I am looking for a proof of this fact, where $H^{1,0} = Ker(d: \mathscr{E}^{1,0} \rightarrow \mathscr{E}^{2})$ and $H^{0,1} = Coker(\overline{\partial}: \mathscr{E} \rightarrow \mathscr{E}^{0,1}$, (These are the dolbeault cohomology groups) where we are working on a riemann-surface $X$. This statement is equivalent to $H^0(\Omega) = H^1(\mathscr{O})$, where the latter two are the sheaf cohomology groups of the sheaf of holomorphic 1-forms and holomorphic differentials, respectively. I have read the proof in Donaldson, in which he first establishes an isomorphism $\alpha \rightarrow \overline{\alpha}$ that takes $H^{1,0}$ to $\overline{H^{0,1}}$, and then composes that with the inner product $\langle \alpha, \beta \rangle = \int_X \alpha \wedge \beta$, but I don't understand how we know that this original conjugation map produces an isomorphism. Another proof I have seen uses the theory of distributions, but was incomplete.
I would greatly appreciate a proof or a reference to a complete proof, preferably one that doesn't use too much heavy machinery (The proofs I've seen only seem to rely on a couple of facts form analysis.