Suppose $(E,p,B;F)$ is a fiber bundle such that $E$ is homeomorphic to $B\times F$, is it true that the fiber bundle is trivial? A non connected counter example has been provided, so I'll ask for E,B and F to be connected (hopefully low dimensional) manifolds.

Consider the pullback $\xi$ of $TS^2$ via the projection of $S^2\times\mathbb R$ onto the first factor. The bundle $\xi$ is a nontrivial $\mathbb R^2$bundle over $S^2\times\mathbb R$ because its pullback under the inclusion $S^2\to S^2\times\mathbb R$ is $TS^2$, which is nontrivial. On the other hand, its total space is $\mathbb R\times TS^2$ which is diffeomorphic to $S^2\times\mathbb R^3$, which is the total space of the trivial $\mathbb R^2$bundle over $S^2\times\mathbb R$. 


No. Let $B=S^1$, $F=\mathbb Z$, $E=S^1\times\mathbb Z$, and let $p$ be a twofold covering on each component: $p(z,n)=z^2$ where $S^1$ is regarded as the unit circle in $\mathbb C$ (for the purposes of computing $z^2$). 

