Topological boundary of open set given as preimage

Question

Let $M$ be a smooth manifold and $f:M\rightarrow \mathbb{R}$ be a smooth function such that $f(M)=[0,1]$. Let $1/2$ be a regular value and suppose we consider the open and non-empty set $U:=f^{-1}(\frac{1}{2},\infty)\subset M$. I would like to show that $f^{-1}(\frac{1}{2})$ must coincide with the topological boundary of $U$, i.e. $\partial U=f^{-1}(\frac{1}{2})$.

I could prove that $\partial U\subset f^{-1}(\frac{1}{2})$. But I have problems to show the opposite inclusion. How can one prove that $ f^{-1}(\frac{1}{2})\subset\partial U$?

Best wishes

Answer 1

By the Rank Theorem (Theorem 4.12 in my Introduction to Smooth Manifolds, 2nd ed.), each point $p\in f^{-1}(\frac 1 2)$ is contained in the domain of a coordinate chart on which $f$ has a coordinate representation of the form $f(x^1,\dots,x^n) = x^n$. Thus any sufficiently small neighborhood of $p$ contains both points where $f>\frac 1 2$ and points where $f<\frac 1 2$.

Answer 2

Let $f: (-\epsilon, 1+\epsilon) \to [0,1]$
$t\mapsto \begin{cases}0 & t\leq 0 \\ \frac{3}{2}t & 0<t<\frac{1}{3}\\ \frac 1 2 & \frac 1 3 \leq t \leq \frac 2 3\\ \frac 3 2 t- \frac 1 2 & 1>t> \frac 2 3 \\ 1 & t\geq 1\end{cases}$

Then the preimage of $\frac 1 2$ is not contained in the boundary of $U$.

Edit: The question was edited to live in the smooth category. If you assume 1/2 to be a regular value of f, then this will be true for the manifold notion of boundary, as a consequence of the regular value theorem.