= Solution
A $d$-closed <differential form of type (p, q)> is $\bar\partial$-closed, and a <Bott-Chern cohomology> coboundary is a <Dolbeault cohomology> coboundary because
$$
\partial\bar\partial\beta=-\bar\partial(\partial\beta).
$$
Hence taking the same representative defines a <complex-linear map> $\phi:H^{p,q}_{BC}(X)\to H^{p,q}_{\bar\partial}(X)$.
If $X$ is compact and Kähler, the <Dolbeault Hodge decomposition> supplies a $\bar\partial$-harmonic representative $\alpha$ of every <Dolbeault cohomology> class. The scalar <Kähler Laplacian identity> is $\Delta_d=2\Delta_{\bar\partial}$, so $\alpha$ is also $d$-harmonic. Consequently
$$
0=\langle\Delta_d\alpha,\alpha\rangle
=\|d\alpha\|^2+\|d^*\alpha\|^2,
$$
and $d\alpha=0$. It therefore defines a <Bott-Chern cohomology> class mapping to the original class. Thus \b[$\boxed{\phi\text{ is surjective on a compact Kähler manifold}.}$]
For clarity, the scalar <Kähler Laplacian identity> follows from the <Kähler identities>: they make the mixed anticommutators $\{\partial,\bar\partial^*\}$ and $\{\bar\partial,\partial^*\}$ zero, while the curvature-zero instance of the identity proved in 4(b) gives $\Delta'=\Delta''$. Expanding $\Delta_d$ yields $\Delta'+\Delta''=2\Delta''$.
Back to article page
