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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!

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    Locally, the submanifold has a tubular neighborhood $U$ which looks like a product of a ball times your submanifold. Then you can take $f$ to be the projection on the ball. (If you want $f$ defined on the whole big manifold, just extend smoothly in any way)2012-05-08

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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.

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    This proof also generalizes to show that codimension doesn't matter: every embedded $k$-submanifold is locally the inverse image of a regular value.2012-05-08
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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 $f$.

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    Dear treble, Even if $M$ is two-sided locally in $\overline{M}$ (i.e. the normal bundle to $M$ is trivial), it may not separate $M$ and so may not be two-sided globally. So your function $f$ may only be definable in some (tubular) neighbourhood of $M$, not on all of $\overline{M}$. Regards,2012-05-08