We define the sheaf of holomorphic functions on $\mathbb C$, hence for any holomorphic function on an open subset we can do differentiation. I want to ask on what kind of open subset $U$ can it be surjective(i.e. every holomorphic function on $U$ has a holomorphic integration on $U$)?
Is simply connected enough? Could you help me prove it or give an another description of such open sets? Thank you!
As @peek-a-boo noted, if $U$ is simply connected the derivative is surjective. I claim that $U$ must be simply connected (if $U$ is connected) in order for the derivative to be surjective, thus giving a complete characterization. Let $U\subsetneq \mathbb{C}$ and $\alpha\in\mathbb{C}-U$. Since$\frac d{dz}$ is surjective, there exist $f: \frac{df}{dz}=\frac{1}{z-\alpha}$. This implies that $[\exp(-f)(z-\alpha)]'=-\exp(f)+\exp(f)=0$ and so $\exp(f(z))=\frac{K}{z-\alpha}$, where $K=\exp(-f(z_0))(z_0-\alpha)$, which we can assume wlog to be $1$. With a similar proof, one can show that, thanks to the existence of a primitive $g$ of $\frac{h'}{h(z)-\alpha}$ (with $\alpha\not\in h(U)$), $\exp(g)=h(z)-\alpha$. So we have proved that the logarithm is well defined on $U$. This also implies that the square root is well defined.
This two results together imply simple connectedness: to see why this is true, recall the proof of Riemann's mapping theorem and notice that the only points in which the simply connectedness of the set is used is to ensure the existence of $\log(z-\alpha)$ and of the square root (at least this is true for some proofs of the result, e.g. the one in Stein Shakarchi's Complex analysis) and so the usual proof of RMT proves $U$ is biholomorphically equivalent to $\mathbb{D}$.
If $U$ is not connected, the usual decomposition of $U$ as a countable disjoint union of open connected sets implies that $U$ must be the countable disjoint union of open simply connected sets, thus characterizing completely the sets for which the derivative is surjective.