- #1

- 35

- 0

I need prove or disprove the following statement:

If f: R^2->R^2 is a differentiable function whose derivative at (0,0) is not invertible, then there does not exist a differentiable function g: R^2->R^2 such that:

(g o f)(X)=X

I've been trying to find counter examples like crazy (f(x,y)=(x+y,x-y), (xy,x+y), etc.) but everything seems to agree. I think this is actually true, but I don't want an entire proof (I can do that work); just a hint or two as to which way I should begin to approach this (basically I just need a starting point).

Nevermind; I found the trick. I realized if (g o f)(X)=X, the derivative of this must be the identity matrix. From there, no problems.

If f: R^2->R^2 is a differentiable function whose derivative at (0,0) is not invertible, then there does not exist a differentiable function g: R^2->R^2 such that:

(g o f)(X)=X

I've been trying to find counter examples like crazy (f(x,y)=(x+y,x-y), (xy,x+y), etc.) but everything seems to agree. I think this is actually true, but I don't want an entire proof (I can do that work); just a hint or two as to which way I should begin to approach this (basically I just need a starting point).

Nevermind; I found the trick. I realized if (g o f)(X)=X, the derivative of this must be the identity matrix. From there, no problems.

Last edited: