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A biholomorphism of the unit disk to itself. Every such map has the form $e^{i\theta }(z-a)/(1-\overline{a}z)$ with $|a|<1$. These maps normalize an extremal univalent function in the proof of the Riemann mapping theorem.

A locally uniform convergence limit of univalent functions on a domain is either constant or univalent. For a nonconstant limit $f$, suppose distinct $x,y$ satisfy $f(x)=f(y)$. Choose a small closed disk around $x$ excluding $y$ with $f-f(y)$ nonzero on its boundary. The Rouche theorem then forces $f_j-f_j(y)$ to have a zero in that disk for large $j$, contradicting injectivity of $f_j$. This supplies the injectivity step in an extremal proof of the Riemann mapping theorem without assuming it as a special lemma.

Use the simply connected domain shown in the diagram. A Riemann mapping theorem map followed by an automorphism of the unit disk sends its marked points to $0$ and $x\in (0,1)$. Split this disc along its real diameter. Map the upper conformal quadrilateral from c to a rectangle $[0,w]\times [0,h]$, with $[-1,0]$ and $[x,1]$ as vertical sides and $(0,x)$ as its bottom. The lower half is a second identical rectangle. The full disc is obtained by gluing the two rectangles along their two vertical sides and their bottom sides; the two top sides form the original outer boundary.

In this doubled rectangle extremal-length duality, loops enclosing both marked bottom corners have shortest flat length $2w$. They must meet both vertical seams, and a closed curve's total horizontal travel is at least $2w$; the doubled horizontal segments attain it. Crosscuts separating the two corners must meet the bottom seam, and run from outer boundary back to outer boundary. Their total vertical travel is at least $2h$, attained by doubled vertical segments. The total flat area is $2wh$.

These flat metrics are extremal, not merely lower bounds. For an arbitrary metric, average the lengths of the doubled horizontal segments over their height and apply the Cauchy-Schwarz inequality to get $L_{\rho }(\Gamma {)}^{2}h\le 2w{A}_{\rho }$. Averaging the doubled vertical segments gives $L_{\rho }({\Gamma }^{\ast }{)}^{2}w\le 2h{A}_{\rho }$. ConsequentlyThe conformal metric on the doubled rectangles pulls back to an admissible metric on the original punctured domain; the two marked corner points have zero area. Here ${\Gamma }^{\ast }$ consists of crosscuts with endpoints on the outer boundary, as drawn. Boundedness alone, without the illustrated domain topology and the specified curve families, would not justify this construction.

A simply connected plane domain carrying a hyperbolic metric is proper. By the Riemann mapping theorem, choose a biholomorphism $\varphi :\Omega \to \mathbb{D}$ sending a basepoint to zero. The same distance formula now givesThus the Blaschke condition holds. In particular the $a_n$ have no interior accumulation point: near any such point the quantities $1-|a_n|$ would stay bounded away from zero.

Construct the Blaschke productHere $m=1$ precisely when the sequence contains zero. For $|\zeta |\le r<1$, its normalized Blaschke factor $b_a$ satisfiesThe Blaschke condition therefore makes the infinite product converge uniformly on every compact subdisk. Away from the prescribed zeros its tail factors tend uniformly to one with summable errors, so their logarithms converge and the product is nonzero. It has exactly the prescribed zeros, and the maximum modulus principle bounds every finite partial product by one, hence $|B|\le 1$. Thus $f=B\circ \varphi$ is the required bounded analytic function, with no extra zeros.

Simple connectivity cannot be dropped. On the punctured disk chooseThe hyperbolic metric on the punctured disk is $|dz|/(|z|\log (1/|z|))$. Its universal covering map $\tau \mapsto e^{i\tau }$ pulls it back to the curvature-minus-one upper-half-plane metric. The imaginary axis is a geodesic; the lift from $i$ to $i{n}^2$ has length $\log n^2$, and any other lift differs by a real deck translation and has at least that distance. ThereforeThe triangle inequality again makes the sum finite for every basepoint. However, any bounded holomorphic function on the punctured disk extends over zero by the Riemann removable singularity theorem. These zeros accumulate at zero, so the identity theorem makes the extension identically zero. It cannot have exactly the prescribed zero set. This example of hyperbolic zero sequences on a punctured disk satisfies the distance condition but not the analytic zero-set condition.

We use the complete hyperbolic metric of curvature $-1$; on the unit disc its length element is $2|dz|/(1-|z{|}^2)$. This normalization matters in the exponential summability criterion. A nonzero bounded holomorphic function on the entire plane is constant by Liouville's theorem, so the prescribed infinite zero set forces $\Omega \ne \mathbb{C}$. The Riemann mapping theorem supplies a biholomorphism $\varphi :\Omega \to \mathbb{D}$, and transports the disk hyperbolic metric to $\Omega$.

Fix $w\in \Omega$ and choose $\varphi (w)=0$. Put $a_n=\varphi (z_n)$ and $F=f\circ {\varphi }^{-1}$. First suppose $F(0)\ne 0$. For radii avoiding its zeros, Jensen's formula givesThe monotone convergence theorem as $r\uparrow 1$ gives ${\sum }_n-\log |a_n|<\infty$, hence ${\sum }_n(1-|a_n|)<\infty$. If $F$ has a zero at zero, factor off its finite order first; the resulting holomorphic function is still bounded, since division by that power is bounded away from zero and extends near zero. Thus the Blaschke condition holds in either case, with the possible origin zero treated as one additional finite term.

The Hyperbolic distance in the Poincare disc satisfiesConsequently the exponential distances are summable. This holds for every basepoint: the triangle inequality bounds the ratio of $e^{-\rho (w,z_n)}$ and $e^{-\rho (w_0,z_n)}$ between $e^{-\rho (w,w_0)}$ and $e^{\rho (w,w_0)}$.

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.

Write $D=\mathbb{H}\setminus K$. By definition of a compact H-hull, $D$ is a proper simply connected domain, and it agrees with the complex upper half-plane outside a sufficiently large disc. The Riemann mapping theorem supplies a conformal isomorphism $f:D\to \mathbb{H}$. Choose its boundary normalization so that the prime end at infinity maps to infinity. This prime end is unambiguous because the domain has an ordinary straight real boundary near infinity.

The Schwarz reflection principle, applied there in the coordinate $1/z$, extends $f$ across the real boundary near infinity. It has a simple pole there, with Laurent seriesThe pole is simple because the reflected map is locally conformal at infinity; its leading coefficient is positive because it maps the upper side to the upper side. Postcomposing with the half-plane automorphism $w\mapsto (w-b)/a$ givesThis is hydrodynamic normalization at infinity.

If $g_1,g_2$ both satisfy it, $g_2\circ g_1^{-1}$ is a conformal automorphism of the upper half-plane. The asymptotics force it to fix infinity, so it has the form $w\mapsto \alpha w+\beta$ with $\alpha >0$ and $\beta \in \mathbb{R}$. The same asymptotics force $\alpha =1$, $\beta =0$. Hence the normalized mapping-out function exists and is unique. The argument requires reflection only outside a large disc, not regularity of the hull's entire boundary.

A compact H-hull is a bounded relatively closed set $K\subset \mathbb{H}$ whose complement $D=\mathbb{H}\setminus K$ is a simply connected domain. Equivalently one can retain its compact closure in $\overline{\mathbb{H}}$; real boundary points do not affect the mapping-out function. The Riemann mapping theorem and Schwarz reflection principle near infinity give a unique conformal map $g_K:D\to \mathbb{H}$ with hydrodynamic normalization at infinity. Its Laurent series isThe half-plane capacity is $\mathrm{hcap}(K)=a_K$. The missing constant term is part of the normalization, not a choice to be made again when comparing different hulls.

Here are the conformal map facts we use to justify positivity and comparison. The inverse $f_K:\mathbb{H}\to D\subset \mathbb{H}$ has expansion $f_K(w)=w-a_K/w+O(w^{-2})$. The upper-half-plane Schwarz-Pick theorem, applied to $w$ and $iY$ and then letting $Y\to \infty$, implies $\mathrm{Im}{f}_K(w)\ge \mathrm{Im}w$. Indeed the inequality between pseudohyperbolic distances givesand multiplying by $Y$ gives the height comparison in the limit. At $w=iY$, the inverse expansion reads $\mathrm{Im}{f}_K(iY)=Y+a_K/Y+O(Y^{-2})$, so $a_K\ge 0$.

For $r>0$, uniqueness of the mapping-out function givesConsequentlyTranslation by a real number $b$ similarly gives $g_{K+b}(z)=b+g_K(z-b)$ and leaves half-plane capacity unchanged.

If $K\subset A$ are compact H-hulls, their mapping-out functions compose as $g_A=g_L\circ g_K$, where $L$ is the filled image of $A\setminus K$. Comparing Laurent series yieldsThis proves monotonicity of half-plane capacity, rather than assuming it from inclusion alone. The filled half-disc $A=\{z\in \mathbb{H}:|z-b|\le R\}$ has mapping-out functionwhich takes its semicircular boundary to the real line and is a conformal bijection onto $\mathbb{H}$. Thus its half-plane capacity is $R^2$. Enclose $K$ in such a half-disc, compare capacities, and minimize over real centres and radii. This proves the sharp boundIf one defines the radius by an infimum, take $R$ strictly larger than that infimum and then let $R$ decrease; no attainment assumption is needed.

The Riemann mapping theorem asserts that a nonempty simply connected domain $\Omega ⊊\mathbb{C}$ is biholomorphic to the unit disk $\mathbb{D}$. More precisely, for each $z_0\in \Omega$ there is a unique biholomorphism $f:\Omega \to \mathbb{D}$ satisfying $f(z_0)=0$ and $f^{\prime }(z_0)>0$. We prove existence through an extremal derivative, deriving the compactness and injectivity steps rather than assuming special lemmas for the theorem.

First construct a bounded univalent function on $\Omega$. Choose $a\notin \Omega$ and a holomorphic logarithm of $z-a$, which exists by the permitted logarithm result on the translated simply connected domain. PutThen $u(z{)}^2=z-a$, so $u$ is univalent and never zero. Its image $G=u(\Omega )$ is open, and $G\cap (-G)=\varnothing$: equality $u(x)=-u(y)$ would imply $x=y$ after squaring and then $u(x)=0$.

Fix $w_0\in G$ and choose $r>0$ with $B(w_0,r)\subset G$. It follows that $B(-w_0,r)\subset -G$ is disjoint from $G$, so $|u(z)+w_0|\ge r$ throughout $\Omega$. Henceis a univalent function with $|h|\le 1/2$. For $b\in \mathbb{D}$, defineThe identityand the inverse $T_b^{-1}(w)=(w+b)/(1+\overline{b}w)$ show directly that $T_b$ is an automorphism of the unit disk. Apply $T_{h(z_0)}$ to $h$ and then a rotation to make its nonzero derivative at $z_0$ positive. The family $\mathcal{A}$ of univalent maps $\Omega \to \mathbb{D}$ with value zero and positive derivative at $z_0$ is therefore nonempty.

LetIt is positive. Choose $\rho >0$ with $\overline{B(z_0,\rho )}\subset \Omega$. The Cauchy estimate gives $v^{\prime }(z_0)\le 1/\rho$ for every $v\in \mathcal{A}$, so $M$ is finite. Take a sequence $v_j\in \mathcal{A}$ with $v_j^{\prime }(z_0)\to M$.

We now justify a locally uniform convergence subsequence. On any compact subset of $\Omega$, the bound $|v_j|\le 1$ and the Cauchy estimate on slightly larger interior disks give a uniform derivative bound. They therefore give equicontinuity. Apply the Arzelà-Ascoli theorem on a compact exhaustion of $\Omega$, then take a diagonal subsequence for locally uniform convergence. Its limit $v$ is holomorphic: pass to the limit in the Cauchy integral formula on every small interior circle. The same formula for derivatives givesThus $v$ is nonconstant. Initially $|v|\le 1$, and the maximum modulus principle improves this to $|v|<1$ everywhere.

For completeness, the needed locally uniform limit of univalent functions argument is short. If $x\ne y$ and $v(x)=v(y)$, the nonconstant holomorphic function $v-v(y)$ has an isolated zero at $x$. Choose a small closed disk around $x$ excluding $y$, with no zero on its boundary. Uniform convergence there gives $v_j-v_j(y)\to v-v(y)$. The Rouche theorem then makes $v_j-v_j(y)$ have a zero inside that disk for all sufficiently large $j$. Its only possible zero, by injectivity of $v_j$, is $y$, outside the disk. This contradiction proves injectivity of $v$. Therefore $v\in \mathcal{A}$ and attains the supremum $M$.

It remains to prove surjectivity. Suppose $c\in \mathbb{D}$ is omitted by $v$; since $v(z_0)=0$, necessarily $c\ne 0$. The domain $T_c(v(\Omega ))$ is simply connected, because both constituent maps are homeomorphisms onto their images, and it excludes zero. The permitted holomorphic logarithm gives a holomorphic square rootThis maps $\Omega$ into $\mathbb{D}$ and is univalent, since equality of its values implies equality after squaring and hence equality of the original arguments. Write $\gamma =g(z_0)$. Then ${\gamma }^2=T_c(0)=-c$ and $|\gamma |=\sqrt{|c|}$. Normalize again with $V=T_{\gamma }\circ g$ and a rotation. Its derivative magnitude isThe strict inequality follows from $(1-\sqrt{|c|}{)}^2>0$. This square-root improvement of a normalized conformal map contradicts extremality. Thus $v(\Omega )=\mathbb{D}$. An injective holomorphic map has nonzero derivative and a holomorphic local inverse; these inverses agree, so the resulting bijection is biholomorphic. We have proved existence.

For uniqueness, two normalized maps give an automorphism of the unit disk $v_2\circ v_1^{-1}$ fixing zero. Apply the Schwarz lemma, proved in the next part, to this automorphism and its inverse. Both inequalities force equality, so the equality case makes it a rotation. Its derivative at zero is the positive real number $v_2^{\prime }(z_0)/v_1^{\prime }(z_0)$, forcing that rotation to be the identity. ThereforeThe proper-domain hypothesis is necessary: a bounded entire holomorphic function has derivative at most $1/R$ on disks of arbitrarily large radius by the Cauchy estimate, and hence is constant. Conversely, a domain biholomorphic to the disk is simply connected because the map is a homeomorphism.

Write $D=\mathbb{H}\setminus K$. The mapping-out function is a conformal map $g_K:D\to \mathbb{H}$ with hydrodynamic normalization at infinityEquivalently, $g_K(z)-z\to 0$. The Riemann mapping theorem gives a map onto the half-plane. Since $K$ is bounded, infinity has an analytic real-boundary neighborhood. After sending its image to infinity, the Schwarz reflection principle gives a Laurent series $az+b+O(z^{-1})$ there, with $a>0$ and $b$ real. A real affine automorphism removes $a$ and $b$, giving the stated normalization. Reflection also makes $a_K$ real.

If $g_1,g_2$ both have this normalization, $g_2{g}_1^{-1}$ is a conformal automorphism of the upper half-plane. Such maps are real Möbius transformations. The expansion forces it to fix infinity, have leading coefficient $1$, and have zero translation. It is therefore the identity. Thus the normalized mapping-out function exists and is unique.

Fix $z_0\in \mathcal{U}$. We prove the Riemann mapping theorem in this case by maximizing a normalized derivative. Choose $a\notin \mathcal{U}$. Since $\mathcal{U}$ is a simply connected domain and $z-a$ never vanishes, it has a holomorphic square root $g$. This $g$ is injective, and $g(\mathcal{U})$ is disjoint from $-g(\mathcal{U})$: equality up to sign would first force the original points to coincide. Choose $b\in g(\mathcal{U})$ and $r>0$ with $B(b,r)\subset g(\mathcal{U})$. The ball $B(-b,r)$ is omitted, so $1/(g+b)$ is a bounded univalent function. Scaling and composing with an automorphism of the unit disk produces an injective holomorphic function $\varphi :\mathcal{U}\to \mathbb{D}$ with $\varphi (z_0)=0$ and ${\varphi }^{\prime }(z_0)>0$.

Let $\mathcal{F}$ be the family of all such normalized univalent functions. A Cauchy estimate in a small disk about $z_0$ bounds their derivatives, so $M={\sup }_{\varphi \in \mathcal{F}}{\varphi }^{\prime }(z_0)$ is finite and positive. Choose a maximizing sequence in this normal family. Montel theorem gives a subsequence converging uniformly on compact subsets to $\varphi$. Its derivative at $z_0$ is $M>0$. The maximum modulus principle puts its image in $\mathbb{D}$, and Hurwitz's theorem implies that a nonconstant limit of injective holomorphic functions is injective. Thus the maximum is attained.

If $a_1\in \mathbb{D}$ is omitted, then $a_1\ne 0$. Put $T_a(w)=(w-a)/(1-\overline{a}w)$. A holomorphic square root $v$ of $T_{a_1}\circ \varphi$ exists, is injective, and takes values in $\mathbb{D}$. Compose $v$ with $T_{v(z_0)}$ and a rotation to normalize it. Writing $s=|a_1|\in (0,1)$, the new derivative has magnitudea contradiction. Therefore $\varphi$ maps onto $\mathbb{D}$. The Cayley transform between the half-plane and disk now givesIts inverse is holomorphic by the holomorphic inverse function theorem, so this is a biholomorphism. The proper-subset hypothesis was used to choose $a\notin \mathcal{U}$; the whole complex plane cannot be mapped this way, by Liouville theorem.

Write $\mathbb{H}=\{z:\mathrm{Im}z>0\}$. A compact H-hull is a bounded, relatively closed set $A\subseteq \mathbb{H}$ for which $D_A=\mathbb{H}\setminus A$ is a simply connected domain. “Compact” refers to its compact closure in $\overline{\mathbb{H}}$; boundary points on $\mathbb{R}$ can be adjoined without changing the remaining domain.

The Riemann mapping theorem and normalization at infinity give a unique mapping-out function of a compact H-hull $g_A:D_A\to \mathbb{H}$ with hydrodynamic normalization at infinity. By the Schwarz reflection principle outside a large disc, it has the Laurent seriesThe half-plane capacity is the coefficientThe empty compact H-hull has $g_A(z)=z$ and half-plane capacity zero. Nonnegativity follows from the Brownian representation of half-plane capacity proved next.

A compact H-hull is a bounded subset $A\subset \mathbb{H}$, closed relative to the complex upper half-plane, such that $\mathbb{H}\setminus A$ is a simply connected domain. The word compact refers to its bounded closure in $\overline{\mathbb{H}}$; a nonempty hull need not be a compact subset of the open half-plane.

The Riemann mapping theorem and normalization at infinity give a unique mapping-out function of a compact H-hull $g_A:\mathbb{H}\setminus A\to \mathbb{H}$ with $g_A(z)-z\to 0$ as $z\to \infty$. This hydrodynamic normalization at infinity yields a Laurent seriesThe coefficient is the half-plane capacity:

The uniformization theorem says that every simply connected Riemann surface is conformally equivalent to exactly one of the Riemann sphere, the complex plane, and the unit disc. A proper simply connected plane domain is noncompact, so it is not the sphere. It cannot be uniformized by the plane: after choosing a point outside the domain, a branch of a square root and then a Möbius transformation would produce a nonconstant bounded entire function, contradicting the Liouville theorem. It is therefore conformally equivalent to the disc, which is the Riemann mapping theorem.

A compact H-hull is a bounded relatively closed set $A\subset \mathbb{H}$ for which $\mathbb{H}\setminus A$ is a simply connected domain. The Riemann mapping theorem and hydrodynamic normalization at infinity give a unique mapping-out function of a compact H-hull $g_A:\mathbb{H}\setminus A\to \mathbb{H}$ withThe coefficient $a\ge 0$ is the half-plane capacity $\mathrm{hcap}(A)$.

For $\lambda >0$ and $x\in \mathbb{R}$, uniqueness of the normalized map givesSubstituting the expansion of $g_A$ yieldsTherefore the scaling and translation of half-plane capacity is

Use the univalent functions on $D$, namely the univalent functions $F:D\to \mathbb{C}$. Each has nonzero derivative, so its image $F(D)$ is a domain conformally equivalent to $D$, and is therefore simply connected. This image is proper: otherwise $F^{-1}:\mathbb{C}\to D$ would be a bounded nonconstant entire function, contrary to the Liouville theorem.

Conversely, every proper simply connected domain $G\subset \mathbb{C}$ is $F(D)$ for some such $F$, by taking the inverse of a Riemann mapping theorem bijection. Thus these functions parameterize all proper simply connected domains by their images. The complex plane itself is the exceptional simply connected domain and must be listed separately.

The parameterization has redundancy: two univalent functions $F_1,F_2$ have the same image if and only if $F_2=F_1\circ \varphi$ for an automorphism of the unit disk $\varphi$. All such automorphisms have the formIf an interior point $w_0\in G$ is prescribed, the normalization $F(0)=w_0$ and $F^{\prime }(0)>0$ real removes this redundancy. Equivalently one can describe these maps through their convergent Taylor series on $D$, imposing univalence on the resulting holomorphic functions.

The Riemann mapping theorem gives a conformal bijection $G\to D$ exactly whengiven the hypothesis that $G$ is a simply connected domain. A domain is nonempty, open, and connected, so no extra nonemptiness qualification is needed.

For necessity, a conformal bijection $\mathbb{C}\to D$ would be a bounded nonconstant entire function, contradicting the Liouville theorem. For sufficiency, $G$ is a proper simply connected domain, exactly the hypothesis of the Riemann mapping theorem.