Am I missing something here? Do I see shapes differently than everyone else?
A Cubohemioctahedron is cited to have a Euler characteristic of negative 2. This is because most texts say its 10 faces + 12 vertices - 24 edges = -2. One example where this is cited: website.
It's normally justified by observing that the interior is actually 4 intersecting hexagons so therefore the interior only contributes 4 faces. I believe this is wrong. First, hexagon is a two a dimensional term and thus would have no faces at all. Second, if you color each of the 4 two dimensional hexagons a different color, you will notice the only way to arrive at 4 internal faces is by counting the front and back of each hexagon as one face. Of course front and back of anything implies three dimensions with both front and back being an independent face. If the hexagons were three dimensional that would make them octahedrons and this would no longer be a Cubohemioctahedron. The 4 hexagons referenced here are simply two dimensional shapes that are clouding people's vision and should be ignored completely.
1) Saying a two dimensional shape has even a single face, as we are doing here, is a mathematical error. Two dimensional shapes have zero faces.
2) A face can't face two different directions at the same time.
3) Counting opposite sides of a 3 dimensional object as one face is also a mathematical error.
Is there a flaw in my reasoning?
The Euler characteristic is defined for any topological space which is formed by gluing together finitely many "cells" of different dimensions. It is defined to be $\sum_{i} (-1)^in_i$, where $n_i$ is the number of cells of dimension $i$. The simplest example is a sphere, which you can get by gluing a 2 dimensional disk onto a point by its boundary. So $n_0=1=n_2$ and the Euler characteristic is $1+1=2$. You can do a similar calculation for the torus which you can think of as taking a square and gluing the top edge to the bottom edge and the left to the right. There is a 1 $2$-cell, the middle of the square. There is one vertex, represented by the square's four corners which are all identified together, and finally two edges. So the Euler characteristic is $1-2+1=0$. You can do a similar calculation for the Klein bottle and get $0$. But here is a crucial point. When you normally draw the Klein bottle, it has a circle of self intersection. But that's not really there in the Klein bottle. It is an artifact of it being drawn in three dimensions. If you were to include the circle of self intersection, the Euler characteristic would be much different. So now we get to the Cubohemioctahedron. In this case the intersections of the hexagons are not really part of the shape. They are an artifact of needing to push it into 3 dimensional space. If you regard the cubohemioctahedron instead as an abstract topological space, then those intersections are not there, and furthermore the fact that they are on the "inside" of the shape no longer is relevant. You need to count all the cells.