Suppose $f$ is a smooth function from smooth manifold $M$ to real numbers. If $f$ is regular on $[a,b]$, prove that $f^{-1}(a)$ is diffeomorphic to $f^{-1}(b)$.
I understand that these are submanifolds since regular. But how to prove it’s diffeomorphic? My guess is we need to use flow of certain vector field to generate the diffeomorophism, but how to formulate this idea?
If the map $f$ is proper, then this follows from Ehresmann's Lemma, that $f$ is a fibre bundle, hence all of $f^{-1}(x)$ are diffeomorphic. For example if $M$ is compact then the map $f$ has to be proper.
For non-proper functions, the claim is false. Consider $f : \mathbb{R}^{2} \setminus \{0\} \rightarrow \mathbb{R}$, defined by $f(x,y) = x$. Then $f$ is smooth and regular, but $f^{-1}(0)$ is not diffeomorphic to $f^{-1}(1)$ since $f^{-1}(1)$ is connected and $f^{-1}(0)$ is not.