An Inportant Example in the Sheaf Morphism

I mainly want to improve the familarity of me to the language of sheaves through this example, in which I will give some detail computation.

Through the basic theory of sheaves, we know that the image of a sheaf morphism $\phi:\mathcal{F}\longrightarrow\mathcal{G}$ is not necessarily a sheaf, just a presheaf, which is so called the presheaf image , the sheafification of the presheaf image is called the image of the morphism $\phi$ , which will be denoted by $\mathrm{im} \phi$ , and which is a subsheaf of $\mathcal{G}$ , moreover, if we have $\mathrm{im}\phi=\mathcal{G}$ , then we say this morphism is surjective, there is a natural view of the “surjectiveness”, that this the morphism between sections $\phi_{U}:\mathcal{F}(U)\longrightarrow\mathcal{G}(U)$ is a surjective one between the Abelian groups, however, this is the wrong view, and what are the counter examples? Next I shall give and explain why it is not the case.

Let $X=\mathbb{C}$ be the complex plane, and we denoted by $(\mathbb{C},\mathscr{O})$ and $(\mathbb{C},\mathscr{O}^*)$ the sheaf of holomorphic functions and the sheaf of non-vanishing holomorphic functions respectively, let $\exp: \mathscr{O}\longrightarrow \mathscr{O}^*$ which sends for each $f\in \mathscr{O}(U)$ to the function $\exp f$ , clearly this defines a morphism between the two sheaves, first of all, I’m going to show the preasheaf image is not a sheaf.

To be convinient, we denoted by the $\mathcal{I}$ the presheaf image, and recall that the section of which is $\mathcal{I}(U)=\exp \mathscr{O}(U)$ , now take an open cover of $\mathbb{C}$ , namely $\mathbb{C}=(\mathbb{C}\setminus\mathbb{R}_{>0})\cup(\mathbb{C}\setminus\mathbb{R}_{<0})\cup\mathbb{D}$ , and denoted by $U_1,U_2,U_3$ in my ordinal, now we take $s_1=(z\mapsto z)\in \mathcal{I}(U_1)$ , and $s_2=(z\mapsto z)\in \mathcal{I}(U_2)$ , then obviously $s_1|_{U_1\cap U_2}=s_2|_{U_1\cap U_2}$ , but we shall find no globally $s\in\mathcal{I}(U)$ , such that $s|_{U_1}=s_1, s|_{U_2}=s_2$ , since if so, at least on $U_1$ , one has $s=(z\mapsto \log z)$ , but the logarithm cannot be extended analytically to the whole plane.

Next, I’m going to show that $\exp$ is surjective, to see this, we need to compute the sheafification of $\mathcal{I}$ , namely $\mathcal{I}^+$ .

First of all, we need to compute the stalks $\mathcal{I}_p$ for each $p\in \mathbb{C}$ , by definition :

$\begin{aligned} \mathcal{I}_p&=\varinjlim_{p\in U}\exp\mathscr{O}(U)=\left(\coprod_{p\in U}\exp\mathscr{O}(U)\right)\big/\sim\\&=\left(\coprod_{p\in U}\left\{e^f:f\in\mathscr{O}(U)\right\}\right)\big/\sim\\&=\left\{\left[e^f\right]_p:f\in\mathscr{O}(U)\right\}\end{aligned}$

Here $\left[e^f\right]_p$ represents for the function germ at $p$ of the functions with the form $e^f$ , next we will deduce the section $\mathcal{I}^+(U)$ , by the definition:

$\begin{aligned}\mathcal{I}^+(U)&=\left\{s\in\mathrm{Map}\left(U;\coprod_{p\in U}\mathcal{I}_p\right)\left.\right|\forall p\in U,\exists V\ni p, \text{s.t.}\forall t\in\mathcal{I}(U), t_p=s(p)\right\}\\&=\left\{s\in\mathrm{Map}\left(U;\coprod_{p\in U}\mathcal{I}_p\right)\left.\right|\mathrm{locally}, s(p)=t_p\right\}\\&=\coprod_{U\supset V\ni p}\left\{s:p\mapsto \left[e^f\right]_p\left.\right|f\in\mathscr{O}(V)\right\}\end{aligned}$

which is isomorphic to the group of all holomorphic funcions on $U$ which locally have continuous logarithm, that is $\mathscr{O}^*(U)$ , hence it is surjective.