Let $N$ be a smooth, connected manifold and $f:N \to \mathbb R$ a smooth, proper and surjective map, transverse to some $k \in \mathbb N$. This means that $M:=f^{-1}(k) \subset N$ is a finite collection of closed codimension 1 submanifolds of $N$, such that $N \setminus M$ consists of finitely many, and at least 2 components. A set $C \subset N$ is called bounded iff its closure $\overline{C}$ in $N$ is compact, otherwise its unbounded. My claim, which I haven't been able to prove, is the following:
All components of $N \setminus M$ are unbounded.
Note that surjectivity of $f$ implies that the sets $f^{-1}(-\infty,k)$ and $f^{-1}(k,\infty)$ are unbounded. If there was a bounded component $K \subset N \setminus M$, then wlog we may assume $K \subset f^{-1}(-\infty,k)$. Further, we would have that $\overline{K}$ is a compact, connected subset of $N$. So $f(\overline{K}) \subset (-\infty,k]$ is also compact and connected, thus we can write $f(\overline{K}) = [a,k]$ for some $a < k$. Now, if I can show that $a = k$, I arrive at an obvious contradiction. However, it doesn't seem to me that there is a simple point-set argument to prove that. Any help is appreciated.
Not necessarily. Just look at a product like $S^1\times \mathbb{R}$ with the trivial product metric and let $f$ be the projection onto the second factor.
Edit: to take into account the extended meaning of the question. Of course there may be a bounded component, but in that case the map will not be transversal everywhere. Just consider a smooth curve $\mathbb{R}\rightarrow \mathbb{R}$ which is the identity outside of $[-1,1]$ and moves from $-1$ to $1$, then back to $-1$ and then back to $1$ again in $(-1,1)$. Clearly this curve will have at least two points with vanishing derivative, but it will be onto, proper and can be chosen as a smooth map wich is transversal to, say $0$. Clearly $f^{-1}(0)$ will consist of at least three points which will subdivide the source into $4$ components (or more), two of which have to be bounded for trivial reasons.