This question seems so obvious to me, that I think there could be a answer to it already. If so, I would appreciate a link!

Let $X$ be a complex manifold. Let $\exp:\mathcal{O}_X \to \mathcal{O}_X ^\ast$ be the exponential map-morphism between the sheaf of holomorphic functions on $X$ and the sheaf of nowhere vanishing holomorphic functions on $X$. Then $\exp$ is not surjective on arbitrary open subsets of $X$: For example there is a holomorphic log-function on $\mathbb{C}\backslash(-\infty,0]$ AND on $\mathbb{C}\backslash[0,+\infty)$ but not on their union. Right?

Q1.: Is my understanding correct, that the coker-presheaf $\text{coker}(\exp)$ is NOT trivial, then?

Q2.: How do I see explicitly that the sheafification of this coker-presheaf is trivial? Is there a good explicit description of this sheafified coker-presheaf? If $\text{coker}^\sharp(\exp)$ and $\text{im}^\sharp(\exp)$ are the sheafified coker, resp. image presheaves. Is there an "easy" connection between them?

Rem.: The sheafification $F^\sharp$ of a sheaf $F:Open(X)\to Rings$ as I know it, is as defined in Iversen: \begin{align} F^\sharp(U):=\{ \varphi_U \in \prod_{x \in U} F_x | \forall y \in U \exists y \in V \subset U:\exists g\in F(V):\varphi_U(y)=(V,g) \in F_y \}. \end{align} The restriction is the restriction to smaller products of stalks, which then still satisfy the property required. I inserted a bit of my own notation since I can think of it beter this way. Is this construction correct? I could replace $\prod$ with $\coprod$, right? Thanks in advance!


1 Answer 1


Q1: Yes, you are correct.

Q2: Calculate on the level of stalks. If we can show that $\mathcal{O}_X\to\mathcal{O}_X^*$ is surjective on stalks, this would show that the stalks of $\operatorname{coker(exp)}$ are all zero, which imply that it is the zero sheaf. Since the stalk is the colimit over all open neighborhoods and the simply-connected open neighborhoods define a cofinal system of these for any manifold, we may calculate using the simply-connected open neighborhoods. But on any simply-connected open set $U$, we can define a logarithm, which shows that $\mathcal{O}_X(U)\to \mathcal{O}_X^*(U)$ is surjective, and therefore the map is surjective on stalks.


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