If I have a function $\phi:\mathbb{R^{2}}\rightarrow\mathbb{R}$ which is $C^{\infty}$ without critical points, can I assure that all the contour are homeomorphic?
2026-04-02 22:07:42.1775167662
contour which can be homeomorphic?
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No. Here's a counterexample: $\phi(x,y) = e^x(y^2-1)$. Its differential is $$d\phi = e^x(y^2-1)dx + 2ye^x\,dy,$$ which is never zero. For $C\ge 0$, the contour $\phi^{-1}(C)$ is a pair of smooth curves $y=\pm\sqrt{Ce^{-x}+1}$, thus disconnected. But for $C<0$, it is a connected curve given by $x= \log (-C/(1-y^2))$, $-1<y<1$.
Here's a picture of the contours: