Hatcher definition algebraic topology elaboration In page 108, hatcher says the boundary map $\partial:C_n(X)\rightarrow C_{n-1}(X)$ is defined by
$\partial_n{\sigma}=\sum_i(-1)^i\sigma|_{[v_0,...,\hat{v_i}..,v_n]}$. He then immediately states
"Implicit in this formula is the canonical identification of $[v_0,....\hat{v_i},...,v_n]$ with $\Delta^{n-1}$, preserving ordering of vertices.."
Can someone elaborate on the quoted text, please?
 A: The singular $n$-chain $\sigma \in C_n(X)$ is a formal $\Bbb Z$-linear combination of maps $\Delta^n \to X$, so the boundary map suffices to be defined on generators (i.e., on a single map $\Delta^n \to X$). Thus, let $\sigma: \Delta^n \to X$ be a singular $n$-simplex in $X$.
$\Delta^n$ can be written as an ordered list of vertices $[v_0, ..., v_n]$ (if you specify $n+1$ vertices, you've specified an $n$-simplex, and the converse works too; you can see pictures of this idea on page 103).
$\partial_n (\sigma)$ needs to be a singular $(n-1)$-chain, so it needs to be a formal $\Bbb Z$-linear combination of singular $(n-1)$-simplices. For each $i$ you want to delete, $[v_0, ... \hat{v_i}, ...,v_n]$ is an ordered list of vertices with $n$ entries, so it is an $(n-1)$-simplex, and since it's a list containing (all but one of) the vertices of $\Delta^n$, it is an $(n-1)$-simplex making up one of the faces of $\Delta^n$ . You can see pictures of this idea on page 105.
Thus it makes sense to talk about the restriction of $\sigma$ to this simplex, so $\sigma|_{[v_0,...,\hat{v_i},...,v_n]}$ is a map $\Delta^{n-1} \to X$. We can add these, so the alternating sum of these as $i$ varies is an element of $C_{n-1}(X)$. The note about preserving the ordering of vertices just ensures that when you delete a vertex from your list and look at the remaining list, you do indeed get the exact face of the original simplex that contains all the vertices but the one you deleted.
