Why is the euler characteristic of a sphere 2?

Question

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.

Answer 1

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 $$\chi(X)=\sum(-1)^n\mathrm{rank}(H_n(X))\,.$$

Answer 2

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.

Answer 3

You can also use the hash addition formula, we have:

$$ \chi(X \# Y) = \chi(X) + \chi(Y) - 2$$

We have $X=Y$ here (both spheres) and know $X \# Y = X$ :

$$ \chi(X) = 2 \chi(X) - 2$$

Rearranging,

$$\chi(X) =2$$