7
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ How does the proof fail? $\endgroup$ – Oscar Randal-Williams Apr 1 '15 at 19:55
  • 2
    $\begingroup$ 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. $\endgroup$ – Ryan Budney Apr 2 '15 at 3:57
  • 1
    $\begingroup$ @RyanBudney Thank you. Do you have a reference for this version of the theorem? $\endgroup$ – Kathrin L. Apr 2 '15 at 8:54
  • $\begingroup$ 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). $\endgroup$ – Ryan Budney Apr 2 '15 at 14:16
  • $\begingroup$ @RyanBudney can $f(\partial M)\subset \partial N$ without $f:\partial M\to\partial N$ being a submersion? $\endgroup$ – Arrow Nov 6 '19 at 22:07
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ Hasn't that just pushed the problem to how to smooth the corners (upstairs and down)? $\endgroup$ – Oscar Randal-Williams Apr 1 '15 at 19:54
  • 1
    $\begingroup$ @Oscar: I must admit, I'm not sure. $\endgroup$ – John Klein Apr 1 '15 at 21:35
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.