Circle bundle with homotopically trivial fiber in the total space

Question

Consider a smooth circle fiber bundle $$ S^1 \to E\to B $$ where $E$ is a smooth 3-manifold and $B$ is a smooth surface. Assuming any $S^1$ fiber in $E$ is homotopically trivial, can we prove that $E$ is homeomorphic to $S^3$?

Answer 1

Yes.

By the homotopy long exact sequence $$\dots \to \pi_2(B) \to \pi_1(S^1) \to \pi_1(E) \to \dots$$ and the fact that the generator of $\pi_1(S^1)$ is homotopically trivial in $B$, we see that $\pi_2(B) \neq 0$, and thus by classification of surfaces that $B$ is $S^2$ or $\mathbb R\mathbb P^2$.

Circle bundles over the sphere are classified by their Chern class, which is the same thing as the connecting map $\pi_2(S^2) \to \pi_1(S^1)$. Since this must be $\pm 1$, the Chern class must be $\pm $ the fundamental class, and either case gives the Hopf fibration - i.e., $S^3$.

For $\mathbb R\mathbb P^2$, there is an obstruction. The easiest way to see it is to use the Leray spectral sequence of the fiber double cover $S^2 \to \mathbb R \mathbb P^2$ gives a double cover of $E$. This double cover must still have surjective $\pi_2(B) \to \pi_1(S^1)$ since the generator of $\pi_2 (\mathbb R \mathbb P ^2)$ lifts to the double cover. So it must be $S^3$. But then the involution of the double cover swaps the orientation of $S^2$, which means it negates the Chern class of the fibration, so it swaps the orientation of the fibers, and thus it fixes the orientation of $S^3$, which implies it has a fixed point, contradicting the fact that it arises from a double cover.