Ehresmann fibration theorem for manifolds with boundary

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?

• How does the proof fail? – Oscar Randal-Williams Apr 1 '15 at 19:55
• Yes a version of it is true if both $M$ and $N$ have boundary. The extra assumption you need is that when you restrict $f$ to the boundary of $M$ it maps that to the boundary of $N$, and that the restriction map $f_{|\partial M} : \partial M \to \partial N$ is a submersion. From these hypothesis, the proof is basically identical to the classical proof. – Ryan Budney Apr 2 '15 at 3:57
• @RyanBudney Thank you. Do you have a reference for this version of the theorem? – Kathrin L. Apr 2 '15 at 8:54
• Take whichever proof you've seen in the manifold without boundary case and adapt it. I think the cleanest proof would be to apply the tubular neighbourhood theorem to the fibres. The fact that the map is a submersion allows you to trivialize the tubular neighbourhoods of the fibres (this also gives you the fibre bundle structure). – Ryan Budney Apr 2 '15 at 14:16

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.