Find values such that f^{-1}(m) and f^{-1}(n) are not diffeomorphic

Question

Suppose that $f: \mathbb R^2 \to \mathbb R$, $f(x,y) = (x^3-1)e^y$. It is not hard to show that it is a surjective smooth submersion. Then $\forall m,n \in \mathbb R$, $f^{-1}(m)$ and $f^{-1}(n)$ are smooth submanifolds of dimension 1. What is the strategy for finding $m,n$ such that $f^{-1}(m)$ and $f^{-1}(n)$ are not diffeomorphic.

Answer 1

This is false, that is, $f^{-1}(m)\cong \mathbb{R}$ independent of $m\in\mathbb{R}$.

First, for $m = 0$, we see $f^{-1}(m) = \{(1,y)\}\cong \mathbb{R}$, so we will focus our attention on the case where $m\neq 0$.

We may solve the equation $(x^3-1)e^y = m$ for $y$, getting $y = \ln\left(\frac{m}{x^3-1}\right)$. This has domain $(1,\infty)$ for $m > 0$ and domain $(-\infty, 1)$ for $m < 0$.

We will now restrict to $m > 0$, the other case begin analogous. In this case, $$f^{-1}(m) = \left\{\left(x,\ln\left(\frac{m}{x^3-1}\right)\right): x > 1\right\}.$$ Let $\pi:\mathbb{R}^2\rightarrow \mathbb{R}$ project onto the first factor. Then $\pi|_{f^{-1}(m)}:f^{-1}(m)\rightarrow (1,\infty)$ is a diffeomorphism with inverse $g:(1,\infty)\rightarrow f^{-1}(m)\subseteq\mathbb{R}^2$ with $g(x) = \left(x, \ln\left(\frac{m}{x^3-1}\right)\right)$. And, of course, $(1,\infty)\cong \mathbb{R}$.

For the case of $m < 0$, the only change is that one uses $(-\infty,1)$ instead of $(1,\infty)$.