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Following example is from Liu's Algebraic Geometry and Arithmetic curves:

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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:

  1. Here $\mathcal{F}(U)$ an abelian group of holomorphic functions on $X$ under addition operation.
  2. Here $\mathcal{G}(U)$ an abelian group of "locally" invertible holomorphic functions on $X$ under addition operation. Since otherwise $\exp(f) \not\in \mathcal{G}(U)$.
  3. A holomorphic function $f$ is "locally" invertible in a neighbourhood of a point $z_0\in X$ if $f'(z_0)\neq 0$.
  4. $\alpha(X)$ 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 $0\not\in\Omega$.
  5. 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 $\mathcal{G}_x$?

rationalbeing
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  • Correct. 2. and 3. locally invertible holomorphic function = invertible holomorphic function = non vanishing holomorphic function (so no need for the locally : if $f$ is holomorphic and does not vanish, $1/f$ is holomorphic. But $\mathcal{G}(U)$ is a sheaf of group for the multiplication ! We have $\exp(f)\in\mathcal{G}(U)$ since it is nowhere vanishing. 4. correct. 5. correct
    • – Roland Apr 03 '18 at 18:01
  • @Roland Thanks for cross-checking my understanding. But how is a locally invertible function on $X$ = invertible function on $X$? Branch of logarithm can't be continuously extended to a continuous function defined in the full punctured plane, but $\exp(z)$ is non-vanishing on the full punctured plane. Can we still call $\exp(z)$ invertible on $X$? Then my understanding of 4. should be wrong. What should be the description of $\mathcal{F}_x$ and $\mathcal{G}_x$ in words? Thanks. – rationalbeing Apr 04 '18 at 02:11
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    Invertible here means invertible for the multiplication (not composition). That is why you only need that $f$ does not vanish. This is also why $\exp$ is a morphism : $\exp(f+g)=\exp(f)\exp(g)$ (and it has values in invertible functions). – Roland Apr 04 '18 at 05:25
  • @Roland Is this the standard definition of "invertible holomorphic function"? Can you point towards a source? I thought it was referring to inverse function theorem for holomorphic functions. – rationalbeing Apr 04 '18 at 05:32
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    This is standard because when we speak about $\mathcal{O}_X$, we meant holomorphic functions : $X\to\mathbb{C}$ together with its ring operation, and not morphisms together with the composition. So invertible for the composition is usually called "isomorphism". Moreover this exponential map exists and is surjective for any complex variety $X$ (not only of dimension one which is the only case where we can speak about inverse for the composition), and you will soon meet it a lot more. Finally, this is also the only option which makes the exponential map a morphism of groups. – Roland Apr 04 '18 at 09:09