# Circle bundle with homotopically trivial fiber in the total space

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 $$S^3$$?

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.