trivial cokernel, but globally, we will have non-trivial cokernel.
#### § Exponential sheaf sequence

$\begin{aligned}
0
\rightarrow 2\pi i \mathbb Z
\xrightarrow{\alpha: \texttt{incl}} \mathfrak O
\xrightarrow{\beta:exp(\cdot)} \mathfrak O^*
\rightarrow 0
\end{aligned}$

- $\mathfrak O$ is the sheaf of the additive group of holomorphic functions. $\mathfrak O^*$ is the sheaf of the group of non-zero holomorphic functions.
- $\alpha$, which embeds $2\pi n \in 2\pi i \mathbb Z$ as a constant function $f_n(\cdot) \equiv 2 \pi i n$ is injective.
- $\beta(\alpha(n)) = e^{2 \pi i n} = 1$. So we have that the composition of the two maps $\beta \circ \alpha$ is the zero map (multiplicative zero), mapping everything in $2\pi i \mathbb Z$ to the identity of $\mathfrak O^*$. Thus,
`d^2 = 0`

, ensuring that this is an exact sequence. - Let us consider the local situation. At each point
`p`

, we want to show that $\beta$ is surjective. Pick any $g \in \mathfrak O^*_p$. We have an open neighbourhood $U_g$where $g \neq 0$, since continuous functions are locally invertible. Take the logarithm of $g$ to pull back $g \in O^*_p$ to $\log g \in O_p$. Thus, $\beta: O \rightarrow O^p$ is surjective at each local point $p$, since every element has a preimage. - On the other hand, the function $h(z) \equiv z$ cannot be in $O^*$ If it were, then there exists a homolorphic function called $l \in O$ [for $\log$] such that $\exp(l(z)) = h(z) = z$ everywhere on the complex plane.
- Assume such a function exists. Then it must be the case that $d/dz exp(l(z)) = d/dz(z) = 1$. Thus, $exp(l(z)) l'(z) = z l'(z) = 1$[use the fact that $exp(l(z)) = z$]. This means that $l'(z) = 1/z$.
- Now, by integrating in a closed loop of $e^{i \theta}$. we have $\oint l'(z) = l(1) - l(1) = 0$.
- We also have that $\oint l'(z) = \oint 1/z = 2\pi i$.
- This implies that $0 = 2\pi i$ which is absurd.
- Hence, we cannot have a function whose exponential gives $h(z) = z$ everywhere.
- Thus, the cokernel is nontrivial globally.