Let $f: B_1(0)\subset\mathbb{R}^2 \to \mathbb{R}$ be a smooth function, and say $c\in \mathbb{R}$ is not a critical value of $f$. Is it true that each connected component $\Gamma$ of $\left\{f\le c\right\}$ has $\partial \Gamma = \emptyset$? Or better, is it possible to exclude the case $\partial \Gamma \approx [0,1]$?
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$\Gamma$ is going to be some kind of 2-manifold, right? Do you mean boundary of $\Gamma$ in the last line? Also is $B_1(0)$ the closed disk? – Tim kinsella May 27 '16 at 10:46
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Yes, exactly! Γ is a 2-manifold, B_1(0) is the unit disk (you can assume it to be closed, I don't know whether it matters or not at the end) and ∂Γ is as usual the boundary of Γ. – user342725 May 27 '16 at 11:04
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So $\Gamma$ won't be homeomorohic to [0,1] since it's a one-manifold – Tim kinsella May 27 '16 at 11:10
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Of course, I meant ∂Γ ≈ [0,1] in the last line. Sorry for the mistake. I modified the question. – user342725 May 27 '16 at 11:42
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If $B_1$ is the open disk, then $\partial \Gamma= f^{-1}(c)$ is a one manifold without boundary. So a union of circles and lines. If $B_1$ is closed, I'm not sure what happens, but I doubt there's much difference, – Tim kinsella May 27 '16 at 12:05
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How can you say that $\partial \Gamma = f^{-1}(c)$ ? It seemed to me to not be true in general! I had in mind something like $f: B_1(0) \to \mathbb{R}$, $f(x, y) = y$. Then every value is a regular value and $f^{-1}(0) = (-1, 1)\times \left{0\right}$. But $f^{-1}((-1, 0)) = B_1(0)^- := B_1(0)\cap \left{y<0\right}$, hence its boundary (which has only one connected component) cannot be written as a level set for $f$ (or, at least, seems so to me). Or am I missing something? – user342725 May 27 '16 at 12:52