This is an exercise from "Calculus on Manifolds" by Michel Spivack (first edition, p.100):
If $c$ is a singular $1$-cube in $\mathbb{R}^2-\{0\}$, with $c(0)=c(1)$, show that there is an integer $n$ such that $c-c_{i,n}=\partial c^2$ for some $2-$chain $c^2$.
Hint : First partition $[0,1]$ so that each $c([t_{i-1},t_i])$ is contained on one side of some line through $0$..
First of all i do not even understand the hint...
There is an exercise just before this one which says the following :
For $R>0$ and $n$ an integer, define the singular $1$ cube $c_{R,n} : [0,1]\rightarrow \mathbb{R}^2-0$ by $c_{R,n}(t)=(R\cos 2\pi n t,R\sin 2\pi nt)$. Show that there exists a singular $2$ cube $c:[0,1]^2\rightarrow \mathbb{R}^2-0$ such that $c_{R_1,n}-c_{R_2,n}=\partial c$.
This exercise was easy.. We have two circles of different radius with same centre.. This suggest me to consider shape of two concentric circles.. the path $sc_{R_1,n}(t)+(1-s)c_{R_2,n}(t)$ gives singular $2$ cube whose boundary is $c_{R_1,n}-c_{R_2,n}$..
I am thinking of using similar idea but with no luck..
The $2$ cube can be as wild as possible.. Hint says to consider partition such that $c([t_{i-1},t_i])$ lies one side of some line passing through origin..
We do not want $2$ cube but a $2$ chain.. So, we can consider some finite number of $2$ cubes and i believe that number has some thing to do with number of elements in the partition..
I have some vague ideas but that is good for nothing..
Help me to solve this...