Calculation kernel of boundary operator. (Easy example in Hatcher) Consider the graph as given in the third page (pg.99) in the link below. 
https://www.math.cornell.edu/~hatcher/AT/ATch2.pdf
We 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 edges minus the vertex at the tail. Thus we have $$ \partial(ka + lb + mc + nd) = (k+l +m +n) y - (k+l + m +n )x$$, and the cycles are precisely the kernel of $\partial$.
Apparently it is a simple calculation to verify that $a-b$, $b-c$, and $c-d$ form a basis for this kernel.
Unfortunately, I do not know what this calculation would be. It would be nice to see an example of a such a calculation in this easy example. 
 A: First, assume that $\sigma = ka + lb + mc + nd$ belongs to the kernel of $\partial$. By what you have written, this is equivalent to $k + l + m + n = 0$. Then, noticing that $$\sigma = k(a-b) + (l + k)(b-c) + (l+k+m)(c-d) + (l + k + m + n)d$$$$ = k(a-b) + (l+k)(b-c) + (l+k+m)(c-d)$$ we see that $\sigma$ lies in the span of $a-b$, $b-c$, and $c-d$.
Second, it is clear that $a-b$, $b-c$, and $c-d$ belong to the kernel of $\partial$, so we can conclude from this and the first part that the kernel of $\partial$ is generated by $a-b$, $b-c$, and $c-d$.
It only remains to see that they form a basis of the kernel. To prove this we must show they are linearly independent. Assume then that $n_1(a-b) + n_2(b-c) + n_3(c-d) =0$. Expanding this out, we have $$n_1a + (n_2 - n_1)b + (n_3-n_2)c - n_3d = 0$$ Since $a$, $b$, $c$, and $d$ are linearly independent in $C_1$, we conclude that $n_1 = 0$, $n_2 - n_1 = 0$, $n_3 - n_2 = 0$, and $n_3 = 0$, and thus that $n_1 = n_2 = n_3 = 0$. This completes the proof.
