# Presheaf and Sheaf Images

## A Presheaf Image that is not a Sheaf Image

### The Global Sections Functor is not Exact

Presheaf morphisms are the same as sheaf morphisms; they’re natural transformations: we get one sequence for each open set. So, it seems odd that exactness can hold globally, but not for each component. After all, an exact sequence is a bunch of objects and morphisms between them. Yet, taking sections of an exact sequence of sheaves does not, in general, yield exact sequences.

How? Because, an exact sequence is not just some objects and morphisms. An exact sequence is also the fact that the image of each morphism is the kernel of the next. The image and the kernel are objects that depend on an ambient abelian category.

Definition 1. Let $\mathcal{A}$ be an abelian category and $f : A \to B$ a morphism is $\mathcal{A}.$ Then, the kernel of $f$ is a map $i : K \to A$ universal with respect to the property that $f \circ i = 0.$

Definition 2. Let $\mathcal{A}$ be an abelian category and $f : A \to B$ a morphism is $\mathcal{A}.$ Then, the cokernel of $f$ is a map $s : B \to C$ such that $s \circ f = 0$ universally. The image of $f$ is $\text{im} f := \text{ker coker} f.$

Let $\varphi : \mathcal{F} \to \mathcal{G}$ be a morphism of presheaves. The kernel of $\varphi$ is a presheaf given by $\text{ker}_\varphi (U) = \text{ker}\varphi_U.$ The image presheaf is defined the same way. Now, the presheaf kernel of a map of sheaves is always a sheaf. In particular, the presheaf kernel is isomorphic to the sheaf kernel. But, the presheaf image of a map of sheaves is in general not a sheaf! Images in the abelian category of sheaves are constructed by sheafifying the presheaf image.

Let $X = \mathbb{C}^\ast.$ Consider the exponential sequence of sheaves on $X$

$0 \to \mathbb{Z} \to \mathcal{O}_X \to \mathcal{O}_X^\ast \to 0,$

with maps given by $\times 2\pi i : \mathbb{Z} \to \mathcal{O}_X$ and $\text{exp} : \mathcal{O}_X \to \mathcal{O} _X^\ast.$ On each open set, it is an exact sequence of abelian groups. So, it is an exact sequence as a sequence of presheaves of abelian groups. Moreover, one checks that exp$: \mathcal{O}_X \to \mathcal{O}_X^\ast$ is surjective on the stalks. It is a fact that exactness of a sequence of sheaves can be checked at the level of stalks. Thus, this is an exact sequence of sheaves.

Taking global sections does not yield an exact sequence because there is no global logarithm. So, the exponential is not surjective at the level of global sections.

As suggested, this is because the presheaf image of the exponential is not a sheaf. Consider the sections $t_i := \text{id}_{|U_i} \in \mathcal{O}^\ast_X(U_i).$ Using the logarithm function, there are sections $s_j \in \mathcal{O}_X(U_j)$ such that $t_i = \text{exp}(s_i).$ Since $t_i{_|U_i \cap U_j} =t_j{_|U_i \cap U_j}$ the sheaf axioms glue the exp$(s_i)$ to get a global section $t$ that restricts to $\text{id}_{|U_i}$ on each $U_i.$ But, the identity is not in the image of exp. So, im$_\text{pre}\text{exp}$ fails the gluing axiom.