TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-31

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The terminal germ of a continuation along a fixed path is chain-independent

Statement

Let $γ : [0, 1] \to Ω$ be a path and let $ξ_{0}$ be a holomorphic germ at $γ (0)$ . If two admissible continuation chains of $ξ_{0}$ along $γ$ exist, then they determine the same terminal germ at $γ (1)$ .

Facts & Assumptions

Given: A path $γ : [0, 1] \to Ω$ , an initial germ $ξ_{0}$ at $γ (0)$ , and two admissible continuation chains of $ξ_{0}$ along $γ$ .

[L1]

Two admissible continuation chains along the same path admit a common refinement whose subinterval images lie in one element of each chain (Two admissible continuation chains along one path admit a common refinement).

[L2]

A holomorphic germ at a point is equality on some neighbourhood of that point, and an admissible continuation chain requires successive representatives to agree as germs at the joining path points (Holomorphic germs at a point, Analytic continuation along a path by admissible chains).

[L3]

If two holomorphic functions on a complex domain agree on a set with an accumulation point in that domain, then they agree on the whole domain (Identity theorem for holomorphic functions).

Proof

technique · direct

1.1

By [L1], refine both chains so that they use the same subdivision $0 = u_{0} < \dots < u_{r} = 1$ , and on each interval $[u_{k}, u_{k + 1}]$ the path image lies in both a function element $(f_{k}, U_{k})$ from the first chain and a function element $(g_{k}, V_{k})$ from the second.

1.2

At the initial point $γ (u_{0}) = γ (0)$ the two first representatives have germ $ξ_{0}$ , so [L2] gives an open neighbourhood of $γ (0)$ on which $f_{0} = g_{0}$ .

givenL2

2.1

Assume inductively that $f_{k}$ and $g_{k}$ have the same germ at the left endpoint $γ (u_{k})$ . The path segment $γ ([u_{k}, u_{k + 1}])$ is connected, so it lies in one connected component $W_{k}$ of $U_{k} \cap V_{k}$ . By [L2] the functions $f_{k}$ and $g_{k}$ agree on a neighbourhood of $γ (u_{k})$ contained in $W_{k}$ , and [L3] therefore gives $f_{k} = g_{k}$ on all of $W_{k}$ . In particular they have the same germ at the right endpoint $γ (u_{k + 1})$ .

L2L3step 1.2cases

3.1

Applying step 2.1 successively for $k = 0, \dots, r - 1$ shows that the two refined chains have the same germ at every subdivision point, hence especially at $γ (1)$ .

step 2.1induction

4.1

The terminal germs of the original chains equal those of the refinements, so the terminal germ depends only on $ξ_{0}$ and $γ$ , not on the chosen admissible chain.

step 1.1step 3.1∎

Depends on

Used by

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Sources

Lars V. Ahlfors, Complex Analysis, 3rd ed., Ch. 8 §1.4 (standard reference, not scraped)
Henry Wilton, Riemann Surfaces lecture notes, §9.2 (standard reference, not scraped)