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On a simply connected domain, pathwise continuation glues to one holomorphic function

Statement

Let ΩC be simply connected, let a0Ω, and let ξ0 be a holomorphic germ at a0 that admits analytic continuation along every path in Ω starting at a0. Then there is a holomorphic function F:ΩC such that for every path γ in Ω starting at a0, the terminal germ of the continuation of ξ0 along γ is exactly the germ of F at γ(1).

Facts & Assumptions

Given: A simply connected complex domain Ω, a base point a0Ω, and a germ ξ0 at a0 that admits continuation along every path from a0.

[L1]

Fixed-endpoint path-homotopic paths give the same terminal germ (Fixed-endpoint homotopic paths give the same analytic continuation).

[L2]

A simply connected space is nonempty, path-connected, and has trivial fundamental group at every basepoint (Simply connected topological spaces).

[L3]

A based loop class is the class of a loop modulo endpoint-fixed path homotopy, and the constant loop is the identity element (Based loops and the fundamental group, Loop classes form the group π1(X,x0) under concatenation).

Proof

technique · direct
1.1

Fix zΩ. Because Ω is path-connected by [L2], there is at least one path γz from a0 to z. Let Tγ denote the terminal germ obtained by continuing ξ0 along a path γ from a0.

L2choose
1.2

If γ and β are two paths from a0 to z, then λ:=γˉβ is a based loop at z. By [L2] and [L3], its loop class is the identity, so there is an endpoint-fixed path homotopy K:[0,1]×[0,1]Ω from λ to the constant loop cz.

L2L3
2.1

Define e:[0,1][0,1]2 by e(t)=(0,3t) for 0t13, e(t)=(3t1,1) for 13t23, and e(t)=(1,33t) for 23t1. For (s,t)[0,1]2, put qt(s):=(1s)(12,0)+se(t) and H(s,t):=K(qt(s)). Then H is continuous, H(0,t)=K(12,0)=λ(12)=a0, and H(1,t)=K(e(t))=z because e(t) lies on the three edges where K is constantly z. Also q0(s)=((1s)/2,0) and q1(s)=((1+s)/2,0), so H(s,0)=λ((1s)/2)=γ(s) and H(s,1)=λ((1+s)/2)=β(s). Thus H is a path homotopy from γ to β relative to the endpoints.

step 1.2algebra
3.1

Fact [L1] applied to the path homotopy of step 2.1 gives Tγ=Tβ. Therefore the value of the terminal germ over z is independent of the chosen path from a0 to z.

L1step 2.1
4.1

Define F(z) to be the value at z of this common germ. This is well defined by step 3.1.

step 3.1
5.1

Let zΩ and choose a path γ from a0 to z. Let (f,U) represent the terminal germ Tγ at z. For every wU, the same function element (f,U) continues that germ from z to w, so step 3.1 forces the terminal germ over w to be [f]w. Hence FU=f, and F is holomorphic on U. Since z was arbitrary, F is holomorphic on all of Ω, and its germ at each point is the continued germ.

step 3.1step 4.1given

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