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 $\sigma : \Delta^n \to X$ from the standard geometric $n$-simplex $\Delta^n$ to $X$, with boundary map $\partial_n = \sum_{i=0}^n (-1)^i d_i$ where $d_i: C_n (X) \to C_{n-1} (X)$ is the $i$th face map ("deleting" the $i$th vertex).
The singular homology groups are then the homology groups of this complex (ie. $H_n(X)=\ker(\partial_n)/\text{im}(\partial_{n+1})$).
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 $C_n(S) = \mathbb{Z}[S_n]$, where $S_n \subset S$ is the set of $n$-simplices in $S$, i.e. the free abelian group generated by $S_n$. This complex has boundary operator $\partial_n = \sum_{i=0}^n (-1)^i d_i$, where $d_i$ is the $i$th face map. The homology groups of this is $H^\Delta_n(S) = \ker(\partial_n) / \text{im}(\partial_{n+1})$.
Now for this to make any sense in a topological framework, we have the realization of $S$, $|S| = \coprod (S_n \times \Delta^n) / (d_i \sigma, y) \sim (\sigma, d^iy)$ for all $(\sigma, y) \in S_n \times \Delta^{n-1}$ and $d^i$ 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 $\Delta^n$ into $X$ before or after you form your homology groups, but there must be something I'm missing…
Best 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 $\Delta^n \to X$ from the standard geometric $n$-simplex $\Delta^n$ into $X$. The maps $d^i: \Delta^{n-1} \to \Delta^{n}$, sends $\Delta^{n-1}$ to the face of $\Delta^n$ sitting opposite the $i$th vertex of $\Delta^n$.
An ordered $n$-simplex is a partially ordered set $n_+ = \{ 0 < 1 < \cdots < n \}$. The $n+1$ elements of $n_+$ is called the vertices of $\sigma$. The subsets of $n_+$ are called the faces of $\sigma$. There are morphisms of simplices $d^i: (n-1)_+ \to n_+$ called coface maps, given by $d^i((n-1)_+) = \{ 0 < 1 < \dots < î < \cdots < n \}$ omitting the $i$th 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 $n$th homology group $H_n(X) = \ker(\partial_n) / \text{im}(\partial_{n+1})$ of this complex is the $n$th 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 $n$th homology groups of this complex $H^\Delta_n(S) = \ker(\partial_n) / \text{im}(\partial_{n+1})$ is the $n$th 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..