"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