# Inverse Function Theorem and Local Invertibility

I have the following function: $$f: \mathbb{R^2}/{(0,0)} \rightarrow \mathbb{R^2}$$ where $$f(x,y) = (\frac{x²-y²}{2}, xy).$$

I've shown that $f$ is a local $C^1$-diffeomorphism for all $(a,b) \in \mathbb{R^2}/{(0,0)}$.

Is f a class $C^1$-homeomorphism?

How can I prove that f is invertible in a neighborhood of $(0,0)$?

I tried using the differentiability of the inverse homeomorphism and the inverse function theorem, but the hypothesis fails because the jacobian matrix on $(0,0)$ is non-invertible.

• The point $(0,0)$ does not belong to the domain of definition of your function. – Tomas Jun 6 '16 at 11:29
• Does that imply I can't take a neighborhood of it where f is invertible? Isn't there a way around it by using an accumulation point? – DrHAL Jun 6 '16 at 11:56
• determinant of Jacobi-Matrix $\ne 0$ => invertible – user90369 Jun 6 '16 at 11:57

$f$ is not invertible in any neighborhood of $(0,0)$. I'm actually going to prove this for the function $g(x,y) = 2 \cdot f(x,y) = (x^2-y^2,2xy),$ but we'll get the result for $f$.
Think of $g$ as a function $\mathbb{C} \to \mathbb{C}$. Then it's clear that $g(z) = z^2$, since $(a+ib)^2 = (a^2-b^2)+i(2ab).$ Hence $g$ is a two-sheeted cover $\mathbb{C} \to \mathbb{C}$ branched at $z = 0$, and cannot be a homeomorphism in a neighborhood of $0$.
Essentially, any neighborhood of $0$ must contain both $z'$ and $-z'$ for some small $z'$, and since $g(z') = g(-z')$ it cannot be a homeomorphism in such a neighborhood. For any nonzero point $z_0 \in \mathbb{C}$ we can find a small neighborhood (take radius less than $|z_0|$) that does not contain any antipodal pairs, and $g$ will be a smooth (in fact holomorphic) homeomorphism on such a neighborhood. But it cannot even be a homeomorphism in any neighborhood of 0.
Note that this is not implied by the vanishing of the Jacobian. The function $h: \mathbb R^2 \to \mathbb R^2$, $h(x,y) = (x^3,y)$ has vanishing Jacobian at $(0,0)$ but is a smooth homeomorphism there. But the Jacobian vanishing generally indicates that things are going wrong.