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The uniformization theorem states that every simply connected Riemann surface is biholomorphic to exactly one of the Riemann sphere $\hat{\mathbb{C}}$, the complex plane $\mathbb{C}$, or the unit disc $\mathbb{D}$. A surface is uniformized by one of these when that surface is its universal cover. The only surface uniformized by $\hat{\mathbb{C}}$ is the sphere itself, because every nonidentity Möbius transformation has a fixed point and therefore cannot act as a deck transformation. The Riemann surfaces uniformized by the complex plane are

Let $U\subseteq \mathbb{C}$ have a complement containing at least two points. Its universal cover cannot be the sphere because $U$ is noncompact. If it were the plane, the covering map $\mathbb{C}\to U\hookrightarrow \mathbb{C}$ would be a nonconstant entire function omitting two values, contrary to the Little Picard theorem. The plane domain with two omitted points is hyperbolic, so

Now put $S=R\setminus \{P_1,\dots ,P_n\}$. The spherical possibility occurs only for $(g,n)=(0,0)$. By the preceding list, the plane-uniformized possibilities arewhich give respectively $2g-2+n=-1,0,0$. Every remaining pair has $2g-2+n>0$ and must have disc universal cover. Thus the uniformization of a punctured compact Riemann surface gives

For the complex torus $X=\mathbb{C}/\Lambda$, any analytic map $f:\mathbb{C}\to X$ lifts through the universal covering map to an entire function $F:\mathbb{C}\to \mathbb{C}$. If $F$ is constant, so is $f$. Otherwise Little Picard says that $F$ omits at most one complex number. For every $x\in X$, its lifts form an infinite coset $z+\Lambda$, so at least one lift is attained by $F$. Hence the holomorphic map from the complex plane to a one-dimensional complex torus satisfies

Finally, $\mathbb{C}$ and $\mathbb{D}$ are homeomorphic byThey are not conformally equivalent: a biholomorphism $\mathbb{C}\to \mathbb{D}$ would be a nonconstant bounded entire function, contradicting the Liouville theorem. This proves the claimed example of complex plane and unit disc are homeomorphic but not conformally equivalent.

The uniformization theorem states that every simply connected Riemann surface is conformally equivalent to exactly one of the Riemann sphere ${\mathbb{C}}_{\infty }$, the complex plane $\mathbb{C}$, or the open unit disc $\mathbb{D}$. Equivalently, every connected Riemann surface is a quotient of one of these by a free properly discontinuous group of conformal automorphisms.

For ${\mathbb{C}}_{\infty }$, every nonidentity Möbius transformation has a fixed point. A free action is therefore trivial, so the only surface uniformized by the sphere is

Every conformal automorphism of $\mathbb{C}$ has the form $z\mapsto az+b$. If $a\ne 1$, it has a fixed point, so a group acting freely consists only of translations. A discrete subgroup of $(\mathbb{C},+)$ has rank zero, one, or two. The corresponding quotients arewhere $\Lambda$ is a rank-two lattice and $\mathbb{C}/\Lambda$ is a complex torus. This proves the completeness of the Riemann surfaces uniformized by the complex plane.

Now let $D\subseteq \mathbb{C}$ be a domain whose complement contains two distinct points. Its universal cover cannot be the sphere because no nonconstant holomorphic map from a compact Riemann surface has image in the noncompact proper domain $D$. If its universal cover were $\mathbb{C}$, the covering map followed by $D\hookrightarrow \mathbb{C}$ would be a nonconstant entire function omitting the two chosen complementary points, contradicting the Little Picard theorem. The uniformization theorem therefore leaves onlywhich is the plane domain with two omitted points is hyperbolic result.

Finally let $j:{\mathbb{C}}^{\ast }\hookrightarrow R$ be a holomorphic embedding with $R$ compact. At each puncture zero and infinity, an essential singularity would, by the Great Picard theorem in a coordinate chart, force values to be taken repeatedly and contradict injectivity. Thus $j$ extends holomorphically toA generic point of the open image $j({\mathbb{C}}^{\ast })$ has exactly one preimage, so $J$ has degree one. A degree-one holomorphic map between compact Riemann surfaces is a biholomorphism; equivalently, the Riemann-Hurwitz formula rules out branching and gives the same conclusion. Therefore the compact Riemann surface containing an embedded punctured plane is