$x_0,x_1$ are points in the path-connected space $X$. Is it true that $[\overline{g*\alpha}] = [\bar{g}*\bar{\alpha}]$, where $g\in \pi_1(X,x_0)$ and $\alpha$ is a path from $x_0$ to $x_1$?
$g*\alpha$ starts at $x_0$ and ends at $x_1$, so $\overline{g*\alpha}$ starts at $x_1$ and ends at $x_0$. I think $\bar{g}*\bar{\alpha}$ is invalid since $\bar{g}$ starts at $x_0$ and ends at $x_0$, while $ \bar{\alpha}$ starts at $x_1$.
The path ${g*\alpha}$ is defined as $$ g*\alpha(s) = \begin{cases} g(2s) & 0 \leq s \leq 1/2 \\ \alpha(2s - 1) & 1/2 \leq s \leq 1. \end{cases} $$ So this first traverses $g$ at double speed, then $\alpha$ at double speed. The path $\overline{g*\alpha}$ traverses this backwards; thus, it is defined as $$ \overline{g*\alpha}(s) = g*\alpha(1 - s) = \begin{cases} g(2 - 2s) & 0 \leq 1 - s \leq 1/2 \\ \alpha(1 - 2s) & 1/2 \leq 1 - s \leq 1. \end{cases} $$ Rewriting the conditionals in the last brackets, we get \begin{align}\overline{g* \alpha}(s) &= \begin{cases} g(2-2s) & 1/2 \leq s \leq 1 \\ \alpha(1-2s), & 0 \leq s \leq 1/2 \end{cases} \\ &=\begin{cases} \alpha(1-2s) & 0 \leq s \leq 1/2 \\ g(2-2s) & 1/2 \leq s \leq 1 \end{cases} \\ &= \begin{cases} \alpha(1-2s) & 0 \leq s \leq 1/2 \\ g(1-(-1+2s)) & 1/2 \leq s \leq 1 \end{cases} \\ &= \begin{cases} \bar{\alpha}(2s) & 0 \leq s \leq 1/2 \\ \bar{g}(2s-1) & 1/2 \leq s \leq 1 \end{cases} \end{align} But by definition of $*$, this is exactly the path $\overline{\alpha}*\overline{g}(s)$.
