(Resources: Exercises from Munkres, Topology and personal notes from long ago.)
A homotopy is a continuous deformation between two continuous functions.
We begin by reviewing the concatenation operation on loops.
Given loops $\alpha, \beta, \gamma$ based at $x_0 \in X$, for topological space $X$, we first prove the following properties of the concatenation operation $\star$:
- (Associativity) $[\alpha] \star ([\beta] \star [\gamma]) =
([\alpha] \star [\beta]) \star [\gamma]$.
- (Right/left identities) $[\alpha] \star [e_{x_0}] = [e_{x_0}] \star [\alpha] = [\alpha]$.
- (Right/left inverses) $[\alpha] \star [\hat{\alpha}] = [\hat{\alpha}] \star [\alpha] = [e_{x_0}]$ where $\hat{\alpha}(t) = \alpha (1 - t)$.
- (Associativity) The concatenation operation $\star$ defined on paths instead of the induced operation on path-homotopy classes is not associative.
Namely,
\begin{align*}
(\alpha \star (\beta \star \gamma))(s) & =
\begin{cases}
\alpha(2s) & s \in [0, \frac{1}{2}]\\
\beta(4s - 2) & s \in [\frac{1}{2}, \frac{3}{4}]\\
\gamma(4s - 3) & s \in [\frac{3}{4}, 1]
\end{cases}\\
((\alpha \star \beta) \star \gamma)(s) & =
\begin{cases}
\alpha(4s) & s \in [0, \frac{1}{4}]\\
\beta(4s - 1) & s \in [\frac{1}{4}, \frac{1}{2}]\\
\gamma(2s - 1) & s \in [\frac{1}{2}, 1].
\end{cases}\\
\end{align*}
However, the two concatenated paths are path-homotopic by defining the following homotopy:
\begin{align*}
H(s, t) & =
\begin{cases}
\alpha (\frac{4s}{2-t}) & s \in [0, \frac{2-t}{4}]\\
\beta ((4s-2)(1-t) + (4s-1)t) & s \in [\frac{2-t}{4}, \frac{3-t}{4}]\\
\gamma (\frac{4s + t -3}{t + 1}) & s \in [\frac{3-t}{4}, 1]
\end{cases}\\
& =
\begin{cases}
\alpha (\frac{4s}{2-t}) & s \in [0, \frac{2-t}{4}]\\
\beta (4s + t - 2) & s \in [\frac{2-t}{4}, \frac{3-t}{4}]\\
\gamma (\frac{4s + t -3}{t + 1}) & s \in [\frac{3-t}{4}, 1]
\end{cases}.
\end{align*}
Note that:
\begin{align*}
H(s, 0) & = (\alpha \star (\beta \star \gamma))(s)\\
H(s, 1) & = ((\alpha \star \beta) \star \gamma)(s)\\
H(0, t) & = ((\alpha \star \beta) \star \gamma)(0) \\
& = (\alpha \star (\beta \star \gamma))(0)\\
& = \alpha(0) = x_0\\
H(1, t) & = ((\alpha \star \beta) \star \gamma)(1) \\
& = (\alpha \star (\beta \star \gamma))(1)\\
& = \gamma(1) = x_0.
\end{align*}
One can plug in the bounds for the $s$ to show that $\frac{4s}{2-t}$ at $s = \frac{2-t}{4}$ is equivalent to $1$, $4s + t - 2$ at $s = \frac{2-t}{4}$ is equivalent to $0$, $4s + t - 2$ at $s = \frac{3-t}{4}$ is equivalent to $1$, and $\frac{4s + t -3}{t + 1}$ at $s = \frac{3-t}{4}$ is equivalent to $0$.
Since $\alpha(1) = \beta(0) = \beta(1) = \gamma(0) = x_0$, the boundary points of the closed intervals for $s$ agree in each case.
The denominators are all bounded away from $0$ since $t \in [0, 1]$, so for each path reparametrization we have a composition of continuous functions which is continuous.
Thus, the pasting lemma can be applied and $H(s, t)$ is continuous.
Hence $H(s, t)$ is a path-homotopy from $(\alpha \star (\beta \star \gamma))(s)$ to $((\alpha \star \beta) \star \gamma)(s)$.
It follows that:
\begin{align*}
[\alpha \star (\beta \star \gamma)] = [(\alpha \star \beta) \star \gamma].
\end{align*}
By definition of the induced operation on path-homotopy classes:
\begin{align*}
[\alpha] \star ([\beta] \star [\gamma]) &= [\alpha \star (\beta \star \gamma)]\\
&= [(\alpha \star \beta) \star \gamma]\\
&= ([\alpha] \star [\beta]) \star [\gamma].
\end{align*}
- (Right/left identities)
Define the path $e_{x_0}(s) = x_0$ for $s \in [0, 1]$.
Note that
\begin{align*}
(\alpha \star e_{x_0}) (s) &=
\begin{cases}
\alpha(2s) & s \in [0, \frac{1}{2}]\\
e_{x_0}(s) & s \in [\frac{1}{2}, 1]
\end{cases}\\
&=
\begin{cases}
\alpha(2s) & s \in [0, \frac{1}{2}]\\
x_0 & s \in [\frac{1}{2}, 1]
\end{cases}.
\end{align*}
\begin{align*}
(e_{x_0} \star \alpha) (s) & =
\begin{cases}
e_{x_0}(2s) & s \in [0, \frac{1}{2}]\\
\alpha(2s - 1) & s \in [\frac{1}{2}, 1]
\end{cases}\\
& =
\begin{cases}
x_0 & s \in [0, \frac{1}{2}]\\
\alpha(2s - 1) & s \in [\frac{1}{2}, 1]
\end{cases}\\
\end{align*}
We prove that $\alpha \star e_{x_0}$ and $e_{x_0} \star \alpha$ are each path-homotopic to $\alpha$.
Define the mapping $H_1(s, t)$ between $\alpha \star e_{x_0}$ and $\alpha$:
\begin{align*}
H_1(s, t) & =
\begin{cases}
\alpha(2s) (1 -t) + \alpha(s) t & s \in [0, \frac{1}{2}]\\
x_0 (1 -t) + \alpha(s) t & s \in [\frac{1}{2}, 1]
\end{cases}
\end{align*}
$H_1(s,t)$ is in fact continuous by first observing we are using a linear reparametrization and then noting in the boundary condition at $s = \frac{1}{2}$ that $2s = 1$ and $\alpha(1) = x_0$ so $\alpha(2s)(1-t) + \alpha(s) t = x_0(1-t) + t \alpha(s)$; hence we can apply the pasting lemma.
$H_1(s, t)$ also satisfies the other part of being a path-homotopy between $\alpha \star e_{x_0}$ and $\alpha$ :
\begin{align*}
H_1(s, 0) & = \alpha \star e_{x_0}\\
H_1(s, 1) & = \alpha \\
H_1(0, t) & = \alpha(0) = x_0 \\
H_1(1, t) & = \alpha(1) = x_0.
\end{align*}
Hence $[\alpha \star e_{x_0}] = [\alpha]$.
Similarly, we can define $H_2(s, t)$ to be:
\begin{align*}
H_2(s, t) & =
\begin{cases}
x_0 (1 -t) + \alpha(s) t & s \in [0, \frac{1}{2}]\\
\alpha(2s - 1) (1 -t) + \alpha(s) t & s \in [\frac{1}{2}, 1]
\end{cases}
\end{align*}
which is a path-homotopy between $e_{x_0} \star \alpha$ an $\alpha$ by the same reasoning above.
Hence $[e_{x_0} \star \alpha] = [\alpha]$.
Therefore,
\begin{align*}
[e_{x_0}] \star [\alpha] & = [e_{x_0} \star \alpha]\\
&= [\alpha]\\
&= [\alpha \star e_{x_0}]\\
&= [\alpha] \star [e_{x_0}].
\end{align*}
- (Right/left inverses) $[\alpha] \star [\hat{\alpha}] = [\hat{\alpha}] \star [\alpha] = [e_{x_0}]$ where $\hat{\alpha}(t) = \alpha (1 - t)$.
\begin{align*}
\alpha \star \hat{\alpha} &=
\begin{cases}
\alpha(2s) & s \in [0, \frac{1}{2}]\\
\hat{\alpha}(2s -1) & s \in [\frac{1}{2}, 1]
\end{cases}\\
&=
\begin{cases}
\alpha(2s) & s \in [0, \frac{1}{2}]\\
\alpha(2 -2s) & s \in [\frac{1}{2}, 1]
\end{cases}\\
\end{align*}
We define the following path-homotopy $G_1$ between $\alpha \star \hat{\alpha}$ and $e_{x_0}$:
\begin{align*}
G_1(s, t) &=
\begin{cases}
\alpha(2s)(1-t) + x_0 t & s \in [0, \frac{1}{2}]\\
\alpha(2-2s)(1-t)+ x_0 t & s \in [\frac{1}{2}, 1]
\end{cases}.
\end{align*}
Note that $G_1(s, t)$ is continuous by the pasting lemma since $\alpha(1) = \hat{\alpha}(0)$ and composition of continuous functions are continuous.
Moreover:
\begin{align*}
G_1(s, 0) &=
\begin{cases}
\alpha(2s) & s \in [0, \frac{1}{2}]\\
\alpha(2-2s) & s \in [\frac{1}{2}, 1]
\end{cases}\\
& = \alpha \star \hat{\alpha}\\
G_1(s, 1) & = x_0\\
G_1(0, t) & = x_0\\
G_1(1, t) & = x_0.
\end{align*}
Hence $G_1(s, t)$ is a path-homotopy from $\alpha \star \hat{\alpha}$ to $e_{x_0}$.
Similarly, we define the following path-homotopy $G_2$ between $\hat{\alpha} \star \alpha$ and $e_{x_0}$:
\begin{align*}
G_2(s, t) &=
\begin{cases}
\alpha(1 - 2s)(1-t) + x_0 t & s \in [0, \frac{1}{2}]\\
\alpha(2s - 1)(1-t)+ x_0 t & s \in [\frac{1}{2}, 1]
\end{cases}.
\end{align*}
Note that
\begin{align*}
G_2(s, 0) &=
\begin{cases}
\alpha(1 - 2s) & s \in [0, \frac{1}{2}]\\
\alpha(2s - 1) & s \in [\frac{1}{2}, 1]
\end{cases}\\
&= \hat{\alpha} \star \alpha\\
G_2(s, 1) &= x_0\\
G_2(0, t) &= x_0\\
G_2(1, t) &= x_0.
\end{align*}
And $G_2$ is continuous for the same reasons given above for $G_1$.
Hence $G_2(s, t)$ is a path-homotopy from $\hat{\alpha} \star \alpha$ to $e_{x_0}$.
Therefore,
\begin{align*}
[\alpha] \star [\hat{\alpha}] & = [\alpha \star \hat{\alpha}]\\
&= [e_{x_0}]\\
&= [\hat{\alpha} \star \alpha]\\
&= [\hat{\alpha}] \star [\alpha]
\end{align*}
Given spaces $X$ and $Y$, let $[X, Y]$ denote the set of homotopy classes of maps of $X$ into $Y$.
Let $I = [0, 1]$. For any $X$, the set $[X ,I]$ has a single element.
We show that give any two continuous functions $f : X \to I$ and $f' : X \to I$, we define a homotopy $H : X \times [0, 1] \to I $ from $f$ to $f'$.
Since $[0, 1]$ is a subset of the real line which is a vector space, the additional and scalar multiplication operations are well-defined.
\begin{align*}
H(s, t) &= f'(s) (1-t) + f(s) t.
\end{align*}
Note that
\begin{align*}
H(s, 0) &= f'(s)\\
H(s, 1) &= f(s)
\end{align*}
and $H$ is continuous as it is a linear combination of continuous functions.
So $H$ is a homotopy, and any two continuous functions $f, f'$ are homotopic.
Hence, the set $[X, I]$ has a single homotopy class.
If $Y$ is path connected, the set $[I, Y]$ has a single element.
Suppose that $Y$ is path connected.
Given any point $x_0 \in Y$, the identity loop given by $e_{x_0} : [0, 1] \to Y$ where $e_{x_0}(s) = x_0$ is homotopic to any path $\alpha$ starting at $x_0$ by the following homotopy:
\begin{align*}
H(s,t) = \alpha(s(1-t)).
\end{align*}
Note that:
\begin{align*}
H(s, 0) &= \alpha(s)\\
H(s, 1) &= \alpha(0) = x_0 = e_{x_0}
\end{align*}
and $s(1-t)$ is continuous in $s, t$ so $H(s, t)$ is continuous by composition.
Since $Y$ is path connected, there exists a path from every point $y \in Y$ to $x_0$ so the identity loop for any $y$ is homotopic with the path from $y$ to $x_0$ which is homotopic by symmetry to the identity loop on $x_0$.
So any path from $x$ to $y$ is homotopic by transitivity to the identity loop on $x_0$ and hence $[I, Y]$ is a single element.