Consider the map $\pi:S^1\times S^3\to S^3\to S^2$, where the first map is projection and the second map is Hopf fibration. This is a torus bundle over $S^2$. I am trying to show the converse: if $p:S^1\times S^3\to \Sigma$ is a smooth fiber bundle then it must be isomorphic to $\pi$.
Here is my attempt: let $F$ be a fiber of $p$. From the homotopy exact sequence of $p$, and using $\chi(F)\chi(\Sigma)=\chi(S^1\times S^3)=0$, I've shown that $p$ must be a torus bundle over $S^2$. But how can we show that $p$ must be isomorphic to $\pi$?