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



Covering seifert manifolds
This is certainly true, though one needs to be sufficiently careful about one's definition of Seifert fibred---it'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 3-manifolds', Bull. LMS 15(5), 1983, pp. 401--487 .


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 fibre-preserving homeomorphism (and hence an isomorphism of Seifert bundles) unless one of the following occurs.
> 
> 
> 1. $M$ is covered by $S^3$ or $S^2\times\mathbb{R}$,
> 2. $M$ is covered by $S^1\times S^1\times S^1$,
> 3. $M$ is $S^1 \times D^2$ or an I-bundle 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 3-manifolds 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). 


