Spivak Inverse Function Theorem Proof

1. Nov 27, 2009

krcmd1

On p. 36 of "Calculus on Manifolds" Spivak writes:

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

This far I understand. However, he next says:

"Therefore we may assume at the outset that $$\lambda$$ is the identity."

I don't understand how this follows, since he previously defined $$\lambda$$ = Df(x).

I would appreciate someone adding a bit more explanation here.

Thank you.

2. Nov 28, 2009

HallsofIvy

Staff Emeritus
What he is saying is that since, whenever some theorem is true for $\lambda^{-1}\circle f$, it is true for f, we can work with f rather than $\lambda^{-1}\circle f$- that is that we can take $\lambda= 1$. What you need to do is look at why "if the theorem is true for $\lambda^{-1}\circle f$, it is clearly true for f.[/itex]

3. Nov 29, 2009

krcmd1

I can understand why

a)if this particular theorem is true for $$\lambda^{-1}\circ$$f it is true for f, but

b) is it true as your posting suggests that any theorem true for $$\lambda^{-1}\circ$$f is true for f?

and

c) how does his proof depend upon (a)? I mean, how does the subsequent argument depend upon (a)? I understand that $$\lambda$$ is a linear transformation with non-zero determinant. Doesn't that already imply that f(x+a) -f(a) <> 0 in some neighborhood of a?

I hate to look a gift horse in the mouth but I'm studying this stuff on my own with no one to talk to - not in a class.

Thank you all.

4. Dec 29, 2011

tjkubo

I know this thread is old, but I wanted to put in my two cents since I had trouble with this at first as well.
Suppose the theorem is true for any function with derivative at a equal to the identity function and suppose we have a function f with λ = Df(a) not necessarily the identity. As the author points out, $\lambda^{-1} \circ f$ is a continuously differentiable function with derivative at a equal to the identity, so the theorem is true for $\lambda^{-1} \circ f$ by the assumption above. Hence, the theorem is true for f.