Every embedded hypersurface is locally a regular surface? I'm reading do Carmo's Riemannian Geometry, in ex6.11 d) he wrote that"every embedded hypersurface $M^{n} \subset \bar{M}^{n+1}$ is locally the inverse image of a regular value". Could anyone comment on how to show this?
To be more specific, let $\bar{M}^{n+1}$ be an $n+1$ dimensional Riemannian manifold, let $M^{n}$ be some $n$ dimensional embedded submanifold of $\bar{M}$, then locally we have that $M=f^{-1}(r)$, where $f: \bar{M}^{n+1} \rightarrow \mathbb{R}$ is a differentiable function and $r$ is a regular value of $f$.
Thank you very much!

By choosing good local coordinates, you can assume that $M = \mathbb{R}^n\subset\mathbb{R}^{n+1} = \overline{M}$. Specifically, assume that $M = \{x\in \mathbb{R}^{n+1} : x_{n+1} = 0\}$. Then $M = f^{-1}(0)$, where $f\colon \mathbb{R}^{n+1}\to \mathbb{R}$ is the map $f(x) = x_{n+1}$. Since $0$ is a regular value for $f$, this is exactly what you want.
