# Relative version of the cohomology product

Let $$f\colon X\to Y$$ be a continuous map of 'nice' topological spaces (e.g. $$f$$ is a smooth map of smooth manifolds; $$f$$ might be assumed to be proper although I am not sure it is relevant). Let $$\underline{\mathbb{F}}_X$$ be the constant sheaf on $$X$$ with coefficients in a field $$\mathbb{F}$$.

Does the obvious morphism of sheaves $$\underline{\mathbb{F}}_X\otimes \underline{\mathbb{F}}_X\to \underline{\mathbb{F}}_X$$ induce a canonical morphism in the derived category $$D^+(Sh_\mathbb{F})$$ of sheaves of $$\mathbb{F}$$-vector spaces $$Rf_*(\underline{\mathbb{F}}_X)\otimes Rf_*(\underline{\mathbb{F}}_X)\to Rf_*(\underline{\mathbb{F}}_X),$$ where $$Rf_*$$ is the push-forward functor. If yes, is this morphism associative?

Remark. When $$Y$$ is a point the required map does exist and is just the multiplication map in cohomology $$H^*(X,\mathbb{F})$$.

• Short answer: since $f^*$ is symmetric monoidal, its right adjoint $Rf_*$ is lax symmetric monoidal and so in particular it sends algebras to algebras. – Denis Nardin Dec 24 '19 at 14:37

Write $$\mathcal O_X = \mathbf F_X$$ and $$\mathcal O_Y = \mathbf F_Y$$. Then $$f \colon (X,\mathcal O_X) \to (Y,\mathcal O_Y)$$ is a morphism of ringed spaces, with $$f^{-1}\mathcal O_Y = \mathcal O_X$$. In particular, $$f^* = f^{-1}$$ is exact, so $$Lf^* = f^*$$. Moreover, since $$\mathbf F$$ is a field, we also have $$(-) \otimes_{\mathcal O_X}^{\mathbf L} (-) = (-) \otimes_{\mathcal O_X} (-)$$ and similarly on $$Y$$.
Then the counit $$\varepsilon \colon f^*Rf_* \to \operatorname{id}$$ of the adjunction $$f^* \dashv Rf_*$$ gives a canonical map $$f^*Rf_* \mathbf F_X \underset{\mathbf F_X}\otimes f^*Rf_* \mathbf F_X \stackrel{\varepsilon \otimes \varepsilon}\longrightarrow \mathbf F_X \underset{\mathbf F_X}\otimes \mathbf F_X.$$ By [Stacks, Tag 079U], the source is canonically isomorphic to $$f^*(Rf_* \mathbf F_X \otimes_{\mathbf F_Y} Rf_* \mathbf F_X)$$. The target simplifies to $$\mathbf F_X$$. Passing through the adjunction then produces the desired map $$c \colon Rf_* \mathbf F_X \underset{\mathbf F_Y}\otimes Rf_* \mathbf F_X \to Rf_* \mathbf F_X.$$ To check associativity [Stacks, Tag 0FP4], we need to show that the diagram $$\begin{array}{ccc}Rf_* \mathbf F_X \underset{\mathbf F_Y}\otimes Rf_* \mathbf F_X \underset{\mathbf F_Y}\otimes Rf_* \mathbf F_X & \overset{1 \otimes c}\to & Rf_* \mathbf F_X \underset{\mathbf F_Y}\otimes Rf_* \mathbf F_X \\ \!\!\!\!\!\!\!\!\!\!\!{\scriptsize c \otimes 1}\downarrow & & \downarrow{\scriptsize c}\!\!\!\! \\ Rf_* \mathbf F_X \underset{\mathbf F_Y}\otimes Rf_* \mathbf F_X & \underset{c}\to & Rf_* \mathbf F_X\end{array}$$ commutes. Both compositions are adjoint to the map $$f^*\left(Rf_* \mathbf F_X \underset{\mathbf F_Y}\otimes Rf_* \mathbf F_X \underset{\mathbf F_Y}\otimes Rf_* \mathbf F_X\right) \to \mathbf F_X$$ given by identifying the left hand side with $$f^*Rf_*\mathbf F_X \otimes_{\mathbf F_X} f^*Rf_* \mathbf F_X \otimes_{\mathbf F_X} f^*Rf_* \mathbf F_X$$, and using associativity $$(\varepsilon \otimes \varepsilon) \otimes \varepsilon = \varepsilon \otimes (\varepsilon \otimes \varepsilon)$$ in $$D(X,\mathbf F_X)$$. $$\square$$