Dolbeault cohomology of a compact riemann surface is finite dimensional

The smooth complex supplied by [F1] has zero incoming space in degree zero and zero outgoing space in degree one. Thus [F2]'s kernel-modulo-image construction gives exactly the displayed groups, and the degree-zero group is $H^0(X,E)$. By [F3] it equals ${\mathcal{H}}^{0,0}(E)$; by [F4] this harmonic subspace of the finite-dimensional total kernel is finite-dimensional.

Let $P_1$ be the degree-one harmonic projection. By [F3], every smooth $u$ has a unique splitting $u=h+{\bar{\mathrm{\partial }}}_{E}f$ with $h\in {\mathcal{H}}^{0,1}(E)$, and $P_{1}u=h$. Hence $P_1$ vanishes on exact forms, so $[u]\mapsto P_{1}u$ is a well-defined complex-linear map from $H^{0,1}(X,E)$. It is surjective because each harmonic $h$ is smooth by [F4] and satisfies $P_{1}h=h$. Its kernel is zero because $P_{1}u=0$ forces $u={\bar{\mathrm{\partial }}}_{E}f$. Inclusion of harmonic forms followed by passage to the quotient is its inverse. Thus the degree-one group is isomorphic to the finite-dimensional space ${\mathcal{H}}^{0,1}(E)$, proving the dimension formula and unique harmonic representation.

The operator ${\bar{\mathrm{\partial }}}_E$ and the smooth form spaces in [F1] are determined by the holomorphic structure, independently of $h,g$. Their fixed kernel and quotient therefore define the same two cohomology vector spaces for every choice of these metrics. Applying steps 1.1–2.1 to each choice identifies its harmonic spaces with these fixed finite-dimensional spaces, so both dimensions agree. Full AC is inherited exactly through [F5].