 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 becauseHence taking the same representative defines a complex-linear map $\varphi :H_{BC}^{p,q}(X)\to H_{\bar{\partial }}^{p,q}(X)$.

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. Consequentlyand $d\alpha =0$. It therefore defines a Bott-Chern cohomology class mapping to the original class. Thus $\boxed{\varphi \text{ is surjective on a compact Ka¨hler manifold}.}$

For clarity, the scalar Kähler Laplacian identity follows from the Kähler identities: they make the mixed anticommutators $\{\partial ,{\bar{\partial }}^{\ast }\}$ and $\{\bar{\partial },{\partial }^{\ast }\}$ zero, while the curvature-zero instance of the identity proved in 4(b) gives ${\Delta }^{\prime }={\Delta }^{\prime \prime }$. Expanding ${\Delta }_d$ yields ${\Delta }^{\prime }+{\Delta }^{\prime \prime }=2{\Delta }^{\prime \prime }$.