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Let $M$ be a smooth manifold with boundary (say of dimension $m$) and let $N$ be a smooth manifold with no boundary (say of dimension $n$ with $m\geq n$). We have the following classical result:

Theorem: Let $f:M\rightarrow N$ be a surjective smooth map. Let $x\in N$ be a regular value of both, $f$ and $f|_{\partial M}$. Then $f^{-1}(x)$ is a manifold with boundary of dimension $m-n$ ( note that $f^{-1}(x)$ is never empty since $f$ is surjective). Furthermore $\partial (f^{-1}(x))=f^{-1}(x)\cap\partial M$.

So I'm especially interested by the last part of the theorem. So if we remove the assumption that $x$ is a regular value of $f|_{\partial M}$ then there is no reason for the theorem to hold true.

Q1: What are interesting families of manifolds (over a fixed $N$ that we always assumed to be without boundary) for which $\partial (f^{-1}(x))\neq (f^{-1}(x)\cap\partial M)$?

Q2: In general do we have some set relationship between $\partial (f^{-1}(x))$ and $(f^{-1}(x)\cap\partial M)$?

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  • $\begingroup$ Take a compact disc, and any real-valued function on it with no critical points. That's a pretty typical example. $\endgroup$
    – Ryan Budney
    Commented Mar 5, 2014 at 6:20
  • $\begingroup$ if you remove the hypothesis that f is regular on the boudary, then $X=f^{-1}(x)\cap M$ is a, not necessarily proper, sub-manifold of the interior of $M$. In general $\partial X$ may empty, defined, or even not being well-defined in the sense of manifolds. If $X$ accumulates to the boundary, the accomulation points belong to $f^{-1}(x)\cap\partial M$, but can be smaller than that (example: take a cylynder, collapse the boundary components to points to get a spehre and map it to R) $\endgroup$
    – user126154
    Commented Mar 5, 2014 at 10:58
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    $\begingroup$ So I thought about the example about the solid cylinder in $R^3$ (of length $1$) along the $x$-axis and center at $(0,0,0)$ where $f$ is the distance between the projection of a point $P$ on $x$ and $(0,0,0)$. But I found this example a bit artificial since in this case, $N$ is really just $[-1/2,1/2]$ since the fibers outside this interval are empty. But now I just realize that one may identify the two end points of that interval to get a circle. Ok, I see so anything can happen. I guess this answers my question. $\endgroup$
    – Hugo Chapdelaine
    Commented Mar 5, 2014 at 14:23

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