Spivak Inverse Function Theorem Proof

[krcmd1]

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

"If the theorem is true for (\lambda^{-1})\circf , 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.

 

[HallsofIvy]

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]

 

[krcmd1]

I can understand why

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

b) is it true as your posting suggests that any theorem true for \lambda^{-1}\circf 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.

 

[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.