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

Question

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.

Answer 1

I have never seen any Euler's characteristic without something involving homological algebra (or generalization of homological algebra such as triangulated categories). Check the part generalization in this wikipedia page (the French one is even clearer).

It is very much not probable that you will find a proof that does not use it. I see two possibles scenarios:

• You use homological algebra with our saying it and you make it look like it does not look like it.
• You use a restriction of the cases you study, like you only consider polyhedra or something like that and in this case you could try a combinatorial proof as Euler did for its proof of Euler-Poincaré formula.

So you should try to study basics of homological algebra (it is very useful in general) and how homology is computed for CW-complexes, and how does the Euler-Poincaré's characteristic can be computed on such sets.

When you have done that, it will become very clear to you why it is a topological invariant.

Answer 2

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.