On Hatcher's proof that $H_0(X)$ is a direct sum of $\mathbb{Z}$s?

Question

I'm confused about part of the proof of Proposition 2.7 in Hatcher.

If $X$ is nonempty and path-connected, then $H_0(X)\simeq\mathbb{Z}$. Hence for any space $X$, $H_0(X)$ is a direct sum of $\mathbb{Z}$s, one for each path-component of $X$.

Proof: By definition, $H_0(X)=C_0(X)/\operatorname{Im}\partial_1$. Define a homomorphism $\epsilon\colon C_0(X)\to\mathbb{Z}$ by $\epsilon(\sum_i n_i\sigma_i)=\sum_i n_i$. This is surjective, so the proof proceeds to show $\operatorname{Im}\partial_1=\ker\epsilon$.

To show $\ker\epsilon\subset\operatorname{Im}\partial_1$, suppose $\epsilon(\sum_i n_i\sigma_i)=0$, so $\sum_i n_i=0$. The $\sigma_i$ are singular $0$-simplicies, which are simply points of $X$. Choose a path $\tau_i\colon I\to X$ from a basepoint $x_0$ to $\sigma_i(v_0)$, and let $\sigma_0$ be the singular $0$-simplex with image $x_0$. We can view $\tau_i$ as a singular $1$-simplex $\tau_i\colon [v_0,v_1]\to X$ and then we have $\partial\tau_i=\sigma_i-\sigma_0$.

This final sentence is confusing to me. I know we can think of $1$-simplicies as line segments, so it sort of makes sense to think of $\tau_i$ as a singular $1$-simplex. But I thought then $\partial\tau_i=\tau_i|[v_1]-\tau_i|[v_0]$, why does Hatcher say $\partial\tau_i=\sigma_i-\sigma_0$?

Answer 1

Because $\tau_i| [v_1]$ is $\tau_i$ evaluated at $v_1$, which is just $\sigma_1$.

Answer 2

You are almost all the way there. You have $\partial \tau_i=\tau_i|[v_1]-\tau_i|[v_0]$, but $\tau_i|[v_j]$ is just mapping $v_j$ to a point of $X$, i.e. it is a zero simplex. By construction, it coincides with $\sigma_i(v_j)$.