All manifolds in consideration may have nonempty boundary and may be disconnected.
Let me fix a definition first. A map between smooth manifolds $M\rightarrow N$ is a fiber bundle, iff it's locally smoothly trivial. I neither assume that all fibers are diffeomorphic nor the map being surjective.
The classical Ehresmann fibration theorem says: If $f\colon M\rightarrow N$ is a proper submersion between smooth manifolds without boundary, then it is a smooth fiber bundle.
Is this also true if $M$ and $N$ have boundary? If not, which natural conditions can one impose on $M$, $N$ or $f$ such that the theorem holds?
Best Answer
Let $D(M)$ be the boundary of $M \times [0,1]$ (by smoothing corners, this can be understood as smooth). Then $f: M \to N$ induces a smooth map $$ D(f): D(M) \to D(N)\, . $$ Further, $D(f)$ is a proper submersion of boundary-less manifolds so it's a smooth fiber bundle. Now pull this back along the inclusion $N \times 0 \subset D(N)$ to conclude that $f$ is a smooth fiber bundle.