Products of Paths Relative to {0, 1} in a Topological Space

# Products of Paths Relative to {0, 1} in a Topological Space

 Definition: Let $X$ be a topological space. A Path in $X$ is a continuous function $\alpha : [0, 1] \to X$.
 Definition: Let $X$ be a topological space and let $\alpha, \beta : [0, 1] \to X$ be paths such that $\alpha(1) = \beta (0)$. The Product of the paths $\alpha$ and $\beta$ is the new path $\alpha \beta : [0, 1] \to X$ defined for all $t \in [0, 1]$ by $\alpha \beta = \left\{\begin{matrix} \alpha(2t) & \mathrm{if} \: 0 \leq t \leq \frac{1}{2}\\ \beta (2t - 1) & \mathrm{if} \: \frac{1}{2} \leq t \leq 1 \end{matrix}\right.$.

Observe that the order of the notation for denoting the path product $\alpha \beta$, is important. We traverse the pathes in the product from left to right. We first traverse $\alpha$ at twice the speed, and hence $\beta$ at twice the speed. Since the terminal point of $\alpha$ is equal to the initial point of $\beta$, $\alpha \beta$ is indeed a continuous function.

Let $X$ be a topological space and let $\alpha : I \to X$ be a path. Observe that $I = [0, 1]$ is a topological space and that $A = \{0, 1 \} \subset I$. From the Homotopic Mappings Relative to a Subset of a Topological Space, we proved that the relation that two continuous functions are homotopic relative to a subset $A$ of the domain is an equivalence relation. We let:

(1)
\begin{align} \quad [\alpha] = \{ \beta : \mathrm{beta} \: \mathrm{is \: a \: path \: and} \: \alpha \simeq_{\{0, 1\}} \beta \} \end{align}

In the following proposition we show that multiplication of such equivalence classes is a well-defined operation.

 Theorem 1: Let $X$ be a topological space and let $\alpha, \beta : I \to X$ be paths such that $\alpha (1) = \beta (0)$. Then $[\alpha][\beta] := [\alpha\beta]$ is well-defined.
• Proof: Let $\alpha, \beta : I \to X$ be paths and let $\alpha', \beta' : I \to X$ also be paths such that:
(2)
\begin{align} \quad \alpha &\simeq_{\{0, 1\}} \beta \\ \quad \alpha' &\simeq_{\{0, 1\}} \beta' \end{align}
• We want to then show that $[\alpha][\beta] = [\alpha'][\beta']$, i.e., $[\alpha\beta] = [\alpha'\beta']$ which shows that the product in the theorem is well-defined and independent of the choice of representative path for the equivalence class.
• Since $\alpha \simeq_{\{0, 1\}} \alpha'$ there exists a continuous function $H : I \times I \to X$ such that $H_0 = \alpha$, $H_1 = \alpha'$, $H_t(0) = \alpha(0) = \alpha'(0)$ and $H_t(1) = \alpha(1) = \alpha'(1)$ for all $t \in I$.
• Similarly, since $\beta \simeq_{\{0, 1\}} \beta'$ there exists a continuous function $H' : I \times I \to X$ such that $H'_0 = \beta$ $H'_1 = \beta'$, $H'_t(0) = \beta(0) = \beta'(0)$ and $H'_t(1) = \beta(1) = \beta'(1)$ for all $t \in I$.
• Define a new function $H^* : I \times I \to X$ by:
(3)
\begin{align} \quad H^*(s, t) = \left\{\begin{matrix} H(2s, t) & 0 \leq s \leq \frac{1}{2}\\ H'(2s - 1, t) & \frac{1}{2} \leq s \leq 1 \end{matrix}\right. \end{align}
• Then $H^*$ is continuous since $H$ and $H'$ are continuous and by The Gluing Lemma. Also:
(4)
\begin{align} \quad H^*_0 = H^*(s, 0) = \left\{\begin{matrix} H(2s, 0) & 0 \leq s \leq \frac{1}{2}\\ H'(2s - 1, 0) & \frac{1}{2} \leq s \leq 1 \end{matrix}\right. = \left\{\begin{matrix} H_0(2s)& 0 \leq s \leq \frac{1}{2}\\ H'_0(2s - 1) & \frac{1}{2} \leq s \leq 1 \end{matrix}\right. \left\{\begin{matrix} \alpha (2s)& 0 \leq s \leq \frac{1}{2}\\ \beta(2s - 1) & \frac{1}{2} \leq s \leq 1 \end{matrix}\right. = \alpha \beta \end{align}
(5)
\begin{align} \quad H^*_1 = H^*(s, 1) = \left\{\begin{matrix} H(2s, 1) & 0 \leq s \leq \frac{1}{2}\\ H'(2s - 1, 1) & \frac{1}{2} \leq s \leq 1 \end{matrix}\right. = \left\{\begin{matrix} H_1(2s)& 0 \leq s \leq \frac{1}{2}\\ H'_1(2s - 1) & \frac{1}{2} \leq s \leq 1 \end{matrix}\right. \left\{\begin{matrix} \alpha' (2s)& 0 \leq s \leq \frac{1}{2}\\ \beta'(2s - 1) & \frac{1}{2} \leq s \leq 1 \end{matrix}\right. = \alpha' \beta' \end{align}
• Therefore $\alpha \beta \simeq_{\{0, 1 \}} \alpha ' \beta '$. So $[\alpha\beta] = [\alpha'\beta']$. So indeed, the product $[\alpha][\beta] := [\alpha\beta]$ is well-defined. $\blacksquare$