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

1.1L1given

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.

2.1step 1.1L2∎

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.

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)