Associativity of Products of Paths in a Topological Space

# Associativity of Products of Paths in a Topological Space

 Proposition 1: Let $X$ be a topological space and let $\alpha, \beta, \gamma: I \to X$ be paths such that $\alpha(1) = \beta(0)$ and $\beta (1) = \gamma (0)$. Then $[\alpha]([\beta][\gamma]) = ([\alpha][\beta])[\gamma]$.
• Proof: Define a function $H : I \times I \to X$ defined by:
(1)
\begin{align} \quad H(s, t) = \left\{\begin{matrix} \alpha \left ( \frac{4s}{2 - t} \right ) & \mathrm{if} \: 0 \leq s \leq \frac{2 - t}{4} \\ \beta \left (4s + t - 2 \right ) & \mathrm{if} \: \frac{2 - t}{4} \leq s \leq \frac{3 - t}{4} \\ \gamma \left ( \frac{4s}{1 + t} - \frac{3 - t}{1 + t} \right ) & \mathrm{if} \: \frac{3 - t}{4} \leq s \leq 1 \end{matrix}\right. \end{align}
• Then $H$ is continuous since $\alpha, \beta, \gamma$ are continuous and the ends match up (and applying The Gluing Lemma). Furthermore:
(2)
\begin{align} H_0(s) = H(s, 0) = \left\{\begin{matrix} \alpha \left ( 2s \right ) & \mathrm{if} \: 0 \leq s \leq \frac{2}{4} \\ \beta \left (4s - 2 \right ) & \mathrm{if} \: \frac{2}{4} \leq s \leq \frac{3}{4} \\ \gamma \left ( 4s - 3 \right ) & \mathrm{if} \: \frac{3}{4} \leq s \leq 1 \end{matrix}\right. = \alpha(\beta \gamma) \end{align}
(3)
\begin{align} H_1(s) = H(s, 1) = \left\{\begin{matrix} \alpha \left ( 4s \right ) & \mathrm{if} \: 0 \leq s \leq \frac{1}{4} \\ \beta \left (4s - 1 \right ) & \mathrm{if} \: \frac{1}{4} \leq s \leq \frac{2}{4} \\ \gamma \left ( 2s - 1 \right ) & \mathrm{if} \: \frac{2}{4} \leq s \leq 1 \end{matrix}\right. = (\alpha \beta)\gamma \end{align}
• And lastly:
(4)
\begin{align} \quad H_t(0) = \alpha(0) = \alpha(\beta \gamma)(0) = (\alpha \beta)\gamma(0) \quad \forall t \in I \end{align}
(5)
\begin{align} \quad H_t(1) = \gamma (1) = \alpha (\beta \gamma)(1) = (\alpha \beta)\gamma (1) \quad \forall t \in I \end{align}
• Therefore $\alpha (\beta \gamma) \simeq_{\{0, 1\}} (\alpha \beta)\gamma$. So $[\alpha]([\beta][\gamma]) = ([\alpha][\beta])[\gamma]$. $\blacksquare$