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I was reading through Lee's Smooth Manifolds on the part regarding orientations and I was wondering if the connected sum preserves the orientability of manifolds. Intuitively it seems to be true, but I'm not sure how to make this rigorous. Is the connected sum of orientable manifolds necessarily orientable and why?

user186571
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  • Is our attention restricted to closed manifolds or no? – Matt Samuel Nov 06 '14 at 02:13
  • We are just considering any manifold in general by the usual definition as patchings of coordinate charts from euclidean space. – user186571 Nov 06 '14 at 02:17
  • If the manifold is compact without boundary homology gives us a way to prove it: http://math.stackexchange.com/questions/187413/computing-the-homology-and-cohomology-of-connected-sum – Matt Samuel Nov 06 '14 at 02:19
  • Thanks, is there a more elementary method to prove this? – user186571 Nov 06 '14 at 02:20
  • Possibly. I'm not personally aware of one, but this is not my specialty. – Matt Samuel Nov 06 '14 at 02:22
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    The connected sum of orientable manifolds is indeed orientable. It's really not so hard to prove it yourself, so you should try! –  Nov 06 '14 at 02:52
  • In what direction/tools should I think about it? I'm not perfectly sure, since I'm slightly uncomfortable with the exact definition of orientability and connected sums. – user186571 Nov 06 '14 at 03:25
  • Maybe think about collar neighborhoods. Sketch: Pick an orientation on the first manifold; there's a unique orientation on the $S^{n-1} \times I$ you attach via a collar neighborhood that agrees with that of the first manifold. Then (attaching this to the second manifold) there's a unique orientation on the second manifold that agrees with that of $S^{n-1} \times I$. So there's a consistent orientation on the whole thing. –  Nov 06 '14 at 07:53

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