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On a standard treatment of time-orientability in Lorentzian spaces, it is usual to see a remark of the form: "If a space-time (M,g) is not time-orientable, then it has a double covering space which is." Then, "the" time-orientable covering space is defined in the natural way: as the manifold whose point set is ${(p,a)}$ with $pin M$ and $a$ a time-orientation at $p$, where the covering map is just the projection down to $M$. The physical significance of this basic observation is usually that time-orientability is "cheap"--- if ever you find yourself dealing with a spacetime that fails to be time-orientable, you can always move canonically to a locally isometric spacetime that does not suffer the same defect. (See Hawking and Ellis, p. 181, for an example of this sort of treatment.)

My question concerns the uniqueness of that canonical move, and I have not found any good leads. In particular, I care about the status of the following proposition:

Question. Given a geodesically complete spacetime $(M,g)$​​ that is not time-orientable, let $(M',g')$​​ and $(M'',g'')$​​ be covering spacetimes of $(M,g)$​​. If $(M',g')$​​ and $(M'',g'')$​​ are each geodesically complete, connected, and time-orientable, then does it follow that $(M',g')$​​ and $(M'',g'')$​​ are isometric?

Edit. Here is where I am now in an attempt to prove the answer "yes".

Let (X,h) be the time-orientable covering space of (M,g) as standardly defined (e.g. as in Hawking and Ellis). Then (X,h) is time-orientable and is connected (and is as well geodesically complete), in which case we can find a universal cover of (X,h) that is time-orientable and geodesically complete (and connected, by definition). This universal cover of (X,h) is also a universal cover of (M,g) by composing the two respective covering maps. So suppose, without loss of generality, that (M',g') is that universal cover. Then it follows that (M',g') is simply connected. Since universal covers are unique, if (M'',g'') is a universal cover as well, then I am done. But suppose that (M'',g'') is not a universal cover. Then it is not simply connected, in which case (M',g') and (M'',g'') are not even homeomorphic.

So I need to show that every geodesically complete, connected, and time-orientable cover of (M,g) is a universal cover. (I'll also note that so far, I haven't seemed to need any of the assumed properties of geodesic completeness.) I can't help shake the intuition that this is false, but I have yet to find a counterexample.

Edit 2. There is a special case, where the time-oriented double cover of (M,g) is the universal cover of the same, in which it would suffice to show that any time-oriented cover of (M,g) is a two-sheeted cover. (Then the standard uniqueness proofs pertaining to 2-sheeted oriented covers should be enough to finish the proof.) I am more optimistic about this case.