In my notes I have that $(g,E)$ is an analytic continuation of $(f,D)$ along a path $\gamma :[0,1]\to U$ if there exists function elements $(f_i,D_i)$ with $i\in \{ 0,\ldots ,n\}$ and $0=t_0<t_1<\ldots t_n=1$ which satisfy
- For $j\in \{ 0,1,\ldots ,n-1\}$ $\gamma([t_j,t_{j-1}])\subset D_j$
- $(f,D)= (f_0,D_0)\sim (f_1,D_1)\sim \ldots \sim (f_n,D_n)=(g,E)$
Where we write $(f,D)\sim (g,E)$ if the two function elements are a direct analytic continuation of eachother. I.e. $D\cap E\neq \emptyset$ and $f,g$ agree on $D\cap E$. For clarity a function element is a pair $(f,D)$ with $f$ an analytic function on a domain $D$.
My question is that I'm unsure why the path condition is needed. If we have a sequence of function elements, with one a direct continuation of the last. Since each domain in the sequence is connected and each has a non-empty intersection with the last surely that means such a path in the definition above must exist anyway?
Not really sure what I'm missing here so any help would be appreciated.