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Math Help - Question on something in: Milnor: Topology from the Differentiable Viewpoint

  1. #1
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    Question on something in: Milnor: Topology from the Differentiable Viewpoint

    Hello,

    since I don't want to write down a whole page of the book, you need to have the book to answer me. I'm aware of that and I hope someone has it.

    On page 8 Milnor shows that #   f^{-1}(y) is locally constant as a function of y.

    That means, that for every regular value y, there exists a neighborhood V of y in N, so that every x in V is a regular value (did i get this right?) and #   f^{-1}(y) =#   f^{-1}(y') for every y' in V.

    In the last line of the section, he shows a possible neighborhood V of y:
     $V=V_1 \cap ... \cap V_k - f(M-U_1-...-U_k)$

    I see that  V_1 \cap ... \cap V_k is a neigborhood of y, but why is it still one, if you remove  $f(M-U_1-...-U_k)$?

    Thanks in advance,
    engmaths
    Last edited by engmaths; November 20th 2011 at 01:31 PM.
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  2. #2
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    Re: Question on something in: Milnor: Topology from the Differentiable Viewpoint

    For simplicity we can suppose the neighborhoods U1,...,Uk are all open. So W=M-U1-...-Uk is closed, so W is compact as a closed subset of a compact space M. So f(W) is compact. Again, since N as a manifold is Hausdorff, so f(W) is closed.
    So V is open, and it's easy to verify that y belongs to V.
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  3. #3
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    Re: Question on something in: Milnor: Topology from the Differentiable Viewpoint

    Thanks, that helped me alot.

    Did I get right, that every x in V is a regular value? So that would mean, that for every regular value, there is a whole neighborhood of regular values.
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    Re: Question on something in: Milnor: Topology from the Differentiable Viewpoint

    The set of all critical values are of 0 measure as Sard's theorem says. But I'm not sure about your statement. Sard's theorem says the set of regular values are dense, but it doesn't say it's open.
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  5. #5
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    Re: Question on something in: Milnor: Topology from the Differentiable Viewpoint

    I think Milnor already says on page 8 that every x in V is a regular value and therefore every regular value has a neighborhood of regular values. I just ask, because I'm no native speaker and not sure if I got it right. (You don't need to give me a proof.)

    edit: V is the neigbhorhood in the starting post.

    edit2:
    I thought of a proof for: "every x in V is a regular value".
    Proof:
     x \in f^{-1}(V)=f^{-1}(U_1 \cap ... \cap U_k - f(M-U_1-...-U_k)) means x \in U_i for a  i  \in \{1,...,k\}. f|_{U_i} is a Diffeom. and therefore D(f|_{U_i})(x) an Isomorphism. Let i be the inclusion map from U_i to N. Then f|_{U_i}=f\circ i and D(f|_{U_i})(x)=D(f)(i(x)) \circ D(i)(x)=D(f)(x) \circ D(i)(x). So D(f)(x) must be surjective => x is a regular value of f.
    Last edited by engmaths; November 20th 2011 at 02:46 PM.
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