# 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
• @RyanBudney can $f(\partial M)\subset \partial N$ without $f:\partial M\to\partial N$ being a submersion? – Arrow Nov 6 '19 at 22:07

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.

• Hasn't that just pushed the problem to how to smooth the corners (upstairs and down)? – Oscar Randal-Williams Apr 1 '15 at 19:54
• @Oscar: I must admit, I'm not sure. – John Klein Apr 1 '15 at 21:35

To think, consider $$N$$ connected boundaryless manifold. I believe that when $$f$$ is a proper map and also the restriction $$f:\partial M\rightarrow N$$ is a submersion, then (M,f,N) become a fiber bundle.

Claim:

Let $$p:E\longrightarrow B$$ a submersion such that $$p$$ is a proper map, $$\partial B=0$$. Then, $$E$$ is the total space of a fiber bundle over $$B$$ with projection $$p$$. When $$\partial E\neq \emptyset$$ this result is true if $$p\vert_{\partial E}:\partial E\longrightarrow B$$ is a submersion.

Proof's Sketch:

Fixe $$x\in B$$ and let $$W$$ be a tubular neighborhood of $$p^{-1}(x)$$ in $$E$$ with smooth retraction $$r:W\longrightarrow p^{-1}(x)$$ (see Hirsch, Differential topology page 109. The hypothesis that $$p\vert_{\partial E}$$ is a submersion ensures that $$p^{-1}(x)$$ is a neat submanifold for each $$x\in B$$, hence there exists tubular neighborhood for $$p^{-1}(x)$$ for each $$x\in B$$). The differential of the map $$p\times r:W\longrightarrow B\times p^{-1}(x)$$ is non singular in each point of $$p^{-1}(x)\subset W$$. Since $$p^{-1}(x)$$ is compact, we can obtain an open neighborhoord $$W^{'}$$ of $$p^{-1}(x)$$ such that $$p\times r:W^{'}\longrightarrow B\times p^{-1}(x)$$ is an embendding. As $$p$$ is a proper map, we can obtain an open set $$U$$ of $$B$$ such that $$p^{-1}(x)\subset p^{-1}(U)$$ and $$p^{-1}(U)\subset W^{'}$$. Thus, $$p\times r:p^{-1}(U)\longrightarrow B\times p^{-1}(x)$$ is a diffeomorphisms satisfying $$\pi_1\circ(r\times p)=p$$. We conclude by showing that given $$y\in B$$, $$p^{-1}(y)$$ is diffeomorphic to $$p^{-1}(x)$$. First, the condition $$p^{-1}(y)$$ is diffeomorphic to $$p^{-1}(x)$$ is an open condition, since that for $$y\in p^{-1}(U)$$, the restriction $$p\times r:p^{-1}(y)\longrightarrow \{y\}\times p^{-1}(x)$$ is a difeomorphism. Now, given $$y\in B$$ such that there exists a sequence $$y_n\longrightarrow y$$ and $$p^{-1}(y_n)$$ which is diffeomorphic to $$p^{-1}(x)$$, then by the early construction, for sufficiently large $$n$$, $$p^{-1}(y_n)$$ is diffeomorphic to $$p^{-1}(y)$$, concluding that the condition $$p^{-1}(y)$$ is diffeomorphic to $$p^{-1}(x)$$ is a closed condition. Since that $$B$$ is a connected space, this proves the claim.

Ps. If $$B$$ is a disconnected manifold, we have that $$p$$ is locally trivial but the fibers can be different.