Talk:Cup product

Error: Not every homology class can be represented by a submanifold

One passage near the end of the article reads as follows:

"When two submanifolds of a smooth manifold intersect transversely, their intersection is again a submanifold. By taking the fundamental homology class of these manifolds, this yields a bilinear product on homology. This product is dual to the cup product, i.e. the homology class of the intersection of two submanifolds is the Poincaré dual of the cup product of their Poincaré duals."

But transveral interesection of submanifolds does not in general directly yield a bilinear product on homology, simply because not every homology class can be represented by an embedded submanifold.

However, according to René Thom's 1954 paper "Quelques proprietés globales . . .", every homology class has an integer multiple which is represented by the fundamental class of a submanifold.

So, if classes u, v have multiples U = K·u, V = L·v that are represented by submanifolds, we may define the intersection number i(u,v) as #(U ∩ V) / (K·L). (Here U and V are assumed to be perturbed if necessary so that they are transverse, and the size of their intersection is counted algebraically.)Daqu (talk) 04:41, 30 January 2013 (UTC)[reply]

exterior product

Being more comfortable with differential topology, I like to think of "cup product" by drawing analogy with exterior product. I guess they will be exactly equivalent in de Rham cohomology theory (which of course works only for smooth manifolds and not general topological spaces unlike cup product). But will it be a good idea to mention that analogy in an "intuitive explanation/analogy" section at the end of this article for people coming from differential topology background? Likewise, cap product can be considered to be analogous to "integration over fibre". - Subh83 (talk | contribs) 23:14, 9 May 2013 (UTC)[reply]

ahh... I see it's already there in the article. My bad! - Subh83 (talk | contribs) 23:16, 9 May 2013 (UTC)[reply]