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Let $F:\mathbb{R}^n \to \mathbb{R}^m$ be a smooth surjective submersion. Are the fibers of $F$ necessarily connected? What if we substitute $\mathbb{R}^n$ and $\mathbb{R}^m$ with open neighborhoods of the origin?

The problem is motivated by an application of Exercise 8.18 (p.202) of Lee's book on smooth manifolds.

a.padoan
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    Hint: First think about the case $n=m=2$. Can you find a counter-example? – Moishe Kohan Feb 10 '21 at 10:51
  • @MoisheKohan I did try to find a counter-example, but couldn't so far :( Intuitively, I understand that this should not be true in general. I would like to understand under which additional assumptions on F this should be true. – a.padoan Feb 11 '21 at 20:38
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    As for sufficient additional assumptions: if $F$ is additionally proper, Ehresmann's theorem implies it's a fiber bundle, but since $\mathbb{R}^m$ is contractible, this bundle has then to be trivial, so $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^m\times\text{Fiber}$ and the fibers have to be connected. – Thorgott Feb 11 '21 at 22:55
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    https://math.stackexchange.com/questions/553299/surjective-immersion-mathbbr2-to-mathbbr2-which-is-not-a-diffeomorphi/553408#553408, also https://mathoverflow.net/questions/147110/surjective-entire-functions-without-critical-points – Moishe Kohan Feb 12 '21 at 04:11
  • Thank you both! – a.padoan Feb 12 '21 at 10:29

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