Intuition for plane graphs

Question

In the graph theory section of 2-Dimensional Categories by Johnson and Yau, they define a plane graph as a graph $G$ together with a topological embedding $m_G:|G|\to\mathbb{C}$ of the geometric realization of $G$ into the complex plane. This is supposed to fix a drawing of the graph, where (unless I'm mistaken) we want edges to be allowed to cross each-other as needed to draw the graph the way we want to draw it, even if these crossings aren't at vertices. For posterity:

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Definition. A graph $G$ is a triplet $(V_G,E_G,\psi_G)$ where $V_G$ is a finite set called the set of vertices, $E_G$ is a finite set called the set of edges, and $\psi_G:E_G\to V_G\times V_G$ is a function called the incidence function. The geometric realization of $G$, denoted $|G|$, is the topological quotient $$|G|=\Big[\big(\coprod_{v\in V_G}\{v\}\big)\coprod\big(\coprod_{c\in E_G}[0,1]_e\big)\Big]\big/\sim$$ where $\{v\}$ is a one point space indexed by a vertex $v\in V_G$, $[0,1]_e$ is a copy of the unit interval indexed by an edge $e\in E_G$, and the identification $\sim$ is generated by $$u\sim0\in[0,1]_e\ni1\sim v\ \ \ \ \text{if}\ \ \ \ \psi_G(e)=(u,v).$$

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This is where I'm confused -- the geometric realization has distinct copies of the unit interval for each edge of $G$, only glued together at endpoints where edges share vertices. If $m_G$ is a topological embedding then it's injective, so we aren't allowed to send two non-endpoints on two separate unit intervals to the same point in $\mathbb{C}$. That is, for $x,y\in(0,1)$ and edges $e,e'\in G$ we have that $$e\neq e'\implies m_G(e,x)\neq m_G(e',y)$$ if $m_G$ is an embedding. But this prevents drawings where edges overlap at non-vertex points, and this would prevent us from even having a geometric realization for non-planar graphs using this definition, correct?

If the above nitpicking is correct, are the authors implicitly restricting attention to planar graphs because all pasting diagrams have planar underlying graphs? Are there any diagrams in (bi-) categories, pasting or otherwise, whose underlying graphs are not planar?

Answer 1

Yes, this definition restricts to planar graphs. I'm not sure exactly what their motivation for doing so is (I haven't read most of the paper), but I imagine it is because their theorems are about how the diagrams they are considering interact with the topology of the plane. For instance, from a quick skim, Theorem 3.3.7 seems to be saying there is a coherent way to compose 2-morphisms by juxtaposing appropriate plane graphs geometrically, corresponding to "gluing together" faces of a plane graph to form larger faces. This is specifically using the geometric configuration of the faces to tell you what 2-morphisms you are composing together, and so it doesn't make sense if you don't have a plane graph with well-defined faces. Another way to think about it is that what's going on is that these diagrams don't really correspond to graphs at all, but instead to 2-dimensional polyhedral complexes, and they are looking at the special case that those polyhedral complexes are embedded in the plane. The graphs are merely the 1-skeletons of these 2-dimensional diagrams.

It's certainly not true that an arbitrary diagram in a category is always a planar graph. For instance, if you take a 5-element linear order and draw all its non-identity morphisms, that forms a diagram whose graph is $K_5$, which is non-planar.