Let $F\to E\to B$ be a fiber bundle of smooth manifolds. Suppose that the fiber $F$ and the base $B$ are parallelizable. Is the total space $E$ necessarily parallelizable? What if $B$ is the circle $S^1$?
It is clear that when the bundle of tangent bundles $TF\to TE\to TB$ is trivial then $E$ is parallelizable. But is this a necessary condition? Is $B=S^1$ a special case?
$\mathbb{S}^1$ is parallelizable but the Klein bottle is a fibre bundle over $\mathbb{S}^1$ with a fiber $\mathbb{S}^1$ and it is not parallelizable because all parallelizable manifolds are oriented.