The Problem: Suppose $I=[0, 1]$, $1\in S^1\subset\mathbb{C}$, $f, g, h: I\to S^1\times I$ are paths defined by $f(s)=(e^{2\pi is}, 1)$, $p(s)=(1, s)$, and $h(s)=(e^{2\pi is}, s)$. Are $p\ast f\ast\bar{p}$ and $h\ast f\ast\bar{h}$ path homotopic? Here $g\ast k$ denotes the path $$\sigma(s)= \begin{cases} g(2s), s\in[0, 1/2] \\[2ex] k(2s-1), s\in[1/2, 1] \end{cases}$$, and $\bar{g}(s)$ denotes the path $g(1-s)$.
My Attempt: I tried to establish the straight-line homotopy, which failed because the points would not all be contained in $S^1\times I$. Then I tried to show that they are not homotopic, by assuming that $p\ast f\ast\bar{p}$ is path homotopic to $h\ast f\ast\bar{h}$ and tried to deduce a contradiction, also to no avail. Now I am stuck-any help would be greatly appreciated.
At the meeting point between $f$ and $\overline{p},\overline{h}$ respectively gives $1 = f(4s-1) = p(1-4s)$ and $1 = f(4s-1) = h(1-4s)$ at $s = 1/4$. and meeting point between $f$ and $p,h$ respectively gives, $1 = f(4s-1) = p(2s-1)$, $1 = f(4s-1) = h(2s-1)$ at $s = 1/2$.
You need to come up with homotopy between $p$ and $h$ which satisfies the above boundary conditions.
Let $h(s)=(h_1(s),h_2(s))$, $f(s)=(f_1(s),f_2(s))$, $p(s) = (p_1(s),p_2(s))$.
Try Homotopy: For $0 \leq s\leq 1/4$, $$F(s,t) = (h_1(1-4s)e^{-j 2\pi ((1-4s)t+(1-t)))},h_2(1-4s)).$$
For $1/4 \leq s\leq 1/2$, $$F(s,t) = (f_1(4s-1)e^{-j 2\pi (1-t)},f_2(4s-1)).$$
For $1/2 \leq s\leq 1$, $$F(s,t) = (h_1(2s-1)e^{-j 2\pi ((2s-1)t+(1-t)))},p_2(2s-1)).$$