I'm looking for a function f in $\mathbb{R}^2$ such that the inverse function therem at some point P = (x,y) does not give an answer of whether the function is invertable in some neighborhood of P, ie the Jacobian is zero but the inverse exists in some neighborhood of P.
I know that the invertibility of f is related to f being 1-1 in some neighborhood of the point P, but I'm not sure how to actually find such point since it would need to be a point at which the Jacobian is zero. Finding points at which the Jacobian of a function is zero isn't that hard, it's just the condition of f needing to also be locally invertible.
Attempt:
The function I've been playing with is $f(x,y) = (x2^y, x2^{-y}) = (u, v) $ since the Jacobian is 0 when x = 0. I'm hoping that $f^{-1}$ exists around (0,0) since nice things always seem to happen at the origin, but I wouldn't be surprised if this was wrong.
Edit: I calculated the inverse to be $f^{-1}(u,v) = (\sqrt{uv}, log_2(\frac{u}{\sqrt{uv}}))$.
Edit: 2 The answer by Mark Joshi is probably the simplest example for this. I am still curious if the example I tried to give (above) works as well.
$f(x,y) = (x^3,y^3)$ at $(0,0).$