Hatcher's Algebraic Topology states on page 108 that singular homology is a special case of simplicial homology:

Though singular homology looks so much more general than simplicial homology, it can actually be regarded as a special case of simplicial homology by means of the following construction. For an arbitrary space $X$, define the singular complex $S(X)$ to be the $\Delta$-complex with one $n$ simplex $\Delta^n_\sigma$ for each singular $n$ simplex $\sigma: \Delta^n \to X$ with $\Delta^n$ attached in the obvious way to the $(n − 1)$ simplices of $S(X)$ that are the restrictions of $\sigma$ to the various $(n − 1)$ simplices in $\partial \Delta^n$ It is clear from the definitions that $H^\Delta_n(X)$ is identical with $H_n(X)$ for all $n$, and in this sense the singular homology group $H_n(X)$ is a special case of a simplicial homology group. One can regard $S(X)$ as a $\Delta$-complex model for $X$, although it is usually an extremely large object compared to $X$.

I would like to know if my understanding of this paragraph is correct:

We know that for $Y$ a $\Delta$-complex, its simplicial homology and its singular homology are the same, i.e. $H_n^\Delta(Y)= H_n(Y)$. This fact is not clear from the definition but can be proven with, for example, the Five Lemma. So the phrase "it's clear from the definition" probably shouldn't be used.

I think that maybe (and an explanation of this would be great) that over the singular complex $S(X)$, the singular homology and the simplicial homology are the same, i.e. $H_n(S(X))=H_n(X)$.

Using those two equalities we would obtain $$ H_n^{\Delta}(S(X))=H_n(S(X))=H_n(X) $$

is this correct?

This answer seems to imply I'm wrong, but I am not sure if I understand the given explanation.


1 Answer 1


Although the statements in your post are all true, none of them are correct interpretations of this passage from Hatcher's book.

Instead, what Hatcher is saying is that for any topological space $X$ there exists a simplicial complex $Y$ such that the singular chain complex $S(X)$ and the simplicial chain complex $\Delta(Y)$ are isomorphic as chain complexes. And then, because of this chain complex isomorphism, it follows that the singular homology of $X$ is isomorphic to the simplicial homology of $Y$.

Furthermore Hatcher's description of the isomorphism is defined by bijections, namely one bijection for each $n \ge 0$, between the basis of $S_n(X)$ (i.e. the set of singular $n$-simplices in the topological space $X$) and the basis of $\Delta_n(Y)$ (i.e. the set of $n$ simplices of the simplicial complex $Y$).

  • $\begingroup$ So $S(X)$ is the complex with objets $C_n(X)$, the sets of singular $n$-chains, while $Y$ is the result of "regularizing/fixing" $S(X)$ to make it a $\Delta$-complex. The result follows by the bijections/isomorphisms you describe. Is that correct? $\endgroup$
    – RyeCatcher
    Feb 22 at 19:59
  • $\begingroup$ and this "fixing/extra structure" was the whole point of Hatcher's paragraph $\endgroup$
    – RyeCatcher
    Feb 22 at 20:06
  • $\begingroup$ I think you're on the right track, but let me correct things slightly. The basis of $S_n(X)$ is the set of singular $n$-simplices of $X$ (i.e. the set of continuous functions from the standard $n$-simplex to $X$). And the set of $n$-simplices of $Y$ is bijectively indexed by the exact same set, namely the set of singular $n$-simplices of $X$. The "regularizing/fitting" procedure is what you use to to fit the simplices of $\Delta(Y)$ together, in just such a way that the bijection between bases extends to an isomorphism of chain complexes $S(X) \approx \Delta(Y)$. $\endgroup$
    – Lee Mosher
    Feb 22 at 21:12

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