How to show a map has the homomorphism property?

Question

Consider the map $f$ defined in the answer from here. I'm trying to understand why $f$ has the homomorphism property. The hint given in the comments is to include the point $1/2\in I$ in the partition. But the points of partition had the property that $\gamma(I_j)$ lies entirely within $U_1$ or $U_2$ (see the question referred to above for the notation). Why should the subintervals corresponding to $1/2$ have this property? Further, even if they have this property, $G$ is still not abelian, which looks like an obstruction to proving the homomorphism property (or maybe I'm overthinking...).

Answer 1

Given closed paths $\gamma_1, \gamma_2$, the path $\gamma = \gamma_1 \cdot \gamma_2$ is defined by $\gamma(t) = \gamma_1(2t)$ for $t \le 1/2$ and $\gamma(t) = \gamma_2(2t-1)$ for $t \ge 1/2$. Forv $k = 1,2$ choose partitions $I^k_j = [x^k_{j-1},x^k_j]$ of $I$, $j = 1,\dots,n_k$, such that $\gamma_k(I^k_j)$ is contained in $U_1$ or $U_2$. Define a partition $I_j = [x_{j-1},x_j]$ of $I$, $j = 1,\dots,n_1+n_2$, by $x_j = \frac{1}{2}x^1_j$ for $j = 0,\dots,n_1$ and $x_j = \frac{1}{2}(x^2_j+1)$ for $j = n_1,\dots, n_1 + n_2$. Then $1/2$ is a partition point. By construction $\gamma(I_j)$ is contained in $U_1$ or $U_2$. Choose paths $\beta_j^k$ in the appropriate $U_{r(k,j)}$ which connect $x$ and $\gamma_k(x^k_j)$, where $\beta_0^k$ and $\beta^k_{n_k}$ are constant. The paths $\beta_j^k$ provide us with paths $\beta_j$ in $U_{r(j)}$ connecting $x$ and $\gamma(x_j)$: Take $\beta_j = \beta_j^1$ for $j = 0,\dots,n_1$ and $\beta_j = \beta^2_{n_1-j}$ for $j = n_1,\dots,n_1+n_2$. But now it is clear that $$f(\gamma) = f_{r(1)}([\Gamma_1]) \dots f_{r(n_1+n_2)}([\Gamma_{n_1+n_2}])$$ $$= f_{r(1)}([\Gamma_1]) \dots f_{r(n_1)}([\Gamma_{n_1}]) f_{r(n_1+1)}([\Gamma_{n_1+1}]) \dots f_{r(n_1+n_2)}([\Gamma_{n_1+n_2}])$$ $$= f_{r(1,1)}([\Gamma^1_1]) \dots f_{r(1,n_1)}([\Gamma^1_{n_1}]) f_{r(2,1)}([\Gamma^2_1]) \dots f_{r(2,n_2)}([\Gamma^2_{n_2}]) = f(\gamma_1) f(\gamma_2) .$$