Let $X$ be a compact, connected complex manifold. I denote by $H$ the space $\ker (\partial \colon H^{1,0}(X) \to H^{2,0}(X))$. I now want to prove that there is an exact sequence $$0 \to H \stackrel{\iota}\to H_{dR}^1(X,\mathbb C) \stackrel{\pi^{0,1}}\to H^{0,1}(X) \stackrel{\partial}\to H^{1,1}(X).$$ I assume that all cohomology groups are taken with respect to $\bar{\partial}$, although this is not written explicitly in the text. I seem to be able to prove that $\iota$ is an inclusion, so its injectivity should be fine.
I am missing something while proving $\ker \pi^{0,1} \subset \mathrm{Im}\iota$ and $\ker \partial \subset \mathrm{Im}\iota$. Let us take the first one for example. The map $\pi^{0,1}$ maps $[\alpha] \in H_{dR}^1(X,\mathbb C)$ to $[\alpha^{0,1}]$, i.e. the class in $H^{0,1}$ of the $(0,1)$-part of $\alpha$. So if $\pi^{0,1}[\alpha] = 0$ then $\alpha$ must be $\bar{\partial}$-exact, hence there must be a complex function $g$ on $X$ such that $\alpha^{0,1} = \bar{\partial}g$. Now $\alpha$ is also $d$-closed by assumption, so \begin{align*} 0 & = \partial \alpha + \bar{\partial}\alpha \\ & = \partial \alpha^{1,0} + \partial \alpha^{0,1} + \bar{\partial} \alpha^{1,0} + \bar{\partial}\alpha^{0,1} \\ & = \partial \alpha^{1,0} + \partial \alpha^{0,1} + \bar{\partial} \alpha^{1,0}. \end{align*} In particular, $\partial \alpha^{1,0} = 0$ and $ \partial \alpha^{0,1} = -\bar{\partial}\alpha^{1,0}$. Here I am stuck. The idea was to take as a preimage of $\alpha$ its $(1,0)$-part, but I cannot really prove it lies in the same equivalence class of $\alpha$. The only thing I have for now is that its $\partial$-derivative is $\bar{\partial}$-closed. Perhaps there is a local argument that helps, but I am not sure how to use it. Is there any way to fix this?
Consider $\alpha-dg = \alpha^{0,1}- \partial g$. This is cohomologous to $\alpha$. But it's obviously in $H$.