Homotopy Group is Group

Theorem
The set of all homotopy classes of continuous mappings $c:[0. . 1]^n \to X$ satisfying $c(\partial [0 . . 1]^n) = x_0$ in a space $X$ at a base point $x_0$, under the operation of concatenation on class members, form a group.

This group is called the $n^{th}$ homotopy group.

Proof
We examine each of the group axioms separately.

G0
The concatenation of any two functions $c_1, c_2: [0. . 1]^n \to X$ from any two (not necessarily distinct) equivalence classes is another function $c_3:[0. . 1]^n \to X$ by the definition of concatenation, which will have its own equivalence class.

G1
Let $c_1, c_2, c_3$ be three functions $[0,1]^n \to X$, selected from three (not necessarily different) equivalence classes.

The concatenation $(c_1 * c_2) * c_3$ is, from the definition of concatenation:

$A(\hat{v}) = \begin{cases} c_1 (4v_1, v_2, \ldots v_n) & : v_1 \in [0. . \dfrac 1 4] \\ c_2 (4v_1-1, \ldots, v_n) & : v_1 \in (\dfrac 1 4 . . \dfrac 1 2) \\ c_3 (2v_1-1, v_2, \ldots v_n) & : v_1 \in [\dfrac 1 2. . 1] \end{cases}$

Likewise, the concatenation $c_1 * (c_2 * c_3)$ is by definition:

$B(\hat{v}) = \begin{cases} c_1 (2v_1, v_2, \ldots v_n) & : v_1 \in [0. . \dfrac 1 2] \\ c_2 (4v_1-2, \ldots, v_n) & : v_1 \in (\dfrac 1 2 . . \dfrac 3 4) \\ c_3 (4v_1-3, v_2, \ldots v_n) & : v_1 \in [\dfrac 3 4. . 1] \end{cases}$

We can construct a homotopy:

$H(\hat{v},t) = \begin{cases} c_1 (\dfrac {4v_1} {1+t}, v_2, \ldots v_n) & : v_1 \in [0. . \dfrac {1+t} 4] \\ c_2 (4v_1-t-1, \ldots ,v_n) & : v_1 \in (\dfrac {1+t} 4 . . \dfrac {2+t} 4) \\ c_3 (\dfrac{4v_1-(2+t)} {2-t}, v_2, \ldots v_n) & : v_1 \in [\dfrac {2+t} 4. .1 \end{cases}$

which satisfies $H(\hat{v},0)=A(\hat{v})$ and $H(\hat{v},1)=B(\hat{v})$.

Therefore $A$ and $B$ are in the same equivalence class.

G2
The identity is simply the function $i:[0. . 1]^n \to X$ defined as $i(\hat{v})=x_0$, where $\hat{v} \in [0,1]^n$.

Given the function $c:[0,1]^n \to X$ and its concatenation with $i$:
 * $(c*i)(\hat{v}) = \begin{cases}

c(2v_1, v_2, \ldots v_n) & : v_1 \in [0. . 1/2] \\ i(\hat{v})=x_0 & : v_1 \in [1/2. . 1] \end{cases}$

we can construct a homotopy:

$H(\hat{v},t) = \begin{cases} c(\dfrac{2v_1}{2-t}, v_2, \ldots v_n) & : v_1 \in [0. . 1 - \dfrac t 2] \\ i(\hat{v}) = x_0 & : v_1 \in [1 - \dfrac t 2. . 1] \end{cases}$

which satisfies $H(\hat{v},0)=c(\hat{v})$ and $H(\hat{v},1)=(c*i)(\hat{v})$.

This shows $c$ and $c*i$ are in the same equivalence class.

G3
For any $c:[0. . 1]^n \to X$, $c^{-1}=c((1,0, \ldots, 0)-\hat{v})$.

Then we can construct a homotopy:

$H(\hat{v},t) = \begin{cases} c(2v_1, v_2, \ldots v_n) & : v_1 \in [0. . \dfrac{1-t} 2] \\ c(1-t, \ldots ,v_n) & : v_1 \in (\dfrac{1-t} 2 . . \dfrac{1+t} 2) \\ c^{-1}(2v_1-1, v_2, \ldots v_n) &: v_1 \in [\dfrac{1+t} 2. . 1] \end{cases}$

which satisfies:

$H(\hat{v},0) = \begin{cases} c(2v_1, v_2, \ldots v_n) & : v_1 \in [0. . 1/2] \\ c^{-1}(2v_1-1, v_2, \ldots v_n) & : v_1 \in [1/2. . 1] \end{cases}$

and:
 * $H(\hat{v},1)=x_0$

Hence $i$ and $c * c^{-1}$ are in the same equivalence class.