Following example is from Liu's Algebraic Geometry and Arithmetic curves:
The purpose of this example is to illustrate that a surjective morphism of presheaves doesn't imply that every group homomorphism $\mathcal{F}(U)\to \mathcal{G}(U)$ is surjective for every open set $U$ of $X$.
I understand the outline of the argument but would like to crosscheck my understanding:
- Here $\mathcal{F}(U)$ an abelian group of holomorphic functions on $X$ under addition operation.
- Here $\mathcal{G}(U)$ an abelian group of "locally" invertible holomorphic functions on $X$ under addition operation. Since otherwise $\exp(f) \not\in \mathcal{G}(U)$.
- A holomorphic function $f$ is "locally" invertible in a neighbourhood of a point $z_0\in X$ if $f'(z_0)\neq 0$.
- $\alpha(X)$ is not surjective because $X$ is not simply connected. We can find a holomorphic branch of logarithm $F(z)$ such that $\exp(F(z))=z$ for all $z\in \Omega$ if and only if $\Omega$ is simply connected with $1\in \Omega$ and $0\not\in\Omega$.
- The induced group homomorphism between the stalk of $\mathcal{F}$ and $\mathcal{G}$ at $x\in X$, i.e. $\alpha_x: \mathcal{F}_x\to \mathcal{G}_x$ is surjective since "locally every open set around a point $x\in X$ is simply connected" and a nowhere vanishing function $f$ (required for it to be invertible) on a simply connected domain $U$ is of the form $f(z)=\exp(g(z))$ for some holomorphic function $g$ on $U$.
I have following doubt:
Given a presheaf $\mathcal{F}$ on $X$, the stalk of $\mathcal{F}$ at $x\in X$ is the group $\mathcal{F}_x = \varinjlim_{U\ni x} \mathcal{F}(U)$. Then what is the explicit description of stalk $\mathcal{F}_x$ and $\mathcal{G}_x$?