Let $M,N$ be compact connected manifolds, $f:M \to N$ a smooth map with $\operatorname{rank}{(df)}=\dim{N}$. Then for all points $p,q \in N$ ; $f^{-1}p$ is diffeomorphic to $f^{-1}q$.

Please help me solve this question, I've no idea.

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    $\begingroup$ Surely you have some thoughts. For example what does the condition 'rank(df)=dim N' imply? $\endgroup$ Commented Jun 24, 2011 at 9:48
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    $\begingroup$ @wild: Good remark of yours. Unfortunately I am not sure one can have some thoughts and at the same time post five questions in a time span of one hour or so. $\endgroup$
    – Did
    Commented Jun 24, 2011 at 9:54
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    $\begingroup$ please make the body of your questions self-contained. It is customary here to state the question in the main body and not only in the title. Also, most of your questions seem to be relating more to differential geometry or differential topology than algebraic topology. Moreover, what kind of homework is this? It would be much easier for us to answer your questions if you provided a little bit of your own background and thoughts. $\endgroup$
    – t.b.
    Commented Jun 24, 2011 at 10:18
  • $\begingroup$ Moved to meta (and let's remove meta-comments from here, please) $\endgroup$
    – Grigory M
    Commented Jun 24, 2011 at 10:30
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    $\begingroup$ In particular, please consult our homework FAQ too see what other information you should include in a question like this to make it easier for other users to help you. $\endgroup$ Commented Jun 24, 2011 at 12:40

1 Answer 1


Fix $p \in N$. Since $N$ is connected, it suffices to show that the set $S := \{q \in N : f^{-1}(p)\simeq f^{-1}(q) \}$ is opened and closed.

From the response of a question on MSE, we know that $S$ is opened. If $q \in \bar{S}$, let $U \subset N$ be a neighborhood of $q$ on which all fibers are diffeomorphic to $f^{-1}(q)$. Since $U$ contains some point in $S$, we conclude that $q \in S$, hence $S$ is closed.


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