I am trying to understand the pictorial representations of homotopies, by writing down the equations for them. I want to prove that $(f\cdot g)\cdot h\simeq f\cdot (g\cdot h)$, where $f,g,h:I\to X$ are paths with $f(1)=g(0)$ and $g(1)=h(0)$. The following picture is from "A Concise Course in Algebraic Topology" by J. P. May.
I have read the explanation in this thread: How should I interpret a homotopy schematic?, but I still can't wrap my head around this. According to the thread, the homotopy is given by
$$
H(s,t)=\begin{cases}
f\Big (\frac{4}{t+1}s \Big )\quad\qquad\qquad\quad\text{ }\text{ } s\in \Big [0,\frac{1}{4}t+\frac{1}{4} \Big] \\
g(4s-t-1)\quad\qquad\quad\text{ }\text{ } s\in \Big [ \frac{1}{4}t+\frac{1}{4},\frac{1}{4}t+\frac{1}{2} \Big ] \\
h\Big (\frac{4}{2-t}(s-1)+1\Big )\quad\quad s\in \Big [\frac{1}{4}t+\frac{1}{2},1\Big ]
\end{cases}$$
It is straightforward to find that the left diagonal is given by the following line $t=4s-1$. So the left-most region is given by $t\geq 4s-1$. It is also straightforward to find that the right diagonal is given by $t=4s-2$, so the middle region should be given by $4s-2 \leq t\leq 4s-1$, and the right region given by $t\leq 4s-2$, I guess?
I also understand at time $t=0$, we would like to walk the path $f$ at $4$ times the speed, while at $t=1$, we would like to walk it just twice the speed. And so forth.
Maybe I am stupid, but I can't seem to figure out what to put inside $f$, $g$ and $h$ by just looking at the picture. I can see that the formula for $H$ works, but I cannot really see how to come up with it. For example:
In $f$, we have $\frac{4}{t+1}s$. My formula for the line is $t=4s-1$. I can manipulate this equation as follows $t=4s-1\iff t+1=4s\iff 1=\frac{4}{t+1}s$. Then I have something which looks like the expression on the right-hand side, but I do not know why I would like to do this manipulation.
In $g$, we have $4s-t-1$. My expression for the line looks as follows $t=4s-1$, by a simple manipulation, we have the following $0=4s-t-1$, and we have the expression on the right-hand side.
In $h$, we have the expression $\frac{4}{2-t}(s-1)+1$. I tried to expand it to see what it looks like, then I got this $\dfrac{4s-2-t}{2-t}$. The equation for my line is $t=4s-2$, which can be rewritten as $0=\frac{4s}{t+2}-1\iff 0=\frac{4s}{t+2}-\frac{t+2}{t+2}=\frac{4s-t-2}{t+2}$, which almost looks like the expression inside $h$.
But I don't know why I want to rewrite the expressions in this way, and I don't know what I am doing, really.
I would be really happy if someone could help me a little how to decode the information encoded in a homotopy diagram. Thanks!
Best Answer
Look at the schematic diagram in your link. The two line segments inside the square are given by the equations $t= 4s -1$ where $s \in [\frac{1}{4},\frac{1}{2}]$ and $t = 4s-2$ where $s \in [\frac{1}{2},\frac{3}{4}]$. They intersect the line segment $I_t = [0,1] \times \{t\}$ in the two points $(s_1,t)$ with $s_1 = \frac{t+1}{4}$ and $(s_2,t)$ with $s_2 = \frac{t+2}{4}$. This splits $I_t$ in the three segments $I^f_t = [0,\frac{t+1}{4}] \times \{t\}$ having length $l^f_t= \frac{t+1}{4}- 0 = \frac{t+1}{4}$, $I^g_t = [\frac{t+1}{4}, \frac{t+2}{4}] \times \{t\}$ having length $l^g_t= \frac{t+2}{4}- \frac{t+1}{4} = \frac{1}{4}$ and $I^h_t = [\frac{t+2}{4}, 1] \times \{t\}$ having length $l^h_t = 1- \frac{t+2}{4} = \frac{2-t}{4}$.
On $I^f_t$ we walk the path $f$ with speed $\dfrac{1}{l^f_t} = \dfrac{4}{t+1}$ and get $H(s,t) = f(\dfrac{4}{t+1}s)$.
On $I^g_t$ we walk the path $g$ with speed $\dfrac{1}{l^g_t} = 4$ and get $H(s,t) = g(4(s-s_1)) = g(4s - t - 1)$.
On $I^h_t$ we walk the path $h$ with speed $\dfrac{1}{l^h_t} = \dfrac{4}{2-t}$ and get $H(s,t) = h(\dfrac{4}{2-t}(s-s_2)) = h(\dfrac{4}{2-t}(s-1) + 1)$.