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Let $\pi:\mathbb R^d\longrightarrow\mathbb R$ be Lipschitz continuous and such that $\|\nabla\pi\|>0$ almost everywhere. Suppose that the level set $\pi^{-1}(0)$ is compact. Can I conclude that the level sets $\pi^{-1}(c)$ are diffeomorphic for all $c\geq0$?

If not, what are the additional conditions for which I can conclude it?

Thank you

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    $\begingroup$ $\pi(x) = |x|$? (More generally, since the empty set is compact, you probably want to assume that $\pi$ is at minimum surjective to $[0,\infty)$ for your conclusion to hold.) $\endgroup$
    – Willie Wong
    Commented Nov 19, 2020 at 17:03
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    $\begingroup$ Also, diffeo is certainly too strong. It shouldn't be that hard to cook up lipschitz functions with some level sets rounds spheres and some level sets the surface of a cube. For what you want (supposing you want the level sets to be $C^1$ diffeos) you need something like $\pi$ being $C^2$, that $[0,\infty)$ contains no critical value, and that $\pi$ is proper. You should look for "Morse Theory" in textbooks. $\endgroup$
    – Willie Wong
    Commented Nov 19, 2020 at 17:16
  • $\begingroup$ Thank you. Could not be sufficient that $\pi\in C^1$ and $\|\nabla\pi\|>0$? $\endgroup$
    – Redeldio
    Commented Nov 19, 2020 at 21:11
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    $\begingroup$ Maybe. Without writing it down I am 100% sure I can prove it with C^2. With C^1 I'd actually have to think. $\endgroup$
    – Willie Wong
    Commented Nov 19, 2020 at 21:36
  • $\begingroup$ The same question was discussed on math.stackexchange: math.stackexchange.com/questions/3914334/… $\endgroup$
    – Voliar
    Commented Feb 20 at 6:53

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