Let $A$ and $B$ are Riemann surfaces. $f:A→ B$ is morphism of Rieman surface, in other word, holomorhic function, and supposed that $f$ is bijective. Then, can we say that inverse of $f$ is also holomorphic?
Thank you in advance.
Asked
Active
Viewed 87 times
2
-
5Does this answer your question? The inverse of a bijective holomorphic function is also holomorphic. For the record: The same applies for bijective holomorphic maps between higher-dimensional complex manifolds, inverses of such maps are again holomorphic. See here. But this is a more difficult result. – Moishe Kohan May 27 '21 at 16:29
1 Answers
4
"Yes".
In suitable coordinates, a holomorphic mapping is locally represented by a convergent power series $z \mapsto w = a_{n}z^{n} + \cdots$ with $n \geq 1$ and $a_{n} \neq 0$. Such a mapping is locally bijective if and only if $n = 1$, if and only if it locally has a holomorphic inverse.
Andrew D. Hwang
- 78,195
-
Can you please write more than one term? I assume that higher terms follow? – Martin Brandenburg May 27 '21 at 14:44
-
@MartinBrandenburg I'm happy to edit if that will clarify, but yes, $a_nz^n$ denotes the lowest-order term of a convergent power series $\sum_{k=n}^{\infty} a_kz^k$ at the origin, and the qualitative local behavior of a holomorphic function is governed by the first non-constant term. That seemed the crucial item to include (and potentially more clear than giving the whole series). – Andrew D. Hwang May 28 '21 at 01:58