Why do the holomorphic functions form a sheaf. Different functions on disjoint open sets can't be glued.

Question

I'm only just learning about sheafs so this is probably silly, but I am trying to work out why the presheaf of holomorphic functions on the complex numbers forms a sheaf.

My problem is that if you consider two disjoint open disks in the complex plane, and pick a random holomorphic function on each ball, their overlap is empty, so the restrictions agree trivially. But there is no guarantee that we can find a single holomorphic function which restricts to the given functions on the two disjoint disks, so we can't glue, and therefore this isn't a sheaf.

What am I missing?

Answer 1

The function that is equal to the first function on the first disk and to the second function on the second disk is indeed holomorphic if the disks are open and disjoint. Recall that being holomorphic means that the function is complex differentiable. This differentiability needs only to be checked in the neighborhood of each point. This is a local condition.

Answer 2

Your “problem” is not a problem at all. If we have homomorphism functions $f, g$ defined on $U, V$ respectively and $U, V$ are disjoint, we can define a holomorphic function $h$ on $U \cup V$ by

$$h(x) = \begin{cases} f(x) & x \in U \\ g(x) & x \in V \end{cases}$$

and $h$ is holomorphic on its domain.