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May 14, 2011 at 21:54 comment added Ryan Budney I wouldn't say (1) is naive, since much of the work of completing the task is left un-done, in that you still have to perturb the homology classes to be transverse. This isn't always easily done. Morse theory can be useful in these situations.
May 14, 2011 at 19:28 answer added Peter May timeline score: 30
May 11, 2011 at 23:41 comment added Sergey Melikhov Strategy (1) does not need $X$ to be homotopy equivalent to a manifold. See R. Fenn, Techniques of geometric topology'', Chapter 1 and S. Buoncristiano, C. P. Rourke, B. J. Sanderson A geometric approach to homology theory'', Chapter 2.
May 11, 2011 at 17:04 comment added user332 In my experience, any manifold that I understand well enough to be able to compute the de Rham algebra of is probably a manifold that I understand well enough to work out the intersection product on. Would you happen to know of an example which doesn't fit this description?
May 11, 2011 at 15:42 comment added Donu Arapura When I actually have to compute anything like this, I tend to use (1) or and sometimes $(\alpha,\beta)\mapsto \int\alpha\wedge \beta$ in de Rham. (2) is likely to lead to a mess.
May 11, 2011 at 14:29 answer added Johannes Ebert timeline score: 10
May 11, 2011 at 13:24 answer added James Cranch timeline score: 8
May 11, 2011 at 13:10 answer added Jeff Strom timeline score: 6
May 11, 2011 at 12:45 history asked user332 CC BY-SA 3.0