Difference between simplicial and singular homology?

I am having some difficulties understanding the difference between simplicial and singular homology. I am aware of the fact that they are isomorphic, i.e. the homology groups are in fact the same (and maybe this doesnt’t help my intuition), but I am having trouble seeing where in the setup they differ.

To my understanding, the singular chain complex on a space X consists of the free abelian groups generated by the sets of n-simplices in X, where an n-simplex in this context is a continuous map σ:ΔnX from the standard geometric n-simplex Δn to X, with boundary map n=ni=0(1)idi where di:Cn(X)Cn1(X) is the ith face map (“deleting” the ith vertex).
The singular homology groups are then the homology groups of this complex (ie. Hn(X)=ker(n)/im(n+1)).

Now for the simplicial homology, we have a simplicial complex S, which is a set of (abstract?) ordered simplices, such that a face of any simplex in S is itself a simplex in S. Then we form the simplicial chain complex where Cn(S)=Z[Sn], where SnS is the set of n-simplices in S, i.e. the free abelian group generated by Sn. This complex has boundary operator n=ni=0(1)idi, where di is the ith face map. The homology groups of this is HΔn(S)=ker(n)/im(n+1).
Now for this to make any sense in a topological framework, we have the realization of S, |S|=(Sn×Δn)/(diσ,y)(σ,diy) for all (σ,y)Sn×Δn1 and di is the coface map.
(As I understand it, S is a blueprint of how to “assemble” the geometric n-simplices to form a space).
And then of course, if you want to talk about a specific space X, you need to find a simplicial complex S, whose realization is homeomorphic to X.

I can see very well that these two are two very different ways of building up the framework, but what I don’t understand is where in practice it differs. Don’t they both require that you find a way to divide X into n-simplices? The only difference I see, is whether you map from Δn into X before or after you form your homology groups, but there must be something I’m missing…

Answer

Just to sum up, mostly for my own reference, but I thought others might find it useful. (I am new to the site, so please excuse me if this shouldn’t be an answer…)

First some preliminary notions:

For a topological space X, an n-simplex in X is a continuous map ΔnX from the standard geometric n-simplex Δn into X. The maps di:Δn1Δn, sends Δn1 to the face of Δn sitting opposite the ith vertex of Δn.

An ordered n-simplex is a partially ordered set n+={0<1<<n}. The n+1 elements of n+ is called the vertices of σ. The subsets of n+ are called the faces of σ. There are morphisms of simplices di:(n1)+n+ called coface maps, given by d^i((n-1)_+) = \{ 0 < 1 < \dots < î < \cdots < n \} omitting the ith vertex of n_+.

Then for the two homologies:

The singular (unreduced) chain complex on a space X, is the chain complex
\cdots \xrightarrow{\partial_{n+1}} C_n(X) \xrightarrow{\partial_n} C_{n-1}(X) \xrightarrow{\partial_{n-1}} \cdots C_1(X) \xrightarrow{\partial_1} C_0(X) \to 0
where C_n(X) is the free abelian group \mathbb{Z}[S_n(X)] generated by the set S_n(X) = \{ \sigma : \Delta^n \to X \} of all n-simplices in X (i.e. the set of all continuous maps \Delta^n \to X). The boundary maps \partial_n : C_n(X) \to C_{n-1}(X) is given by \partial_n (\sigma) = \sum_{i=0}^{n}(-1)^i \sigma d^i : \Delta^{n-1} \to \Delta^n \to X.

The nth homology group H_n(X) = \ker(\partial_n) / \text{im}(\partial_{n+1}) of this complex is the nth singular homology group of X.

A simplicial complex S is a set S = \bigcup_{n=0}^{\infty} S_n where S_n = S(n_+) being a set of ordered n-simplices, such that a face of any simplex in S is itself a simplex in S. The simplicial chain complex
\cdots \xrightarrow{\partial_{n+1}} C_n(S) \xrightarrow{\partial_n} C_{n-1}(S) \xrightarrow{\partial_{n-1}} \cdots C_1(S) \xrightarrow{\partial_1} C_0(S) \to 0
consists of the free abelian groups C_n(S) = \mathbb{Z}[S_n] generated by the n-simplices. The boundary map \partial_n : C_n(S) \to C_{n-1}(S) is given by \partial_n(\sigma) = \sum_{i=0}^n (-1)^i d_i \sigma where d_i = S(d^i) : S_n \to S_{n-1} is the face maps d_i(\sigma) = \sigma \circ d^i.

The nth homology groups of this complex H^\Delta_n(S) = \ker(\partial_n) / \text{im}(\partial_{n+1}) is the nth simplicial homology group of S.

Lastly we have the realization of S, |S| = \coprod (S_n \times \Delta^n) / \left((d_i \sigma, y) \sim (\sigma, d^iy) \right) for all (\sigma, y) \in S_n \times \Delta^{n-1}, where d_i \sigma \times \Delta^{n-1} is identified with the i'th face of \sigma \times \Delta^n.

Then if you want to say something about a specific space X, you need to find a simplicial complex S, whose realization is homeomorphic to X (i.e. you triangulate X and find the homology groups of the resulting simplicial complex).

NOTE: Feel free to edit any mistakes and clarify where you find it necessary. I'm still not 100% comfortable with it yet..

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