I came across the following definition for functions on a Riemann surface:
A nonconstant analytic function on a Riemann surface, $f_{1}:X_{1} \rightarrow \mathbb{C}$ extends $f_{0}:X_{0} \rightarrow \mathbb{C}$ if there is an embedding $j:X_{0} \rightarrow X_{1}$ such that $f_{0}=f_{1} \circ j$.
I am curious: if we have two extensions of $(X_{0},f_{0})$, then is there a third pair $(X_{3},f_{3})$ that extends both? This seems reasonable to me, but how might we prove this? Also, what is the use of this notion of extending a function on a Riemann surface?
Consider a torus $T$ and a cover $\{X_1, X_2, X_3\}$ of $T$ such that each $X_i$ is homeomorphic to $S^1\times (0,1)$. You can choose the $X_i$ as tubes, such that also $X_i\cap X_j$ as well as $X_i\cup X_j$ is also homeomorphic to $S^1\times (0,1)$. Now if you analytically extend from $X_1$ to $ X_1\cup X_2$ and simultaneously to $ X_1\cup X_3 $ you need to check whether the extension agree on $X_3$, which they will, in general, not.
This kind of extension is useful to explore the topology of Riemann surfaces, cause it allows you to construct (universal) coverings in a canonical way.