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Inverse Function Theorem in Spivak

  1. Jun 14, 2009 #1
    In his proof of the IFT, on p. 36 of "Calculus on Manifolds," Spivak states: "If the theorem is true for [tex]\lambda[/tex][tex]^{-1}[/tex] [tex]\circ[/tex]f, it is clearly true for f. Therefore we may assume at the outset that [tex]\lambda[/tex] is the identity.

    I don't understand why we may assume that.

    thanks for your help!

    Ken Cohen
     
  2. jcsd
  3. Jun 14, 2009 #2

    Office_Shredder

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    For those of us without the textbook handy, can you post the context of what lambda is?
     
  4. Jun 14, 2009 #3
    Is there a way to scan a page and post it?
     
  5. Jun 14, 2009 #4
    "Suppose that f: R[tex]^{n}[/tex] -> R[tex]^{m}[/tex] is continuously differentiable in an open set containing a, and det f'(a) [tex]\neq[/tex] 0. Then there is an open set V containing a and an open set W containing f(a) such that f: V -> W has a continuous inverse f[tex]^{-1}[/tex]: W -> V which is differentiable and for all y [tex]\in[/tex] W satisfies


    (f[tex]^{-1}[/tex])'(y) = [f'(f[tex]^{-1}[/tex](y))][tex]^{-1}[/tex].

    Proof. Let [tex]\lambda[/tex] be the linear transformation Df(a). Then [tex]\lambda[/tex] is non-singular, since det f'(a) [tex]\neq[/tex] 0. Now D([tex]\lambda[/tex][tex]\circ[/tex]f)(a) = D([tex]\lambda[/tex][tex]^{-1}[/tex])(f(a) = [tex]\lambda[/tex][tex]^{-1}[/tex][tex]\circ[/tex]Df(a) is the identity linear transformation."


    This much I think I follow.

    "If the theorem is true for [tex]\lambda[/tex][tex]^{-1}[/tex][tex]\circ[/tex]f, it is clearly true for f."

    I think I understand this as well.

    "Therefore we may assume at the ouset that [tex]\lambda[/tex] is the identity"

    That I don't understand. Since [tex]\lambda[/tex] = Df(a), making it the identity seems a very severe condition on f(a).

    It was easier that I thought to type this in with the Latex Reference. Thank you to whoever programmed that!

    Ken Cohen
     
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