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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

Write a point of the band $X$ as $(u,t)$ with $u\in S^2$, $|t|\le 1/2$, after rescaling the radius of the first three coordinates. The antipodal identification isThe homotopy $(u,t)\mapsto (u,(1-s)t)$ is equivariant under this identification, so it descends to a deformation retraction of $Y$ onto the central sliceThusThe two boundary spheres of $X$ are exchanged by the antipodal map, so $\partial Y\cong S^2$. This is the mapping-cylinder model of punctured real projective three-space.

The standard cellular chain complex for ${\mathbb{RP}}^3$, with one cell in each dimension zero through three, has boundary maps alternating between multiplication by two and zero:Therefore the integral homology of real projective three-space isIn particular,

Apply the Mayer–Vietoris theorem toThe relevant pieces of the long exact sequence areandThe last map is injective, so

Since $\partial Y\cong S^2$ is simply connected and ${\pi }_1(Y)\cong {\pi }_1({\mathbb{RP}}^2)\cong \mathbb{Z}/2$, the Seifert-van Kampen theorem givesthe infinite dihedral group. Topologically, $Z$ is the double of punctured real projective three-space, namely ${\mathbb{RP}}^3\#{\mathbb{RP}}^3$.

The universal cover of each copy of $Y$ is $S^3$ with two disjoint open balls removed, homeomorphic to $S^2\times [0,1]$. The universal cover of the double strings infinitely many such cylinders together according to the Cayley line of $D_{\infty }$. Hence the familiar covering space is