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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 basepoint 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).