The uniformization theorem states that every simply connected Riemann surface is biholomorphic to exactly one of the Riemann sphere , the complex plane , or the unit disc . A surface is uniformized by one of these when that surface is its universal cover. The only surface uniformized by 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 have a complement containing at least two points. Its universal cover cannot be the sphere because is noncompact. If it were the plane, the covering map 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 . The spherical possibility occurs only for . By the preceding list, the plane-uniformized possibilities are
which give respectively . Every remaining pair has and must have disc universal cover. Thus the uniformization of a punctured compact Riemann surface gives
For the complex torus , any analytic map lifts through the universal covering map to an entire function . If is constant, so is . Otherwise Little Picard says that omits at most one complex number. For every , its lifts form an infinite coset , so at least one lift is attained by . Hence the holomorphic map from the complex plane to a one-dimensional complex torus satisfies
Finally, and are homeomorphic by
They are not conformally equivalent: a biholomorphism 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.