## § Cokernel is not sheafy

I wanted to understand why the Cokernel is not a sheafy condition. I found an explanation in Ravi Vakil's homework solutions which I am expanding on here.

#### § Core idea

We will show that there will be an exact sequence which is surjective at each stalk, but not globally surjective. So, locally, we wil have 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 $f_n(\cdot) \equiv 2 \pi i n$ isinjective.
• $\beta(\alpha(n)) = e^{2 \pi i n} = 1$. So we have that the composition of the two maps $\beta \circ \alpha$ isthe 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 showthat $\beta$ 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.