I'm trying to prove the first part of Proposition 5.47 of Lee's Smooth Manifolds, which is left to the reader. It says

Suppose $M^m$ is a smooth manifold, and $f\colon M\to\mathbb{R}$ smooth. For each regular value $b$ of $f$, the sublevel set $f^{-1}(-\infty,b]$ is a regular domain, that is, a properly embedded codimension $0$ submanifold with boundary.

First, $f^{-1}(\infty,b)$ is open, hence an embedded submanifold of codimension $0$. Also, $f^{-1}(-\infty,b]$ is closed in $M$, so if $f^{-1}(-\infty,b]$ is a embedded submanifold, it is in fact a properly embedded submanifold of codimension $0$.

I want to show $S:=f^{-1}(-\infty,b]$ satisfies the local $m$-slice condition. If $p\in f^{-1}(-\infty,b)$, then since this set is open, we can find a chart $(U,\varphi)$ around $p$ in $S$. But then $\varphi(S\cap U)=\varphi(U)$, so $(U,\varphi)$ is an $m$-slice chart around $p$.

I suspect $f^{-1}(b)$ is the boundary of $S$. Since $f^{-1}(b)$ is a regular level set, it is a properly embedded submanifold of dimension $m-1$ in $M$. I could then find an $m-1$ slice chart $(U,\varphi)$ in $M$ for $f^{-1}(b)$, so that $$ \varphi(f^{-1}(b)\cap U)=\{(x^1,\dots,x^m)\in\varphi(U):x^m=0\} $$

I want to try to modify it somehow to a chart such that $$ \varphi(U\cap S)=\{(x^1,\dots,x^m)\in\varphi(U):x^m\geq 0\} $$ to show it is an $m$-dimensional half slice. Is there maybe a way to restrict to a precompact open set, so that the coordinate functions achieve a mimnimum, and then just shift the coordinate map so the last coordinate is always nonnegative?


2 Answers 2


By submersion theorem, after suitable coordinate transformation, your map should look like: $$ f:(x^1,\dots,x^m)\mapsto x^m $$ in a neighborhood for each regular point in $f^{-1}(b)$. And the claimed conclusion is obvious: you only need to worry about two types of points, the interior points and the boundary points. For an interior point $x\in M$,i.e. $f(x)<b$, everything is fine: $f^{-1}((f(x)-\varepsilon,f(x)+\varepsilon))$ is the desired neighborhood for $x$, which (possibly after shrinking) is homeomorphic to $\mathbb R^m$. For a boundary point, on the other hand, you need to show the existence of a neighborhood homeomorphic to $\mathbb H^m=\{x^m\leq0\}$, which is also automatic under the aforementioned canonical form.

  • $\begingroup$ I think I see now, thanks. $\endgroup$
    – Clara
    Aug 15, 2014 at 23:51

I dont think upstair is right, because he doesn't shown that it is with a boundary.


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