Platonic solids - topological and geometrical conditions With V, E, F as the numbers of vertices, edges and faces of a given polyhedron and based on Euler‘s polyhedron formula
$$ V - E + F = 2 $$
it is quite simple to derive a necessary topological condition for Platonic solids.
One uses p-sided polygons and q-valent vertices to calculate V and E. Inserting this in Euler‘s polyhedron formula yields
$$ F_{p,q} = \frac{4q}{2(p+q) - pq} $$
from which all allowed positive integers F can be calculated.
My question is, if and why this is sufficient geometrically.
It is never mentioned that this pure topological condition could result in „fake objects“ that cannot be constructed geometrically, especially b/c the angles at the vertices are not considered. Of course we know that the formula does not generate such „fake objects“ but I do not see how this follows without any geometrical reasoning.
 A: The condition you state is necessary but it is not sufficient. Consider $p=q=4$. Then $$2(p+q)−pq=0$$ so $$F_{p,q} = \infty$$
Which is the plane tiling of squares.
Similarly you can get a finite answer for $p=7, q=3$, which is a hyperbolic tiling.
However in neither case is $V−E+F=2$. In the first instance these values are now infinite. What you can do is build out a tiling and show that for say the square tiling, the value converges towards $0$.
So you also need to assume that $V−E+F=2$ and any solutions which do not meet this condition must be rejected. This is equivalent to assuming that the polyhedron is a topological ball. But even here, there are oddities. For example if you take a vertex of the regular icosahedron and push it inwards to form a dimple, it meets all the above conditions but is not Platonic.
So there is also your unspoken assumption that all faces are congruent, all vertices are congruent and all edges are equal. For any topological ball, this ensures that the polyhedron will be convex.
The reason why it all works is basically a result from graph theory. It is true for any Eulerian graph, and a convex polyhedron may be treated as such a graph drawn on a ball.
The classic work on the subject is HSM Coxeter's Regular Polytopes.
Historical ramble: Poinsot derived the Platonic solids by assembling regular faces around vertices. Given regular p-gonal faces and q-valent vertices, Schläfli derived his symbols {p q} for the regular polyhedra. These were later generalised to plane and hyperbolic tilings; the examples above are respectively {4 4} and {7 3}. They can be further generalised to other topological surfaces such as the torus or the projective plane, but in these cases the symbol is no longer unique. For example {4 4} may be drawn on a torus manifold with any number of squares (given a suitable intrinsic metric) in each direction, and any toroidal polyhedron has $V−E+F=2$. Constructing a "cube" {4 3} on the (finite) projective plane yields the hemicube, with only half the numbers of faces, edges and vertices - any projective polyhedron has $V−E+F=1$. This number is called the Euler number $\chi$ of the surface. So for a ball, i.e. any convex polyhedron, $\chi=2$.
