I am currently studying simplicial topology and got stuck with a homotopy construction: $$ \alpha(s) = \begin{cases} e^{4\pi is} \\ e^{4\pi i(2s-1)} \\ e^{8\pi i(1-s)} \end{cases} \simeq \beta(s) = e^{2\pi is} $$
Now one approach is to view the above loops as particles traveling with different speed and the task reduces to construct a relation between the two particles as illustrated below.
Now the text constructs the homotopy as, $$ F(s, t) = \begin{cases} \alpha(\frac s {t+1}), 0 \leq s \leq \frac {t+1} 2\\ \alpha(s-\frac t 2), \frac{t+1}2 \leq s \leq \frac {t + 3} 4 \\ \alpha(s), o.w.\end{cases}$$
So just how is this related to the graph, I am stuck. I understand the range of $s, t$ in three different areas but failed to construct such a explicit homotopy just as the text did. However, following a hint, I can construct $f$ as: $$ f = \begin{cases} 2s, 0 \leq s \leq \frac 1 2 \\ 4s-2, \frac 1 2 \leq s \leq \frac 3 4 \\ 4 - 4s, \frac 3 4 \leq s \leq 1 \end{cases}. $$ Therefore $\alpha = \beta \circ f$ and since $f: I \to I$, and $I$ is convex, it follows $f \simeq id$ by straight line homotopy.
My question is how does one construct any of the two version above using the illustration.