In one of the exercise sheets for my complex analysis course we are given the following task
Prove that the set of all meromorphic functions on $\mathbb{C}$ defines a (pre)sheaf.
showing that they form a presheaf is fairly trivial. In order to check the second sheaf Axiom (the gluing Axiom - at least Wikipedia calls is so) the solution suggests that one has to use Mittag-Leffler and the identity theorem.
However I think I can provide an "Elementary" proof: Let $\mathcal{M}$ denote the Sheaf of meromorphic functions, $U \subset \mathbb{C}$ be open and $\{U_i\}_{i \in I}$ be an open cover of $U$.
I have to show that if a familiy $f_i \in \mathcal{M}(U_i)$ satisfies $f_i|_{U_i \cap U_j} = f_j|_{U_i \cap U_j}$ for all $i, j \in I$ then there exists an $f \in \mathcal{M}(U)$ such that $\forall i \in I: f|_{U_i} = f_i$.
Now as fas as I can see I can construct a function $f: U \to \mathbb{C} \cup \{\infty\}$ which maps any $x \in U$ to an $f_i(x)$ where $i$ is chosen such that $x \in U_i$. This function satisfies the aforementioned property so we only have to check that it is in fact meromorphic. For every $x$ there is an $U_i$ which contains $x$ and therefore also a neighborhood $V$ of $x$. Since $f_i|_V$ is meromorphic it follows, that $f|_V = f_i|_V$ is meromorphic. Since meromorphy is a local property $f$ as a whole is meromorphic.
Is this correct?
I am not at all familiar with Sheafs so please point out if there is any wrong use of notation in my question.