Legitimate homotopy between $(\gamma_1 * \gamma_2) *\gamma_3$ and $\gamma_1*(\gamma_2 * \gamma_3)$?

Question

I want to find a homotopy between curves $(\gamma_1 * \gamma_2) *\gamma_3$ and $\gamma_1*(\gamma_2 * \gamma_3)$ that are closed loops in some topological space $X.$

I found $H(t,s):=$

$ \gamma_1 (2t(2-s)),t \in [0, \frac{1+s}{4} ]$

$\gamma_2(4t-1-s) , t \in [\frac{1+s}{4}, \frac{2+s}{4}]$ and

$\gamma_3(2(1+s)t-1-2s), t \in [\frac{2+s}{4},1].$

This map certainly fixed the end-points and $H(t,0) = (\gamma_1 * \gamma_2) *\gamma_3(t)$ as well as $H(t,1)= \gamma_1*(\gamma_2 * \gamma_3)(t)$ is satisfied. My question is if this already means that $H$ is a homotopy or if there are further things to show?

Answer 1

The argument $2t(2-s)$ in the expression $\gamma_1(2t(2-s))$ is incorrect.

The argument you should have in that expression is the increasing function of constant derivative which maps the interval $\bigl[0,\frac{1+s}{4}\bigr]$ to the interval $[0,1]$. That function is $t \mapsto \frac{4t}{1+s}$. So the expression should be $\gamma_1\bigl(\frac{4t}{1+s}\bigr)$.

It looks like the arguments of $\gamma_2$ and of $\gamma_3$ in your homotopy need similar corrections.

Answer 2

The only thing missing is that $H$ is a continous map $H \colon I^2 \to X$ (where $I = [0,1]$).