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Can someone help me with this problem?

I have a $C^1$ function $G\colon\mathbb{R}^n\rightarrow \mathbb{R}^m$, where $k=n-m> 0$. If $M$ is the set of points $x\in G^{-1}(0)$ such that $(DG)_x$ has rank $m$, then $M$ is a smooth manifold of dimension $k$.

If someone could give me a starting kick it would be great!

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    [Theorem 8.8, page 182](http://books.google.fr/books?id=eqfgZtjQceYC&printsec=frontcover&hl=fr#v=onepage&q&f=false) Google lets you read that very page!2012-05-21
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    @GeorgesElencwajg Not quite the same thing. Gustavo does not assume that the rank of differential is constant; he defines $M$ as the set of points where the rank is $m$.2012-05-21
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    Dear @Leonid: no, Gustavo defines $M$ as the set of points where the rank of $DG_x$ is $m$ **and** $G=0$ .You must apply Theorem 8.8 with the source manifold in the book being the open subset $U\subset \mathbb R^n$ where the rank of $DG_x$ is $m$.Then $M=(G\mid U)^{-1}(0)$ and 8.8 applies.2012-05-21
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    Dear @Georges, thanks for the clarification. I did not notice that $m$ was the maximal rank.2012-05-21
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    Nice blog you have, by the way, Леонид !2012-05-21
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    Now I think my teacher is going too fast with exercises and too slow with class notes. The question is on a list about manifolds but this problem seems so advanced (it is talking about embedded manifolds and things I didn't see yet). I should probably include a full-text demonstration of this theorem on the solution just to get things straight.2012-05-22

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