Technical Question, re: Le: Tensor of Cochain Complexes, and Isomorphism of Complexes

zhentil

I'm not sure quite sure what this means. In the case of a Kunneth theorem, you would have this, but in general there's no map from a fiber bundle to the fiber. This may be misunderstanding your question, but if you mean a Kunneth-style argument where you pull back cohomology classes based on two projections, that can't work in a nontrivial fiber bundle. Think of the proof of Thom isomorphism in de Rham cohomology: you don't get the fiber cohomology class for free, i.e. it's not pulled back from anything.

lavinia

no but since there is a mapping that respects the product structure on might think that there is a mapping of chain complexes. that was the original question I think. No one was expecting a Kunneth argument. Still there definitely are non-trivial bundles where the fiber cohomology is pulled back from classes in the total space of the bundle.