 Solution

The set ${\mathcal{Q}}_n$ consists of rooted planar maps with $n$ faces, every face having degree four. The set ${\mathcal{Q}}_n^{\bullet }$ consists of these quadrangulations with an additional distinguished vertex.

For the trivial bijection between planar maps and quadrangulations, start from a rooted planar map with $n$ edges. Put a new vertex in every face and join it to the original vertex at every incident corner. Delete the original edges. The two endpoints of each deleted edge and the new vertices in its two adjacent faces bound a quadrangular face, so the result lies in ${\mathcal{Q}}_n$. Its bipartition distinguishes old from new vertices and reconstructs the original map. With the standard root convention this is a bijection, and hence