For given path-connected topological spaces $X$, $x_0 , x_1 \in X$, and given loops $\alpha$, $\beta$ $: I \rightarrow X$ with base point at $x_1$ and a path $p: I \rightarrow X$ such that $p(0) = x_0$ and $p(1)=x_1$, I want to show, If $\alpha \simeq_{x_1} \beta$ then $ \bar{p} * \alpha * p \simeq_{x_0} \bar{p} * \beta * p$. where $\bar{p}(t) = p(1-t)$ is a path with opposite direction.
My trial is using the definition of path product $\alpha* \beta(t)= \alpha(2t)$ for $0\leq t\leq \frac{1}{2}$ and $=\beta(2t-1)$ for $\frac{1}{2} \leq t \leq 1$. but i am confused with handling of $\bar{p} * \alpha * p(t)$.
I figure out what i was wrong.
apparently \begin{align} \bar{p} * (\alpha * p)(t) = \begin{cases} \bar{p} (2t), & 0 \leq t \leq \frac{1}{2} \\ \alpha * p (2t-1), & \frac{1}{2} \leq t\leq 1 \end{cases} = \begin{cases} p (1-2t), & 0 \leq t \leq \frac{1}{2} \\ \alpha(4t-2) & \frac{1}{2} \leq t \leq \frac{3}{4} \\ p(4t-3) & \frac{3}{4} \leq t \leq 1 \end{cases} \end{align} and \begin{align} \bar{p} * (\beta * p)(t) = \begin{cases} \bar{p} (2t), & 0 \leq t \leq \frac{1}{2} \\ \beta * p (2t-1), & \frac{1}{2} \leq t\leq 1 \end{cases} = \begin{cases} p (1-2t), & 0 \leq t \leq \frac{1}{2} \\ \beta(4t-2) & \frac{1}{2} \leq t \leq \frac{3}{4} \\ p(4t-3) & \frac{3}{4} \leq t \leq 1 \end{cases} \end{align} we know a path is homotopy to itself, $p \simeq p$, the leftover parts is the region $\frac{1}{2} \leq t \leq \frac{3}{4}$. $i.e$, $\alpha(4t-2)$ and $\beta(4t-2)$ from our assumption $\alpha \simeq_{x_1} \beta$ it is guaranteed. Thus $ \bar{p} * \alpha * p \simeq_{x_0} \bar{p} * \beta * p$