Ehresmann fibration theorem for manifolds with boundary

Question

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?

Answer 1

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.

Answer 2

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.