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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 $H^*(X,\mathbb{F})$.

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  • $\begingroup$ Short answer: since $f^*$ is symmetric monoidal, its right adjoint $Rf_*$ is lax symmetric monoidal and so in particular it sends algebras to algebras. $\endgroup$ – Denis Nardin Dec 24 '19 at 14:37
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This answer is a translation of the much more general [Stacks, Tag 0B68] to this setup:

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 $D(X,\mathbf F_X)$. $\square$

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