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Basically I am following this question:$M \times N$ orientable if and only if $M, N$ orientable

The confusing part is $\widetilde{M \times \mathbb{R}^k} \cong \widetilde{M} \times \mathbb{R}^k$. Is this very obvious? As set, they are obviously bijective. But how to prove they are homeomorphic?

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  • why do you need this in order to complete your original proof? I mean if you want, I can tell you some alternative proof – Anubhav Mukherjee Mar 06 '16 at 19:10
  • @Anubhav.K If you can tell me another proof it will be great! By the way,I am finding proof without assumption that the manifold is smooth. –  Mar 06 '16 at 20:18
  • We would be assuming that $M$ is closed if we use the global homology characterization of orientability, @PVAL. – iwriteonbananas Mar 08 '16 at 13:23
  • Better if you solve the question by your own... Ill give you hints...Try to use kunneth formula for one direction, for other direction use kunneth formula with some trick... Think, if you failed, I'll give you more hints – Anubhav Mukherjee Mar 08 '16 at 18:11
  • @Anubhav.K Initially I attempt to do this by kunneth formula and use the top homology group. But actually I am not sure to what extent kunneth formula is valid for manifold. The version I know is for CW complex. –  Mar 08 '16 at 21:12
  • That is a valid point, I've not thought about this before, for smooth manifold we can certainly use kunneth, because Morse theory says us that it will have a CW structure, ow not sure – Anubhav Mukherjee Mar 08 '16 at 21:15

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