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Let $M$ be a manifold and $C_0,C_1$ be two disjoint closed subsets of $M$ , then smooth Urysohn Lemma says that there exists a smooth function $f:M\rightarrow [0,1]$ such that $f(C_0)=\{0\},f(C_1)=\{1\}$.

Is it always possible to choose such a smooth Urysohn function such that $0$ is a regular value of $f$ i.e. at each point of $x\in f^{-1}(0)$ the derivative map $df_x:T_x(M)\rightarrow T_0([0,1])=\Bbb R$ is a surjective.

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  • $\begingroup$ What happens when $C_0$ is a closed interval? $\endgroup$ Oct 21, 2018 at 16:18
  • $\begingroup$ I think I have done . If I choose $M=\Bbb R$ and $C_0=[1,2]$ then $f|_{C_0}$ will be a submersion but $dim([1,2])=1$ hence $f|_{C_0}$ will be an immersion also hence derivative will be isomorphism and now inverse function theorem will gives the contradiction. Am I right. $\endgroup$
    – Sumanta
    Oct 21, 2018 at 17:26

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Actually, it's never possible (except in the trivial case when $C_0=\emptyset$). If $x$ is any point of $C_0$, the fact that $df_x$ is surjective implies that $x$ has a neighborhood $U\subseteq M$ on which $f\colon M\to\mathbb R$ is a submersion. Since every submersion is an open map, the image of $f$ would have to contain a neighborhood of $0$, which is impossible since $f$ takes its values in $[0,1]$.

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  • $\begingroup$ Actually I was doing the problem : If $M$ is a manifold with boundary then there exists a smooth non-negative function $f$ on $M$ such that $\partial M=f^{-1}(0)$ and $0$ is a regular value of $f$. That's why this question arises in mind. Can give me some hints for this problem. $\endgroup$
    – Sumanta
    Oct 21, 2018 at 17:34
  • $\begingroup$ @UserD: Ah, if $M$ is a manifold with boundary, it's a whole different story. Can you see how to do it locally in a coordinate neighborhood? $\endgroup$
    – Jack Lee
    Oct 21, 2018 at 18:29

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