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The following question comes from Introduction to Smooth Manifolds by Lee:

Suppose $\widetilde{M}$ smoothly covers $M$ where $M$ is orientable. Show that $\widetilde{M}$ is orientable.

I think the following proof works:

Orientability is equivalent to the existence of a nowhere vanishing continuous top form on $M$, so let $\Omega$ be any such form on $M$. Then the pullback $\pi^*\Omega$ is a top form on $\widetilde{M}$ which cannot vanish ($\pi$ here denotes the smooth covering map). For if it did that would imply that $\Omega$ vanished somewhere.

This proof concerns me since I am nowhere using the fact that $\widetilde{M}$ is covering space other than to know that $\pi^*\Omega$ is a top form on $\widetilde{M}$ since $\pi$ is a local diffeomorphism. Have I proven too much here?

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You used one more thing that you're not explicitly making note of: the existence of a nowhere vanishing form on $M$. –  whuber Mar 1 '13 at 23:30

1 Answer 1

up vote 7 down vote accepted

What you say is perfectly correct.
Indeed, if $\pi:\tilde M\to M$ is a local diffeomorphism between $n$-dimensional differential manifolds and $M$ is oriented by the nowhere zero top form $\omega\in \Omega^n(M)$, then you obtain an orientation on $\tilde M$ by lifting $\omega$ to the nowhere zero top form $\pi^*\omega\in \Omega^n(\tilde M)$ .
In other words, the stronger assumption that $\pi$ be a covering map is irrelevant.

Remark 1
The converse is false: in the universal covering $\pi:S^2\to \mathbb P^2(\mathbb R)$ of the projective plane by the 2-sphere, the sphere is orientable but the projective plane is not.
Actually, any manifold admits of an orientable covering map: its universal cover, but this says nothing about the orientability of the manifold.

Remark 2
The same result holds more generally for local homeomorphisms between topological manifolds. The above proof obviously doesn't work since you cannot talk about differential forms on topological manifolds: even the definition of orientability must be changed.
The key to the solution of these problems is the Algebraic Topology concept of relative homology: see for example Greenberg-Harper's Algebraic Topology, Chapter 22, page 157.

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Thanks! I guess Lee uses the covering assumption to make it clear what to do in subsequent questions, where for example one shows real projective space of odd dimension is orientable. –  rondo9 Mar 2 '13 at 16:32
    
Dear rond09, Lee writes extremely well and I am sure he had very good reasons to state the theorem the way he did: it is not necessarily a good idea to state results in their maximum generality. It just so happens that I like local diffeomorphisms and their analogue in topology, complex analysis and algebraic geometry (where Grothendieck gave them the beautiful name étale maps), so I wrote an answer emphasizing them. –  Georges Elencwajg Mar 2 '13 at 16:54

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