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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: $x\mapsto x^3$ –  Georges Elencwajg Jun 24 '12 at 5:02
    
@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. –  Potato Jun 24 '12 at 5:37
    
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? –  Potato Jun 24 '12 at 5:52
    
Use the Open Mapping Theorem for non-constant analytic functions. –  Robert Israel Jun 24 '12 at 7:09
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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 $f'=0$). –  Robert Israel Jun 24 '12 at 8:08
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1 Answer 1

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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Dear @GeorgesElencwajg I didn't understand how your answer prove the statement of my question –  Jr. Jun 25 '12 at 3:12
    
Dear 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 ? –  Georges Elencwajg Jun 25 '12 at 5:30
    
wouldn't it be "$z \in U$"? –  Jr. Jun 25 '12 at 16:55
    
Yes, there was a typo: I corrected it by replacing $z$ by $a$ in the fifth line.Thanks. –  Georges Elencwajg Jun 25 '12 at 17:26
    
How does one know that $f$ can be written in the form $f(z)=cz^n(1+zg(z))$? –  Jr. Jun 26 '12 at 17:50
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