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 $S^3$?



Seifert fiber space with homotopically trivial generic fiber
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.


