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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!

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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 $1$.

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  • $\begingroup$ I am not working in $\mathbb{C}$, and this problem was not part of complex analysis, so would you mind explaining why the "multiplicity" would be greater than $1$? $\endgroup$ – Ryker Nov 2 '14 at 6:14
  • $\begingroup$ @Ryker If your function is continuous, the Looman-Menchoff theorem says that your function can be viewed as a holomorphic function from $\mathbb{C}$ to $\mathbb{C}$ and thus is complex-analytic, so Prof Israel's solution goes through immediately. If you want to avoid complex analytic methods please note it in your question. $\endgroup$ – guest Nov 2 '14 at 6:21
  • $\begingroup$ @guest, I edited the original post and stated that then. And, yes, as I said, I cannot use any notions from complex analysis whatsoever. $\endgroup$ – Ryker Nov 2 '14 at 6:24

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