Prove: Euler’s formula. For any simple polyhedron

            V - E + F = 2

where

            V is the number of vertices

            E is the number of edges

            F is the number of faces

Proof. Imagine the given polyhedron to be hollow with a surface made of thin rubber. Now let us cut out one of the faces of the polyhedron and then stretch the remaining surface out flat on a plane. Of course, doing this will cause gross distortions in the faces, angles, etc. but the network of vertices and edges in the plane will contain the same number of vertices and edges as the original polyhedron and the number of polygons will be one less than in the original polyhedron, since one face was removed. We shall now show that for the plane network, V - E + F = 1, so that if the removed face is counted, the result is V - E + F = 2 for the original polyhedron. We do this by performing a succession of alterations on our plane network. We will show that each alteration that we make causes no change in the value of the quantity V - E + F. After all alterations are made we will end up with a single triangle for which V - E + F = 1.

1. First set of alterations. In the first set of alterations we change the entire network into a network of triangles by repeated use of the following operation:

●          In any polygon of the network that is not already a triangle, draw a diagonal.

The effect of this alteration in the network is to increase both E and F by 1. The value of V - E + F is thus not changed by the alteration. We continue drawing diagonals until the entire network consists entirely of triangles. The value of V - E + F for the network of triangles that we end up with has the same value as for the network we started with (the values of E and F have changed but the quantity V - E + F has not changed).

2. Second set of alterations. We now use a series of alterations to eliminate boundary triangles of the network until all triangles have been eliminated except for a single triangle. The boundary triangles may have one edge on the network boundary or two edges. We use one alteration for those with one edge on the boundary and another for those with two edges on the boundary.

1] Alteration for triangles with one edge on the boundary. Remove the boundary edge and the face. This alteration decreases E and F by 1, while V is unaffected. Thus the quantity V - E + F is unchanged.

2] Alteration for triangles with two edges on the boundary. Remove the two boundary edges, the vertex, and the face. This alteration decreases E by 2, V by 1, and F by 1. Thus the quantity V - E + F is unchanged.

Making these alterations, we work our way repeatedly around the network boundary until all but a single triangle remains. Fig. 1 shows an example of the procedure in which we have started with a rectangular solid (1). We imagine it to be hollow and constructed of thin rubber. We cut out the top face and stretch the remaining surface out flat on a plane to get the plane network (2). We then draw in diagonals and convert it to a network of triangles (3). Now we eliminate boundary triangles to get (4), (5) and finally (6). Triangle ABC in (3) is an example of a triangle with one edge on the boundary. Triangle DEF in (4) is an example of a triangle with two edges on the boundary.

The triangle we end up with has 3 edges, 3 vertices and one face. Thus the value of the quantity V - E + F for this triangle is 3 - 3 + 1 = 1. Hence in the original plane network the value of V - E + F is 1 and in the original polyhedron with one face missing the value of V - E + F is 1. Consequently, V - E + F = 2 for the original polyhedron. End of proof.