# Inverse Function Theorem in Spivak

1. Jun 14, 2009

### krcmd1

In his proof of the IFT, on p. 36 of "Calculus on Manifolds," Spivak states: "If the theorem is true for $$\lambda$$$$^{-1}$$ $$\circ$$f, 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.

Ken Cohen

2. Jun 14, 2009

### Office_Shredder

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

3. Jun 14, 2009

### krcmd1

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

4. Jun 14, 2009

### krcmd1

"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($$\lambda$$$$\circ$$f)(a) = D($$\lambda$$$$^{-1}$$)(f(a) = $$\lambda$$$$^{-1}$$$$\circ$$Df(a) is the identity linear transformation."

This much I think I follow.

"If the theorem is true for $$\lambda$$$$^{-1}$$$$\circ$$f, 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