Boundary of a $2$-chain If $c$ is a singular $1$-cube in $\mathbb{R}^2 \backslash \left\{ (0,0)\right\}$ with $c(0)=c(1)$, show that there is an integer $n$ such that: $c-c_{(1,n)} = \partial (c^2)$ for some $2$-chain $c^2$.

Ok, so, as Jesse says, this is problem 4-24 from Little Spivak. Maybe you should look at problem 4-23, where he explains what $c_{(1,n)}$ is: a parametrization of the circumference of radius $1$, centered at the origin, that goes round the origin $n$ times
$$
c_{(1,n)} (t) = (\cos 2\pi nt, \sin 2\pi nt) \ .
$$
So the problem says that, given any closed parametrized curve $c(t)$, there is an integer $n$ such that this $c$ together with $c_{(1,n)}$ are the boundary of some $2$-chain $c^2$.
If I'm not wrong, this $n$ is the winding number of $c$; that is, the number of times $c$ goes round the origin. So, if $c$ goes round the origin one time, you need the circumference $c_{(1,1)}$ that goes round the origin once too; if $c$ goes round twice, you need $c_{(1,2)}$...
Maybe you should try first problem 4-23, where Spivak asks a similar question, but where $c$ is also a circumference (of different radius). There you see too that you need both circumferences to go round the origin the same number of times in order they are the boundary of some $2$-chain. For instance, if $n=1$ (just once round the origin), these circumferences are the boundary of a regular anulus.
Anyway, you may try to take a look at Little Spivak solutions. They are quite right. For instance, in the solution to 4-24, Ken Kubota (the author of these solutions), begins computing $L = \int_0^1 c^*(d\theta)$, which is the winding number of $c$ multiplied by $2\pi$.
