad \tilde\eta : \mathcal{O}_ X[-j] \to M$

Combining this with the functoriality of the derived tensor product we obtain

$\mathcal{O}_ X[-i - j] = \mathcal{O}_ X[-i] \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{O}_ X[-j] \xrightarrow {\tilde\xi \otimes \tilde\eta } K \otimes _{\mathcal{O}_ X}^\mathbf {L} M$

which by the same token as above is an element of $H^{i + j}(X, K \otimes _{\mathcal{O}_ X}^\mathbf {L} M)$.

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. $\square$

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

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

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

Lemma 20.31.2. In the situation above the following diagram commutes

$\xymatrix{ f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_ Y}^\mathbf {L} f_*\mathcal{M}^\bullet \ar[r] \ar[d] & Rf_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_ Y}^\mathbf {L} Rf_*\mathcal{M}^\bullet \ar[d]^{\text{Remark 0B68}} \\ \text{Tot}( f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_ Y} f_*\mathcal{M}^\bullet ) \ar[d]_{\text{naive cup product}} & Rf_*(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{M}^\bullet ) \ar[d] \\ f_*\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}^\bullet ) \ar[r] & Rf_*\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}^\bullet ) }$

Proof. By the construction in Remark 20.28.7 we see that going around the diagram clockwise the map

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

\begin{align*} Lf^*(f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_ Y}^\mathbf {L} f_*\mathcal{M}^\bullet ) & = Lf^*f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_ Y}^\mathbf {L} Lf^*f_*\mathcal{M}^\bullet \\ & \to Lf^*Rf_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_ Y}^\mathbf {L} Lf^*Rf_*\mathcal{M}^\bullet \\ & \to \mathcal{K}^\bullet \otimes _{\mathcal{O}_ Y}^\mathbf {L} \mathcal{M}^\bullet \\ & \to \text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}^\bullet ) \end{align*}

By Lemma 20.28.6 this is also equal to

\begin{align*} Lf^*(f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_ Y}^\mathbf {L} f_*\mathcal{M}^\bullet ) & = Lf^*f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_ Y}^\mathbf {L} Lf^*f_*\mathcal{M}^\bullet \\ & \to f^*f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_ Y}^\mathbf {L} f^*f_*\mathcal{M}^\bullet \\ & \to \mathcal{K}^\bullet \otimes _{\mathcal{O}_ Y}^\mathbf {L} \mathcal{M}^\bullet \\ & \to \text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \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}_ Y}^\mathbf {L} f_*\mathcal{M}^\bullet ) & \to Lf^*\text{Tot}( f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_ Y} f_*\mathcal{M}^\bullet ) \\ & \to Lf^*f_*\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}^\bullet ) \\ & \to Lf^*Rf_*\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}^\bullet ) \\ & \to \text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}^\bullet ) \end{align*}

By Lemma 20.28.6 this is also equal to

\begin{align*} Lf^*(f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_ Y}^\mathbf {L} f_*\mathcal{M}^\bullet ) & \to Lf^*\text{Tot}( f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_ Y} f_*\mathcal{M}^\bullet ) \\ & \to Lf^*f_*\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}^\bullet ) \\ & \to f^*f_*\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}^\bullet ) \\ & \to \text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}^\bullet ) \end{align*}

Now the proof is finished by a contemplation of the diagram

$\xymatrix{ Lf^*(f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_ Y}^\mathbf {L} f_*\mathcal{M}^\bullet ) \ar[d] \ar[rr] & & Lf^*f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X}^\mathbf {L} Lf^*f_*\mathcal{M}^\bullet \ar[d] \\ Lf^*\text{Tot}( f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_ Y} f_*\mathcal{M}^\bullet ) \ar[d]_{naive} \ar[r] & f^*\text{Tot}( f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_ Y} f_*\mathcal{M}^\bullet ) \ar[ldd]^{naive} \ar[dd] & f^*f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X}^\mathbf {L} f^*f_*\mathcal{M}^\bullet \ar[dd] \ar[ldd] \\ Lf^*f_*\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}^\bullet ) \ar[d] \\ f^*f_*\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}^\bullet ) \ar[rd] & \text{Tot}(f^*f_*\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} f^*f_*\mathcal{M}^\bullet ) \ar[d] & \mathcal{K}^\bullet \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{M}^\bullet \ar[ld] \\ & \text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}^\bullet ) }$

All of the polygons in this diagram commute. The top one commutes by Lemma 20.27.5. 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. $\square$

Let $(X, \mathcal{O}_ X)$ be a ring space. Let $\mathcal{K}^\bullet$ and $\mathcal{M}^\bullet$ be complexes of $\mathcal{O}_ X$-modules. Then we have a “naive” cup product

$\mu ' : \text{Tot}( \Gamma (X, \mathcal{K}^\bullet ) \otimes _ A \Gamma (X, \mathcal{M}^\bullet )) \longrightarrow \Gamma (X, \text{Tot}( \mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}^\bullet ))$

By Lemma 20.31.2 applied to the morphism $(X, \mathcal{O}_ X) \to (pt, A)$ this naive cup product is related to the cup product $\mu$ defined in the first paragraph of this section by the following commutative diagram

$\xymatrix{ \Gamma (X, \mathcal{K}^\bullet ) \otimes _ A^\mathbf {L} \Gamma (X, \mathcal{M}^\bullet ) \ar[d] \ar[r] & R\Gamma (X, \mathcal{K}^\bullet ) \otimes _ A^\mathbf {L} R\Gamma (X, \mathcal{M}^\bullet ) \ar[d]^-\mu \\ \text{Tot}(\Gamma (X, \mathcal{K}^\bullet ) \otimes _ A \Gamma (X, \mathcal{M}^\bullet )) \ar[d]_-{\mu '} & R\Gamma (X, \mathcal{K}^\bullet \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{M}^\bullet ) \ar[d] \\ \Gamma (X, \text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}^\bullet )) \ar[r] & R\Gamma (X, \text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}^\bullet )) }$

in $D(A)$. On cohomology we obtain the commutative diagram

$\xymatrix{ H^ i(\Gamma (X, \mathcal{K}^\bullet )) \times H^ j(\Gamma (X, \mathcal{M}^\bullet )) \ar[d] \ar[r] & H^{i + j}(X, \text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{M}^\bullet )) \\ H^ i(X, \mathcal{K}^\bullet ) \times H^ j(X, \mathcal{M}^\bullet ) \ar[r]^\cup & H^{i + j}(X, \mathcal{K}^\bullet \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{M}^\bullet ) \ar[u] }$

relating the naive cup product with the actual cuproduct.

Lemma 20.31.3. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{K}^\bullet$ and $\mathcal{M}^\bullet$ be bounded below complexes of $\mathcal{O}_ X$-modules. Let $\mathcal{U} : X = \bigcup _{i \in I} U_ i$ be an open covering Then

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

where the horizontal arrows are the ones in Lemma 20.25.1 commutes in $D(A)$.

Proof. 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 of the lemma involving (20.25.3.2) follows from the already proven commutativity of Lemma 20.31.2 where $f$ is the morphism to a point (see discussion following Lemma 20.31.2). $\square$

Lemma 20.31.4. Let $f : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces. The relative cup product of Remark 20.28.7 is associative in the sense that the diagram

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

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

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

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

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

Lemma 20.31.5. Let $f : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces. The relative cup product of Remark 20.28.7 is commutative in the sense that the diagram

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

is commutative in $D(\mathcal{O}_ Y)$ for all $K, L$ in $D(\mathcal{O}_ X)$. Here $\psi$ is the commutativity constraint on the derived category (Lemma 20.46.5).

Proof. Omitted. $\square$

Lemma 20.31.6. Let $f : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ and $g : (Y, \mathcal{O}_ Y) \to (Z, \mathcal{O}_ Z)$ be morphisms of ringed spaces. The relative cup product of Remark 20.28.7 is compatible with compositions in the sense that the diagram

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

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

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

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

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

$L(g \circ f)^*R(g \circ f)_* = Lf^* Lg^* Rg_* Rf_* \to Lf^* Rf_* \to \text{id}$

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

 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 $\mathcal{O}_ X[-i - j] = \mathcal{O}_ X[-i] \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{O}_ X[-j]$.

Comment #4921 by Cailan Li on

Maybe change the $R$ in $H^{i+j}( R\Gamma(X, K)\otimes_R^L R\Gamma(X, M))$ to an $A$ in the 9th line?

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