Let \(U\subset\widehat{\mathbb C}\) be a region such that \[ |\widehat{\mathbb C}\setminus U|\ge3. \] Prove that the universal cover of \(U\) is conformally equivalent to the unit disk \(\Delta\).
Solution
Let \[ \pi:(X,p)\longrightarrow(U,u_0) \] be the abstract universal cover. We construct a covering map \(X\to\Delta\) by an extremal argument and then prove that it is surjective.
1. There exists a nonconstant holomorphic map\[ F:(X,p)\longrightarrow(\Delta,0) \]which is a covering map onto its image.
Choose three distinct points \[ a,b,c\in\widehat{\mathbb C}\setminus U. \] Set \[ Y=\widehat{\mathbb C}\setminus\{a,b,c\}. \] The thrice-punctured sphere \(Y\) has the disk as universal cover; the modular function \(\lambda\), precomposed with a Möbius map \(\Delta\to\mathbb H\) and postcomposed with a Möbius map sending \(\{0,1,\infty\}\) to \(\{a,b,c\}\), gives a universal covering \[ q:\Delta\longrightarrow Y. \]
The inclusion \(j:U\hookrightarrow Y\) composed with \(\pi\) gives a map \[ j\circ\pi:X\to Y. \] Since \(X\) is simply connected, this map lifts through \(q\) to a holomorphic map \[ F:X\to\Delta \] with \(F(p)=0\) after choosing a suitable point of the fiber \(q^{-1}(u_0)\) and postcomposing by a disk automorphism if necessary.
More concretely, \(F(X)\) is the connected component of \(q^{-1}(U)\) containing \(F(p)\). The restriction of \(q\) to that component is a covering of \(U\), and the usual lifting property shows that \(F:X\to F(X)\) is its universal covering. Thus \(F\) is a covering map onto its image.
2. Let \(\mathcal F\) be the family of all holomorphic maps\[ F:(X,p)\to(\Delta,0) \]which are covering maps onto their images. Then \(\mathcal F\) is a normalfamily, and a nonconstant locally uniform limit of members of \(\mathcal F\)again belongs to \(\mathcal F\).
Normality is immediate locally: every member of \(\mathcal F\) takes values in the bounded domain \(\Delta\), so Montel’s theorem applies in every coordinate disk of \(X\); a diagonal argument gives subsequential convergence on compact subsets of \(X\).
For the closure statement, let \(F_n\in\mathcal F\) converge locally uniformly to a nonconstant \(F\). Since each covering map has no critical points, \(F_n'\) never vanishes in local coordinates. Hurwitz’s theorem applied to the derivatives shows that \(F'\) never vanishes either. Thus \(F\) is a local biholomorphism.
The covering property is stable under this normalized nonconstant limit: inverse germs of the \(F_n\) are univalent, and on every compact chain of coordinate disks they form normal families. Their limits give analytic continuation of every inverse germ of \(F\) along every path in \(F(X)\); Hurwitz prevents two inverse branches from colliding. The path-lifting criterion for coverings therefore shows that \[ F:X\to F(X) \] is a covering map. Hence \(F\in\mathcal F\).
3. Fix a local coordinate \(\zeta\) near \(p\) with \(\zeta(p)=0\). There isan extremal map \(F\in\mathcal F\) satisfying\[ |F'(p)|=\sup_{H\in\mathcal F}|H'(p)|>0, \]where derivatives are taken in the coordinate \(\zeta\).
The family is nonempty by 1, and every member is a local biholomorphism, so the supremum is positive.
Choose a small coordinate disk \(D\) about \(p\) whose closure is compact in \(X\). Since every \(H\in\mathcal F\) satisfies \(|H|<1\), Cauchy’s estimate on \(D\) gives a uniform upper bound on \(|H'(p)|\). Hence the supremum is finite.
Choose \(F_n\in\mathcal F\) with \(|F_n'(p)|\) tending to the supremum. By normality, pass to a locally uniformly convergent subsequence \(F_n\to F\). Derivatives converge locally as well, so \(|F'(p)|\) equals the positive supremum. In particular \(F\) is nonconstant, and by the closure statement in 2 we have \(F\in\mathcal F\).
4. The extremal map \(F\) from 3 is surjective onto \(\Delta\).
Suppose instead that some \(a\in\Delta\) is omitted by \(F\). Since \(F(p)=0\), we have \(a\neq0\).
Let \[ \phi_a(w)=\frac{w-a}{1-\overline a\,w}, \] so \(\phi_a\) is a disk automorphism sending \(a\) to \(0\). The map \[ F_1=\phi_a\circ F \] omits \(0\). Since \(X\) is simply connected, \(F_1\) has a holomorphic square root \[ h:X\to\Delta, \qquad h^2=F_1. \] Equivalently, \(h\) is the lift of \(F_1\) through the covering \(s(w)=w^2\) of \(\Delta^*\). Because \(F\) is a covering onto its image, the lifting property shows that \(h\) is also a covering onto its image.
Put \[ u=h(p). \] Then \[ |u|^2=|F_1(p)|=|\phi_a(0)|=|a|. \] Set \(r=|u|=\sqrt{|a|}\), so \(0<r<1\), and let \[ \psi_u(w)=\frac{w-u}{1-\overline u\,w}. \] The normalized map \[ H=\psi_u\circ h \] belongs to \(\mathcal F\) and satisfies \(H(p)=0\).
We now compare derivatives at \(p\). Since \[ |\phi_a'(0)|=1-|a|^2=1-r^4, \] the identity \(h^2=F_1\) gives \[ 2|u|\,|h'(p)|=(1-r^4)|F'(p)|. \] Also \[ |\psi_u'(u)|=\frac1{1-|u|^2}=\frac1{1-r^2}. \] Therefore \[ \begin{aligned} |H'(p)| &=\frac{|h'(p)|}{1-r^2}\\ &=\frac{1-r^4}{2r(1-r^2)}|F'(p)|\\ &=\frac{1+r^2}{2r}|F'(p)|. \end{aligned} \] Since \(0<r<1\), \[ 1+r^2>2r, \] so \[ |H'(p)|>|F'(p)|. \] This contradicts the extremality of \(F\). Hence \(F\) omits no point of \(\Delta\), and therefore \[ F(X)=\Delta. \]
5. The universal cover \(X\) is conformally equivalent to \(\Delta\).
By 2 and 4, \[ F:X\to\Delta \] is a covering map. The target \(\Delta\) is simply connected. A connected covering of a simply connected space has one sheet, so \(F\) is a homeomorphism. Since \(F\) is holomorphic and locally biholomorphic, its inverse is holomorphic. Thus \[ X\cong\Delta \] conformally, as required.