Associativity of group operation in the fundamental group

Question

I am trying to understand Hatcher's argument as to why the binary operation on the homotopy classes paths is associative, where we define $[f][g] = [f \cdot g]$ with $f \cdot g$ is the concatenation of two paths sharing an endpoint. He argues that $(f \cdot g) \cdot h$ is a reparametrization of $f \cdot (g \cdot h)$ with $\phi$ as the piecewise linear function shown in the figure below:

I have tried to write out the explicit formulations for $f \cdot (g \cdot h), (f \cdot g) \cdot h),$ and $\phi$ as follows:

$$ ((f \cdot g) \cdot h) (s) = \begin{cases} f(4s), & 0 \leq s \leq \frac{1}{4}\\ g(4s-1), & \frac{1}{4} \leq s \leq \frac{1}{2} \\ h(2s-1), & \frac{1}{2} \leq s \leq 1. \end{cases} $$

$$ (f \cdot (g \cdot h)) (s) = \begin{cases} f(2s), & 0 \leq s \leq \frac{1}{2}\\ g(4s-2), & \frac{1}{2} \leq s \leq \frac{3}{4} \\ h(4s-3), & \frac{3}{4} \leq s \leq 1. \end{cases} $$

$$ \phi(s) = \begin{cases} \frac{1}{2}s & 0 \leq s \leq \frac{1}{2}\\ s-\frac{1}{4}, & \frac{1}{2} \leq s \leq \frac{3}{4} \\ 2s-1, & \frac{3}{4} \leq s \leq 1. \end{cases} $$

But clearly, $((f \cdot g) \cdot h) \phi$ and $(f \cdot (g \cdot h))$ are not the same function, for instance, on the interval $[\frac{1}{4}, \frac{1}{2}]$, the former equals to $g(2s-1)$, while the latter is $f(2s)$.

Where is the error?

Answer 1

When I plug in $s=3/8$, for example, I get $f(3/4)$ for both. You seem to agree for the second composition. For the first: $\phi(3/8) = 3/16$, so $((f \cdot g) \cdot h) \phi(3/8) = ((f \cdot g) \cdot h)(3/16)$, and since $3/16 \leq 1/4$, this equals $f(4 \cdot 3/16) = f(3/4)$. The same works for any $s \in [0, 1/2]$: $\phi$ takes the interval $[0,1/2]$ to the interval $[0, 1/4]$, and so into the part of the path given by $f$.

Answer 2

The concatenation $f\cdot g$ does $f$ in $[0,1/2]$ and $g$ in $[1/2,1]$, both 2 times the original path-speed. Thus $(f\cdot g)\cdot h$ does $f$ in the first $1/4$ of $[0,1]$, $g$ in the second $1/4$ both at 4-times the original speed, and $h$ at 2 times the original speed in the final $1/2$ of $[0,1]$.

On the other hand $f\cdot(g\cdot h)$ does $f$ at 2 times the original speed in the first half-interval, and then $g$ and $h$ in the remaining two $1/4$s at 4 times the original speed.

Thus the speeds over the images of $f,g$ and $h$ as a multiple of the original path-speeds are $(4,4,2)$ and $(2,4,4)$ over the intervals $[0,1/4],[1/4,1/2],[1/2,1]$ and $[0,1/2],[1/2,3/4],[3/4,1]$ respectively. The reparametrization should thus have speeds $1/2,1,2$ over $[0,1/2]$, $[1/2,3/4]$ and $[3/4,1]$ respectively, just as Hatcher describes.

Update: to provide more details on how the speeds lead to the reparametrizations. Knowing the speeds of the two parametrizations, given that you know they are both traversing the same paths, suffices to determines the reparametrization: since the speeds for $(fg)h$ and $f(gh)$ are $(4,4,2)$ and $(2,4,4)$ respectively, if we seek a reparametrization which, when precomposed with $((fg)h)$ will yield the path $(f(gh))$, then the speeds along the reparameterized version of $((fg)h)$ must match those of $(f(gh))$. But then since in $(f(gh))$ the path of $f$ is traversed at twice the initial speed rather than 4 times that speed in $((fg)h)$, the reparametrization must halve the speed of a particle traversing the path using the $((fg)h)$ parametrization, showing that this part of the reparametrization is given by $t\mapsto \frac{1}{2}t$ on $[0,1/2]$. Then $g$ is traversed at the same speed in both, so we can identify their schedule, and, correspondingly, the reparametrization of $(fg)h$ is just the identity map shifted by $-1/4$ over $[1/2,3/4]$, and finally, since $(f(gh))$ traverses the path of $h$ at 4 times the speed of the original path, while $((fg)h)$ traverses it at twice the speed, we see that the reparametrization must double the speed in that interval, so that in $[3/4,1]$ it is $1/2+2(s-1/2) = 1+2s$.