Let $X$ be a complex manifold, there is a natural map $f:H^2(X,\mathbb C)\to H^2(X,\mathcal O)$ induced by the inclusion map $\mathbb C\hookrightarrow \mathcal O$ which coincides with the natural projection map $\pi^{0,2}:H^2_{dR}(X,\mathbb C)\to H^2(X,\mathcal O),[\omega]\mapsto [\omega^{0,2}]$, where $\pi^{0,2}:A^2(X)\to A^{0,2}(X)$ take a $2$-form to its $(0,2)$ component (see Huybrechts' book *complex geometry: an introduction* p.132 Lemma 3.3.1).

When $X$ is a compact Kähler manifold, it is easy to conclude that the natural map $f:H^2(X,\mathbb C)\to H^2(X,\mathcal O)$ is surjective. Since for any $\Delta_{\bar\partial}$-harmonic representative $\omega^{p,q}$ of $H^{p,q}_{\bar\partial}(X)$, we can add $\omega^{p,q}$ together and get $\omega=\sum\limits_{p+q=k}\omega^{p,q}$, which is $\Delta$-harmonic by Kähler identity: $\Delta=2\Delta_{\bar\partial}$.

But for a non-Kähler manifold, for example, the $\partial\bar\partial$-manifold: a compact complex manifold with any $\partial$-,$\bar\partial$-closed, $d$-exact $(p,q)$ form being $\partial\bar\partial$-exact, is the map $f$ surjective?

I guess that this map is also surjective. Since its Frolicher spectral sequence degenerates at $E_1$ (see remark 5.21 of Deligne, Griffiths, Morgan, Sullivan, Real homotopy theory of Kähler manifolds), so we have $H^2(X)\cong H^{2,0}_{\bar\partial}(X)\oplus H^{1,1}_{\bar\partial}(X)\oplus H^{0,2}_{\bar\partial}(X)$, and there is an isomorphism between Bott-Chern cohomology and Dolbeault cohomology $H_{BC}^{p,q}(X)\cong H^{p,q}_{\bar\partial}(X)$ for any $\partial\bar\partial$-manifold. But for any $\bar\partial$-close $(0,2)$ form $\alpha$, how can we conclude there must be a $d$-closed form $\omega$ with its $(0,2)$ component $\omega^{0,2}$ equals to $\alpha$?