$\mathbb{S}^2$ as a fibre bundle

Question

I know, by the Hopf fibration, that $\mathbb{S}^3$ is an $\mathbb{S}^1$-fibre bundle over $\mathbb{S}^2$.

Can $\mathbb{S}^2$ be an $\mathbb{S}^1$-fibre bundle over some manifold $M$?

Answer 1

If $F\to E\to B$ is a fiber bundle, $\chi(E)=\chi(F)\cdot\chi(B)$. But $\chi(S^1)=0$ and $\chi(S^2)=2\ne0$, so there is no $S^1$ bundle with total space $S^2$ (or with any even-dimensional sphere for that matter).

Answer 2

The answer is no. Suppose $S^1\rightarrow S^2\rightarrow M$ is a fiber bundle sequence with $M$ a manifold. By the local product property, $M$ has dimension $1$ as a manifold and so is either $\mathbb{R}$ or $S^1$. However, by the long exact sequence in homotopy of this fiber bundle, we have a short exact sequence: $$0\rightarrow\mathbb{Z}\rightarrow\pi_2(M)\rightarrow\mathbb{Z}\rightarrow 0$$ and so $\pi_2(M)$ is non-trivial, hence $M$ can not be the real line or the circle.