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Let $X$ be Hausdorff. Suppose further that it is triangulable. Let $K$ and $L$ be two simplicial complexes such that their underlying space $|K|=|L|=X$. It is a lot of work to show (see Chapter 2 of Munkres' Elements of Algebraic Topology) that the simplicial homology groups are the same, i.e., independent of the simplicial structure and only dependent on the underlying space.

Compare the situation with a CW-complex. The fact that the cellular homology is independent of the filtration of the space is proved in only two pages by Hatcher as the result that $H^{CW}_n(X) \simeq H_n(X)$. The proof is not only much shorter but works for a wider category.

My question, which may seem strange, is:

Where is all the work going?

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1 Answer 1

The key idea in proving the topological invariance of the simplicial homology groups as outlined in Munkres' textbook is that it suffices to prove the following correspondence is functorial: the association of a simplicial chain complex $C(K)$ to a simplicial complex $K$ and the association of a sequence of homomorphisms $(f_{*})_p:H_p(K)\to H_p(L)$ in simplicial homology for all nonnegative integers $p$ to a continuous function $f:\left|K\right|\to \left|L\right|$ between polyhedra.

An alternate way to prove the topological invariance of the simplicial homology groups is to construct an isomorphism between simplicial homology theory and singular homology theory. In other words, one must construct for each simplicial complex $K$ a chain map $\eta:C(K)\to S(\left|K\right|)$ that induces an isomorphism in homology in all dimensions and is natural (i.e., commutes with homomorphisms in homology induced by continuous functions). In order for this to be an isomorphism of theories, one needs also that $\eta$ commutes with the homology boundary homomorphism in the long exact homology sequence.

In fact, the proof outlined in the second paragraph above is much simpler (it is carried out in pages 190 - 193 inclusive of Munkres' textbook). The idea is to choose a partial ordering of the vertices of a simplicial complex $K$ that induces a linear ordering on the vertices of each simplex of $K$. We can then define a chain map $\eta:C(K)\to S(\left|K\right|)$ by the rule $\eta([v_0,\dots,v_p])=l(v_0,\dots,v_p)$ where $[v_0,\dots,v_p]$ is an oriented $p$-simplex of $K$ such that $v_0<\cdots<v_p$ in the given ordering and $l(v_0,\dots,v_p)$ is the linear singular $p$-simplex in $\left|K\right|$ mapping the vertices of the standard $p$-simplex onto the vertices $v_0,\dots,v_p$ of $K$ (in the given order).

In this case, one can prove with some effort that $\eta$ induces isomorphisms in homology in all nonnegative dimensions; one does not need the theory of subdivisions as in the proof outlined in the first paragraph. It is interesting to note that the homomorphisms on homology induced by $\eta$ do not depend on the given partial ordering of the vertices of $K$ (but, of course, the chain map $\eta$ does). One can also prove that the homomorphism on homology $\eta_{*}$ is natural with respect to homomorphisms on homology induced by simplicial maps. Unfortunately, one cannot easily prove the naturality of $\eta_{*}$ with respect to arbitrary continuous maps for one needs to know that we can associate a homomorphism in homology to an arbitrary continuous map between polyhedra (and at least one proof of this fact requires the theory of subdivisions).

In particular, the topological invariance of the simplicial homology groups follows without the use of the theory of subdivisions. In fact, one can fairly easily prove that singular homology and cellular homology are isomorphic for a CW-complex (as you have pointed out).

In summary, the reason the proof of the topological invariance of the simplicial homology groups is lengthy in Munkres' exposition is Munkres seeks to prove a stronger fact (from which the topological invariance of the simplicial homology groups follows as a corollary); namely, that one can associate a homomorphism in homology to any continuous map between polyhedra and that this association is functorial. The proof of this given in Munkres requires the theory of subdivisions which is lengthy. On the other hand, a much simpler proof of the topological invariance follows from the isomorphism between simplicial and singular homology theory. The disadvantage of this proof, however, is that one cannot derive the result that a continuous map between polyhedra induces a homomorphism in homology from the techniques used in this proof.

In fact, I quote a paragraph in the preface of Munkres' textbook:

The instructor who wishes to do so can abbreviate the treatment of simplicial homology by omitting Chapter 2. With this approach the topological invariance of the simplicial homology groups is proved, not directly via simplicial approximations as in Chapter 2, but as a consequence of the isomorphism between simplicial and singular theory (section 34).

I hope this helps!

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Once you have a functorial chain homotopy between two functorial chain complexes, doesn't the isomorphism of homology functors automatically commute with the connecting maps in the long exact sequence? –  Zhen Lin Dec 26 '11 at 14:03
    
Hi. Thanks for your reply. But, I am sorry. What you've written, which is very informative though not unknown to me, is not really the kind of answer I am looking for. However, reading your answer, I have a feeling that all that 'missing' work may been subsumed into establishing excision. I will try to make this clearer in my head. –  doofus Dec 26 '11 at 14:18
    
@doofus The proof that simplicial homology theory satisfies the excision axiom is trivial. On the other hand, the proof that singular homology theory satisfies the excision axiom requires a little work. I believe one needs to prove that if $\cal A$ is a collection of subsets of a topological space $X$ whose interiors cover $X$, then the inclusion map $S^{\cal A}(X)\to S(X)$ induces an isomorphism in homology (both ordinary and reduced). In this case, $S^{\cal A}(X)$ is the subchain complex of the singular chain complex of $X$ consisting of the ``$\cal A$-small chains" of $X$ ... –  Amitesh Datta Dec 26 '11 at 14:39
    
@doofus ... However, the case of the excision axiom used in the proof of the existence of an isomorphism between the simplicial and singular homology theories is trivial. On the other hand, a non-trivial case of the homotopy axiom of the singular homology theory is used in this proof. The proof of the homotopy axiom requires care (and can be derived from the the acyclic model theorem). However, even this proof is not as involved as the totality of those constituting the theory of subdivisions of simplicial complexes. –  Amitesh Datta Dec 26 '11 at 14:42
    
@Zhen I am not sure what you mean; could you please clarify? I agree that the map $\eta$ of my answer commutes with the homology boundary homomorphism in the long exact homology sequence and that this is an easy consequence of the naturality of the zig-zag lemma (the naturality of $\eta$ implies, in particular, that $\eta$ commutes with inclusions) if this is what you meant. –  Amitesh Datta Dec 26 '11 at 14:45
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