Inverse Function Theorem in Spivak

**krcmd1** · Jun14-09, 06:13 PM

In his proof of the IFT, on p. 36 of "Calculus on Manifolds," Spivak states: "If the theorem is true for $LaTeX Code: \\lambda$ $LaTeX Code: ^{-1}$ $LaTeX Code: \\circ$ f, it is clearly true for f. Therefore we may assume at the outset that $LaTeX Code: \\lambda$ is the identity.

I don't understand why we may assume that.

thanks for your help!

Ken Cohen

**Office_Shredder** · Jun14-09, 06:31 PM

For those of us without the textbook handy, can you post the context of what lambda is?

**krcmd1** · Jun14-09, 07:10 PM

Is there a way to scan a page and post it?

**krcmd1** · Jun14-09, 07:28 PM

"Suppose that f: R $LaTeX Code: ^{n}$ -> R $LaTeX Code: ^{m}$ is continuously differentiable in an open set containing a, and det f'(a) $LaTeX Code: \\neq$ 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 $LaTeX Code: ^{-1}$ : W -> V which is differentiable and for all y $LaTeX Code: \\in$ W satisfies

(f $LaTeX Code: ^{-1}$ )'(y) = [f'(f $LaTeX Code: ^{-1}$ (y))] $LaTeX Code: ^{-1}$ .

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

This much I think I follow.

"If the theorem is true for $LaTeX Code: \\lambda$ $LaTeX Code: ^{-1}$ $LaTeX Code: \\circ$ f, it is clearly true for f."

I think I understand this as well.

"Therefore we may assume at the ouset that $LaTeX Code: \\lambda$ is the identity"

That I don't understand. Since $LaTeX Code: \\lambda$ = 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