Simplicial homology and homeomorphisms

Question

In Hatcher's book, in the introduction page of singular homology, he mentions that "it is obvious that homeomorphic spaces have the same singular homology, in contrast to simplicial homology". However I thought that this was also true for simplicial homology (and looking at the construction I don't see why this would not be true for simplicial homology).

What does he mean by this statement? Does he point at the fact that not all spaces are triangulable and hence do not admit the construction of simplicial homology?

Answer 1

As Max points out in the comments, the same space might have different triangulations.

To elaborate, the construction of singular homology relies only on continuous maps, so it's "obvious" that singular homology is invariant under homeomorphism. However, the construction of simplicial homology relies on the additional structure of a triangulation of your space. You could have multiple triangulations of a space which might -- conceivably! -- produce different homology.

Because of this dependence in the construction of simplicial homology, you need to prove that simplicial homology is independent of triangulation. That extra work is why Hatcher says it is not obvious that simplicial homology is a homeomorphism invariant.

Answer 2

Neal's answer explains the problem of different triangulations. Nevertheless it turns out that the simplicial homology $H_*(\mathcal{T})$, where $\mathcal{T}$ is a simplicial complex triangulating $X$, is a topological invariant of $X$. That is, if $X_1, X_2$ are homeomorphic and $\mathcal{T}_i$ are triangulations of $X_i$, then $H_*(\mathcal{T}_1) \approx H_*(\mathcal{T}_2)$. The standard proof relies on "identifying" the simplicial homology of a triangulation of $X$ with the singular homology of $X$. This is a genuine topological proof.

Historically, the need for a topological proof was not so obviuos. In the "early days" mathematicians conjectured that a combinatorial proof was possible. It is a simple observation that if $\mathcal{T}$ is a triangulation of $X$ and $\mathcal{T}'$ is a subdivision of $\mathcal{T}$, then $H_*(\mathcal{T}) \approx H_*(\mathcal{T}')$. Now the so-called Hauptvermutung said that any two triangulations of a triangulable space have a common subdivision. This would obviuosly prove that $H_*(\mathcal{T})$ is a topological invariant of $X$.

Unfortunately the Hauptvermutung fails as was shown by John Milnor in 1961. Therefore there is no combinatorial proof.

See for example https://en.wikipedia.org/wiki/Hauptvermutung.