When calculating the Euler Characteristic of any regular polyhedron the value is 2. Since a sphere is homoeomorphic to all regular polyhedrons, the sphere ought to have a Euler Characteristic of 2 as well.
So:
$V-E+F=2$
holds true
A sphere obviously do not have vertices nor edges, which ought to mean they have 2 faces, which i assume are the inside and outside.
If that is the case, why dont you count the inside and outside as two seperate faces on any of the other regular polyhedrons? A tetrahedron for example only has 4 faces.
If not, then where is the other face.
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$\begingroup$For any triangulation of the sphere, it is true that $V-E+F=2$, where $V$ is the number of vertices in the triangulation, $E$ the number of edges in the triangulation and $F$ the number of faces in the triangulation. For example, consider the triangulation below:
There are $6$ vertices, $12$ edges and $8$ faces, so $V-E+F=6-12+8=2$.
There are also more complicated definitions of the Euler Characteristic in terms of homology or number of cells in each dimension in a CW complex. It can be defined as$$\chi(X)=\sum(-1)^n\mathrm{rank}(H_n(X))\,.$$
$\endgroup$ $\begingroup$Consider a sequence of simple configurations on the surface of a sphere. Start with two points on the surface and connect them by two edges. This divides the surface of the sphere into two faces. Thus $\,2-2+2=2.\,$ Now remove one of the edges. This leaves only one face on the sphere. Thus$\,2-1+1=2\,$. Now move the two points together until they merge into only a single point and there is no edge. Thus $\,1-0+1=2\,$ again.
NOTE: Topology enters at the first step with the two edges. On a sphere, they divide the surface into two faces which are topologically disks, but for other topologically different surfaces this may not be the case. For example, a torus requires more edges to be able to form faces which are topologically disks.
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