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Let $M$ and $N$ be two smooth manifolds and $f$ a surjective smooth map from $M$ to $N$ which is a submersion. If for any $p \in N$, $f^{-1}(p)$ is compact, then is $f$ necessarily proper, that is, the preimage of compact set is compact?

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Counter-example: Let us identify $\Bbb R^2$ with $\Bbb C$ for convenience. Let $M=\Bbb C\setminus(-\infty,0]$ and let $N=\Bbb C\setminus\{0\}$, which are considered as open submanifolds of $\Bbb R^2$. Then for $$f: M \to N,\quad f(z)=z^2,$$ it satisfies all of your assumptions, but for $D:=\{z\in N: |z-2|\le 1\}$, $f^{-1}(D)$ is not compact.

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