# 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!

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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) –  Mariano Suárez-Alvarez May 8 '12 at 20:06

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. –  Jesse Madnick May 8 '12 at 20:31
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 $f$.
Dear treble, you have to be a little bit careful using the distance function. E.g. if $M = \{0\} \subset \mathbb R = \overline{M}$, then your function $f(x)$ is just $|x|$, which is not differentiable at $0$, making it a little tricky to talk about $0$ being a "regular value for $f$". Regards, –  Matt E May 8 '12 at 21:42
This difficulty is superficial. You merely need to decide "what side" of $M$ you are on. This is equivalent to choosing a normal vector (i.e. how to define $\nabla f$) in some region of $M$, or in your example, a choice of $f = x$ or $f = -x$. The above argument then applies to either choice. But you are right, I could have clarified this in the original answer. –  treble May 8 '12 at 22:00
I will further remark that if $\overline M = \mathbb R^3$ and if $(t, x, y)$ are coordinates as in my answer, then $\partial_t, \partial_x, \partial_y$ form what is known as a "Darboux Frame." Moreover, in higher dimensions the well-known Fermi coordinates are an example of the construction in my answer. I am just remarking that, in particular, these useful coordinates also answer affirmatively the OP's question. –  treble May 8 '12 at 22:15
Dear treble, Sounds good! (Although for this to work globally I guess you need a hypothesis to the effect that $M$ separates $\overline{M}$.) Cheers, –  Matt E May 8 '12 at 23:24
Thanks for the comments :) You are right, I was really assuming that $M$ was oriented when I was thinking that the argument should work globally. –  treble May 8 '12 at 23:35