A question about the proof of $\pi_1(G,e)$ being abelian

Question

I have seen a lot of posts dealing with the fundamental group over a topological group. In almost every proof I’ve seen of it being abelian, we need to show that if I have two paths $\gamma_1$,$\gamma_2$ based at e, the identity of G, then

$\gamma_1(s)*\gamma_2(s) \cong (\gamma_1 \circ \gamma_2)(s)$

where $\circ $ denotes concatenation, $*$ denotes the product from the group, and $\cong$ denotes path homotopy equivalence. None of the solutions I’ve seen show how to prove this homotopy equivalence. Even things like the Eckmann-Hilton argument seem to be taking this for granted. Is there some trivial homotopy that I’m just not seeing? If so, what is it?

Answer 1

Let $I = [0,1]$ and $K = I \times \{0\} \cup \{1\} \times I \subset I^2 = I \times I$ and let $i : K \hookrightarrow I^2$ denote the inclusion map. Define

1. $u : I \to K, u(t) = (2t,0)$ for $t \le 1/2$ and $u(t) = (1,2t-1)$ for $t \ge 1/2$.

2. $\gamma_{12} : K \to G, \gamma_{12}(x,0) = \gamma_1(x), \gamma_{12}(1,y) = \gamma_2(y)$.

3. $\Delta : I \to I^2, \Delta(t) = (t,t)$. This is the diagonal map.

4. $\Gamma : I^2 \to G, \Gamma(x,y) = \gamma_1(x) * \gamma_2(y)$.

These are continuous maps; clearly $\gamma_1 \circ \gamma_2 = \gamma_{12} u$ and $\gamma_1 * \gamma_2 = \Gamma \Delta$. Moreover we have $$\Gamma i = \gamma_{12} .$$ In fact, $\Gamma i(x,0) = \Gamma(x,0) = \gamma_1(x) * \gamma_2(0) = \gamma_1(x) * e = \gamma_1(x) = \gamma_{12}(x,0)$ and $\Gamma i(1,y) = \Gamma(1,y) = \gamma_1(1) * \gamma_2(y) = e * \gamma_2(y) = \gamma_2(y) = \gamma_{12}(1,y0)$.

Therefore $\gamma_1 \circ \gamma_2 = \gamma_{12} u = \Gamma i u$. But the paths $iu : I \to I^2$ and $\Delta : I \to I^2$ are homotopic via $$H : I \times I \to I^2, H(t,s) = s\Delta(t) + (1-s) iu (t) = \begin{cases} (2t -st),st) & t \le 1/2 \\(st+1-s,2t -st +s -1) & t \ge 1/2 \end{cases}.$$ Hence $\bar H = \Gamma H : I \times I \to G$ is a homotopy between the paths $\gamma_1 \circ \gamma_2$ and $\gamma_1 * \gamma_2$. Explicitly we have $$\bar H(t,s) = \begin{cases} \gamma_1(2t -st) * \gamma_2(st ) & t \le 1/2 \\ \gamma_1(st+1-s) * \gamma_2(2t -st +s -1) & t \ge 1/2 \end{cases}$$

The above construction is an adaption of the approach in A confusion on $\Omega$ and $\Sigma$ functors.