Homotopic Paths Implies Homotopic Composition

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $f_1, f_2, g_1, g_2: \left[{0 \,.\,.\, 1}\right] \to S$ be paths in $T$.

Let $f_1$ be homotopic to $f_2$ and $g_1$ be homotopic to $g_2$.

Then the concatenated paths $f_1 * g_1$ and $f_2 * g_2$ are homotopic.

Proof
Let $F: \left[{0 \,.\,.\, 1}\right] \times \left[{0 \,.\,.\, 1}\right] \to S$ be a homotopy between $f_1$ and $f_2$.

Let $G: \left[{0 \,.\,.\, 1}\right] \times \left[{0 \,.\,.\, 1}\right] \to S$ be a homotopy between $g_1$ and $g_2$.

Define $H: \left[{0 \,.\,.\, 1}\right] \times \left[{0 \,.\,.\, 1}\right] \to S$ by:
 * $H \left({s, t}\right) = \begin{cases} F \left({2 s, t}\right) & : s \in \left[{0 \,.\,.\, \dfrac 1 2}\right] \\

G \left({2 s - 1, t}\right) & : s \in \left[{\dfrac 1 2 \,.\,.\, 1}\right]\end{cases}$

By Continuous Mapping on Finite Union of Closed Sets, $H$ is continuous.

By definition of concatenation of paths, $H$ is a homotopy between $f_1 * g_1$ and $f_2 * g_2$.

Also see

 * Definition:Multiplication of Homotopy Classes of Paths