Simplicial homology

Question

Im studying algebric topology and after defining $H_n^{sinp}(X) $ there is a theorem that states that $H_*^{sinp}(X) $ is independent of simplicial structure of $X$. My question is, where can i find a proof for this theorem? Also ive head that $H_*^{sinp}(X) $ depends only on geometric realisation of $X$. This is what this theorem is saying? Im a bit confused, a little explanation would help. Thanks

Answer 1

Generally speaking (and this is sort of a point that Hatcher emphasizes), simplicial homology is "fairly easy" to compute directly without knowledge about the homology of any other spaces. This allows you to get started computing the homology of some relatively simple spaces, even before knowing any of the basic properties a homology theory ought to have.

Most of the useful properties of homology (like homotopy invariance, excision and the long exact sequence) are easier to prove using singular homology. The fact that the two are naturally isomorphic (when restricted to "nice spaces") allows you to deduce these properties for simplicial homology. In particular, it implies the properties that you're asking about.

I'm not going to prove the isomorphism here, I'll leave that to Hatcher (as @Berni Waterman says, thats theorem 2.27). If you don't know singular homology, its worthwhile to learn that first before jumping right to the proof.

Answer 2

As others have mentioned, the most common proof is via the isomorphism between singular and simplicial homology. But if you want a purely simplicial proof, see Corollary 18.2 in James Munkres's Elements of Algebraic Topology.

The idea is to subdivide both triangulations enough that the identity map has a simplicial approximation, which therefore induces an isomorphism on simplicial homology.