I'm obviously missing something obvious here. Euler's formula for planar graphs is: $$v-e+f = 2$$ Consider a triangle. It has three vertices and three edges. Substituting this into the formula above gives $f = 2$. Maybe this is because the region inside the triangle and that outside is counted as two faces. Now, for a Tetrahedron, the number of vertices is four and edges is six. Substituting this above we get $f = 4$. But this is exactly the number of faces a Tetrahedron has. We didn't have to split it into two regions. What am I missing?
In fact the "outside" of your graph could be considered to be wrapped around a ball. Thus your graph becomes the edge skeleton of a polyhedron, the regions then are the faces. Therefore Euler's formula likewise is valide for ball like polyhedra in general too.
And that result can even be expanded to other dimensions as well. Let X_i be the count of i-dimensional elements of a polytope which is hyperball like, where X_(-1) is the (unique) empty set and X_n is the (unique) hyperbody of the polytope, then you'd have
sum (-1)^i X_i = 0
For n=3, just as for every odd dimension, the partial sum of -X_(-1) and (-1)^n X_n would add to -2. Whereas for any even dimension n those add to 0.