Problem: Let $X$ be a compact Riemann surface. Show that $$\Omega^1(X) \to \operatorname{Rh}^1(X) = \frac{\ker (d : \mathcal{E}^{(1)}(X) \to \mathcal{E}^{(2)}(X))}{\operatorname{im}( d: \mathcal{E}(X) \to \mathcal{E}^{(1)}(X))}$$ is injective.
My try: I already know that the map is well-defined. Let $\omega \in \Omega^1(X)$ and suppose its equivalence class in the de Rham group is $[\omega] = 0$. I try to show that $\omega = 0$.
We may write $\omega = f dz$ for some holomorphic $f$. On the other hand $\omega \in d\mathcal{E}(X)$, so $\omega = dg$ for some $g \in \mathcal{E}(X)$. It follows that $\omega = fdz = dg = \frac{\partial g}{\partial z} dz$. I think it should follow that $f = 0$ (so that $\omega = 0$), but I cannot prove that.
Suppose $\omega = dg$ for some smooth function $g : X \to \mathbb{C}$. As $dg = \partial g + \bar{\partial g}$ and $\omega$ is a $(1, 0)$-form, $\bar{\partial g} = 0$. That is, $g$ is holomorphic and hence constant. Therefore $\omega = dg = 0$.