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Problem says:

Assume that $f:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ is smooth and C-R equations

$$\frac{\partial f_{1}}{\partial x} = \frac{\partial f_{2}}{\partial y} $$

$$\frac{\partial f_{1}}{\partial y} = -\frac{\partial f_{2}}{\partial x}$$

holds. Suppose that $f$ is locally invertible at $x_{0}$ . Show that $Df(x_{0})\neq0$ . Find a counter-example when you don't have C-R equations. .

Clearly, by inverse function theorem, $\Rightarrow $ direction holds. But showing that the other direction hold under C-R equation seems to be hard since the theorem (implicit, inverse function theorem) only says about ($\Rightarrow $) direction. How could I show that the reverse holds?

Chris Taylor
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Darae-Uri
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1 Answers1

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Identifying $\mathbb{R}^2\simeq \mathbb{C}$. Now let $U$ be the neighbourhood of $x_0$ where $f$ is bijective. Using the open mapping theorem we can deduce that $f^{-1}$ is continous on $f(U)$. The injectivity of $f$ gives you that $f'$ is nowhere constant $0$ in $U$ and indeed, using the identity theorem you can deduce that the set of zeros, call it $N$, of $f'$ is discret (i.e. there is a open ball around every element where $f'$ is non-vanishing). Using the inverse function theorem on $f(U)\setminus f(N)$ (note that $f(N)$ is still discret) you can deduce that $f$ is holomorphic on $f(U)\setminus f(N)$. Now use Riemanns theorem for removable singularitys to deduce that $f^{-1}$ is indeed holomorphic on $f(U)$. The rule for the derivative of the inverse gives you that $f'$ is nonzero on $U$.

As a counterexample when C-R Relations are not available take

$$(x,y)\mapsto (x^3,y^3). $$

Maik Pickl
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