# Trivial nature of orientable fibre bundle with cylinder base space

Given a bundle whose base space is a "cylinder", i.e. $\mathbb{S}^1 \times \mathbb{B}^n$ where $\mathbb{B}^n$ is an $n$-ball, is orientability (of the total space) enough to ensure that the bundle is trivial, and how does one go about showing this.

I would basically like to understand the next simplest case after that of bundles with contractible base space, which are all trivial.

It feels like there should be a simple answer, that doesn't involve the diffoemorphism class of the fibres, possibly using some homotopy-type argument to reduce the question to a bundle with base space $\mathbb{S}^1$.

Also, are there any good references I should be looking at?

Thanks!

• The classic reference is Steenrod's Topology of Fiber Bundles. – Ted Shifrin Sep 18 '14 at 14:55

Theorem 11.4: If $X$ is a $C_\sigma$-space [normal and locally compact], then any bundle $\mathcal{B}'$ over the base space $X \times \mathbb{B}^1$ is equivalent to a bundle of the form $\mathcal{B} \times \mathbb{B}^1$.