Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 :(

share|improve this question

1 Answer 1

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.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.