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Given a local diffeomorphism $f: N \to M$ with $M$ orientable. Why is $N$ orientable? My professor wrote this in class without giving a proof and said "you should try to prove this for fun :)". I am clueless, please help :(

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up vote 6 down vote accepted

Do you know about the connection between differential forms and orientability? That $M$ is orientable means precisely that $M$ has a non-vanishing volume form $\omega$. To show $N$ is orientable, we must show $N$ also admits a non-vanishing volume form. It seems natural to consider the pullback of $\omega$, and you can check directly that it is non-vanishing at each point because $f$ is a local diffeomorphism.

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All right that is quite helpful. I am now trying to show that $f^* \omega$ is non-vanishing. During this process I am examining the determinant of Df. I can tell that this determinant must be not be zero since f is local diffeo. However, I do not follow why? –  Leo Spencer Apr 28 '13 at 1:46
    
Actually, I think I got that now. If the determinant was zero, then Df would not be invertible which means f could not be a diffeomorphism. –  Leo Spencer Apr 28 '13 at 2:01
    
@LeoSpencer Precisely. –  Potato Apr 28 '13 at 2:19
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