# Explicit quasi-inverse of Künneth-isomorphism?

With $A_X$ the complex of $\mathbb{R}$-differential forms on $X$, the Künneth theorem states that \begin{align*} A_X \otimes A_Y &\to A_{X \times Y}, \\ (\omega,\eta) &\mapsto {\rm pr}_X^\ast \omega \wedge {\rm pr}_Y^\ast \eta \end{align*} gives an isomorphism on cohomology. Is there any explicit formula describing a quasi-inverse of this map? I.e. a map $$A_{X\times Y} \to A_X \otimes A_Y$$ inducing an inverse on cohomology groups ?

Thanks

• Generally speaking, even if a canonically defined map happens to be invertible, there won't be a nice way to describe the inverse. For example, already the inverse of the map $V \to V^{\ast \ast}$ from a finite-dimensional vector space to its double dual doesn't have a nice description (somehow, if it did, it wouldn't be sensitive to the fact that $V$ is finite-dimensional). – Qiaochu Yuan Sep 10 '12 at 20:09