I'm trying to study for a quiz. I think I'm on the right track with this problem, however, I'm having a difficult time formalizing it.
Prove that if graph $G$ is a 3-connected planar graph then its dual must be simple.
By simple we mean no self-loops nor multiple edges between the same pair of vertices. The problem doesn't specify whether we're discussing directed graphs or not, I would assume not though.
My idea is that if $G$ is 3-connected (In other words, the graph may be disconnected by removing a minimum of 3 edges), then the edges which make up the cutset would form 2 internal faces (excluding the infinite face). So when we take the dual of $G$, these two faces become vertices as shown with my crude Paint skills below.
Now supposing G was not 3-connected, but instead 2-connected. Then we would only have one internal face and two edges running between the internal face and the infinite face, which creates a general graph, not a simple one, as there are multiple edges between the same vertices.
So perhaps proof by contradiction would be best here I'm assuming?