I have been asked to show that the dual of the dual, $(G^*)^*$, of a planar graph $G$ is isomorphic to $G$ if and only if $G$ is connected.
I understand the reasoning in the answer one but that cannot be the whole proof, right? I also know that $|E(G)|=2|V(G)|-2$ which is mentioned in the comments of answer two. My idea is that I should try to show that any neighboring vertices in $G$ will also be neighboring vertices in $(G^*)^*$ (like in answer one), non-neighboring vertices in $G$ will be non-neighboring vertices in $(G^*)^*$ and that there will be the same number of vertices in both $G$ and $G^*$ but I cant really see what the conectedness gives. Is this correct and enough? Anyone who can help me with this?