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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?

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    $\begingroup$ The inverse image, not the inverse. The stalk at $z$ is $f^{-1}(\{z\}) = $ the set of all germs locally defined at $z$. $\endgroup$ Mar 20, 2021 at 4:14
  • $\begingroup$ Thanks Alex @AlexKruckman $\endgroup$
    – Seth
    Mar 20, 2021 at 5:10

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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.

https://en.wikipedia.org/wiki/Sheaf_(mathematics)

https://en.wikipedia.org/wiki/Germ_(mathematics)

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