In the course of research, I made use of the following theorem:
Any fibre bundle over a contractible CW-complex is trivial.
As a tangent to what I was actually using this for, I came to the following conclusion:
Let $S$ be a differentiable manifold not homeomorphic to one of the form $A \times \mathbb R$. Then there is no continuous map $f : S \rightarrow \mathbb R$ which is free of stationary points.
I would like my reasoning verified, as well as a non-trivial example where this might be a useful result.
The Proof:
If we put a continuous scalar map on any differentiable manifold... $$f : S \rightarrow \mathbb R$$ ...then restrict the codomain to the range of $f$, we have a continuous surjection... $$f : S \rightarrow \mathrm{Range}(f)$$ ...which is a submersion iff $f$ has no stationary points. For simplicity, let's assume that $S$ has a single path-component. This also implies that $\mathrm{Range}(f)$ has a single path-component due to continuity.
Now if $f$ is a submersion, then $S$ can be considered a fibre bundle... $$F \rightarrow S \rightarrow \mathrm{Range}(f)$$ ...where the fibre $F$ is (up to homeomorphism) the preimage of any point in $\mathrm{Range}(f)$. The theorem mentioned above implies that if $f$ is indeed a submersion, then $S$ is homeomorphic to $\mathrm{Range}(f) \times F$. Since $S$ is a manifold (without boundary, as this is not technically part of the definition), this can only be possible if $\mathrm{Range}(f)$ is an open interval. (Side note: I think this rules out all compact $S$ automatically since by continuity, these map into closed intervals.)
So it seems that the only case in which we can put a scalar function $f$ on $S$ free of stationary points is for $S$ which are homeomorphic to some product space of the form $A \times \mathbb R$. For $S$ with more than one path-component: well, if we can't have a map free of stationary points for even a single path-component, then we can't do it for the whole of $S$. It is only possible when each path component is homeomorphic to something of the form $A_i \times \mathbb R$, in which case the whole manifold $S$ is homeomorphic to $A \times \mathbb R$.
Seeking an Example:
I cannot think of an example of a non-compact differentiable manifold that is not of the form $A \times \mathbb R$! Perhaps something of infinite genus?