Barycentric Subdivision of General Chains. Define $S:C_n(X) \to C_n(X)$ by setting $S\sigma = \sigma_\# S\Delta^n$ for a singular $n$-simplex $\sigma: \Delta^n \to X$.
How are we using $S$ in the definition of $S$?
Is this the $S$ from the previous part of the proof?
Is $\Delta^n$ now a map?
I know:
Since $\sigma: \Delta^n \to X$ is a $n$-simplex, then $\sigma_\#: C_n(\Delta^n) \to C_n(X)$ is the chain homomorphism.
In this formula $\Delta^n$ refers to the identity map on the space $\Delta^n$, considered as a linear $n$-chain on $\Delta^n$. So $S\Delta^n$ is defined as in part (2) of the proof as another linear $n$-chain on $\Delta^n$ (as the next line of the proof says, it's just a sum of all the simplices in the barycentric subdivision of $\Delta^n$ with appropriate signs). Then, $\sigma_\# S\Delta^n$ is the $n$-chain on $X$ obtained by applying $\sigma$ to $S\Delta^n$.