Let $X$ be $\mathbb{C}$ with the classical topology, let $\mathscr{O}_X$ be the sheaf of holomorphic functions and let $\mathscr{F}$ be the presheaf of functions admitting a holomorphic logarithm. Show that $$0 \longrightarrow \underline{\mathbb{Z}} \longrightarrow \mathscr{O}_X \longrightarrow \mathscr{F} \longrightarrow 0$$ is exact where $\underline{\mathbb{Z}} \longrightarrow \mathscr{O}_X$ is given by the natural inclusion and $\mathscr{O}_X \longrightarrow \mathscr{F}$ is given by $f \mapsto \exp(2\pi if)$. Show also that $\mathscr{F}$ is not a sheaf. (Hint: The problem is that there are functions that don't have a logarithm but locally have a logarithm.)
So let $i : \underline{\mathbb{Z}} \to \mathscr{O}_X$ denote the inclusion and $j: \mathscr{O}_X \to \mathscr{F}$. The first question is that how is this $j$ defined? I think they mean that for $U$ open in $X$ we consider $$\begin{align*} j(U): \mathscr{O}_X(U) &\to \mathscr{F}(U) \\\\ j(U)(f) &= \exp(2\pi i f) \end{align*}$$ but does the rhs of $$j(U)(f) = \exp(2\pi i f)$$ make sense since $f$ is a map from $U$ to some ring?
I think I can only proceed with the problem when I know how this $j$ is supposed to behave as I need to show next that $$\ker(j) = \operatorname{im}(i)$$ to verify exactness.