> [!definition] > > Let $X$ be a [[Topological Space|topological space]]. A **path** in $X$ is a [[Continuity|continuous]] map $f: [0, 1] \to X$. > [!definition] > > Let $f_0, f_1$ be paths in $X$ such that $f_0(0) = f_1(0)$ and $f_0(1) = f_1(1)$. A **homotopy of paths** connecting $f_0$ and $f_1$ is a continuous map $F: [0, 1]^2 \to X$ such that > 1. $F(s, 0) = f_0(s)$, $F(s, 1) = f_1(s)$. > 2. $F(0, t) = f_0(0) = f_1(0)$ and $F(1, t) = f_0(1) = f_1(1)$ for all $t \in [0, 1]$. > > For each $t \in [0, 1]$, denote $f_t(s) = F(s, t)$. > > If there exists a homotopy connecting $f_0$ and $f_1$, then $f_0 \simeq f_1$ are **homotopic**. > [!theorem] > > Let $f$ be a path in $X$ and $\varphi: [0, 1] \to [0, 1]$ be a continuous map such that $\varphi(0) = 0$ and $\varphi(1) = 1$, then $f \simeq f \circ \varphi$, and $f \circ \varphi$ is a **reparametrisation** of $f$. > > *Proof*. Let $F(s, t) = f((1 - t)s + t\varphi(s))$, then $F$ is continuous with $F_0 = f$ and $F_1 = f \circ \varphi$.