Does there exist a submersion $f:\mathbb{R}^{3}\setminus\{0\} \to \mathbb{R}$ for which there are $c_1$ and $c_2$ in $\mathbb{R}$ such that $f^{-1}(c_1)$ is compact and $f^{-1}(c_2)$ is non-compact.
Does there exist a submersion $f:\mathbb{R}^{3}\setminus\{0\} \to \mathbb{R}$ with one pre-image compact and another pre-image non-compact?
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analysis
differential-geometry
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0I can imagine one when you look for submersions $f:\mathbb{R}^3\setminus\lbrace -1,+1\rbrace\rightarrow\mathbb{R}$, but not when you only take out one point... I think the answer may be no. – 2011-07-25
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0One could try to look for a function $\mathbb{R}^3\setminus\lbrace0\rbrace\to\mathbb{R}^*_+$ whose level sets would look like spheres for small $0
and become gradually more elongated as $c\to 1$, so that they "hug" an infinite vertical cylinder, be that infinite vertical cylinder for $c=1$, and be vertical cylinders of increasing radii as $c\to\infty.$ So it might be possible after all... – 2011-07-25 -
0Something like $f(x,y,z)=(x^2+y^2)$ when $x^2+y^2\geq 1$, and $\frac{x^2+y^2+z^2(1+z^2)\varphi(x^2+y^2))}{x^2+y^2+z^2(1+z^2)}$ where $\varphi:\mathbb{R}\to\mathbb{R}$ is a smooth function that is increasing, $\geq 0$, constantly $=0$ on some neighborhood of $0$ and $\varphi(x)=1+\lambda\times (1-x)$ on some neighborhood of $1$. One would need to do some refinements, maybe add some constants here and there and modify $f$ and $\varphi$ to make $f$ globally smooth, but this should essentially work. – 2011-07-25
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0Oh no xD, the formula I gave doesn't do the job... I'll think about it :D. – 2011-07-25