TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30

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Every proper homologically simply connected plane domain is conformally equivalent to the unit disc

Statement

Assume the Axiom of Choice. Let $Ω ⊊ C$ be a homologically simply connected complex domain and let $z_{0} \in Ω$ . Then there is a biholomorphic map $f : Ω \to D$ such that

$f (z_{0}) = 0, f^{'} (z_{0}) > 0.$

Facts & Assumptions

Given: The Axiom of Choice, a proper homologically simply connected complex domain $Ω ⊊ C$ , and a point $z_{0} \in Ω$ .

[A1]

The Axiom of Choice is used by the extremal-attainment lemma (The Axiom of Choice).

[L1]

Under the Axiom of Choice, the extremal family is nonempty and the supremal derivative is attained by a holomorphic map $f : Ω \to D$ (A proper homologically simply connected plane domain has a bounded univalent competitor, A maximizing sequence has a locally uniform limit with extremal derivative).

[L2]

That extremal map is univalent and surjective onto $D$ (The extremal limit is univalent, An extremizer onto a proper subdomain of the disc can be enlarged).

[L3]

An injective holomorphic map on a complex domain is biholomorphic onto its open image (An injective holomorphic map has no critical point and is biholomorphic onto its image).

Proof

technique · direct

1.1

By [A1] and [L1], choose a holomorphic map $f : Ω \to D$ with $f (z_{0}) = 0$ and extremal derivative $f^{'} (z_{0}) > 0$ .

A1L1givenchoose

2.1

Fact [L2] makes this map injective and gives $f (Ω) = D$ . Since $Ω$ is a complex domain by the given data, [L3] makes $f$ biholomorphic onto its image, which is exactly $D$ .

L2L3step 1.1

3.1

The map of step 2.1 has the required normalization, so it is the desired conformal equivalence.

step 1.1step 2.1∎

Depends on

Used by

The normalized Riemann map is unique Corollary
A normalized Riemann map for a horizontal strip Example
A normalized Riemann map for a sector with an explicit branch choice Example
A normalized Riemann map for the slit plane Example
The normalized Riemann map from the upper half-plane sending i to 0 Example
FALSE: the Riemann map is unique without normalization False statement

Dependency tree · two levels

25 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Matthias Weber, Complex Analysis, Theorem 5.2.6 (standard reference, not scraped)
Walter Rudin, Real and Complex Analysis, Theorem 14.9 (standard reference, not scraped)