The following is an exercise from Hatcher's Algebraic Topology:
Suppose we build $\mathbb S^2$ from a finite collection of polygons by identifying edges in pairs. Show that in the resulting CW structure on $\mathbb S^2$ the $1$-skeleton cannot be either of the two graphs $K_5$ (the complete graph with $5$ vertices) or $K_{3,3}$ (the complete bipartite graph between sets of $3$ vertices).
This question is answered here and here, but both solutions make a crucial assumption: we do not identify edges from the same polygon. So for example, the classic CW structure on the torus with one face (a square), two edges (opposite sides of the square), and one vertex, resulting in the $1$-skeleton $\mathbb S^1 \vee \mathbb S^1$, would not be permitted.
My question: Why can't we construct a CW structure on $\mathbb S^2$ with edges of the same face identified, and for which the resulting $1$- skeleton is $K_5$ or $K_{3,3}$?
If $K_5$ is the $1$-skeleton of a weird CW complex on $\mathbb S^2$, then it would have to have $7$ faces by considering the Euler characteristic. There are certain kinds of polygons with self-identified edges that wouldn't be allowed; a triangle with an edge identified, for example, would include self-loops or vertices connected with two edges, which don't exist in $K_5$. But the casework needed to look at each type of self-identification is incredibly tedious, and I'm not sure how I'd do that systematically anyways (especially if we identify vertices and not edges, creating "parachutes" as sub-CW-complexes).
Any thoughts?