Lemma 1.3
The relations $\simeq$ and $\simeq_P$ are equivalence relations.
Proof: We need to show that the relations $\simeq$ and $\simeq_P$ are reflexive, symmetric, and transitive. For the first two, we will look at $\simeq$ only, leaving out $\simeq_P$ since it will follow by the same argument. For transitivity, we will need to address the extra condition of $\simeq_P$ which requires that the endpoints stay the same.
Reflexivity: Let
$$f: X \longrightarrow Y$$ be a continuous map.
$f \simeq f$ with the trivial homotopy
$$F(x,t) = f(x) \quad \forall t \in [0,1]$$
Symmetry: Define $f$ as before and let Let
$$f': X \longrightarrow Y$$
be another continuous function such that $f \simeq f'$. Let $F(x,t)$ be a homotopy between $f$ and $f'$, meaning that
$$F(x,0) = f(x) \quad \text{and} \quad F(x,1) = f'(x)$$
Then, we can define $G(x,t) = F(x,1-t)$, which is also a homotopy between $f$ and $f'$. We can think of $G$ as $F$ going backwards, since $t$ is starting at one and going to zero. So we have
$$G(x,0) = F(x,1) = f'(x)$$
and
$$G(x,1) = F(x,0) = f(x)$$
which gives us
$f \simeq f'$.
Transitivity: Introduce a third path $$f'': X \longrightarrow Y$$ such that $f \simeq f'$ and $f' \simeq f''$. Let $F$ be a homotopy between $f$ and $f'$ and let $F'$ be a homotopy between $f'$ and $f''$. We need to find a homotopy between $f$ and $f''$.
Define
$$G: X \times I \longrightarrow Y$$
by
$$G(x,t) = \begin{cases}
F(x,2t) & t \in \left[ 0,\frac{1}{2}\right]\\
F'(x,2t-1) & t \in \left[\frac{1}{2}, 1\right]
\end{cases}$$
While $t$ only goes from zero to one, this transformation takes place over 2 "seconds", since its components go from 0 to 1 one after another. It's clear that this maps $f$ to $f''$, since it does the transformation of $F$ followed by the transformation of $F'$. We can even check:
$$G(x,0) = F(x,2\cdot 0) = F(x,0) = f(x)$$
$$G(x,1) = F'(x,2(1) - 1) = F'(x,1) = f''(x)$$
So now we only need to show that it is continuous to meet the requirements of a homotopy.
This map is indeed continuous, since at $t= \frac{1}{2}$ we have
$$F(x,2t) = f'(x) = F'(x,2t-1)$$
Therefore, by the Pasting Lemma, $G$ s continuous on $X \times I$
and is the required homotopy between $f$ and $f''$.
Let's also look at transitivity for path homotopies. Let:
$$F: I \times I \longrightarrow X$$
$$F': I \times I \longrightarrow X$$
be path homotopies between $f$ and $f'$, and $f'$ and $f''$ respectively, and define $G$ nearly the same as before, only with variable $s$
$$G(s,t) = \begin{cases}
F(s,2t) & t \in \left[ 0,\frac{1}{2}\right]\\
F'(s,2t-1) & t \in \left[\frac{1}{2}, 1\right]
\end{cases}$$
Since $f$ and $f'$ are both path homotopies, we have that for all $t \in I$
$$F(0,t) = x_0, \quad F(1,t) = x_1$$
and
$$F'(0,t) = x_0, \quad F'(1,t) = x_1$$
So we have
$$G(0,t) = \begin{cases}
F(0,2t) & t \in \left[ 0,\frac{1}{2}\right]\\
F'(0,2t-1) & t \in \left[\frac{1}{2}, 1\right]
\end{cases} \, = x_0 \quad \forall t\in [0,1]$$
and
$$G(1,t) = \begin{cases}
F(1,2t) & t \in \left[ 0,\frac{1}{2}\right]\\
F'(1,2t-1) & t \in \left[\frac{1}{2}, 1\right]
\end{cases} \, = x_1 \quad \forall t\in [0,1]$$
Therefore, $G$ is a path homotopy between $f$ and $f'$.
Definition 1.4
We say that a set $X$ is convex if every line segment connecting two points in $X$ fully contained in $X$.
Theorem 1.5
Let $f$ and $g$ be any two maps of a space $X$ into $\mathbb{R}^2$. Then, $f$ and $g$ are homotopic and
$$F(x,t) = (1-t) f(x) + t g(x)$$
is a homotopy between them.
The proof of this comes straight from the definition of homotopy.
For an example, let
$$f: x \mapsto \left(x,x^2\right)$$
$$g: x \mapsto \left(x, \frac{x}{5} \right)$$
Below, we can see the homotopy $$F(x,t) = (1-t)\left(x,x^2\right) + t \left(x, \frac{x}{5} \right)$$
transforming $f$ into $g$ as $t$ goes from 0 to 1.
$F$ in this case is called a straight line homotopy. The straight line homotopy will work for any convex subset $A$ of $\mathbb{R}^n$ and
$$f: X \longrightarrow A$$
$$g: X \longrightarrow A$$
Example 1.6
Let $X = \mathbb{R}^2 - \{0\}$. (A plane with a hole in the middle) The paths
$$f(s) = (\cos \pi s, \sin \pi s)$$
$$g(s) = (\cos \pi s, 2\sin \pi s) $$
are path homotopic in $X$.
But if we let
$$h(s) = (\cos \pi s, -\sin \pi s) $$
$f$ and $h$ are not path homotopic. Why? Transforming $f$ into $h$ would require passing through $(0,0)$—where the hole is! Therefore the transformation cannot be continuous, which is required of a homotopy.
Definition 1.7
An equivalence class under homotopy is called a homotopy class. The homotopy class consisting of all maps homotopic to $f$ is denoted $[f]$.
Definition 1.8
Let $f$ be a path in $X$ from $x_0$ to $x_1$, and let $g$ be a path in $X$ from $x_1$ to $x_2$. We call these paths composable, since $f(1) = g(0)$.
We define the product
$$h(s) = (f * g)(s) = \begin{cases}
f(2s) & s \in \left[ 0,\frac{1}{2}\right]\\
g(2s-1) & s \in \left[\frac{1}{2}, 1\right]
\end{cases}$$
This function is also sometimes called concatenation, and is well defined and continuous by the Pasting Lemma. This functions represents the path created by traveling along $f$ from $x_0$ to $x_1$, and then along $g$ from $x_1$ to $x_2$.
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Theorem 1.9
Let $f$ and $f'$ be two paths from $x_0$ to $x_1$ and let $g$ and $g'$ be two paths from $x_1$ to $x_2$. If $F$ is a path homotopy between between $f$ and $f'$ and $G$ is a path homotopy between $g$ and $g'$, the product $H = F * G$ is a homotopy between $f * g$ and $f' * g'$.
Proof: By the definition of the product we have that
$$H(s,t) = (F * G)(s,t) = \begin{cases}
F(2s,t) & s \in \left[ 0,\frac{1}{2}\right]\\
G(2s-1,t) & s \in \left[\frac{1}{2}, 1\right]
\end{cases}$$
This map is also well-defined and continuous, since at $t=\frac{1}{2}$ we have $F,(1,t) = G(0,t) = x_1$ for all $t$.
Now, if we let $t=0$ we get
$$H(s,0) = \begin{cases}
F(2s,0) & s \in \left[ 0,\frac{1}{2}\right]\\
G(2s-1,0) & s \in \left[\frac{1}{2}, 1\right]
\end{cases} $$
$$= \begin{cases}
f(2s) & s \in \left[ 0,\frac{1}{2}\right]\\
g(2s-1) & s \in \left[\frac{1}{2}, 1\right]
\end{cases} \, = (f * g)(s)$$
Similarly, for $t=1$ we have
$$H(s,1) = \begin{cases}
f'(2s) & s \in \left[ 0,\frac{1}{2}\right]\\
g'(2s-1) & s \in \left[\frac{1}{2}, 1\right]
\end{cases} \, = (f' * g')(s)$$
Therefore, $H$ is a homotopy between $f * g$ and $f' * g'$. In other words, the product of the homotopies, is the homotopy of the products!
Corollary 1.10
For homotopy classes $[f]$ and $[g]$, we have the property
$$[f] * [g] = [f * g]$$
Theorem 1.11
The operation * has the following properties:
Before we can prove this, we need the following lemma:
Lemma 1.12
Proof of Lemma:
(a) Since $F$ be a path homotopy, it is a continuous map $$F: X \times I \longrightarrow Y$$ such that
$$F(s,0) = f(s), \quad \quad F(s,1) = f'(s)$$
and
$$F(0,t) = x_0, \quad \quad F(1,t) = x_1$$
Now if we compose $k$ with $F$, we get a continuous map
$$k \circ F : I \times I \longrightarrow Y$$
such that
$$(k \circ F) (s,0) = k(F(s,0)) = k(f(s)) = (k\circ f)(s)$$
$$(k \circ F) (s,1) = k(F(s,1)) = k(f'(s)) = (k\circ f')(s)$$
Additionally we have that
$$(k \circ F) (0,t) = k(F(0,t)) = k(x_0) $$
$$(k \circ F) (1,t) = k(F(1,t)) = k(x_1) $$
which, if $x_0$ and $x_1$ are the endpoints of $f$ and $f'$, will be the endpoints of the paths $k \circ f$ and $k \circ f'$.
Therefore $k \circ F$ is a path homotopy between $k \circ f$ and $k \circ f'$. 
(b) Using the definition of the product:
$$k \circ (f * g)(s) = k \circ \begin{cases}
f(2s) & s \in \left[ 0,\frac{1}{2}\right]\\
g(2s-1) & s \in \left[\frac{1}{2}, 1\right]
\end{cases}$$ $$= \begin{cases}
k \circ f(2s) & s \in \left[ 0,\frac{1}{2}\right]\\
k \circ g(2s-1) & s \in \left[\frac{1}{2}, 1\right]
\end{cases}$$
$$=(k \circ f) * (k \circ g)(s) $$
Now we are ready to prove the theorem.
Proof of Theorem 1.11: We will do these out of order since associativity is a bit trickier than the rest.
II. The Identity Property: First we will show the first half of the statement, that $[e_{x_0}] * [f] = [f]$. To do this, we will define two functions and show they are homotopic. Then, we will compose each with $f$, which by part (a) of our lemma will give us our result.
First, define the function $$e_0 : I \longrightarrow I$$ by
$$e_0(s) = 0 \quad \quad \forall s \in [0,1]$$
and the function $$i : I \longrightarrow I$$ by
$$i(s) = s \quad \quad \forall s \in [0,1]$$
Now, take the product
$$(e_0 * i)(s) = \begin{cases}
e_0(2s) & s \in \left[ 0,\frac{1}{2}\right]\\
i(2s-1) & s \in \left[\frac{1}{2}, 1\right]
\end{cases}$$
$$ = \begin{cases}
0 & s \in \left[ 0,\frac{1}{2}\right]\\
2s-1 & s \in \left[\frac{1}{2}, 1\right]
\end{cases}$$
Since $I$ is convex, by Theorem 1.5 we can define $$G: I \times I \longrightarrow I$$ by
$$G (s,t) = (1-t)i(s) + t (e_0 * i)(s)$$
which is a homotopy of $i$ and $e_0 * i$.
Now we will use our lemma. We know that $f \circ G: I \times I \longrightarrow I$ is continuous and:
$$(f \circ G)(s,0) = f(G(s,0)) = f(s)$$
$$(f \circ G)(s,1) = f(G(s,1)) = f(e_0 * i)(s) = \big( (f \circ e_0) * (f \circ i) \big)(s)$$
$$= \begin{cases}
f \circ e_0(s) & s \in \left[ 0,\frac{1}{2}\right]\\
f \circ i(2s-1) & s \in \left[\frac{1}{2}, 1\right]
\end{cases}$$
$$= \begin{cases}
x_0 & s \in \left[ 0,\frac{1}{2}\right]\\
f(2s-1) & s \in \left[\frac{1}{2}, 1\right]
\end{cases}$$
$$= \left(e_{x_0} * f \right) (s)$$
This gives us $f \simeq_p e_{x_0} * f$ and our result
$$[f] = \left[e_{x_0} * f \right] = [e_{x_0}] * [f]$$
To show that $[f] * [e_{x_1}] = [f]$, we will use a similar method. Define $$e_1: I \longrightarrow I$$ by
$$e_1(s) = 1 \quad \quad \forall s \in [0,1]$$ and keep $i$ the same.
Now we have the product $i * e_1: I \longrightarrow I$ given by
$$(i * e_1)(s) = \begin{cases}
2s & s \in \left[ 0,\frac{1}{2}\right]\\
1 & s \in \left[\frac{1}{2}, 1\right]
\end{cases}$$
Just like in the first part, we can define $H: I \times I \longrightarrow I$ to be a homotopy between $i$ and $i * e_1$.
Therefore, $f \circ H$ is a homotopy between $f = f \circ i$ and $(f \circ i) * (f \circ e_1) = f * e_{x_1}$, and we have our result
$$[f] \simeq_p [f] * [e_{x_1}]$$
III. The Inverse Property: Using the same identity map $i$ as before, we can define its reverse by $$\tilde{i}(s) =1-s$$
$i * \tilde{i}$ will be a path in $I$ from 0 to 0, and therefore we have that
$$e_0 \simeq_p i * \tilde{i} $$
By our lemma, this gives us
$$f \circ e_0 \simeq_p (f \circ i) * (f \circ \tilde{i})$$
where $f \circ e_0 = x_0$ and
$$(f \circ i) * (f \circ \tilde{i}) = \begin{cases}
f \circ i(2s) & s \in \left[ 0,\frac{1}{2}\right]\\
f \circ \tilde{i}(2s-1) & s \in \left[\frac{1}{2}, 1\right]
\end{cases}$$
$$=\begin{cases}
f(2s) & s \in \left[ 0,\frac{1}{2}\right]\\
f(-2s) & s \in \left[\frac{1}{2}, 1\right]
\end{cases}$$
$$= f * \tilde{f}$$
is a path that starts and ends at $x_0$.
Therefore we have
$$f * \tilde{f} \simeq_p e_{x_0}$$
and our intended result
$$[f] * [\tilde{f}] = [e_{x_0}]$$
Through the same process we can show that
$$[\tilde{f}] * [f] = [e_{x_0}]$$
With that, the time for associativity has come.
I. The Associative Property To begin this proof, we have to establish the algebraic fact which goes as follows: Let $[a,b]$ and $[c,d]$ are two intervals in $\mathbb{R}$. There will always be a unique positive linear map
$$p: [a,b] \longrightarrow [c,d]$$
which transforms $[a,b]$ into $[c,d]$ and has the form $y = mx + b$. Positive here means that the map will always have a positive slope, which makes sense because we need to preserve order. For both intervals, as $x$ increases, $y$ increases.
For instance, the positive linear map from $[1,2]$ to $[-5,-1]$ is given by
$p(x) = 4x - 9$.
Now, let $f,g,h$ be three composable functions. That is:
We can now consider the triple product of these three functions. \\ \\
First, consider the product $f * (g * h)$, which is given by
$$\big(f * (g * h)\big)(s) = \begin{cases}
f(2s) & s \in \left[ 0,\frac{1}{2}\right]\\
(g * h)(2s-1) & s \in \left[\frac{1}{2}, 1\right]
\end{cases}$$
$$= \begin{cases}
f(2s) & s \in \left[ 0,\frac{1}{2}\right]\\
g(4s-2) & s \in \left[\frac{1}{2}, \frac{3}{4}\right] \\
h(4s-3) & s \in \left[\frac{3}{4}, 1\right]
\end{cases} $$ \\
Now what if we moved the parentheses? We would get
$$\big((f * g) * h\big)(s) = \begin{cases}
(f * g)(2s) & s \in \left[ 0,\frac{1}{2}\right]\\
h(2s-1) & s \in \left[\frac{1}{2}, 1\right]
\end{cases}$$
$$= \begin{cases}
f(4s) & s \in \left[ 0,\frac{1}{4}\right]\\
g(4s-1) & s \in \left[\frac{1}{4}, \frac{1}{2}\right] \\
h(2s-1) & s \in \left[\frac{1}{2}, 1\right]
\end{cases} $$
Our challenge is to prove that these are equal. To do this, we will look at the product as a series of compositions involving positive linear maps.
Looking at $f * (g * h)$, the first component of the product maps the interval $[0, \frac{1}{2}$ to $[0,1]$ and then plugs it into $f$. The second maps $[\frac{1}{2}, \frac{3}{4}]$ to $[0,1]$ and then plugs it into to $g$, and similarly for the third component.
Let's define a path to represent these three linear maps. Let $a$ and $b$ be points such that $ 0 < a < b < 1$. Define a path $k_{a,b}$ as follows:
The way we defined this path, it could represent either $f * (g * h)$ or $(f * g) * h$.
We can show that if we replace $a$ and $b$ with two different points $c$ and $d$ where $0 < c < d < 1$, then $k_{a,b} \simeq_P k_{c,d}$.
Let $p: I \longrightarrow I$ be a map where $p(a) = c$ and $p(b) = d$. 
$p$ is made up of the positive linear maps that map $[0,a]$, $[a,b]$, and $[b,1]$ to $[0,c]$, $[c,d]$, and $[d,1]$ respectively, and we have
$$k_{a,b} \circ p = k_{c,d}$$
Since $p$ turns $a$ into $c$ and $b$ into $d$.
Now, since $p$ maps $I$ to $I$ and so does the identity map,
$i: I \longrightarrow I$, there must exist a path homotopy $P$ between them. By Lemma 1.12, this means that $k_{a,b} \circ P$ is a path homotopy between $k_{a,b}$ and $k_{c,d}$.
Hence, $f * (g * h)$ and $(f * g) * h$ are in the same homotopy class. This gives us our result
$$[f] * \big([g] * [h] \big) = \big([f] *[g] \big) * [h] $$
In fact, this extends to associativity for any finite product of paths!
Theorem 1.13
Let $f$ be a path in $X$ and let
$$0 = a_0 < a_1 < \dots < a_n = 1$$
be a partition of the interval $[0,1]$ and let
$$f_i : I \longrightarrow X \quad i = 1,2,...,n$$
be the path that represents the positive linear map of $I$ onto $[a_{i-1},a_i]$ plugged into $f_i$.
Then,
$$[f] = [f_1] * [f_2] * \dots * [f_n]$$
Proof: Consider the set of paths $$p_i: I \longrightarrow I$$ given by
$$p_i(s) = (1-s)a_{i-1} + sa_i \quad \quad 0 \leq s \leq 1, \quad i=1,2,...,n $$
Since $I$ is convex, all of these paths are fully contained in $I$. All of these paths together are essentially connecting the dots $a_0, a_1, ...,a_n$.
Now define $P: I \longrightarrow I$ by
$$p = p_1 * p_2 * \dots *p_n$$
Since all of these together create a path from 0 to 1, we have that
$$[p] = [p_1] * [p_2] * \dots * [p_n] = [i]$$
where $i$ is the identity map in $I$,
and
$$p(0) = p_1(0) = 0 = i(0)$$
$$p(1) = p_n(1) = 1 = i(1)$$
Therefore, we can say that there is a path homotopy $F: I \times I \longrightarrow I$ between $p$ and $i$ which satisfies
$$F(s,0) = p(s) \quad \quad F(s,1) = i(s)$$
$$F(0,t) = 0 \quad \quad F(1,t) = 1$$
Finally, we can compose each with $f$ and get
$$f \circ i \simeq (f \circ p_1) * (f \circ p_2) * \dots
*(f \circ p_n) $$
which gives us the intended result
$$[f] = [f_1] * [f_2] * \dots * [f_n] $$
Problem 1
Show that if $$h,h': X \longrightarrow Y$$ are homotopic and
$$h,k': Y \longrightarrow Z$$ are also homotopic, then $k \circ h \text{ and } k' \circ h' $
are homotopic.
Proof: Since we know $h \simeq h'$, there exists a homotopy
$$H: X \times I \longrightarrow Y$$
$$H(x,0) = h(x), \quad H(x,1) = h'(x)$$
Similarly, since $k \simeq k'$ we have the homotopy
$$K: Y \times I \longrightarrow Z$$
$$K(y,0) = h(y), \quad K(y,1) = h'(y)$$
Now, to define a homotopy between $k \circ h \text{ and } k' \circ h' $, we need to build a composition map. So define a new function
$$\Phi: X \times I \longrightarrow Y \times I $$
given by
$$\Phi(x,t) = \big(H(x,t),t \big)$$
and define
$$F: X \times I \longrightarrow Z$$
to be a continuous map.
Now we can build a complete diagram to connect the three spaces: 
This shows us visually that $F = K \circ \Phi$. So we have
$$F(x,0) = (K \circ \Phi)(x,0) = K\big(H(x,0),0 \big)$$
$$ = K\big( h(x), 0 \big) = (k \circ h)(x)$$
and
$$F(x,1) = (K \circ \Phi)(x,1) = K\big(H(x,1),1 \big)$$
$$ = K\big( h'(x), 1\big) = (k' \circ h')(x)$$
Therefore $F$ is the required homotopy which gives us
$$k \circ h \simeq k' \circ h'$$
Problem 2
Given the spaces $X$ and $Y$, let $[X,Y]$ denote the set of homotopy classes of maps of $X$ into $Y$.
Proof: \\
(a) To show that the set of homotopy classes has only one element, we need to show that if $f$ and $g$ are two maps between $X$ and $I$, it must be true that $f \simeq g$. So define
$$f,g: X \longrightarrow I$$
If $I$ is convex, we can define the straight line homotopy between them
$$F(x,t) = (1-t)g(x) + tf(x)$$
We've been treating $I$ as convex throughout this section, but it doesn't hurt to prove it here. To show that a set is convex, we need to show that for any two points $a, b \in I$, the straight line connecting them is also in $I$.
To do this, let $a \leq b$ be points in $I$. The line between them is given the same way that the homotopy is, for $t \in [0,1]$ we have that
$$c(t) = (1-t)a+ tb$$
connects $a$ and $b$, and furthermore, given how we restrict $t$ every point on that line is in $I$. Hence, $I$ is convex and for any $f, g$ that map $X$ to $I$
$$F(x,t) = (1-t)g(x) + tf(x)$$
must be a homotopy between them.
Therefore $[X,Y]$ contains only one element.
(b) For this problem, we define the map $$e_a: I \longrightarrow Y $$
by the constant function
$$e_a (s) = a \quad \quad \forall s \in [0,1]$$
where $a \in Y$ is a fixed point. To prove that $[I,Y]$ has only one element, we will prove that any other continuous map $$g: I \longrightarrow Y$$ must be homotopic to $e_a$.
We can show that $g \simeq e_{g(0)}$ by defining the homotopy
$$G: I \times I \longrightarrow Y$$ by
$$G(x,t) = g\big((1-t) x \big)$$
which gives us the required
$$G(x,0) = g(0), \quad \quad G(x,1) = g(x)$$
However, this is not enough to say $[I,Y] = \left\{ \left[ e_{g(0)}\right] \right\}$ because the value of $e_{g(0)}$ depends on $g$ and we want this to work for any continuous function from $I$ to $Y$.
However, if we can show that $e_a \simeq e_b$ for all $a,b \in Y$, we can use $g \simeq e_{g(0)}$ to show that $g \simeq e_a$.
Since $Y$ is path connected, for any $a,b \in Y$, there exists a path
$$\varphi: I \longrightarrow Y$$
such that
$$\varphi(0) = a, \quad \quad \varphi(1) = b$$
We can use this path to give us the required homotopy between $e_a$ and $e_b$. Consider the map
$$E: I \times I \longrightarrow Y$$
defined by
$$E(s,t) =\varphi(t)$$
We have
$$E(s,0) = \varphi(0) = a = e_a(s)$$
$$E(s,1) = \varphi(1) = b = e_b(s)$$
Therefore $e_a \simeq e_b$.
So, for any path $g: I \longrightarrow Y$ we know that
and now we know $g \simeq e_{g(0)}$, and we now know that
$$g \simeq e_{g(0)} \text{ and } e_{g(0)} \simeq e_a$$
which together give us
$$g \simeq e_a \Rightarrow [g] = [e_a]$$
Hence, the set $[I,Y]$ has a single element.
Problem 3
A space $X$ is said to be contractibleif the identity map is null homotopic. Show that:
Proof:
(a) Now this certainly feels like it should be true, looking at a picture for the identity map in $\mathbb{R}$. 
Let's start with the proof for $I$. Let $i_I: I \longrightarrow I$ denote the identity function in $I$. We only need to prove that the identity map is equal to some constant map, so we'll use the easiest one:
$$0: I \longrightarrow I $$ where
$$0(s) =0\quad \quad \forall s \in I$$ Consider
$$F: I \times I \longrightarrow I$$
where $$F(s,t) = st$$ This gives us
$$F(s,0) = 0 \cdot s = 0(s)$$
$$F(x,1) = 1 \cdot s = i_I(s)$$
Therefore, $i_I \simeq 0$.
Now, for $\mathbb{R}$, we redefine our 0 function as $$0: \mathbb{R} \longrightarrow \mathbb{R}$$ $$0(s) =0 \quad \quad \forall s \in \mathbb{R}$$ and adjust our $F$ to be
$$F: \mathbb{R} \times I \longrightarrow \mathbb{R}$$
where $$F(s,t) = st$$
And just as before we have
$$F(s,0) = 0 \cdot s = 0(s)$$
$$F(x,1) = 1 \cdot s = i_{\mathbb{R}}(s)$$
which gives us our result of $i_{\mathbb{R}} \simeq 0$.
(b) Let $X$ be a contractible space. We need to show that for all $a, b \in X$, there exists a path $\varphi:[0,1] \longrightarrow X$ which connects them.
Let $i_X : X \longrightarrow X$ be the identity function in $X$ and define $$e_a: I \longrightarrow X $$
by
$$e_a (s) = a \quad \quad \forall s \in [0,1]$$ where $a \in X$ is fixed.
Since $X$ is contractible, we have $i_X \simeq e_a$. Therefore there exists a continuous map $$F: X \times I \longrightarrow X$$ such that $$F(x,0) = e_a(x) = a, \quad \quad F(x,1) = i_X(x) = x$$
This is almost enough. We just need adjust our function so it can map $a$ to any point $b \in X$.
To do this, define $$\varphi:I \longrightarrow X$$ by $$\varphi(t) = F(b,t)$$
This gives us exactly what we want
$$\varphi(0) = F(b,0) = e_a(b) = a$$
$$\varphi(1) = F(b,1) = i_X (b) = b$$
Since we can choose any $a,b \in X$ here that we'd like, we have our result that $X$ is path connected.
%3c
(c) Let $$f: X \longrightarrow Y$$ be a continuous map. Just like in Problem 2b, we will get our result by showing that $f$ is homotopic to a constant map.
Since $Y$ is contractible, there exists a continuous map $F: Y \times I \longrightarrow Y$ such that
$$F(y,0) = e_a(x) = a$$
$$F(y,1) = i_Y(y) = y$$
To use this knowledge to say anything about maps from $X \longrightarrow Y$, we need do to some function composition. So define the function $$\Phi: X \times I \longrightarrow Y \times I$$ by
$$\Phi(x,t) = \big(f(x),t \big)$$ We now have the composition diagram
which shows us that
$$G(x,t) = (F \circ \Phi)(x,t) = F\big( f(x),t \big) $$
Therefore, if $F$ and $\Phi$ are continuous, $G$ must be continuous and
$$G(x,0) = F\big( f(x),0 \big) = e_a(x) = a$$
$$G(x,1) = F\big( f(x),1 \big) = f(x)$$
Hence, $f \simeq a$ and the set $[X,Y]$ has only one element.
(d) Let $X$ be a contractible space. Thus we know, for any $p \in X$ that there exists a continuous $H: X \times I \longrightarrow X$ such that
$$H(x,0) = I_X(x) = x$$
$$H(x,1) = p $$
Now, pick a point $q \in Y$ and define the constant map
$$f_q: X \longrightarrow Y$$
by
$$f_q(x) = q \quad \quad \forall x \in X$$
Once again, we want to show that for any continuous function,
$$g: X \longrightarrow Y$$
$g$ must be homotopic to $f_q$.
Given $q$ and $g(p)$ are both points in $Y$, by path-connectedness there must exist a map
$$\varphi: I \longrightarrow Y$$
such that
$$\varphi(0) = g(p), \quad \quad \varphi(1) = q$$
Now define the function
$$ \Phi (x,t) = \begin{cases}
g\big(H(x,2t) \big) & t \in \left[ 0,\frac{1}{2}\right]\\
\varphi(2t-1)& s \in \left[\frac{1}{2}, 1\right]
\end{cases}$$
$\Phi$ is continuous an well defined at $t=\frac{1}{2}$ since $g\big(H(x,1)\big) = g(p) = \varphi(0)$, and since
$$\Phi (x, 0 ) = g\big(H,x,0) \big) = g(x)$$
and
$$\Phi (x, 1 ) = \varphi(1) = f_q(x)$$
we have that $g \simeq f_q$. Therefore, $[X,Y]$ has one element.
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