# Singular Homology is a special case of Simplicial Homology

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.

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 $$Y$$).
• 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? Feb 22 at 19:59
• 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 $S(X) \approx \Delta(Y)$. Feb 22 at 21:12