The projection from the time-orientable double cover preserves topological properties.

Question

In Relativity and Singularities, Natário states that

A connected time-orientable Lorentzian manifold admits a nonvanishing vector field, and hence is either noncompact or has zero Euler characteristic. The same is true for a non-time-orientable Lorentzian manifold, for it must be true for its time-orientable double cover.

I absolutely cannot see the reason why the projection $\pi:\tilde{M}\to M$ perserves topological properties of $\tilde{M}$ (compactness, non-compactness, $\chi(M) = 0$). It's a local isometry, and thus local homeomorphism, sure, but why does the above follow?

Answer 1

In general, $\pi$ will not preserve topological properties of $\tilde{M}$ because it is not a homeomorphism. However,

Me: Let $M$ be a non-time orientable Lorentzian manifold and $\tilde{M}$ its orientable double cover. It seems that the projection $\pi:\tilde{M}\to M$ preserves topological properties such as non-compactness, or compact and zero Euler characteristic. If this is correct, why?

You don't need a Lorentz metric to make sense of the orientation double cover; it can be done in great generality. Let $M$ be a topological manifold; then, every point has a neighborhood homeomorphic to $\mathbb{R}^n$, which has two orientations. So at every point p we have two orientations of a neighborhood, and for nearby points q we can make sense of whether an orientation near p and an orientation near q are the same. Thus the set whose elements are points p and orientations near p is a double cover of M (and hence is also a topological manifold).

Of course, you don't need to know the full generality to answer your questions; the point is that the problem is purely topological, and the answer is too.

preserves topological properties such as [compactness or] non-compactness

This one is kind of following your nose: an open cover of M induces one on M' with twice as many open sets; an open cover on $\tilde{M}$ induces an open cover on $M$ by composing with the covering map $\tilde{M}\to M$. You can check that if an open cover of $\tilde{M}$ has a finite subcover, so does the induced cover on $M$, and if an open cover on M has a finite subcover, so does the induced one on $\tilde{M}$.

Euler characteristic is zero

One way to do this is to use the fact that Euler characteristic is the sum of $(-1)^i \dim H^i(M; Q)$ (which is finite because $M$ is a manifold), and a finite cover induces an isomorphism on rational cohomology. If $M$ is smooth, you can also put a simplicial structure on it; then the Euler characteristic is vertices minus edges plus faces minus ... as usual. But this induces a simplicial structure on M' with twice as many simplices, so if $V - E + F - ...$ is zero, then so is $2V - 2E + 2F - ...$.