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Let M be a manifold without boundary and let , $g:M\to \mathbb R$ has $0$ as a regular value.

Question:

Than the set of $x$ in $M$ with $g(x) \geq 0$ is a smooth manifold with boundary equal to $g^{-1}(0)$.

I am trying to prove this Lemma from Milnor's book 'Topology from a differentiable viewpoint(page 12).

Thank you.

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    $\begingroup$ What is your definition of manifold? This follows from the inverse function theorem. $\endgroup$ – Alex Youcis Oct 27 '13 at 8:30
  • $\begingroup$ for every $x \in M$ there is a neighborhood that is diffeomorphic to an open set i $H^{n}$. $\endgroup$ – Eli Elizirov Oct 27 '13 at 8:36
  • $\begingroup$ i cant see how this follows from the inverse function theorem, i understand that $g^{-1}(0)$ is a smooth manifold , but why $x$ is a manifold. $\endgroup$ – Eli Elizirov Oct 27 '13 at 8:42
  • $\begingroup$ $x$ is not claimed to be a manifold $\endgroup$ – Hagen von Eitzen Oct 27 '13 at 9:23
  • $\begingroup$ The lemma says that x is a manifold with boundary $g^{-1}(X)$ $\endgroup$ – Eli Elizirov Oct 27 '13 at 10:05
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Let $\dim M = m$. By Lemma 1 (of the same book), $g^{-1}(0)$ is a smooth manifold of dimension $m-1$. Let $M' = \{x \in M ; g(x)\geq 0\}$. Then $g$ maps $M'$ to $\mathbb H =\{y \in \mathbb R; y \geq 0\}$. Now the boundary $\partial M'$ is the set of all points in $M'$ that correspond to points of $\partial \mathbb H = \{0\}$ under $g$, that is, $\partial M' = g^{-1}(0)$.

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  • $\begingroup$ Hey, but $g$ may not be a diffeomorphism. $\endgroup$ – mathemather Jul 14 '18 at 7:58

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