OH SNAP!
So I was wrong after all saying this "The implicit f(a)= b is C^1 and well defined in a neighbourhood around those points...?". This follows from b-f(a)= 0 as we've defined g(a,b) and the other result follows from implicit function theorem.
THANKS MAN! YOU ARE THE BEST!
The neighbourhoods are not the same, how did you just switch the letters a and b? And even so... Okay F(b) = b but how did f(a) become just a? What we want to show is that there is an inverse function of f in my previous post (f(a)=b), s.t. f^-1 (b)=a. This is why I don't get? How do we get...
Inverse
f:R^n->R^n,
if f is C^1, f(a)=b and differential is invertible at a, then the inverse is well defined and C^1 on some open subsets U and V where a is in U and b is in V
Implicit
f:R^p+n-->R^n , f(a,b)=0, f is C^1 and the partials with respect to the variables from p onwards form and...
Implicit => inverse function theorem (urgent due to exam, please help)
Homework Statement
Prove the inverse function theorem, knowing the implicit function theorem.
Homework Equations
The statements of both theorems... Can't think of much else.
The Attempt at a Solution...