$a - b, b-c, c-d$ form a basis for this kernel. Hatcher P99 last paragraph:

Define a homomorphism $\partial: C_1 \to C_0$ by sending each basis element $a,b,c,d$ to $y-x$, the vertex at the head of the edge minus the vertex at the tail. Thus $$\partial (ka + lb + mc + nd) = (k + l + m + n)y - (k + l + m + n)x,$$
  and the cycles are precisely the kernel of $\partial$, and $a - b, b-c, c-d$ form a basis for this kernel.

But I can't see how $a - b, b-c, c-d$ form a basis for the kernel? I tried to show that $$r_1(a-b) + r_2(b-c) + r_3(c-d) = 0,$$
but it won't come out?
http://www.math.cornell.edu/~hatcher/AT/ATch2.pdf
 A: First, in that example $\;C_1\cong\Bbb Z^4\;$ = the free abelian group of rank $\;4\;$ , and it's clear that $\;\partial(C_1)\cong\Bbb Z\;$ , so by the first isomorphism theorem
$$C_1/\ker\partial\cong\partial(C_1)\cong\Bbb Z\implies \ker\partial\cong\Bbb Z^3$$
and we're indeed looking for three free (abelian) generators.
Now, as written there, we have that
$$ka+lb+mc+nd\in\ker\partial\iff k+l+m+n=0\implies\;\;\text{we can actually write:}$$
$$ka+lb+mc+nd=\alpha(a-b)+\beta(b-c)+\gamma(c-d)=\alpha a+(\beta-\alpha)b+(\gamma-\beta)c-\gamma d$$
$$\iff \begin{align*}k&=\alpha\\l&=\beta-\alpha\\m&=\gamma-\beta\\n&=-\gamma\end{align*}\iff\begin{cases}\alpha=k\\\beta=k+l\\\gamma=-n\end{cases}$$
For example:
$$3a-5b-2c+4d=3(a-b)-2(b-c)-4(c-d)$$
A: This doesn't specifically answer why $a-b, b-c, c-d$ form a basis, however, I think there is a nicer and easier way to see that a basis consists of three similar elements, namely: $$a-b, a-c, a-d.$$

First, those elements map to $0$ by definition of $\partial$. So, all are in $\ker \partial$.
Next, as written in the book: if $\partial (ka+\ell b+mc+nd)=0$, then $k+\ell+m+n=0$. So, just write $k=-\ell-m-n$. Then we have
\begin{align*}
ka+\ell b + mc +nd &= (-\ell -m-n)a+\ell b +mc +nd\\
&= -\ell(a-b)-m(a-c)-n(a-d).
\end{align*}
So, $ka+\ell b + mc +nd \in \langle a-b, a-c, a-d \rangle$.
Therefore, $$\ker \partial = \langle a-b, a-c, a-d \rangle.$$
