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As far as I know, a germ is a local representation of an analytic function and a sheaf is a collection of germs. We're also told that there is a function which maps a given germ to the point for which it is locally defined. We then define the stalk of $z\in\mathbb{C}$ to be the inverse of such a function (at $z$).
My question, how is this (i.e., the inverse) well-defined? Does a sheaf relate to one function ('equivalency'), or are there not multiple functions defined to a single point?
Question: "My question, how is this (i.e., the inverse) well-defined? Does a sheaf relate to one function ('equivalency'), or are there not multiple functions defined to a single point?"
Answer: If $(X, \mathcal{O})$ is a complex manifold and $x\in X$ is a point, you define the stalk of the sheaf $\mathcal{O}$ at $x$ (denoted $\mathcal{O}_x$) as follows:
$$\mathcal{O}_x:=\{(U,s):x\in U, s\in \mathcal{O}(U)\}/\cong$$
where two pairs $(U,s),(V,t)$ are equivalent iff $s_{U\cap V}=t_{U \cap V}$. With this definition it follows $\mathcal{O}_x$ is a local ring with a maximal ideal $\mathfrak{m}_x$, consisting of equivalence classes of pairs $(U,s)$ with the property that there is an open set $x\in V \subseteq U$ with $s_V=0$ in $\mathcal{O}(V)$.
Note: Here $\mathcal{O}(U)$ is the ring of complex holomorphic functions on the open set $U \subseteq X$. For each open subset $V \subseteq U$ you get trivially a restriction map
$$\rho_{UV}: \mathcal{O}(U) \rightarrow \mathcal{O}(V)$$
and you may verify that $\mathcal{O}$ is a sheaf of rings on $X$.
You get an "evaluation map"
$$ev_x: \mathcal{O}_x \rightarrow \mathcal{O}_x/\mathfrak{m}_x \cong \mathbb{C}$$
defined by $ev_x(U,s):=\overline{(U,s)} \in \mathcal{O}_x/\mathfrak{m}_x$. Since $ev_x$ is surjective with kernel $\mathfrak{m}_x$, it follows $\mathfrak{m}_x$ is a maximal ideal.
If the germ $(U,s)$ has non-zero value at $x$ under the evaluation map, it follows there is an open neigborhood $x\in V \subseteq U$ where $s$ is non-zero, hence on $V$ it follows $s$ has an inverse $t$. It follows $(U,s)$ is a unit in $\mathcal{O}_x$ with multiplicative inverse $(V,t)$. Hence any element $(U,s)\in \mathcal{O}_x- \mathfrak{m}_x$ is a unit and it follows $\mathcal{O}_x$ is a local ring.
The ring $\mathcal{O}_x$ is a Noetherian regular local ring in the sense of "commutative algebra" and for this reason people use methods from commutative algebra in the study of these rings.
