Theorem 1. All cell decompositions of a sphere $S$ have Euler characteristic 2.
This is well-known, but I had this idea for an intuitive proof: for any cell decomposition $\Gamma$ with $V$ vertices, $E$ edges and $F$ faces, let $G(\Gamma) = 1 – (V – E + F)/2$. $G(\Gamma)$ is the genus of the surface represented by $\Gamma$ (if the surface is closed and orientable), but suppose we don't know that yet.
Suppose we have two cell decompositions $\Gamma$ and $\Gamma'$, and we glue them together by removing one face from each and gluing the exposed edges together, like the connected sum $\#$. It can be verified that $G(\Gamma\#\Gamma') = G(\Gamma) + G(\Gamma')$.
Pardon the lack of rigor but I had the vague idea of using the fact that $G(S) = G(S\# S) = 2G(S)$ thus $G(S) = 0$, and Theorem 1 immediately follows. This is an intuitive proof because I allow the assumption that $S\# S = S$. But even so this seems to be circular because I first have to show that $G$ depends on the surface, not the cell decomposition. I have no idea how to do that without proving Theorem 1 beforehand.
Does this vague idea have any value?
Best Answer
The only way I know in order to show that the Euler characteristic we compute does not depend on the cell decomposition is to use homology groups: it shows that $\chi(S)$ depends only on the homotopy type of $S$. Afterwards, it is sufficient to chose a nice cell decomposition, namely a triangulation with just four vertices, and check that the associated Euler characteristic is 2.
However, if you are just interested in triangulations rather than cell decompositions, thanks to a well-chosen stereographic projection, the property you mention is equivalent to Euler's formula on planar graphs. More details may be found here for example.