Take the simplicial complex with vertices {a,b,c} and edges {ab, ac, bc}. In other words, a circle. If I build a chain complex, and make the matrix of my differential, I get that the kernel of $C_0\to0$ mod image of $C_1 \to C_0$ is
$\displaystyle \frac{\mathbb{Z}[a] \times \mathbb{Z}[b] \times \mathbb{Z}[c]} {\mathbb{Z}[a-c] \times \mathbb{Z}[b+c]}$
where the bracket denotes a generator of the group.
Now from basic dimension arguments (and lots of other ways) this has to be just one copy of $\mathbb{Z}$. But how can I see or figure out what will be the generator of my quotient group in terms of my other generators?
If $c=d$, then $C_0=\langle a,b,c\rangle$ and the image of $C_1$ is $H=\langle a-b,a-c,b-c\rangle$. Since $b-c=(a-c)-(a-b)$, $H$ is actually generated by two elements, $H=\langle a-b,a-c\rangle$.
So the quotient group, $C_0/H$, is generated by $a+H$. That's because $b=a+(-(a-b))\in a+H$, $c=a+(-(a-c))\in a+H$. so $a+H=b+H=c+H$.