In Angella and Tomassini's paper p.75, there is an exact sequence:

$\cdots\to B^{\bullet,\bullet}\to H_{\bar\partial}^{\bullet,\bullet}\to H_A^{\bullet,\bullet}\to \cdots$

where $B^{\bullet,\bullet}:=\frac{\ker\bar\partial\cap\text{im }\partial}{\text{im }\partial\bar\partial}$, and recall that Dolbeault cohomology $H_{\bar\partial}^{\bullet,\bullet}:=\frac{\ker\bar\partial}{\text{im }\bar\partial}$, and Aeppli cohomology $H_A^{\bullet,\bullet}:=\frac{\ker\partial\bar\partial}{\text{im }\partial+\text{im }\bar\partial}$, obviously, there are natural maps of $f:B^{\bullet,\bullet}\to H_{\bar\partial}^{\bullet,\bullet}$ and $g:H_{\bar\partial}^{\bullet,\bullet}\to H_A^{\bullet,\bullet}$, but I wonder how do we get $\text{im } f=\ker g$?

My thinking process is like that: In order to get the expression of $B^{\bullet,\bullet}$, we should compute the kernel of the map $g$, for a $\bar\partial$-closed form $\alpha$, let $g(\alpha)=0\in H_A^{\bullet,\bullet}$, then we get $\alpha=\partial\beta+\bar\partial\gamma$, since $\alpha$ is $\bar\partial$-closed, we get $\bar\partial\partial\beta=0$, thus $\alpha\in\ker\bar\partial\cap\text{im }\partial+\text{im }\bar\partial$, then I get stuck, can anyone help me to get $B^{\bullet,\bullet}=\frac{\ker\bar\partial\cap\text{im }\partial}{\text{im }\partial\bar\partial}$?