What's wrong with my map of the hemicube?

Question

Reading from The Foundations of Topological Graph Theory by Bonnington and Little, a map is defined as a set $X$ with two permutations $\pi$ and $\varphi$ such that the orbits of $\pi$ are all of size two.

From their description I made the map for a tetrahedron:

Blue edges represent $\pi$ and red edges represent $\varphi$.

And if I check all the faces of this map are triangles. I also wanted to make the hemi-cube (or Petrial tetrahedron). The Hemi-cube has the same 1-skeleton as the tetrahedron so it should simply be a matter of reversing some of the vertices in the tetrahedron.

If I reverse only 1 vertex it will not work, the result is still orientable, and the triangle which didn't touch that vertex will remain unchanged. If I reverse all vertices I will get the mirror image of the existing diagram and it will just be the tetrahedron. If I reverse 3 vertices I will get the mirror image of reversing 1 vertex so that can't be it either.

So I must reverse two vertices. This will give a non-orientable map (which the hemi-cube is) and it will change every face. By the symmetry of the tetrahedron it does not matter which two vertices I reverse so I reverse two arbitrarily.

But when I draw in the faces I have an issue. There's one square, but the other two squares seem to have been fused into a single octagon which borders itself.

Please excuse the crudeness of this final drawing.

Where have I gone wrong? What is the map for the hemicube (in terms of the permutation definition)?

Answer 1

The misstep here is in the definition. The book The Foundations of Topological Graph Theory is a little confusing in how it uses the term "map". They initially define map as presented in the question, however this is not analogous to the usual sense of map (A graph embedding with a compliment of disconnected discs). Instead this definition is equivalent to an orientable map, or rotation system.

In the next section they define a "map imbedding" (with an "i") which is analogous to the more standard definition of map. The most confusing part is that soonafter they define map imbedding, they begin to just call map imbeddings "maps".

So the issue in the question is that it confuses maps with orientable maps. The Hemi-cube is not orientable, so it has no orientable map.

Using the map imbedding system described the hemicube can be made.

Simply bar all the edges of the tetrahedron. In general barring all the edges of a orientable map imbedding gives the Petrie dual of that map.