Covering seifert manifolds 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?
 A: 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.
  
  
*
  
*$M$ 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 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). 
A: 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.
