Proof. Let \((f_t, D_t)\) and \((g_t, E_t)\) both continue \((f_0, D_0)\) along \(\gamma \), and let \[S = \{ t \in [0,1] : f_t = g_t \text { on a neighbourhood of } \gamma (t)\}.\] Then \(0 \in S\), so \(S \neq \emptyset \). Condition (iii), applied to both families at a point \(t\), shows that membership of \(S\) propagates to all \(s\) near \(t\): so \(S\) is open in \([0,1]\). If \(t\) is a limit point of \(S\), then \(f_t\) and \(g_t\) are both analytic near \(\gamma (t)\) and agree on a set accumulating at \(\gamma (t)\), so by the uniqueness theorem above they agree near \(\gamma (t)\) and \(t \in S\): so \(S\) is closed. A non-empty subset of the connected set \([0,1]\) that is both open and closed is all of \([0,1]\). Hence \(1 \in S\). \(\blacksquare \)

So the outcome of continuing along a path depends on the path but not on the particular chain of discs used to walk along it. The next question is the one that matters: how much does it depend on the path? □