Why is the euler characteristic of a sphere 2? 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.
 A: 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
$$\chi(X)=\sum(-1)^n\mathrm{rank}(H_n(X))\,.$$
A: 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.
A: 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,
$$\chi(X) =2$$
