In Angella and Tomassini's [paper](https://link.springer.com/article/10.1007/s00222-012-0406-3) 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 $B^{\bullet,\bullet}=\frac{\ker\bar\partial\cap\text{im }\partial}{\text{im }\partial\bar\partial}$?



The kernel of $H^{\bullet,\bullet}_{\bar\partial}(X)\to H_A^{\bullet,\bullet}(X)$
Let $\alpha$ be a $\bar{\partial}$-closed form. Denote its Dolbeault cohomology class by $[\alpha]\_{\bar{\partial}}$ and its Aeppli cohomology class by $[\alpha]\_A$; note that the map $g$ is given by $g([\alpha]\_{\bar{\partial}}) = [\alpha]\_A$. Likewise, if $\alpha' \in \ker\bar{\partial}\cap\operatorname{im}\partial$, denote the corresponding element in $B^{\bullet,\bullet}$ by $[\alpha']\_B$; note that the map $f$ is given by $f([\alpha']\_B) = [\alpha']\_{\bar{\partial}}$.


As you noted, if $[\alpha]\_{\bar{\partial}} \in \ker g$, then $\alpha = \partial\beta + \bar{\partial}\gamma$, so $[\alpha]\_{\bar{\partial}} = [\partial\beta + \bar{\partial}\gamma]\_{\bar{\partial}} = [\partial\beta]\_{\bar{\partial}}$. Moreover, since $\bar{\partial}\alpha = 0$, we see that $\bar{\partial}\partial\beta = 0$ and hence $\partial\beta \in \ker\bar{\partial}\cap\operatorname{im}\partial$. Therefore, we can form the element $[\partial\beta]\_B$ which satisfies $f([\partial\beta]\_B) = [\partial\beta]\_{\bar{\partial}} = [\alpha]\_{\bar{\partial}}$, so $[\alpha]\_{\bar{\partial}} \in \operatorname{im}f$ and hence $\ker g \subseteq \operatorname{im}f$.


Suppose now that $[\alpha]\_{\bar{\partial}} \in \operatorname{im}f$, then there is $\alpha' \in \ker\bar{\partial}\cap\operatorname{im}\partial$ with $[\alpha']\_{\bar{\partial}} = [\alpha]\_{\bar{\partial}}$. As $\alpha' \in \operatorname{im}\partial + \operatorname{im}\bar{\partial}$, we see that $g([\alpha]\_{\bar{\partial}}) = g([\alpha']\_{\bar{\partial}}) = [\alpha']\_A = 0$, so $[\alpha]\_{\bar{\partial}} \in \ker g$ and hence $\operatorname{im}f \subseteq \ker g$.


