Triviality of a differentiable sphere bundle I have read in some lecture notes in the internet (without reference) the following result:

Let $E$ be a differentiable sphere bundle whose base $B$ has dimension $n\geq 2$ and whose fibers $F$ have dimension $n-1$. If $B$ is non-compact, then it admits a global smooth cross-section.

The proof is somewhat technical (involves sections with isolate singularities and how to remove them by sending them to infinity) and seems to be correct, but I would like to know a reference in the published literature for this result. In fact, the case $n=2$ would be enough. Thanks!
 A: I am guessing that you misunderstand the claim.
Here are some facts that seem to be in the right ballpark: a bundle whose fibers are spheres of dimension $n$ over a base whose dimension is $n$ or less will certainly have a section. And if the base $B$ has dimension $n+1$ then it will have a section if and only if the Euler class is zero; this is a cohomology class in $H^{n+1}(B)$, or a twisted version of this if the bundle is nonorientable. If the $(n+1)$-manifold $B$ is connected and noncompact then this cohomology group is zero, so a section exists. 
So if you assume $n$ is meant to be dimension of fiber rather than total space then the claim makes sense but with $n+1$ rather than $n-1$ for the base. If the sphere bundle is the unit sphere bundle of a vector bundle and $n$ is the fiber dimension of the vector bundle, then you can change the $n+1$ to $n$, but that's still off by $1$.
A: This is NOT an answer but since the OP's origional statement is not clear, here is the theorem in the book of Greub, W., Halperin, S. and Vanstone that OP mentioned in his own answer:

Theorem III: Every sphere bundle with fibre dimension $n - 1$ over a connected base manifold of dimension $n\ge 2$ admits a cross-section with a single singularity. If the base is not compact, then the bundle admits a cross-section without singularities.

A: Note that the bundles on a product  $\mathbb{R}^m\times S^1 $ are isomorphic to bundles obtained by pullback from  bundles on $S^1$. 
Bundles over $S^1$ with fiber a smooth manifold $F$ are classified  buy the connected components of ${\rm Diffeo}\;(F)$. (Think monodromy.)  The group  ${\rm Diffeo}\; (S^n)$ has at least two components  because the orientation preserving diffeomorphism are not  homotopic to  orientation reversing diffeomorphisms. Hence there are at least two  topological types of sphere bundles over $S^1$, and thus at least two topological types on noncompact manifolds  of product type $\mathbb{R}^m\times S^1$.
A: The question has been solved because I found the reference after about 10 hours of searching!

Greub, W., Halperin, S. and Vanstone, R. Conections, Curvature and Cohomology.

It is Theorem III in Chapter VIII, Section 5.
Thanks, anyway!
