After Hatcher presents the augmented chain complex:
I'm trying to make sense of the following comment in Hatcher's Algebraic Topology:
Formally, one can think of the extra $\mathbb{Z}$ in the augmented chain complex as generated by the unique map $[\emptyset]\to X$ where $[\emptyset]$ is the empty simplex, with no vertices. The augmentation map $\epsilon$ is then the usual boundary map since $\partial [v_0] = [\hat{v}_0] = [\emptyset]$.
I'm having a bit of trouble reconciling the last statement, specifically: "The augmentation map $\epsilon$ is then the usual boundary map...".
Does he mean that, formally, we would define $\tilde{\epsilon}=\phi\big(\epsilon(\sum_i n_i\sigma_i)\big)$ where $\phi$ is the guaranteed isomorphism between $\mathbb{Z}$ and the free abelian group generated by the one map $[\emptyset]\to X$ (which is the empty map)? If so, is my below attempt at showing that $\tilde{\epsilon}$ is the boundary map correct?
Let $\sigma : \Delta^0 \to X$, i.e. $\sigma : [v_0] \to X$. Then:
$$ \phi(\epsilon(\sigma))=\phi(1)=\sigma_\emptyset$$ where $\sigma_\emptyset$ is the empty map.
But $\sigma_\emptyset$ is the same as $\partial(\sigma)$ since $\sigma$ restricted to $[\emptyset]$ can be defined in only one way, which is the empty map, hence equal $\sigma_\emptyset$. So this (hopefully correctly) proves the statement (?)
Also, just to verify, in the statement: $\partial [v_0] = [\hat{v}_0]$, the notations $[v_0]$ and $[\hat{v}_0]$ are just used as shorthand for $\sigma: \Delta^0 \to X$ and its associated restriction, respectively, right?
