Getting rid of exceptional fibers by passing to finite covers? Consider a Seifert fiber space. Is it always possible to find a finite cover that is a circle bundle and the preimage of any fiber is a finite union of circles?
 A: If the Seifert fiber space is compact, then this is true, as long as the base orbifold is "good", which means that it has a finite-sheeted manifold cover, which is a compact surface. This induces a cover of the Seifert fiber space which is a circle bundle over the surface. If the base orbifold is bad, then no such covering will exist. This can happen for a Seifert fibering of $S^3$ over a football orbifold with distinct orders of torsion points, or over a teardrop orbifold. 
If the Seifert fiber space is non-compact, then there may be infinitely many exceptional fibers, and the base orbifold might have torsion of arbitrarily large order, so there is no hope of finding a finite-index cover which is a circle bundle.  
See the draft of Thurston's book for more information on orbifolds and Seifert fibered spaces. Exercise 5.7.10 is on the Seifert fibering of $S^3$ over bad orbifolds. 
A: I believe the answer is yes (although I haven't actually checked). Here's the idea:  Given a Seifert-fibred space you can think of it as being fibred over a $2$-orbifold.  You can de-singularise that $2$-orbifold by taking the appropriate branched cover.  Pulling back the Seifert-fibering gives you a genuine $S^1$-bundle.  This skirts the issue of whether or not you can de-singularize the $2$-orbifold by an appropriate cover but I believe it's not hard to show such "bad" 2-orbifolds never occur as the base space to a Seifert-fibred $3$-manifold.  Right, they're classified here: http://en.wikipedia.org/wiki/Orbifold and you can compare that to the base orbifolds of Seifert-fibred 3-manifolds to answer your question. 
