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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3Surely you have some thoughts. For example what does the condition 'rank(df)=dim N' imply? – wildildildlife Jun 24 '11 at 09:48
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6@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. – Did Jun 24 '11 at 09:54
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1please 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. – t.b. Jun 24 '11 at 10:18
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Moved to meta (and let's remove meta-comments from here, please) – Grigory M Jun 24 '11 at 10:30
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1In 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. – Willie Wong Jun 24 '11 at 12:40
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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.
