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Let $M$ and $N$ be two smooth manifolds, and $f: M \to N$ be a submersion , ${{f}^{-1}}(y)$ is compact for all $y$ in $N$. Then prove for any $x$ in $N$ there is an open neighborhood $U$ of $x$ such that ${{f}^{-1}}(U)$ is diffeomorphic to $U\times {{f}^{-1}}(x)$. I've thought this problem for a long time, but I don't know to use which method. Cobordism can help to solve this question?

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    $\begingroup$ I don't see how cobordism would help. I would try using the implicit function theorem instead. $\endgroup$
    – Neal
    Feb 23, 2012 at 14:10
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    $\begingroup$ show that $f$ is proper and look at en.wikipedia.org/wiki/Ehresmann%27s_fibration_theorem $\endgroup$
    – Blah
    Feb 23, 2012 at 16:59
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    $\begingroup$ You are missing some hypothesis to conclude that $f$ is proper. For example, without any additional hypothesis, we have a counterexample $M=(0,2)$, $N=S^1$ (the unit circle in the complex plane), and $f(t)=e^{2\pi i t}$. $\endgroup$ Apr 28, 2014 at 0:27

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As Blah said in the comments, if we can show that $f$ is proper (and in fact it is!), your statement is just Ehresmann fibration theorem.

To give a proof of Ehresmann fibration theorem, we will use theorem of tubular neighborhoods. This result says that a neighborhood $W$ of $M_y := f^{-1}(y)$ in $M$ is diffeomorphic to a neighborhood of $M_y$ in its normal bundle, and in particular that there exists a differentiable retraction $r : W \to M_y$. But then , consider the map $$(r,f):W \to M_y \times N.$$ This map has a differential (which is invertible along $M_y$). As $M_y$ is compact, there exists an open set $W' \subset W$ containing $M_y$ such that $(r,f)_{|W'}$ is an embedding. Finally, as $f$ is proper, $W'$ contains an open set $W''$ of the form $f^{-1}(U)$, where $U \subset N$ is a neighborhood of $y$. Then clearly $(r,f)(W'') = M_y \times U$ and we have shown that $(r,f)$ is a diffeomorphism from $f^{-1}(U)$ to $M_y \times U$.

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  • $\begingroup$ @ Jacob:Thanks!But I'll make a completment.I've seen somewhere that ${{f}^{-1}}(x)\ and\ {{f}^{-1}}(y)$ is also diffeomorphism.This is a stronger conclusion.How can we get this? $\endgroup$
    – henry
    Feb 24, 2012 at 17:04
  • $\begingroup$ This is the link:link $\endgroup$
    – henry
    Feb 24, 2012 at 17:15
  • $\begingroup$ I've just responded you here. math.stackexchange.com/questions/47373/… $\endgroup$
    – JacobI
    Feb 24, 2012 at 17:36
  • $\begingroup$ @ Jacob: Thank you!I've understood.But where have you learnt these beautiful theorem like Ehresmann fibration theorem?Can you recommended me some modern books about differetial topology?I've read some books which are a little basic that they don't involve some advanced topics. $\endgroup$
    – henry
    Feb 25, 2012 at 3:24
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    $\begingroup$ Dear Henry: I read this theorem on Claire Voisin's book "Hodge theory and complex algebraic geometry" chapter $9$, when dealing with family of complex manifolds. By the way, it would be nice if you accept the answer that you consider helpful ;) $\endgroup$
    – JacobI
    Feb 25, 2012 at 8:19

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