Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 9 / 3 / Solution

For a plane domain, the following are equivalent descriptions of a simply connected domain: every based loop contracts in the domain with its basepoint fixed; any two paths with the same endpoints are homotopic while those endpoints stay fixed; its complement in the Riemann sphere is connected; every holomorphic function has a single-valued primitive; and every nowhere-zero holomorphic function has a holomorphic logarithm. The complement must be taken in the extended plane, including infinity. An equivalent winding number formulation is that every closed curve has index zero about every point outside the domain.

Here is a direct equivalence proof for the first two topological formulations. If all paths with given endpoints are homotopic relative to those endpoints, apply this to a loop based at $a$ and the constant path at $a$, obtaining a contraction. Conversely, let ${\gamma }_0,{\gamma }_1$ join $a$ to $b$. The loop ${\gamma }_0\ast {\gamma }_1^{-1}$ contracts with basepoint fixed. Concatenating this contraction with ${\gamma }_1$ produces a homotopy from ${\gamma }_0\ast {\gamma }_1^{-1}\ast {\gamma }_1$ to ${\gamma }_1$, relative to $a,b$. The middle retracing excursion contracts by shortening its traversal in both directions, giving a homotopy from that first path to ${\gamma }_0$. Hence ${\gamma }_0$ and ${\gamma }_1$ are homotopic relative to their endpoints. Planar domains are path-connected, so no extra path-component condition is needed.

The analytic descriptions follow from path independence of the holomorphic integral. Local primitives exist on discs by integrating power series; a homotopy can be subdivided into pieces lying in such discs, and local path integrals cancel along its common edges. Thus integrals over homotopic paths agree, giving a global primitive. Applied to $h^{\prime }/h$ for a nowhere-zero function, a primitive $H$ makes $h{e}^{-H}$ constant and hence supplies a holomorphic logarithm. These observations are also the tools used in the mapping proof below.

The Riemann mapping theorem states that every nonempty simply connected proper domain $D⊊\mathbb{C}$ admits a conformal bijection onto the unit disc. Given $z_0\in D$, there is exactly one such map with $f(z_0)=0$ and $f^{\prime }(z_0)>0$. The exclusion of the whole plane is necessary: an entire map into the disc would be constant by Liouville's theorem.

An extremal proof has the following concrete steps. First construct an injective bounded holomorphic map. Choose $a\notin D$ and a holomorphic square root $h$ of $z-a$. It is injective, since equality of its values implies equality of their squares. Its image is disjoint from its negative: $h(z)=-h(w)$ would imply $z=w$ and then $h(z)=0$, impossible. If $b=h(z_0)$, choose $r>0$ with $B(b,r)\subset h(D)$; the disc $B(-b,r)$ is omitted. Therefore $1/(h+b)$ is injective and bounded. Scaling it and applying an automorphism of the unit disk gives a map into the disc taking $z_0$ to zero; a rotation makes its derivative positive.

Let $\mathcal{U}$ consist of all injective maps $D\to \mathbb{D}$ with this normalization. It is nonempty. The derivative at $z_0$ is bounded above by the Cauchy integral formula on a small disc compactly contained in $D$. Choose $f_n\in \mathcal{U}$ with derivatives tending to their positive supremum $L$. Boundedness, the Cauchy estimates and a diagonal Arzelà-Ascoli theorem argument give a subsequence converging locally uniformly to $f$; this is the bounded-family form of the Montel theorem. Then $f(z_0)=0$, $f^{\prime }(z_0)=L>0$, and the maximum modulus principle puts its image in the open disc. The limit is injective: if distinct $z,w$ had $f(z)=f(w)$, apply Hurwitz's theorem near $w$ to $f_n(\cdot )-f_n(z)$, which has no zero there; the limit would vanish identically, contradicting $f^{\prime }(z_0)>0$.

Finally suppose $a\in \mathbb{D}$ is omitted by $f$. It is nonzero since $f(z_0)=0$. Let $T_a(w)=(w-a)/(1-\overline{a}w)$ and choose a holomorphic square root $h$ of $T_a\circ f$. It is injective and takes values in the disc. Put $b=h(z_0)$, so $|b|=\sqrt{|a|}$, and normalize $T_b\circ h$ by a rotation. The new map's derivative magnitude at $z_0$ isThis square-root improvement of a normalized conformal map contradicts extremality. Thus $f$ is onto. Two normalized bijections differ by a disc automorphism fixing zero; Schwarz lemma applied to it and its inverse makes it a rotation, and positivity of the normalized derivatives forces that rotation to be the identity. This completes the requested proof outline.