I'm learning more about plane graphs (Chap 4.6) using Diestel's Graph Theory, and there is an exercise which states that every connected plane multigraph has a plane dual, by that it means that there are 2 plane multigraphs $G=(V,E)$ and $G^*=(V^*,E^*)$ and bijections from $F\to V^*$ ($f\mapsto v^*(f)$), $E\to E^*$ ($e\mapsto e^*$) and $V\to F^*$ ($v\mapsto f^* (v)$) which satisfy
- $v^* (f) \in f$ for all faces $f\in F$
- $e^* \cap G = (e^*)^o \cap e^o =e\cap G^*$ consists of a unique point for all $e \in E$
- $v\in f^*(v)$ for all $v\in V$
Here, $F,F^*$are the faces (or connected components of $\mathbb{R}^2 \backslash G,\mathbb{R}^2 \backslash G^*$).
Given a connected graph $G$, I can perform the usual construction by choosing an arbitrary point $v^*(f)$ in each face $f\in F$ to obtain the vertex set $V^*$, then inductively obtain an edge set $E^*$ which satisfy property 2, and thus we can define a plane multigraph $G^* = (V^*,E^*)$.
The natural mapping $V\mapsto F^*$ is just let $f^*(v)$ denote the unique face containing $v$. It's not too hard to show that the mapping is surjective, but I can't seem to easily show that it is injective using the connected property of $G$.
Any help would be appreciated.