Why is the Euler characteristic a topological invariant that is independent of the triangulation

general-topologygeometric-topology

In Do Carmo's "Differential Geometry of Curves and Surfaces", in the proof of the Gauss-Bonnet theorem he gives 2 propositions he does not prove, namely:

Let $S \subset \mathbb{R}^3$ be a compact connected surface, then
$\chi(S)$ assumes one of the values $2,0,-2,…$. Additionally, if
$S'\subset \mathbb{R}^3$ is another compact surface with
$\chi(S)=\chi(S')$, then $S$ is homeomorphic to $S'$.

and

If $R \subset S$ is a regular region of a surface $S$, then the
Euler-Poincaré characterestic does not depend on the triangulation of
$R$. Hence, it can be denoted as $\chi(R)$.

I struggle to find proofs of these two statements that do not involve homology theory, can any of you maybe recommend one that is fairly "elementary"? I assume even a general outline of the proof goes beyond the scope of a comment here.

Best Answer

For the special case of a compact surface $S$, one can prove invariance of Euler characteristic by the following steps.

Step 1: Prove that the Euler charactersitic of a triangulation of $S$ is invariant under subdivision of that triangulation.

Step 2: Prove that any two triangulations of $S$ have subdivisions which are simplicially isomorphic.

Step 1 is not too hard to do by some kind of induction argument.

Step 2 on the other hand is quite hard. It is more-or-less equivalent in difficulty to Rado's Theorem saying that every surface has a triangulation.

There are other special cases where this same 2-step method works. For example it works for compact 3-dimensional manifolds, using a much, much harder theorem of Moise which generalizes Rado's Theorem.

Unfortunately, there are examples of topological spaces for which Step 2 fails completely. So, in general, there is no hope for it: you'll just have to lock yourself in a room and learn singular homology theory.