We examine each of the group axioms separately. === $\text G 0$: Closure === The concatenation of any two mappings: :$c_1, c_2: \closedint 0 1^n \to X$ from any two (not necessarily distinct) equivalence classes is another mapping: :$c_3: \closedint 0 1^n \to X$ by the definition of concatenation, which will have its own equivalence class. === $\text G 1$: Associativity ===  Let $c_1, c_2, c_3$ be three functions $\closedint 0 1^n \to X$, selected from three (not necessarily different) equivalence classes. The concatenation $\paren {c_1 * c_2} * c_3$ is, from the definition of concatenation: :$\map A {\hat v} = \begin {cases} \map {c_1} {4 v_1, v_2, \ldots, v_n} & : v_1 \in \closedint 0 {\dfrac 1 4} \\ \map {c_2} {4 v_1 - 1, v_2, \ldots, v_n} & : v_1 \in \openint {\dfrac 1 4} {\dfrac 1 2} \\ \map {c_3} {2 v_1 - 1, v_2, \ldots, v_n} & : v_1 \in \closedint {\dfrac 1 2} 1 \end{cases}$ Likewise, the concatenation $c_1 * \paren {c_2 * c_3}$ is by definition: :$\map B {\hat v} = \begin {cases} \map {c_1} {2 v_1, v_2, \ldots, v_n} & : v_1 \in \closedint 0 {\dfrac 1 2} \\ \map {c_2} {4 v_1 - 2, v_2, \ldots, v_n} & : v_1 \in \openint {\dfrac 1 2} {\dfrac 3 4} \\ \map {c_3} {4 v_1 - 3, v_2, \ldots, v_n} & : v_1 \in \closedint {\dfrac 3 4} 1 \end {cases}$ We construct a homotopy: :$\map H {\hat v, t} = \begin {cases} \map {c_1} {\dfrac {4 v_1} {1 + t}, v_2, \ldots, v_n} & : v_1 \in \closedint 0 {\dfrac {1 + t} 4} \\ \map {c_2} {4 v_1 - t - 1, v_2, \ldots, v_n} & : v_1 \in \openint {\dfrac {1 + t} 4} {\dfrac {2 + t} 4} \\ \map {c_3} {\dfrac {4 v_1 - \paren {2 + t} } {2 - t}, v_2, \ldots, v_n} & : v_1 \in \closedint {\dfrac {2 + t} 4} 1 \end {cases}$ We observe it satisfies $\map H {\hat v, 0} = \map A {\hat v}$ and $\map H {\hat v, 1} = \map B {\hat v}$. Therefore $A$ and $B$ are in the same equivalence class. === $\text G 2$: Identity ===  The identity is simply the function $I: \closedint 0 1^n \to X$ defined as $\map I {\hat v} = x_0$, where $\hat v \in \closedint 0 1^n$. Suppose we are given the function $c: \closedint 0 1^n \to X$ and its concatenation with $i$: :$\map {\paren {c*i} } {\hat v} = \begin {cases} \map c {2 v_1, v_2, \ldots, v_n} & : v_1 \in \closedint 0 {\dfrac 1 2} \\ \map I {\hat v} = x_0 & : v_1 \in \closedint {\dfrac 1 2} 1 \end {cases}$ We construct a homotopy: :$\map H {\hat v, t} = \begin {cases} \map c {\dfrac {2 v_1} {2 - t}, v_2, \ldots, v_n} & : v_1 \in \closedint 0 {1 - \dfrac {1 - t} 2} \\ \map I {\hat v} = x_0 & : v_1 \in \closedint {1 - \dfrac {1 - t} 2} 1 \end{cases}$  which satisfies $\map H {\hat v, 0} = \map c {\hat v}$ and $\map H {\hat v, 1} = \map {\paren {c*i} } {\hat v}$. This shows $c$ and $c*i$ are in the same equivalence class. === $\text G 3$: Inverses === For any $c: \closedint 0 1^n \to X$, $\map {c^{-1} } {\hat v} = \map c {\tuple {1, 0, \ldots, 0} - \hat v}$. Then we can construct a homotopy: :$\map H {\hat v, t} = \begin {cases} \map c {2 v_1, v_2, \ldots, v_n} & : v_1 \in \closedint 0 {\dfrac {1 - t} 2} \\ \map c {1 - t, v_2, \ldots, v_n} & : v_1 \in \openint {\dfrac {1 - t} 2} {\dfrac {1 + t} 2} \\ \map {c^{-1} } {2 v_1 - 1, v_2, \ldots v_n} & : v_1 \in \closedint {\dfrac {1 + t} 2} 1 \end {cases}$ We observe it satisfies: :$\map H {\hat v, 0} = \begin {cases} \map c {2 v_1, v_2, \ldots, v_n} & : v_1 \in \closedint 0 {\dfrac 1 2} \\ \map {c^{-1} } {2 v_1 - 1, v_2, \ldots, v_n} & : v_1 \in \closedint {\dfrac 1 2} 1 \end {cases}$ and: :$\map H {\hat v, 1} = x_0$ Hence $I$ and $c * c^{-1}$ are in the same equivalence class. {{qed}}