Let $\sigma: \Delta^n\rightarrow X$ be continuous and $n>0$. The boundary is defined to be the map
$\partial_n\sigma=\sum_{i=0}^k(-1)^i(\sigma \circ \epsilon_i^n)$, where $\epsilon_i^n$ is the ith face map.
Now, I just would like someone to verify that my reasoning is correct as a sanity check:
The reason $\partial_n\sigma\in S_{n-1}(X)$ , where $S_{n-1}(X)$ is the free abelian group of singular $n$ chains, where the basis is the set $S=\{ \sigma: \Delta^n\rightarrow X: \sigma \text{ continuous} \}$is because
For each $n$, $S_n(X)=\{ \sum_{i=0}^kn_i\sigma: n_i\in \mathbb{Z}$, $k\in \mathbb{N}\}$, where per usual, I am identifying $S\subseteq S_n(X)$ in the standard way.
Is this correct?