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Inverse Function Theorem in R

  1. Mar 8, 2008 #1
    Aye...title should say in R^2, sorry about that.
    I'm hitting somewhat of a wall in my understanding of a theorem (or rather a special case of a theorem). The theorem as stated in the book is as follows.

    The Inverse FUnction Theorem in the Plane
    Let O be an open subset of the plane R^2 and suppose that the mapping F: O[tex]\rightarrow[/tex]R[tex]^{2}[/tex] is continuously differentiable. Let (x, y) be a point in O at which the derivative matrix DF(x,y) is invertible.

    Then there is a neighborhood U of the point (x,y) and a neighborhood V of its image F(x,y) such that F: U[tex]\rightarrow[/tex]V is one to one and onto.

    The theorem goes on to talk about the inverse functions, but that's not where I'm getting stuck. My problem is this. Consider the function F(r,[tex]\theta[/tex]) = (r*cos[tex]\theta[/tex], r*sin[tex]\theta[/tex]).

    The determinant of the derivative matrix of this function is just r, so the theorem seems that it should only break down at r=0. However consider the point (r, 2*pi)It does not seem to me that it's 1-1 in a neighborhood around this point, which seems to contradict the theorem.

    Where am I misunderstanding things. Is the mapping actually 1-1 here? It seems not to me as it seems to me if you just let theta go to infinity it circles the same ring in the image. I understand that if you restrict it to a ring then it invalidates the theorem as the neighborhood then isn't open due to the ring being thin. But I don't see any way to get around the looping problem.
     
  2. jcsd
  3. Mar 8, 2008 #2

    mathwonk

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    the theorem says the function is 1-1 on SOME nbhd, not on ALL nbhds.
     
  4. Mar 8, 2008 #3
    So let me see if I understand the restriction correctly, as I don't think the book spells this out very well.

    We can only state that it's 1-1 if V is some, perhaps restricted, neighborhood of the image of U. That much I see. However we also must restrict U to the neighborhood in which F:U->V is invertible? Otherwise we have that F[tex]^{-1}[/tex](1,0) can be equal to (1, 2*pi*n) for any n. Do we have to choose a range for the inverse from the outset in order to preserve 1-1? Is that correct?
     
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