Calculus on Manifolds Problem 4-24 Problem $4\text{-}24$: 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 $c - c_{1,n} = \partial c^2$ 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 $0$.
Based on a previous problem where $c_{R,n}$ is defined to be $(R \cos 2\pi nt, R \sin 2\pi nt)$, we define $c_{1,n} = (\cos 2\pi nt, \sin 2\pi nt)$. I'm having trouble understanding what the hint means and how it's going to help in solving the problem. From my understanding, the interval $c([t_{i-1}, t_i])$ is a small section of the curve $c([0,1])$. Does having "each $c([t_{i-1}, t_i])$ contained on one side of some line through $0$" merely mean having a line through $0$ intersecting $c([t_{i-1}, t_i])$ at a single point?
 A: After a rotation if necessary, we may assume that the line is the $x$-axis, and $c(0)$ is a point on the positive $x$-axis. Moreover, let us also assume that $c$ travels in counterclockwise direction.
For $t>0$ close to $0$, the curve lies on the upper half plane $\{y\geq 0\}$. Since the curve is closed, it must eventually turn around back to $x$-axis in order to connect back to $c(0)$, so there will be a first time, denoted as $t_1$, such that $c(t_1)$ is a point on the $x$-axis. Thus the arc $c([t_0,t_1])$ lies entirely in the upper half plane; i.e. on the one side of the line which is the $x$-axis. Note that $0<t_1<1$.
Now it's easy to see how the things go on. As the curve continues the journey, there will be another time $t_2>t_1$ where $c(t_2)$ is again a point on the $x$-axis. The arc $c([t_1,t_2])$ lies entirely in the lower half plane $\{y\leq 0\}$. Of course, it is possible that $t_2=1$, in which case the curve $c$ is simple closed. Otherwise, we continue the same process.
Eventually we obtain a finite number of points
\begin{align}
0=:t_0<t_1<\cdots<t_N:=1
\end{align}
such that for each $1\leq i\leq N$, the arc $c([t_{i-1},t_i])$ is contained on one side of the $x$-axis. In fact, $N$ must be an even number, says $N=2n$. Drawing some pictures by yourself should be able to convince you on this.

To see how the points $\{t_i\}$ above help on the problem, firstly the $n=N/2$ above is the desired $n$ as in $c_{1,n}$. Recall from Problem 4-23 that $c_{1,n}$ wraps the unit circle $n$ times. Now the idea is that

the arc $c([t_{2i-2},t_{2i}])$ should correspond to a single wrap of the circle,

and we do the same for each of the $n$ arcs.
The 2-chain $c^2$ is a function of two variables, says $t$ and $r$. The parameter $t$ plays the role of wrapping the circle, while the parameter $r$ serves to deform the arc $c([t_{2i-2},t_{2i}])$ to the circle $c_{1,n}\left(\left[\frac{i-1}{n},\frac{i}{n}\right]\right)$ (same thing as in the case of annulus in Problem 4-23, except that now the shapes may be quite complicated to imagine). Therefore, we can define $c^2:[0,1]^2\to\mathbb{R}^2-\{0\}$ by
\begin{align}
c\bigg|_{\left[\frac{i-1}{n},\frac{i}{n}\right]\times[0,1]}(t,r)
:=(1-r)\cdot c(\tau_i(t))+r\cdot c_{1,n}(t)
\end{align}
where $\tau_i(t)$ is the point in $[t_{2i-2},t_{2i}]$ satisfying
\begin{align}
\frac{\tau_i(t)-t_{2i-2}}{t_{2i}-t_{2i-2}}
=\frac{t-\frac{i-1}{n}}{\frac{i}{n}-\frac{i-1}{n}}
\end{align}
I'll leave it for you to check that the $c^2$ here is the desired $2$-chain. (Perhaps the sign may not work well, but then you can just choose the minus of my $c^2$ as the answer).

A small remark: Note that in the definition of $c^2$ above, the term $c(\tau_i(t))$ appears but not $c(t)$. It thus seems to me that the statement $\partial c^2=c-c_{1,n}$ is true only up to a reparametrization of $c$. To be pedantic, one perhaps should add that there exists a reparametrization $\phi:[0,1]\to[0,1]$ such that
\begin{align}
\partial c^2=c\circ\phi-c_{1,n}
\end{align}
In our case above, the reparametrization $\phi$ is given precisely by
\begin{align}
\phi\bigg|_{\left[\frac{i-1}{n},\frac{i}{n}\right]}(t)=\tau_i(t)
\end{align}
Since $\tau_i(t)$ depends linearly on $t$, the map $\phi$ is indeed continuous.
