Let $S$ be an $n$-surface in $\mathbb R^{n+k}$ i.e, there is an open set $U \subseteq \mathbb R^{n+k}$ and a smooth map $f : U \to \mathbb R^k$ such that $S=f^{-1}(\{0\})$ and $rank Df(a)=k , \forall a \in S$ , then is it true that any path connected componenet of $S$ is again an $n$ -surface ?
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Do you know about the normal bundle of a submanifold? If so, first prove that a submanifold in $R^{n+k}$ is an $n$-surface in your sense if and only if its normal bundle is trivial. – Moishe Kohan Mar 10 '17 at 14:11
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@MoisheCohen : no I don't know that . I would not like to use it if possible – Mar 10 '17 at 14:13
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Take a look: http://math.stackexchange.com/questions/40419/normal-bundle-of-level-set and http://math.stackexchange.com/questions/23764/is-every-embedded-submanifold-globally-a-level-set?rq=1. I will get back to you when I have more time. – Moishe Kohan Mar 10 '17 at 15:19
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If you read the answer by Manuel Araújo to this MSE question, you learn that a submanifold $M^n\subset {\mathbb R}^{n+k}$ is the regular level set of a smooth map $f: {\mathbb R}^{n+k}\to {\mathbb R}^{k}$ is and only if the normal bundle $\nu_M$ of $M$ is trivial. If $U\subset M$ is an open subset then the restriction of the normal bundle $\nu_M$ to $U$ is the normal bundle $\nu_U$ of $U$. If $\nu_M$ is trivial, so is $\nu_U$. Therefore, if $M$ is the regular level set of a smooth map $f: {\mathbb R}^{n+k}\to {\mathbb R}^{k}$, so is every connected component $U$ of $M$.
Moishe Kohan
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