Constant Paths in a Topological Space

Recall from the Products of Paths Relative to {0, 1} in a Topological Space page that if $X$ is a topological space and $\alpha, \beta : I \to X$ are paths such that $\alpha(1) = \beta(0)$ then the product path $\alpha\beta : I \to X$ is defined by starting at the initial point of $\alpha$, traversing $\alpha$ twice as fast, traversing $\beta$ twice as fast, and ending at the terminal point of $\beta$.

We let:

(1)

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

We proved that the operation of multiplication of these equivalence classes given by:

(2)

\begin{align} \quad [\alpha][\beta] = [\alpha \beta] \end{align}

is well-defined. We now move on to showing the existence of a "left-inverse" and "right-inverse" constant paths.

Definition: Let $X$ be a topological space. For any $x \in X$ we denote $c_x : I \to X$ to be the Constant Path at $x$ defined for all $t \in I$ by $c_x(t) = x$.

Now let $\alpha : I \to X$ be a path such that $\alpha(0) = x$ and $\alpha(1) = y$. We will prove that $[c_y][\alpha] = [\alpha]$ and $[\alpha][c_x] = [\alpha]$ so that $[c_y]$ is a left-handed identity for $[\alpha]$ and $[c_x]$ is a right-handed identity for $[\alpha]$.

Proposition 1: Let $X$ be a topological space and let $\alpha : I \to X$ be a path such that $\alpha(0) = x$ and $\alpha(1) = y$. Then: a) $[c_y][\alpha] = [\alpha]$. b) $[\alpha][c_x] = [\alpha]$.

• Proof of a) To show that $[c_y][\alpha] = [\alpha]$ we must show that $\alpha \simeq_{\{0, 1\}} \alpha c_y$. Define a function $H : I \times I \to X$ by:

(3)

\begin{align} \quad H(s, t) = \left\{\begin{matrix} \alpha \left ( \frac{2s}{t + 1} \right ) & 0 \leq s \leq \frac{t + 1}{2} \\ y & \frac{t+1}{2} \leq s \leq 1 \end{matrix}\right. \end{align}

• Then $H$ is continuous since $\alpha$ is continuous and by The Gluing Lemma. Furthermore:

(4)

\begin{align} \quad H_0(s) = H(s, 0) = \left\{\begin{matrix} \alpha \left (2s \right ) & 0 \leq s \leq \frac{1}{2} \\ c_y & \frac{1}{2} \leq s \leq 1 \end{matrix}\right. = \alpha c_y \end{align}

(5)

\begin{align} \quad H_1(s) = H(s, 1) = \left\{\begin{matrix} \alpha (s) & 0 \leq s \leq 1 \\ y & 1 \leq s \leq 1 \end{matrix}\right. = \alpha \end{align}

• Lastly $H_t(0) = \alpha(0) = x$ for all $t \in I$ and $H_t(1) = y = \alpha(1)$ for all $t \in I$. So indeed, $\alpha \simeq_{\{0, 1\}} \alpha c_y$. So $[\alpha c_y] = [\alpha$, that is:

(6)

\begin{align} \quad [\alpha][c_y] = [\alpha] \quad \blacksquare \end{align}

• Proof of b) Analogous to (a).