This question is similar to Do chains and cochains know the same thing about the manifold? in the sence that both deal with a natural "comparison" quasi-isomorphism that does not preserve the ring structure.

Let $M$ be a smooth manifold. There is a natural comparison map $Comp$ from the differential forms on $M$ to the smooth singular cochains of $M$ (i.e. the linear dual of the vector space spanned by smooth singular simplices). It is defined as follows: take a form $\omega$ of degree $p$ and set $Comp(\omega)$ to be the cochain $\sigma\mapsto \int_\triangle \sigma^*\omega$ where $\triangle$ is the standard $p$-dimensional simplex and $\sigma:\triangle\to M$ is a smooth singular simplex.

$Comp$ is a map of complexes (Stokes' theorem) and moreover, a quasi-isomorphism (the de Rham theorem). But as simple examples show, it does not preserve the ring structure. However it is natural to ask whether the ring structures, up to quasi-isomorphism, of the differential forms and of the cochains contain the same information about $M$. This translates into the following questions.

Can $Comp$ be completed to a morphism of $A_\infty$-algebras?

If the answer to 1. is positive (it presumably is), what about the $E_\infty$ case?

These questions also have natural rational versions. Namely, we can take an arbitrary polyhedron $X$ instead of $M$ and consider Sullivan's $\mathbf{Q}$-polynomial forms. There is a comparison quasi-isomorphism similar to the one above that will go from the $\mathbf{Q}$-polynomial forms of $X$ to the piecewise linear $\mathbf{Q}$-cochains. Can it be completed to a map of $A_\infty$ or $E_\infty$ algebras?