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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.

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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
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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
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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
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