Let $M$ be a 3manifold with boundary. If $M$ has an orientable finite cover that is a Seifert fiber space, then is $M$ also a Seifert fiber space?

4$\begingroup$ I think you must have scrambled the question. Since every manifold has an orientable doublecover, it appears that right now you are asking if every 3manifold is seifertfibered (clearly false). $\endgroup$ – Andy Putman Oct 5 '13 at 5:28

$\begingroup$ Sorry, M have finite covering orientable that is seifert then M is 3manifold seifert $\endgroup$ – jhoel Oct 5 '13 at 12:58

$\begingroup$ A 3manifold is a Seifert fibration if and only if the center of its fundamental group contains an infinite cyclic subgroup (CassonJungreis, Gabai). So you would have to check that this cyclic group is still central in the group extension which is the fundamental group of your original manifold, but this seems not obvious. $\endgroup$ – ThiKu Oct 8 '13 at 5:24
This is certainly true, though one needs to be sufficiently careful about one's definition of Seifert fibredit's important to allow fibres with a neighbourhood that looks like a fibred solid Klein bottle. See p. 429 of P. Scott, 'The geometries of 3manifolds', Bull. LMS 15(5), 1983, pp. 401487 .
I don't think I know a reference in the literature, but one can prove it using the following result (see Theorem 3.9 of Scott's paper).
Theorem: Let $M$ be a compact Seifert fibre space and let $f: M \to N$ be a homeomorphism. Then $f$ is homotopic to a fibrepreserving homeomorphism (and hence an isomorphism of Seifert bundles) unless one of the following occurs.
 $M$ is covered by $S^3$ or $S^2\times\mathbb{R}$,
 $M$ is covered by $S^1\times S^1\times S^1$,
 $M$ is $S^1 \times D^2$ or an Ibundle over the torus or Klein bottle.
It follows that, except in the above three cases, the Seifert structure on the orientable double cover is invariant under the covering transformation, and hence descends.
All closed 3manifolds with finite fundamental group are orientable (by work of Epstein), so there are only finitely many manifolds left to check. I'll leave them as an exercise (which I confess I've never done myself).

$\begingroup$ Thanks Henry, I'm studying his work Residually Free 3Manifolds, I have some questions :). would you be kind enough to answer $\endgroup$ – jhoel Oct 5 '13 at 19:34

$\begingroup$ @jhoel  sure. Please feel free to contact me by email. $\endgroup$ – HJRW Oct 5 '13 at 20:51

2$\begingroup$ One should add here that one needs the Spherical Space Form part of Perelman's geometrization theorem in order to make the conclusion in the spherical case. (The exercise left to the reader.) $\endgroup$ – Misha Oct 6 '13 at 5:52

$\begingroup$ @Misha  I don't think this is necessary. Epstein proved that every compact, nonorientable 3manifold with finite fundamental group is homotopy equivalent to $\mathbb{P}^2\times I$. (The Poincar\'e Conjecture implies that the homotopy equivalence can be upgraded to homeomorphism, but we don't need that here.) In particular, it follows that there are no closed, nonorientable 3manifolds with finite fundamental group, independent of Geometrization. $\endgroup$ – HJRW Oct 6 '13 at 8:15

$\begingroup$ Henry: Maybe I misunderstood the question, but I do not think the manifold was assumed to be nonorientable. $\endgroup$ – Misha Oct 6 '13 at 12:08
You need geometrisation to prove this fact. See Corollary 12.9.5 here for a reference.
You can't prove this without Perelman, at least with our present knowledge. For instance, if the orientable cover is $S^3$, then you must ensure that $M$ be elliptic, and that's precisely the space form conjecture, which is "one third" of geometrisation. But even when the finite cover is some other Seifert space, I don't see an easy argument to conclude without using geometrisation.
Edit. I overlooked the "with boundary" hypothesis. In that case Thurston's proof of geometrisation suffices.

$\begingroup$ I think that, with a bit of care, in the nonspherical case one can get by with just using Nielsen Realization (and similar results for the Euclidean case). But I'm not sure this has been carefully written down anywhere. $\endgroup$ – HJRW Jan 20 '17 at 13:11

$\begingroup$ I am perhaps missing something here, but the question is asking about a 3manifold with boundary. Doesn't that allow us to use Thurston's geometrization of cusped 3manifolds? I don't see how a manifold with boundary can be covered by $S^3$. $\endgroup$ – Neil Hoffman Jan 20 '17 at 16:28