Transition functions are holomorphic

Question

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?

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.