# Homotopy Group is Group

## Theorem

The set of all homotopy classes of continuous mappings:

$c: \left[{0 \,.\,.\, 1}\right]^n \to X$

satisfying:

$c \left({\partial \left[{0 \,.\,.\, 1}\right]^n }\right) = x_0$

in a space $X$ at a base point $x_0$, under the operation of concatenation on class members, forms a group.

This group is called the $n$th homotopy group.

## Proof

We examine each of the group axioms separately.

### G0: Closure

The concatenation of any two functions:

$c_1, c_2: \left[{0 \,.\,.\, 1}\right]^n \to X$

from any two (not necessarily distinct) equivalence classes is another function:

$c_3: \left[{0 \,.\,.\, 1}\right]^n \to X$

by the definition of concatenation, which will have its own equivalence class.

### G1: Associativity

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

The concatenation $\left({ c_1 * c_2 }\right) * c_3$ is, from the definition of concatenation:

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

Likewise, the concatenation $c_1 * \left({ c_2 * c_3 }\right)$ is by definition:

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

We construct a homotopy:

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

We observe it satisfies $H \left({ \hat v, 0 }\right) = A \left({ \hat v }\right)$ and $H \left({ \hat v, 1 }\right) = B \left({ \hat v }\right)$.

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

### G2: Identity

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

Suppose we are given the function $c: \left[{0 \,.\,.\, 1}\right]^n \to X$ and its concatenation with $i$:

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

We construct a homotopy:

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

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

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

### G3: Inverses

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

Then we can construct a homotopy:

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

We observe it satisfies:

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

and:

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

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

$\blacksquare$