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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?

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    $\begingroup$ He means that for simplicial homology it's not obvious, because you may have different triangulations for a given space $\endgroup$ Sep 17, 2018 at 16:13
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    $\begingroup$ Rather a triangulable space will have many different triangulations, and it is not obvious that each of these triangulations will have isomorphic homology. $\endgroup$ Sep 17, 2018 at 16:14

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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.

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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.

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  • $\begingroup$ So in this spirit it is only obvious that two homeomorphic spaces have the same simplicial homology with respect to a certain triangulation (which is how i looked at it)? $\endgroup$
    – NDewolf
    Sep 17, 2018 at 17:04
  • $\begingroup$ @NDewolf Not quite; if two triangulated spaces have a homeomorphism between them that respects the triangulation, then it is "obvious" that they have isomorphic simplicial homology. Key is that you need to take triangulations into account. $\endgroup$
    – Neal
    Sep 17, 2018 at 18:59
  • $\begingroup$ @NDewolf ... the important point here is that simplicial homology is a priori constructed on a tuple $(\mbox{topological space}, \mbox{triangulation})$. Even when you only consider one space $X$, you have to do work to show that simplicial homology doesn't change when you change the triangulation. Why should $H_*(X,\mathcal{T}) \sim H_*(X,\mathcal{S})$ for two different triangulations $\mathcal{T}$, $\mathcal{S}$? $\endgroup$
    – Neal
    Sep 17, 2018 at 19:07
  • $\begingroup$ But isn't a triangulated space just a topological space homeomorphic to a simplicial complex. So given a triangulated space $(X, \mathcal{T})$ and a space $Y$ homeomorphic to $X$, isn't $Y$ trivially also triangulated by $\mathcal{T}$? (Just by composing the homeomorphisms) $\endgroup$
    – NDewolf
    Sep 17, 2018 at 21:12
  • $\begingroup$ @NDewolf yes- but what if $Y$ has a different triangulation? No guarantee the homology constructed from that same triangulation is the same as the homology constructed from the pushed-forward triangulation. $\endgroup$
    – Neal
    Sep 17, 2018 at 21:33

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