NoteNextra-origin/content/Math4202/Math4202_L9.md

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.

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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

1. If f\simeq_p f_1, g\simeq_p g_1, then f*g\simeq_p f_1*g_1. (Product is well-defined)
2. ([f]*[g])*[h]=[f]*([g]*[h]). (Associativity)
3. 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)
4. 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.

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Continue next time.

Definition for the fundamental group

The fundamental group of X at x is defined to be

(\Pi_1(X,x),*)