# Sheafification of coker-presheaf of exp-map

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!

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.