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 $c^2$.
Definitions: A singular $n-$cube in $A\subset \mathbb{R}^n$ is a continous fuction $\alpha:[0,1]^k\to A$. A singular $n-$chain is a sum $\sum^k_{j=1} r_j\alpha_j$, where $\alpha_j$ is a singular $n-$cube and $r_j\in \mathbb{R}$.
So, in this case, $c$ is a curve in $\mathbb{R}^2$ and $c_{1,n}$ is a circumference with radio $1$ and a total of $n$ laps around the origin.
I can consider a line that intersects the origin, and if $\{U_{+}, U_{-}\}$ is an open cover of $\mathbb{R^2}-\{0\}$, then $\{c^{-1}(U_{+}), c^{-1}(U_{-})\}$ is an open cover of $[0,1]$.
I have problems to properly write the proof.
Is this a correct idea? How can I end the proof? Thanks.
Here's a quick proof if you know that $\pi_1(S^1,x)\simeq \mathbb{Z}$.
Since $\mathbb{R}^2-\{0\}$ is homotopy equivalent to the circle, after choosing a basepoint $x$, we have $\pi_1(\mathbb{R}^2-\{0\},x)\simeq \mathbb{Z}$. We can take the curve $c_{i,1}$ as a generator for this group. By definition of the group structure on $\pi_1$, we have $$c_{i,n}=nc_{i,1}\;.$$ Now choose any loop $c$. Then $c$ must be in some homotopy class of loops, which implies that it is homotopic to one of the $c_{i,n}$ and the homotopy between the two gives the desired 2-chain.