Is this the correct homotopy rel $\{0,1\}$? Concerning the problem circled in red:
Would this work as the homotopy rel $\dot I$?
$F(t_1, t_2) =
\begin{cases}
\sigma(e_1),  & \text{if $t_1=0,1$ or $t_2=1$} \\
(\sigma_0 * \sigma_1^{-1}) * \sigma_2[(1-t_2)t_1 + t_2], & \text{otherwise}
\end{cases}$
And also, how would you use theorem $1.6$ here to show this?  I can see by part three I can show that it's nullhomotopic rel $\{1\}$ if we define $f ': S \rightarrow X$ by $f'(e^{2 \pi i t_1}) = f(t_1)$, where $f(t)= (\sigma_0 * \sigma_1^{-1}) * \sigma_2(t)$ and the homotopy as $G(t_1, t_2)$ as $(e^{2 \pi i t_1}, t_2) \rightarrow f'((e^{2 \pi i t_1})^{(1-t_2)})$, but that doesn't show it's rel $\{0,1\}$.




Your homotopy is not a homotopy. Its image always lies on $\partial \Delta^2 \cong S^1$, and $S^1$ is not contractible. More specifically, we have for any $0<t_2<1$,
$$\lim_{t_1 \to 0} F(t_1,t_2) = \sigma_0\sigma_1^{-1}\sigma_2(t_2) \not= \sigma(e_1) = F(0,t_2).$$
I think you have the answer, but you just don't recognize it. Theorem 1.6 tells you have a nullhomotopy of $\sigma:\partial \Delta^2 \to X$ rel $\{e_1\}$. We can then think of $f$ as a composition of maps
$$[0,1] \xrightarrow{g} \partial \Delta^2 \xrightarrow{\sigma} X,$$
where $g(t)=(\epsilon_0\cdot\epsilon_1^{-1}\cdot\epsilon_2)(t)$. Since $g(0)=g(1)=e_1$, the homotopy $\sigma \simeq (\sigma e_1)$ rel $e_1$ gives a homotopy $\sigma \circ g = f \simeq (\sigma(e_1))$ rel $\{0,1\}$.
