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?
Transition functions are holomorphic
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2@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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0@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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0It 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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0Use the Open Mapping Theorem for non-constant analytic functions. – 2012-06-24
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1Sorry, 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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0@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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0As 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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0Dear @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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0@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
1 Answers
Holomorphic functions in one variable are so rigid that they admit of a complete local description, an exceptionally pleasant situation .
Namely, if $f:U\to V$ is a non-constant holomorphic map between connected open subsets of $\mathbb C$ (or between Riemann surfaces), then for $a\in U$ we can write $f(z)=\phi(z)^n$ on an open neigbourhood $W$ of $a$, with $\phi: W\stackrel {\cong}{\to} W' \subset \mathbb C$ a holomorphic isomorphism .
[The proof is not so difficult : suppose $a=f(a)=0$ and write $f(z)=cz^n(1+zg(z))$, with $c\neq 0\in \mathbb C$ and $g$ holomorphic.
Since $c(1+zg(z))$ is non zero near $z=0$, we can extract a holomorphic $n$-th root of it, i.e. find on some neighbourhood $W$ of $a$ a function $h\in \mathcal O(W)$ such that $h(z)^n=c(1+zg(z))$.
The required isomorphism is then $\phi(z)= z\cdot h(z)$ ]
Using this local description it is then clear that injectivity of $f$ forces $n=1$ (because $f(z)=\phi(z)^n$ and $\phi$ is bijective) and thus the bijective function $f$ is an isomorphism because locally it is one: $f \mid W=\phi$.
Notice that the structure theorem also implies that non constant holomorphic functions are open, a non trivial result evoked by Robert in his comment.
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0Dear @GeorgesElencwajg I didn't understand how your answer prove the statement of my question – 2012-06-25
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0Dear Jr.,you know that $f=x\circ y^{-1}$ is holomorphic and you know that it is a homeomorphism, so in particular that it is bijective.The result I proved in the answer then says that $f$ is a holomorphic isomorphism. Hence $f^{-1}=y\circ x^{-1}$ is also holomorphic. Isn't that what you wanted ? – 2012-06-25
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0wouldn't it be "$z \in U$"? – 2012-06-25
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0Yes, there was a typo: I corrected it by replacing $z$ by $a$ in the fifth line.Thanks. – 2012-06-25
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0How does one know that $f$ can be written in the form $f(z)=cz^n(1+zg(z))$? – 2012-06-26
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0Jr., Robert and I have given you a lot of help on a completely standard result. It's time for you to try and do some work yourself. – 2012-06-26