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!

Seems like you should be able to do this globally.
Define $f: \overline M \to \mathbb R$ by 
$$f(x) = \textrm{dist}(x, M)$$
Then the function $\phi:\mathbb R \times M \to \overline M$ given by 
$$(t,p) \mapsto \exp_p(t\nabla f)$$
is, for some small $t$, a diffeomorphism (this follows from existence and uniqueness results for ODE since by definition $exp_p$ gives the geodesic from $p$ in the normal direction $\nabla f$) and therefore $0$ is a regular value for $f$.
