If an analytic function $f: \mathbb{R}^2 \to \mathbb{R}^2$ is locally invertible at $(x_0, y_0)$, then $Df(x_0,y_0) \not = 0$.

Question

I am trying to show that if an analytic function $f: \mathbb{R}^2 \to \mathbb{R}^2$ (i.e. $f$ satisfies the Cauchy-Riemann equations) is locally invertible at $(x_0, y_0)$, then $Df(x_0,y_0) \not = 0$, but am having a lot of trouble with it. I have already shown that $Jf(x_0,y_0) = 0$ iff $Df(x_0,y_0) = 0$, but why local invertibility implies $Df(x_0,y_0) \not = 0$ escapes me. If the inverse was necessarily differentiable, I could do it via the chain rule, but I don't see why this would be the case.

Any help would be greatly appreciated, because I know it's probably a really simple argument and me not getting it is driving me nuts.

Oh, and to add. I cannot use any notions from complex analysis whatsoever, as this is a strictly real analysis problem. Thanks!

Answer 1

Why are you calling it $\mathbb R^2$ instead of $\mathbb C$?

Hint: The number of zeros of $f(z) - p$ inside a circle, counted by multiplicity, is locally constant in $p$. If $f'(z_0) = 0$, the zero of $f(z) - f(z_0)$ at $z_0$ has multiplicity greater than $1$.