The usual proof of the Kunneth formula (say for either the homology or cohomology of manifolds) is essentially pure homological algebra. I was wondering if there was a more geometric proof, i.e., one that does not go through Kunneth for the tensor product of complexes.

This is a bit vague, so here are two possible versions:

- Given smooth manifold $X$ and $Y$ and a de Rham cohomology class $\alpha$ on $X \times Y$, is there some analytic way to find a closed form on $X \times Y$ representing $\alpha$ that is a sum of products of closed forms on $X$ and $Y$?

I would be interested even in the case of Kahler manifolds.

- Given manifolds $X$ and $Y$ and an (integral) homology class $\beta \in H_*(X \times Y)$, is there a multiple of $\beta$ that is represented by a submanifold $M \subseteq X \times Y$, and a geometric procedure for degenerating $M$ to a sum of submanifolds that are products.

Taking a multiple of $\beta$ is necessary because general homology classes need not be represented by submanifolds, and it also deals with the issue that the Kunneth formula is more complicated with integer coefficients.

Such degenerations sometimes show up, e.g. when $X = Y$ is a toric variety and $M \subseteq \Delta(X)$, then one can degenerate $M$ by acting with a generic $1$-parameter subgroup.

My motivation is the Kunneth usually fails in algebraic geometry (say for Chow groups), and when it holds, the proof is very non-formal. I realize that the proof of Kunneth for topological spaces is not very hard, but it would be nice to have a more geometric proof.

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