# Bordism as a generalized cohomology theory.

Can you tell me, where I can find a proof of the following fact: The bordism functor is a generalized cohomology theory, i.e. we can find suitable connecting homomorphisms to obtain long exact cohomology sequences.

-
I'd suggest looking at the book by Rudyak (amazon.com/…). Or the old book by Stong. In any event, the main point is to use transversality to identify (co)bordism (of a point...) with the homotopy groups of a Thom spectrum, which is then the representing spectrum for the desired cohomology theory. – Dan Ramras Apr 23 '12 at 5:03
Also, I think these days one calls the cohomology theory "cobordism." It's confusing, but there is a dual homology theory and it just seems better to call that one bordism. – Dan Ramras Apr 23 '12 at 5:04
Any proof that constructs an appropriate spectrum is the same as the proof that something is a cohomology theory. You go back and forth using Brown Representability. I suggest you try and find the map in the LES using this as a hint as to where to look. – Sean Tilson May 1 '12 at 20:02
Try Davis-Kirk, Chapter 8. – Neal Jun 28 '12 at 0:56
The connecting homomorphism for the pairs of spaces is one of the easier parts of the proof that it's a (co)homology theory. The proof that bordism is a homology theory (I prefer to say homology functor as "theory" IMO is inaccurate and pretentious) is mostly fairly easy -- the part that requires the most care is the proof that it satisfies the excision axiom. – Ryan Budney Jul 30 '12 at 0:07