Seifert fiber space with homotopically trivial generic fiber Let $X$ be a Seifert fiber space, that is, a 3-manifold which is a circle bundle over a 2-orbifold. Suppose all generic fiber of $X$ is homotopically trivial, can we prove that the universal cover of $X$ is homeomorphic to $S^3$?
 A: Yes, this follows from classification of Seifert fibered spaces.  In fact, you can change the hypothesis to "the fiber is torsion in $\pi_1(X)$" and still get the result. $\newcommand{\RR}{\mathbb{R}}$
Suppose that $X$ is reducible.  Thus $X$ has $S^2 \times \RR$ geometry and the fiber is not torsion.
Suppose that $X$ has boundary.  The boundary components are tori.  By "one-half lives, one-half dies" there is a unique slope $\gamma$ that dies in homology.  If the fiber is torsion, a power dies in homology, so in fact $\gamma$ is the fiber slope.  The disk theorem tells us that $\gamma$ bounds a disk.  Irreducibility tells us that $X$ is a solid torus.  We appeal to the classification of fiberings of the solid torus and find that the fiber is not torsion, a contradiction.
Suppose that $X$ is algebraically toroidal.  If the immersed torus is a union of fibers then the fibers are not torsion.  If the torus is transverse to the fibers then it is embedded.  In this case $X$ has $\mathbb{E}^3$ geometry, and the fibers are not torsion.
So, $X$ is Seifert fibered, closed, irreducible, and (algebraically) atoroidal.  Thus it has $S^3$ geometry.
