LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Disc functions omitting 0 and 1 admit holomorphic logarithms for f and 1-f

Statement

Let $f : D \to C$ be holomorphic and omit $0$ and $1$ . Then there exist holomorphic functions $F, G : D \to C$ such that

$e^{F} = f, e^{G} = 1 - f .$

Facts & Assumptions

Given: A holomorphic map $f : D \to C ∖ {0, 1}$ .

[L1]

The unit disc is homologically simply connected (Star-shaped plane domains are homologically simply connected).

[L2]

On a homologically simply connected complex domain, every holomorphic nowhere-zero function has a holomorphic logarithm (A nonvanishing holomorphic function on a homologically simply connected domain has a holomorphic logarithm).

Proof

technique · direct

1.1L1given

Both $f$ and $1 - f$ are holomorphic and nowhere zero on $D$ . Fact [L1] makes $D$ homologically simply connected.

2.1L2step 1.1∎

Applying [L2] first to $f$ and then to $1 - f$ gives holomorphic logarithms $F$ and $G$ with $e^{F} = f$ and $e^{G} = 1 - f$ .

Depends on

Used by

Schottky's theorem Theorem

Dependency tree · two levels

34 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

Aleksander Simonic, The Ahlfors lemma and Picard's theorems, §6.3 (standard reference, not scraped)