Explain the sheaf of holomorphic functions defined over $\mathbb{C}$.

Question

How explain the sheaf of holomorphic functions defined over $\mathbb{C}$. ( $\mathbb{C} \supset A \to F(A)$ is holomorphic function, where $A$ is open set)

I know definition of sheaf and presheaf. But I don't know how I can show it... Could you explain me? Thank you!

Let $X$ be a topological space. A presheaf of sets $F$ on $X$ is a rule associating to each open set $A \subseteq X$ some type of an algebraic object $F(A)$ if right be an abelian group/ a ring $\mathbb{K} $ -algebra ...

$A \mapsto F(A)$:

1. $F(\emptyset) = 0$
2. $A, B$-open set. If $A \subset B $ then there is a homomorphism $\rho_{AB}: F(B) \to F(A) $ -restriction
3. $\rho_{AA} =id_{F(A)}$ $F(A)$ is called the space of section of $F$ $s \in F(A) $, $\rho_{AB}(s)= s_{|B}$

And definition sheaf: A presheaf $F$ on $X$ is a sheaf if for any open set $A \subseteq X$ and open cover $\{A_i\}$ of $A$ such that:

1. if $s \in F(A)$ such that $S_{|A_i}=0$ for all i, then $s \equiv 0$ in $F(A)$.
2. for $f_i \in F(A_i) $ and $f_j \in F(U_j)$ such that $f_i|A_i \cap A_j =f_j|A_i \cap A_j $ then there exists a section $f \in F(A_i \cup A_j )$ such that $f|A_i =f_i$ and $f|A_j = f_j$

Answer 1

The sheaf $F$ (Note: it is usually denoted by $\mathcal{O}$) associates for an open set $A\subset\mathbb{C}$ with the set $F(A)$ of holomorphic functions on $A$. This is a sheaf of commutative rings. Restriction homomorphisms are in fact the restriction of domain of functions.

Suppose $A=\cup_i B_i$ is an open covering. If $f\in F(A)$ restricts to zero functions on every $B_i$, then clearly $f=0$. If we are given $f_i\in F(B_i)$ for each $i$ that agrees on every intersection, then they together define a function $f$ on $A$, which is holomorphic since holomorphicity can be determined only using a small neighborhood. Thus, $F$ is a sheaf.