general topology - Topological boundary of open set given as preimage

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

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 $f<\frac 1 2$.