TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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Analytic continuation along a fixed path is unique whenever it exists

Statement

Let $γ : [0, 1] \to Ω$ be a path and let $ξ_{0}$ be a holomorphic germ at $γ (0)$ . If $ξ_{0}$ admits analytic continuation along $γ$ , then the terminal germ at $γ (1)$ is unique.

Facts & Assumptions

Given: A path $γ : [0, 1] \to Ω$ and a holomorphic germ $ξ_{0}$ at $γ (0)$ .

[L1]

Admitting analytic continuation along $γ$ means admitting at least one admissible continuation chain along $γ$ (Analytic continuation along a path by admissible chains).

[L2]

Any two admissible continuation chains along the same path have the same terminal germ (The terminal germ of a continuation along a fixed path is chain-independent).

Proof

technique · direct

1.1

By [L1], any analytic continuation of $ξ_{0}$ along $γ$ is represented by an admissible continuation chain whose successive representatives agree at the joining points of the subdivision.

L1given

2.1

If two such continuations existed with different terminal germs, their underlying admissible chains would contradict [L2]. Therefore the terminal germ is unique whenever continuation along $γ$ exists.

step 1.1L2∎

Depends on

Used by

Fixed-endpoint homotopic paths give the same analytic continuation Theorem

Dependency tree · two levels

7 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

Lars V. Ahlfors, Complex Analysis, 3rd ed., Ch. 8 §1.4 (standard reference, not scraped)