Homotopy Equivalence Class

Reflexivity Let $\gamma : [0,1] \to X$ be a path. Define $$ H(t,s) = \gamma(t). $$ Then $H$ is continuous and satisfies $$ H(t,0) = \gamma(t), \quad H(t,1) = \gamma(t), $$ and $$ H(0,s) = \gamma(0), \quad H(1,s) = \gamma(1). $$ Thus $\gamma \sim \gamma$. Symmetry Suppose $\gamma_0 \sim \gamma_1$ via a homotopy $H(t,s)$. Define $$ G(t,s) = H(t, 1 - s). $$ Then $G$ is continuous and satisfies $$ G(t,0) = H(t,1) = \gamma_1(t), \quad G(t,1) = H(t,0) = \gamma_0(t), $$ and $$ G(0,s) = \gamma_0(0), \quad G(1,s) = \gamma_0(1). $$ Thus $\gamma_1 \sim \gamma_0$. Transitivity Suppose $\gamma_0 \sim \gamma_1$ via $H$ and $\gamma_1 \sim \gamma_2$ via $K$. Define $$ G(t,s) = \begin{cases} H(t, 2s) & if 0 \le s \le \tfrac{1}{2}, \\ K(t, 2s - 1) & if \tfrac{1}{2} \le s \le 1. \end{cases} $$ Then $G$ is continuous (by the pasting lemma), since at $s = \tfrac{1}{2}$ we have $$ H(t,1) = \gamma_1(t) = K(t,0). $$ Moreover, $$ G(t,0) = H(t,0) = \gamma_0(t), \quad G(t,1) = K(t,1) = \gamma_2(t), $$ and $$ G(0,s) = \gamma_0(0), \quad G(1,s) = \gamma_0(1). $$ Thus $\gamma_0 \sim \gamma_2$.