In what follows, a graph is always a finite multigraph with loops. A geometric dual of a graph is formed by embedding that graph in the plane (taking a set of points for the vertices and curves with non-intersecting interiors between them for the edges), assigning to each face (connected component of the complement of the embedding) a vertex as a point in the face, and connecting vertices with an edge passing through the edge connecting their faces (and intersecting nothing else in its interior), or adding self-loops in the case of an face that borders itself. An abstract dual of a graph $G$ is a graph $G^*$ with a bijection between the set of edges of $G$ and that of $G^*$ such that a set $E$ of edges of $G$ is a cycle if and only if its image is a cutset of $G^*$.
I know that a graph has an abstract dual if and only if it has a geometric dual, but I was wondering if the notions were more closely related. We have a relation $R$ on graphs, such that for graphs $G$ and $H$, $H\in R[G]$ if and only if $H$ is an abstract dual of $G$. I think I’ve shown to myself that $R^3=R$, where $R^3$ is referring to the composition of relations.
Now, take a graph $G$ and start with $S_0[G]=\{G\}$. Take all embeddings of $G$ into the plane, take all graphs corresponding geometric duals of these embeddings, and all graphs with embeddings of which the embedding of $G$ is the geometric dual. Thus, we obtain a new set $S_1[G]$. Apply the previous construction to every member of $S_1$ and take the union of sets obtained in this manner to form $S_2[G]$. Repeat to obtain $S_n[G]$ for all $n\in\mathbb{N}$. Take the union $S[G]$ of $S_n[G]$ for all odd $n$. Let us continue with this notation, and speak of the relation $S$ on finite graphs.
It is clear using $R^3=R$ that for all graphs $G$, $S[G]⊆R[G]$. However, I was wondering if $S[G]=R[G]$. In other, less precise, words, every abstract dual arises from geometric duals. Furthermore, is there some $n\in\mathbb{N}$ such that $S_n[G]=S[G]$, for all $G$? Can it be independent of $G$?
