Inverse Paths in a Topological Spaces

Definition: Let $X$ be a topological space and let $\alpha : I \to X$ be a path. The Inverse Path of $\alpha$ is the path $\alpha^{-1} : I \to X$ defined for all $t \in I$ by $\alpha^{-1}(t) = \alpha(1 - t)$.

If $\alpha : I \to X$ is a path that starts at $x$ and ends at $y$ then $\alpha^{-1} : I \to X$ is a path that starts at $y$ and ends at $x$.

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

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

(1)

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

• Then $H$ is continuous since $\alpha$ and $\alpha^{-1}$ are continuous and by The Gluing Lemma. Also:

(2)

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

(3)

\begin{align} \quad H_1(s) = H(s, 1) = \left\{\begin{matrix} x & 0 \leq s \leq \frac{1}{2} \\ \alpha (2s - 1) & \frac{1}{2} \leq s \leq \frac{1}{2}\\ \alpha^{-1}(2s) & \frac{1}{2} \leq s \leq \frac{1}{2}\\ x & \frac{1}{2} \leq s \leq 1 \\ \end{matrix}\right. = c_x \end{align}

• Furthermore:

(4)

\begin{align} \quad H_t(0) = x \alpha\alpha^{-1}(s) = c_x(s) \quad \forall t \in I, \forall s \in I \end{align}

(5)

\begin{align} \quad H_t(1) = x \alpha\alpha^{-1}(s) = c_x(s) \quad \forall t \in I, \forall s \in I \end{align}

• So indeed $\alpha\alpha^{-1} \simeq_{\{0, 1\}} c_x$. So $[\alpha\alpha^{-1}] = [c_x$, i.e.:

(6)

\begin{align} \quad [\alpha][\alpha^{-1}] = [c_x] \quad \blacksquare \end{align}

• Proof of b) Analogous to (a).