Math4202 Topology II (Lecture 9)

Algebraic Topology

Path homotopy

Consider the space of paths up to homotopy equivalence.

\operatorname{Path}/\simeq_p(X) =\Pi_1(X)

We want to impose some group structure on $\operatorname{Path}/\simeq_p(X)$ .

Consider the $*$ operation on $\operatorname{Path}/\simeq_p(X)$ .

Let $f,g:[0,1]\to X$ be two paths, where $f(0)=a$ , $f(1)=g(0)=b$ and $g(1)=c$ .

f*g:[0,1]\to X,\quad f*g(t)=\begin{cases} f(2t) & 0\leq t\leq \frac{1}{2}\\ g(2t-1) & \frac{1}{2}\leq t\leq 1 \end{cases}

This connects our two paths.

Definition for product of paths

Given $f$ a path in $X$ from $x_0$ to $x_1$ and $g$ a path in $X$ from $x_1$ to $x_2$ .

Define the product $f*g$ of $f$ and $g$ to be the map $h:[0,1]\to X$ .

Definition for equivalent classes of paths

$\Pi_1(X,x)$ is the equivalent classes of paths starting and ending at $x$ .

On $\Pi_1(X,x)$ ,, we define $\forall [f],[g],[f]*[g]=[f*g]$ .

[f]\coloneqq \{f_i:[0,1]\to X|f_0(0)=f(0),f_i(1)=f(1)\}

Lemma

If we have some path $k:X\to Y$ is a continuous map, and if $F$ is path homotopy between $f$ and $f'$ in $X$ , then $k\circ F$ is path homotopy between $k\circ f$ and $k\circ f'$ in $Y$ .

If $k:X\to Y$ is a continuous map, and $f,g$ are two paths in $X$ with $f(1)=g(0)$ , then

(k\circ f)*(k\circ g)=k\circ(f*g)

Proof

We check the definition of path homotopy.

$k\circ F:I\times I\to Y$ is continuous.

$k\circ F(s,0)=k(F(s,0))=k(f(s))=k\circ f(s)$ .

$k\circ F(s,1)=k(F(s,1))=k(f'(s))=k\circ f'(s)$ .

$k\circ F(0,t)=k(F(0,t))=k(f(0))=k(x_0$ .

$k\circ F(1,t)=k(F(1,t))=k(f'(1))=k(x_1)$ .

Therefore $k\circ F$ is path homotopy between $k\circ f$ and $k\circ f'$ in $Y$ .

For the second part of the lemma, we proceed from the definition.

(k\circ f)*(k\circ g)(t)=\begin{cases} k\circ f(2t) & 0\leq t\leq \frac{1}{2}\\ k\circ g(2t-1) & \frac{1}{2}\leq t\leq 1 \end{cases}

and

k\circ(f*g)=k(f*g(t))=k\left(\begin{cases} f(2t) & 0\leq t\leq \frac{1}{2}\\ g(2t-1) & \frac{1}{2}\leq t\leq 1 \end{cases}\right)=\begin{cases} k(f(2t))=k\circ f(2t) & 0\leq t\leq \frac{1}{2}\\ k(g(2t-1))=k\circ g(2t-1) & \frac{1}{2}\leq t\leq 1 \end{cases}

Theorem for properties of product of paths

If $f\simeq_p f_1, g\simeq_p g_1$ , then $f*g\simeq_p f_1*g_1$ . (Product is well-defined)
$([f]*[g])*[h]=[f]*([g]*[h])$ . (Associativity)
Let $e_{x_0}$ be the constant path from $x_0$ to $x_0$ , $e_{x_1}$ be the constant path from $x_1$ to $x_1$ . Suppose $f$ is a path from $x_0$ to $x_1$ . $[e_{x_0}]*[f]=[f],\quad [f]*[e_{x_1}]=[f]$ (Right and left identity)
Given $f$ in $X$ a path from $x_0$ to $x_1$ , we define $\bar{f}$ to be the path from $x_1$ to $x_0$ where $\bar{f}(t)=f(1-t)$ . $f*\bar{f}=e_{x_0},\quad \bar{f}*f=e_{x_1}$ $[f]*[\bar{f}]=[e_{x_0}],\quad [\bar{f}]*[f]=[e_{x_1}]$

Proof

(1) If $f\simeq_p f_1$ , $g\simeq_p g_1$ , then $f*g\simeq_p f_1*g_1$ .

Let $F$ be homotopy between $f$ and $f_1$ , $G$ be homotopy between $g$ and $g_1$ .

We can define

F*G:[0,1]\times [0,1]\to X,\quad F*G(s,t)=\left(F(-,t)*G(-,t)\right)(s)=\begin{cases} F(2s,t) & 0\leq s\leq \frac{1}{2}\\ G(2s-1,t) & \frac{1}{2}\leq s\leq 1 \end{cases}

$F*G$ is a homotopy between $f*g$ and $f_1*g_1$ .

We can check this by enumerating the cases from definition of homotopy.

Continue next time.

Definition for the fundamental group

The fundamental group of $X$ at $x$ is defined to be

(\Pi_1(X,x),*)