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Let $M$ and $N$ be two differential manifolds and there is a surjective submersion $f$ from $M$ to $N$ such that $f^{-1}(p)$ is compact and connected for any $p$ on $N$. Can we conclude that $f$ is proper, that is, the preimage of a compact set is compact?

Also posted on https://mathoverflow.net/q/186764/16323

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Summer
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1 Answers1

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Let $K \subseteq N$ be compact, take a sequence $p_n \in f^{-1}(K)$. Then $q_n = f(p_n)$ admits a subsequence (which we will keep on calling $q_n$) converging to some $q \in K$.

Since $f$ is a submersion, by the regular value theorem $f^{-1}(q)$ is a closed manifold (nonempty because $f$ is surjective). Therefore it admits a tubular neighborhood $U$, and endowing it (or rather, the normal bundle) with a metric, we call $D \subseteq U$ the set of all points (vectors in the normal bundle) with norm $\leq 1$. $D$ is a neighborhood of $f^{-1}(q)$ which is compact because $f^{-1}(q)$ is such, so $\partial D$ is compact as well.

Now, submersions are open maps: a submersion $f \colon M \to N$ can locally be written as a projection $\mathbb{R}^m \to \mathbb{R}^n$ onto the first $n$ coordinates, which is open, so $f$ maps a basis of open sets of $M$ to open sets in $N$. This implies that $f(D)$ is a neighborhood of $q$, so $q_n$ must lie in $f(D)$ for $n$ larger than some $\overline{n}$.

Call $g = f|_{D} \colon D \to f(D)$, then $g$ is onto, so there is a $p'_n \in g^{-1}(q_n) = D \cap f^{-1}(q_n)$ for $n > \overline{n}$. Now suppose that for arbitrarily large $n$ we could find a $p''_n \in f^{-1}(q_n) \setminus D$: since $f^{-1}(q_n)$ is (path-) connected, we would have a path $\alpha_n$ in $f^{-1}(q_n)$ from $p'_n$ to $p''_n$, which would cross $\partial D$ at a point $r_n$ (define a map $\phi \colon M \to \mathbb{R}$ assigning to points in $U$ of norm $\leq 2$ their norm, and for all other points the value $2$: then $\phi(p'_n) \leq 1 < \phi(p''_n)$, so $\phi \circ \alpha_n$ must take the value $1$). By compactness of $\partial D$, $r_n$ admits a subsequence $r_{n_k}$ converging to some $r \in \partial D$, but then $$q \neq f(r) = \text{lim} f(r_{n_k}) = \text{lim} q_{n_k} = q $$ contradiction. Therefore $f^{-1}(q_n)$ must be definitively contained in $D$, so $p_n$ lies in the compact set $D$ from a certain point onwards: further extracting a convergent subsequence, we are done.

Emilio Ferrucci
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    I think $p_{n_k}$ need not to be in $U$. – Summer Nov 03 '14 at 21:12
  • @Hezudao: How about this: we may assume $f$ is a Riemannian submersion. Embed the normal bundle $\nu$ of $f^{-1}(q)$ into $M$ as ${\exp_p (v)}$ where $p\in f^{-1}(q)$ and $v\in T_p M$ has length less that $\epsilon$. By compactness, there is a universal choice of $\epsilon$ which works for all points $p$. Then $f(\nu)$ is the ball of radius $\epsilon$ around $q$. More importantly, $f^{-1}(f(\nu)) = \nu$. So, pick $U = \nu$ in Emilio's argument. – Jason DeVito - on hiatus Nov 03 '14 at 21:24
  • Thanks, edited. I think my original $U$ is ok, what was needed was to justify that $f(U)$ is a nbhd of $q$, using that $f$ is a submersion. Let me know if you think this works :) – Emilio Ferrucci Nov 03 '14 at 22:22
  • @EmilioFerrucci: $f(U)$ is indeed open, but how can you show $f^{-1}(f(U))=U$? – Summer Nov 03 '14 at 22:32
  • @JasonDeVito: Can you explain why do we have $f^{-1}(f(v))=v$? – Summer Nov 03 '14 at 22:34
  • Ah... now I realize connectedness of preimages is needed. Otherwise just take the disjoint union of $(0,1)$ and $(0,2)$ and map to $(0,2)$ in the obvious way. Then the preimage of $[1/2,3/2]$ is not compact. I edited the answer to include where connectedness should be used in my opinion. I'll think a bit more about this, but hopefully someone else will come up with something better (maybe Jason's argument avoids the trouble)... sorry for the confusion! – Emilio Ferrucci Nov 04 '14 at 01:36
  • Your counterexample is also a counterexample to my proposed solution. I'll have to keep thinking about it. – Jason DeVito - on hiatus Nov 04 '14 at 15:26
  • I've changed the proof, this time I used that fibers are connected. Some notations are different. Let me know what you think – Emilio Ferrucci Nov 05 '14 at 11:14
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    Excellent proof. As far as I can see, your final version of proof is correct. This guarantees that Kodaira's definition of complex analytic family is a family with proper projection map, and then Ehressmann fibration lemma applies. – user188722 Sep 26 '17 at 20:59
  • This gap indeed exists in Kodaira's celebrated book "Complex Manifolds and Deformation of Complex Structures". On page 66 he just claims that the integral flow exists globally for $t\in (-r,r)$ and induces the desired trivialization, but I'm stuck with the details. In fact properness is needed to ensure the global existence of flow, and the latter is essential in proving trivialization. Your argument and the proof of Ehressmann theorem (using integral flow) together fill this gap. – user188722 Sep 26 '17 at 21:00