Math4201 Topology II (Lecture 11)

Algebraic topology

Fundamental group

The $*$ operation has the following properties:

Properties for the path product operation

Let $[f],[g]\in \Pi_1(X)$ , for $[f]\in \Pi_1(X)$ , let $s:\Pi_1(X)\to X, [f]\mapsto f(0)$ and $t:\Pi_1(X)\to X, [f]\mapsto f(1)$ .

Note that $t([f])=s([g])$ , $[f]*[g]=[f*g]\in \Pi_1(X)$ .

This also satisfies the associativity. $([f]*[g])*[h]=[f]*([g]*[h])$ .

We have left and right identity. $[f]*[e_{t(f)}]=[f], [e_{s(f)}]*[f]=[f]$ .

We have inverse. $[f]*[\bar{x}]=[e_{s(f)}], [\bar{x}]*[f]=[e_{t(f)}]$

Definition for Groupoid

Let $f,g$ be paths where $g,f:[0,1]\to X$ , and consider the function of all pathes in $G$ , denoted as $\mathcal{G}$ ,

Set $t:\mathcal{G}\to X$ be the source map, for this case $t(f)=f(0)$ , and $s:\mathcal{G}\to X$ be the target map, for this case $s(f)=f(1)$ .

We define

\mathcal{G}^{(2)}=\{(f,g)\in \mathcal{G}\times \mathcal{G}|t(f)=s(g)\}

And we define the operation $*$ on $\mathcal{G}^{(2)}$ as the path product.

This satisfies the following properties:

Associativity: $(f*g)*h=f*(g*h)$

Consider the function $\eta:X\to \mathcal{G}$ , for this case $\eta(x)=e_{x}$ .

We have left and right identity: $\eta(t(f))*f=f, f*\eta(s(f))=f$
Inverse: $\forall g\in \mathcal{G}, \exists g^{-1}\in \mathcal{G}, g*g^{-1}=\eta(s(g))$ , $g^{-1}*g=\eta(t(g))$

Definition for loop

Let $x_0\in X$ . A path starting and ending at $x_0$ is called a loop based at $x_0$ .

Definition for the fundamental group

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

(\Pi_1(X,x),*)

where $*$ is the product operation, and $\Pi_1(X,x)$ is the set o homotopy classes of loops in $X$ based at $x$ .

Example of fundamental group

Consider $X=[0,1]$ , with subspace topology from standard topology in $\mathbb{R}$ .

$\Pi_1(X,0)=\{e\}$ , (constant function at $0$ ) since we can build homotopy for all loops based at $0$ as follows $H(s,t)=(1-t)f(s)+t$ .

And $\Pi_1(X,1)=\{e\}$ , (constant function at $1$ .)

Let $X=\{1,2\}$ with discrete topology.

$\Pi_1(X,1)=\{e\}$ , (constant function at $1$ .)

$\Pi_1(X,2)=\{e\}$ , (constant function at $2$ .)

Let $X=S^1$ be the circle.

$\Pi_1(X,1)=\mathbb{Z}$ (related to winding numbers, prove next week).

A natural question is, will the fundamental group depends on the base point $x$ ?

Definition for $\hat{\alpha}$

Let $\alpha$ be a path in $X$ from $x_0$ to $x_1$ . $\alpha:[0,1]\to X$ such that $\alpha(0)=x_0$ and $\alpha(1)=x_1$ . Define $\hat{\alpha}:\Pi_1(X,x_0)\to \Pi_1(X,x_1)$ as follows:

\hat{\alpha}(\beta)=[\bar{\alpha}]*[f]*[\alpha]

$\hat{\alpha}$ is a group homomorphism

$\hat{\alpha}$ is a group homomorphism between $(\Pi_1(X,x_0),*)$ and $(\Pi_1(X,x_1),*)$

Proof

Let $f,g\in \Pi_1(X,x_0)$ , then $\hat{\alpha}(f*g)=\hat{\alpha}(f)\hat{\alpha}(g)$

\begin{aligned} \hat{\alpha}(f*g)&=[\bar{\alpha}]*[f]*[g]*[\alpha]\\ &=[\bar{\alpha}]*[f]*[e_{x_0}]*[g]*[\alpha]\\ &=[\bar{\alpha}]*[f]*[\alpha]*[\bar{\alpha}]*[g]*[\alpha]\\ &=([\bar{\alpha}]*[f]*[\alpha])*([\bar{\alpha}]*[g]*[\alpha])\\ &=(\hat{\alpha}(f))*(\hat{\alpha}(g)) \end{aligned}

Next, we will show that $\hat{\alpha}\circ \hat{\bar{\alpha}}([f])=[f]$ , and $\hat{\bar{\alpha}}\circ \hat{\alpha}([f])=[f]$ .

\begin{aligned} \hat{\alpha}\circ \hat{\bar{\alpha}}([f])&=\hat{\alpha}([\bar{\alpha}]*[f]*[\alpha])\\ &=[\alpha]*[\bar{\alpha}]*[f]*[\alpha]*[\bar{\alpha}]\\ &=[e_{x_0}]*[f]*[e_{x_1}]\\ &=[f] \end{aligned}

The other case is the same

Corollary of fundamental group

If $X$ is path-connected and $x_0,x_1\in X$ , then $\Pi_1(X,x_0)$ is isomorphic to $\Pi_1(X,x_1)$ .