Construct homotopy from $(\alpha \cdot \beta) \cdot \gamma$ to $\alpha \cdot (\beta \cdot \gamma)$ explicitly

See the picture below.

The idea is that, for any choice of $s$, the loop $t \mapsto F(t, s)$ consists of walking along $\alpha$, $\beta$ and $\gamma$, in that order. The difference between the various choices of $s$ is in the "time schedule": each choice of $s$ allocates different amounts of time to $\alpha$, $\beta$ and $\gamma$.

• If $s = 0$, you walk path $\alpha$ in time $[0, \tfrac 1 4]$, then walk $\beta$ in time $[\tfrac 1 4, \tfrac 1 2 ]$, then walk $\gamma$ in time $[\tfrac 1 2 , 1]$.

• If $s = 1$, you walk path $\alpha$ in time $[0, \tfrac 1 2]$, then walk $\beta$ in time $[\tfrac 1 2, \tfrac 3 4]$, then walk $\gamma$ in time $[\tfrac 3 4 , 1]$.

For intermediate choices of $s$, the schedule is given by interpolation.

So for example, if $s = \tfrac 1 2$, you walk $\alpha$ in time $[0, \tfrac 3 8]$, then walk $\beta$ in time $[\tfrac 3 8, \tfrac 5 8]$, then walk $\gamma$ in time $[\tfrac 5 8, 1]$. And so on.

This is basically a reparametrization of the curve, but the parametrization is changing continuously. We may come up with the homotopy in two steps.

Let $u:I\to X$ be a curve, let $\phi:I\to I$ be a continuous map with $\phi(0)=0$ and $\phi(1)=1$, then $u$ is homotopic to $u\circ\phi$.

Proof: define $H(s, t)=u((1-s)t+s\phi(t))$, then $H(0,t)=u(t)$, $H(1,t)=u(\phi(t))$ and $H(s, 0)=u(0)$, $H(s, 1)=u(1)$. Essentially this is just mapping a homotopy between $\operatorname{Id}$ and $\phi$ in $I$ to the homotopy in $X$ by $u$.

We could construct $\phi:I\to I$ by rescaling the speed according to the time of travel through each curve. More explicitly, we have $[\alpha\cdot(\beta\cdot\gamma)] (t) = [(\alpha\cdot\beta) \cdot\gamma] (\phi(t)) $ if we define: $$\phi(t) =\begin{cases}\frac 12 t, & t\in[0,\frac 12]\\ t-\frac14, & t\in [\frac12, \frac34] \\ 2t-1, & t\in [\frac34, 1]\end{cases} $$

However if you apply the formula above, while we still get a homotopy, but the formula is not as clean (as in we need to divide into 5 cases). To obtain the formula you stated, let's look at the graphs of the intermediate reparametrizations, for each fixed $s_0$, the graph of $\phi_{s_0}(t) =1-s_0)t + s_0 \phi(t) $ would look like a "sheared up" version of $\phi(t)$. So image of the points at which $\phi_{s_0}$ are piecewisely defined do not match with the points at which $(\alpha\cdot\beta) \cdot\gamma$ is piecewisely defined. To make them match we need to "shear" $\phi(t)$ to the left.

So we should take $\tilde\phi_s(t)$ the inverse function of $(1-s)t+s\phi^{-1}(t)$, i.e. $\tilde\phi_s(t)$ is the inverse of $$\begin{cases}(1+s)t, & t\in[0,\frac14] \\ t+\frac s4, & t\in[\frac 14,\frac 12]\\ \frac 12(2-s)t +\frac s2, & t\in[\frac 12,1]\end{cases}$$

Which is given by: $$\tilde\phi_s(t) =\begin{cases} \frac t{1+s}, & t\in[0,\frac {s+1}4]\\ t-\frac s4, & t\in[\frac{s+1}4,\frac{s+2}4]\\ \frac 2{2-s}t - \frac s{2-s}, & t\in[\frac {s+2}4,1]\end{cases} $$

The homotopy you provided in the post is then $H(s, t) =[(\alpha\cdot \beta)\cdot \gamma](\tilde\phi_s(t)) $.

I have checked the definition and all axioms of the first homotopy group in this note and in this I also explain the exact homotopy you describe. I hope it helps you...