Structure of level sets of a noncritical point of a smooth function on a two dimensional domain

Question

Let $\psi$ be a smooth function on a two dimensional simply connected domain $\Omega$ such that $\psi=0$ on the boundary $\partial \Omega$. Suppose $\rho$ is not a critical value of $\psi$ then it is claimed that the level set $\psi= \rho$ consists of finitely many closed disjoint curves. How does one prove / see this? Also, if we were working on the two torus rather than a simply connected domain would this still be true?

Answer 1

1. You need $\Omega$ to be bounded. Otherwise the halfplane $\Omega=\{(x,y)\in\mathbb R^2:x>0\}$ with $\psi (x,y)=\sin x$ and $\rho=1/2$ is a counterexample. (Infinitely many level curves, not closed)
2. You need $\rho\ne 0$. Otherwise $\Omega=\{(x,y)\in\mathbb R^2:x^2+y^2<1 \}$ with $\psi (x,y)=x(1-x^2-y^2) $ and $\rho=0$ is a counterexample. (Level curve is not closed).

With the above assumptions, the statement is true, and you don't even need $\Omega$ to be simply connected. Indeed, let $M=\{(x,y)\in \Omega:\psi(x,y)=\rho\}$. This set is closed in $\Omega$ and has no limit points on $\partial \Omega$. Therefore, $M$ is compact. Since $\nabla \psi\ne 0$ on $M$, the implicit function theorem implies that $M$ is a $1$-dimensional manifold (not necessarily connected). A $1$-manifold can be covered by open sets each of which is homeomorphic to a line interval. By compactness, finitely many such sets are enough. It follows that $M$ has finitely many connected component. Each component is homeomorphic to $S^1$ by the classification of $1$-manifolds.

The above works just as well when $\Omega$ is a bounded open subset of a $2$-dimensional manifold (torus, etc.)