Inverse Function Theorem in Spivak

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The discussion centers on a statement from Spivak's proof of the Inverse Function Theorem (IFT) regarding the assumption that the linear transformation λ can be treated as the identity. Participants seek clarification on why this assumption is valid, given that λ represents the derivative Df(a) and its designation as the identity seems overly restrictive. The context provided explains that if the theorem holds for the inverse transformation λ⁻¹∘f, it should also hold for f itself. The conversation highlights the importance of understanding the implications of this assumption within the proof. Overall, the discussion aims to clarify the reasoning behind Spivak's approach in the proof of the IFT.
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In his proof of the IFT, on p. 36 of "Calculus on Manifolds," Spivak states: "If the theorem is true for \lambda^{-1} \circf, it is clearly true for f. Therefore we may assume at the outset that \lambda is the identity.

I don't understand why we may assume that.

thanks for your help!

Ken Cohen
 
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For those of us without the textbook handy, can you post the context of what lambda is?
 
Is there a way to scan a page and post it?
 
"Suppose that f: R^{n} -> R^{m} is continuously differentiable in an open set containing a, and det f'(a) \neq 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^{-1}: W -> V which is differentiable and for all y \in W satisfies


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

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


This much I think I follow.

"If the theorem is true for \lambda^{-1}\circf, it is clearly true for f."

I think I understand this as well.

"Therefore we may assume at the ouset that \lambda is the identity"

That I don't understand. Since \lambda = 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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