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, 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,

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

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

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

We can construct a homotopy

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

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 $$i$$,

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

we can construct a homotopy

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

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

which satisfies

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

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

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