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

(2)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:

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

- 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:

**Proof of b)**Analogous to (a).