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The fibers $f^{-1}(y)$ of a smooth map $f:\mathbb{R}^n\to\mathbb{R}^m$ at any regular value $y\in\mathbb{R}^n$ are smooth submanifolds by the preimage theorem.

Under which conditions on $f$ are its fibers $f^{-1}(y)$ compact?

Euler_Salter
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    This works if $f$ is proper. – Didier Oct 13 '22 at 14:56
  • @Didier True! However, proper is just the definition of the condition above, so it is just a tautology. I would like to find sufficiency conditions for this to be true – Euler_Salter Oct 13 '22 at 15:06
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    This condition is weaker than $f$ being proper. Properness is equivalent to having compact fibers and being a closed map in this case. You can also have compact fibers and not be a closed map, e.g. consider the composition of a homeomorphism $\mathbb{R}\cong(-1,1)$ and the inclusion $(-1,1)\subseteq\mathbb{R}$. – Thorgott Oct 13 '22 at 15:19
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    @Euler_Salter Proper means that the preimage of a compact is compact. What you consider means that preimage of singletons are compact: this is not at all the same thing – Didier Oct 13 '22 at 20:41
  • @Didier ah you're right! I suppose I am just looking for some condition on $f$ that would imply that the preimage of singletons are compact – Euler_Salter Oct 14 '22 at 09:05
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    @Euler_Salter This is related – Didier Oct 14 '22 at 09:06

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You have the following characterisation of proper maps:

A continuous map $f\colon\mathbf{R}^m\to\mathbf{R}^n$ is proper if, and only if, $\lim\limits_{\|x\|\to +\infty}\|f(x)\|=+\infty$.

The condition ensures that the reverse image of bounded subsets are bounded.

C. Falcon
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