# Unique Complex Structure Making Local Homeomorphism Holomorphic

I'm trying to prove if $f:X\rightarrow Y$ is a local homeomorphism with $Y$ a Riemann surface then there is a unique complex structure on $X$ such that $f$ is holomorphic.

I've proved the existence part and for uniqueness I suppose that $\mathcal{A}=\{(U_x, \phi_x)\}$, $\mathcal{C}=\{(W_a,\chi_a)\}$ are complex structures on $X$, with $\mathcal{A}$ being the one I proved to exist, which comes straight out of the definition of local homeomorphism. Moreover I write $\mathcal{B}=\{(V_{\alpha},\psi_{\alpha})\}$ for the complex structure on $Y$.

I need to prove that the transition functions from $\mathcal{A}$ to $\mathcal{C}$ are homeomorphic. For $x\in W_a$ I consider $U_x\cap W_a$. It clearly suffices to prove that $\chi_a\phi_x^{-1}$ and $\phi_x\chi_a^{-1}$ are holomorphic on the appropriate images of $U_x\cap W_a$.

Now $f|_{U_x\cap W_a}$ is a homemorphism, so in particular I know that $\psi_{\alpha}f\chi_a^{-1}$ and $\psi_{\alpha}f\phi_x^{-1}$ are holomorphic and invertible. I now need only to prove that the inverses of these are themselves holomorphic and I'm done.

I can't see how to do this however! Has anyone got any good ideas? Many thanks!

-
I found a clear proof in Forster book. –  angy Jun 19 '12 at 3:40

Since holomorphicity is a local property, it suffices to work with domains in $\mathbb C$. So, we want to prove that if $f\colon D\to D'$ is a holomorphic bijection, then $f^{-1}$ is also holomorphic.

Step 1: $g:=f^{-1}$ is $C^1$. This follows from the Inverse Function Theorem (from real analysis), by writing $z=x+iy$ etc.

Step 2: $g_{\bar z}=0$. Inverse Function Theorem shows this as well: the derivative matrix of $g$ is of the form $\begin{pmatrix}a & b \\ -b & a\end{pmatrix}$. Alternatively, use the Chain Rule: $0=(f\circ g)_{\bar z}=(f'\circ g)\,g_{\bar z}$, hence $g_{\bar z}=0$.

-
Ah - so any holomorphic homeomorphism is a biholomorphism then? –  Edward Hughes May 29 '12 at 20:16
@hgbreton Yes, that is correct. Even in more than one variable. Even if Wikipedis says [citation needed]. :) –  user31373 May 29 '12 at 20:20
That's marvellous. Thanks very much! –  Edward Hughes May 29 '12 at 20:25
Waiittt...I thought the only hard part is showing that $f'$ is nonvanishing (well, jacobian is nonsingular), to invoke inverse function theorem. –  uncookedfalcon May 30 '12 at 2:00
That is, that's the part that involves complex analysis (for example, $f(x) = x^3$ is injective but $f'$ vanishes at 0, for $f: \mathbb{R} \rightarrow \mathbb{R}$, so it's not true in real analysis that $f$ injective implies $Df$ nonsingular). I've posted an answer which I think addresses this, lemme know what you think. –  uncookedfalcon May 30 '12 at 2:24

1) If Y is a holomorphic manifold of dimension $n$ and if $f:X\to Y$ is a local homeomorphism from the topological space $X$ to $Y$, then there is a canonical structure of holomorphic manifold of dimension $n$ on $X$ making $f$ a holomorphic local isomorphism .

The most efficient way to see this is to forget about charts and atlases: just say that you endow the topological space $X$ with the sheaf of local rings $\mathcal O_X=f^{-1}(\mathcal O_Y)$ .

This presupposes that you are comfortable with the definition of holomorphic manifolds as locally ringed spaces: it is a nice and powerful point of view, unifying geometric categories like differential manifolds, complex analytic spaces, schemes, ....
If you don't yet know this technique , learning it would not be a bad investment.

2) Concerning the comments to Leonid's answer, let me state the theorem:
Any injective holomorphic map $g:U\to V$ between holomorphic manifolds of the same dimension is open and the corestricted morphism $g:U\stackrel {\cong}{\to} g(U)$ is a holomorphic isomorphism.

You will find the proof of this impressively strong result in the brothers Kaup's Holomorphic Functions of Several Variables , Proposition 46.A.1.

-

You need that $f$ holomorphic, injective implies $f: \Omega \rightarrow \mathbb{C}$ biholomorphic (working locally, this really just a question of complex analysis). Well, we have a set theoretic inverse $g: f(\Omega) \rightarrow \Omega$. That it's continuous is the open mapping theorem: $f$ open iff $g$ cts. To see it's holomorphic at $z_0$, we compute for $z_i \rightarrow z$: $$\lim_i \frac{g(z_0) - g(z_i)}{z_0 - z_i} = \frac{1}{\frac{fg(z_0) - fg(z_i)}{g(z_0) - g(z_i)}}$$By continuity, $g(z_i) \rightarrow g(z_0)$, so this tends to $\frac{1}{f'(g(z_0))}$, so for this to converge is that same as $f'(g(z_0)) \neq 0$.

So we need that $f$ holomorphic, injective implies $f'$ is nonvanishing. Suppose not, if $f'$ vanishes at some $z'$, then by Rouche's theorem $f$ is locally $\geqslant 2:1$ with multiplicity. To see that it can't be the case that $f$ is set theoretically $1:1$ but only $\geqslant 2:1$ with multiplicity on a whole open set, note that $f$ is $\geqslant 2:1$ at a point with multiplicity iff $f'$ vanishes there (think about taylor expansion/polynomials), and zeroes of holomorphic functions are isolated.

-