Prove that if $X$ is a nonempty path connected space, then $H_0(X) \approx \Bbb Z$, where $H_0(X) = Z_0(X) / B_0(X)$.
If $B_0(X) = \{\sum m_xx \in S_0(X) : \sum m_x = 0\}$,
then by the first homomorphism theorem the map $\theta : Z_o(X) \rightarrow \Bbb Z$ by $\sum m_xx \mapsto \sum m_x$ is the required isomorphism.
But I'm having trouble showing that $B_0(X) = \{\sum m_xx \in S_0(X) : \sum m_x = 0\}$ by showing an element in the first set is an element of the second set.
I can see that by definition an element of $B_0(X)$ is of the form $\sum_i m_i [\sigma_i(\{(1,0)\})-\sigma_i(\{(0,1)\})]$ or $\sum_i m_i [x_{i, (1,0)}-x_{i, (0,1)}]$, but from here I can't see how to go any further.
Anyone have any ideas? I'm self learning this, so as clear an explanation as possible would be helpful.
Fix $x_0\in X$
$B_0(X)$ is generated by the $[x]-[y]$ where $x$ and $y$ are endpoints of a path in $X$. Therefore it is generated by the $[x]-[y]$ for all $x$ and $y$ in $X$. The augmentation map $\varepsilon:[x]\mapsto1$ vanishes on $B_0(X)$, and each element of $\ker\varepsilon$ has the form $\sum m_x[x]$ where $\sum_x m_x=0$ and then $\sum_x m_x[x]=\sum_x m_x([x]-[x_0])\in B_0(X)$.