Let $X$ be a topological space. Let $H:X\times I\to X$ a homotopy such that $H(x,0)=H(x,1)=x\ \forall x\in X$. We define the path $\gamma$ in $x_0$ as $\gamma(t)=H(x_0,t)$. Show that $[\gamma]$ commutes with every element of $\pi_1(X,x_0)$.
By definiton of the product in the fundamental group, if $[\beta]\in\pi_1(X,x_0)$ then $[\gamma][\beta]=[\gamma * \beta]$, where $*$ denotes the loop product. I've been trying to find a homotopy $G:I\times I\to X$ between $\gamma * \beta$ and $\beta * \gamma$ relative to $\{0,1\}$ without success. Could you please help me find it? Is there another method to show the result?
You can use $H$ to "slide" $\beta$ along $\gamma$. $$G(t) = \gamma|_{[0,t]}*(H_t\circ\beta)*\gamma|_{[t,1]}$$ will do, using the convention of following paths left-to-right with the path product $*$. When $t=0$, it's $\beta*\gamma$, and when $t=1$ it's $\gamma*\beta$. For intermediate $t$, you can check that the three paths are actually composable.