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



Relative version of the cohomology product
This answer is a translation of the much more general [Stacks, [Tag 0B68](https://stacks.math.columbia.edu/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](https://stacks.math.columbia.edu/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](https://stacks.math.columbia.edu/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 $D(X,\mathbf F\_X)$. $\square$


