Applcation of the inverse function theorem

[Kate2010]

Homework Statement

Let f $$\in$$ C1(Rn) be a function such that f(0) = 0 and $$\delta$$1f(0) is nonzero. ($$\delta$$1 means partial derivative with respect to x1)
Show that there exist neighbourhoods U and V of x=0$$\in$$ Rn and a diffeomorphism g:U->V such that f(g(x)) = x1 for all x = (x1,...,xn) $$\in$$ U.

Homework Equations

Inverse function theorem:
Let S$$\subseteq$$Rn be open, f $$\in$$C1(S,Rn) and x0$$\in$$S. If Df(x0) is invertible then there exists an open neighbourhood U of x0 such that f(U) is open and f: U -> f(U) is a diffeomorphism. Also Df^-1(f(x0)) = (D(f(x0))^-1.

The Attempt at a Solution

I have a hint that says to apply the inverse function theorem to some function F:Rn -> Rn.
But, I really have no idea what this should be. I think it should involve f but I'm not sure how. Also I noticed g(x) = f^-1(x1) if f^-1 exists but then I'm not sure this makes sense as f is defined on Rn not R. So any help with finding this F would be great!

 

[micromass]

Hi Kate2010!

You're correct that you can't apply the theorem directly to $f:\mathbb{R}^n\rightarrow \mathbb{R}$. But maybe you can extend f so that the codomain is good. I mean, your function F should be of the form

$$F(x_1,...,x_n)=(f(x_1,...,x_n),?(x_1,...,x_n),...,?(x_1,...,x_n))$$

where ? are the functions you still need to find. Don't look too far, the ?-functions are very easy (but make sure that the condition DF(0) is invertible).

You should use [ itex] and [ /itex] to make LaTeX. The [ tex]-brackets automatically start a new line.

 

[Kate2010]

Ok, thanks!
How about F(x1,...,xn) = (f(x1,...,xn), x2,..., xn)? Then DF(0) is a matrix that is upper triangular, with 1s on the diagonal except for the first diagonal which we know is non-zero, then the det is non-zero, so it is invertible. Also we know that F(0,...,0) = (0,...,0). So by the inverse function theorem F is a diffeomorphism.

But I'm still not sure about how this helps me with g? If we look at the first component, the inverse of F will map f(x1,...,xn) to x1, so I can kind of see that it is true, but I don't know how to write this properly.

 

[micromass]

What if we just choose g=F^-1, won't that work?

 

[Kate2010]

But then doesn't that give g(f(x)) = x1 instead of f(g(x)) = x1?