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Let $X$ be a topological space and $U,V \subset X$ open subsets. Let $x:U \to \mathbb{C}$ and $y:V \to \mathbb{C}$ be homeomorphisms. If $x\circ y ^{-1}$ is holomorphic how do I show that $y\circ x^{-1}$ is holomorphic?

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    @potato: there is nothing to look up because the result is false over $\mathbb R$. Counterexample: $x\mapsto x^3$2012-06-24
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    @GeorgesElencwajg You're right; my mistake. My advice to prove the more general statement that "if $f$ is bijective and holomorphic then so is $f^{-1}$" stands though.2012-06-24
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    It has been a while, but I believe you just invert the power series at an arbitrary point to show the inverse is holomorphic there?2012-06-24
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    Use the Open Mapping Theorem for non-constant analytic functions.2012-06-24
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    Sorry, I wasn't thinking. The Open Mapping Theorem ensures that a nonconstant one-to-one analytic function is a homeomorphism onto its image. The fact that the inverse is analytic is from the formula for the derivative of an inverse function: $(f^{-1})'(z) = 1/f'(f^{-1}(z))$ (and that $f$ is not one-to-one in a neighbourhood of a point where $f'=0$).2012-06-24
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    @RobertIsrael I think the formula you've provided is valid only if $f'(z) \neq 0$. My question lies on this, to ensure that $f^{-1}$ is holomorphic we have to have two hypothesis: $f^{-1}$ is continuous and $f'(z)\neq 0$ in the image of $f^{-1}$, by the way, $f^{-1}$ is holomorphic if and only if these two conditions are virified. The problem here is why $(x\circ y^{-1})'(z) \neq 0$2012-06-24
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    As I said, if $f$ is holomorphic and $f'(z_0) = 0$, then $f$ is not one-to-one in any neighbourhood of $z_0$. In fact, if $d$ is the least positive integer $k$ such that $f^{(k)}(z_0) \ne 0$, $f$ is exactly $d$-to-one in some deleted neighbourhood of $z_0$. This follows from the Argument Principle.2012-06-24
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    Dear @RobertIsrael , I still don't understand why how the result you've mentioned follow from the argument principle.2012-06-25
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    @Jr.: Let $\Gamma$ be a simple closed contour surrounding $z_0$, but with all other zeros of $f(z) - f(z_0)$ and of $f'(z)$ outside $\Gamma$. Let $m(w)$ be the number of zeros (counted by multiplicity) of $f(z)-w$ inside $\Gamma$. Then $m(w) = \frac{1}{2\pi i} \int_\Gamma \frac{f'(z)}{f(z)-w}\ dz$ is constant on a neighbourhood of $f(z_0)$, and $m(f(z_0)) = d$.2012-06-25

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