Does someone know how to proof the following statement:
Let X be a complex manifold. Define the Bott-Chern cohomology as $$H^{p,q}_{BC}(X) := \frac{ \{ \alpha \in \mathcal{A}^{p,q} | d\alpha = 0 \}}{\partial \overline{\partial}\mathcal{A}^{p-1,q-1}(X)}.$$ Show that there are natural maps $$H^{p,q}_{BC}(X) \rightarrow H^{p,q}(X) \text{ and }H^{p,q}_{BC} \rightarrow H^{p+q}(X, \mathbb{C}).$$
This is also part of an exercise in Daniel Huybrechts' "Complex Geometry: An Introduction" (2.6.7) and I already made myself clear that the definition of the Bott-Chern cohomology makes sense.
So does someone know a proof of this statement? I would be very interested in that!
I'll assume $H^{p,q}(X)$ is the Dolbeault cohomology of $X$, and that $H^{p+q}(X)$ is the De Rham cohomology.
We'll only need the notions of linear independence, and that if $U \subset V \subset W$ are vector spaces, then there is a natural linear morphism $W/U \to W/V$.
If $\alpha$ is a smooth $(p,q)$-form, then $d\alpha$ is the sum of the $(p+1,q)$-form $\partial \alpha$ and the $(p,q+1)$-form $\bar\partial \alpha$. If $d\alpha = 0$, we thus conclude that $\bar\partial \alpha = 0$ by the linear independence of $(p+1,q)$ and $(p,q+1)$-forms. There is also a clear inclusion $\partial \mathcal{A}^{p-1,q-1}(X) \subset \mathcal{A}^{p,q-1}(X)$. We thus get two morphisms $$ H^{p,q}_{BC}(X) \to \frac{\operatorname{Ker}(\bar\partial :\mathcal{A}^{p,q}(X) \to \mathcal{A}^{p,q+1})}{\partial\bar\partial \mathcal{A}^{p-1,q-1}(X)} \to \frac{\operatorname{Ker}(\bar\partial :\mathcal{A}^{p,q}(X) \to \mathcal{A}^{p,q+1})}{\operatorname{Im}(\bar\partial : \mathcal{A}^{p,q-1}(X) \to \mathcal{A}^{p,q})} = H^{p,q}(X), $$ the composite of which is the morphism we're looking for.
Similarly, we see that if $v = \partial \bar\partial u$ for a $(p-1,q-1)$-form $u$, then $v = d(\frac12 \bar\partial u - \frac12 \partial u)$, so $v = dw$ for a smooth $(p+q-1)$-form $w$. Thus we get the map $$ H^{p,q}_{BC}(X) \to H^{p+q}(X). $$