NoteNextra-origin/content/Math4202/Math4202_L17.md

Math4202 Topology II (Lecture 17)

Algebraic Topology

Fundamental group of the circle

Recall from previous lecture, we have unique lift for covering map.

Lemma for unique lifting homotopy for covering map

Let p: E\to B be a covering map, and e_0\in E and p(e_0)=b_0. Let F:I\times I\to B be continuous with F(0,0)=b_0. There is a unique lifting of F to a continuous map \tilde{F}:T\times I\to E, such that \tilde{F}(0,0)=e_0.

Further more, if F is a path homotopy, then \tilde{F} is a path homotopy.

Theorem for induced homotopy for fundamental groups

Suppose f,g are two paths in B, and suppose f and g are path homotopy (f(0)=g(0)=b_0, and f(1)=g(1)=b_1, b_0,b_1\in B), then \hat{f}:\pi_1(B,b_0)\to \pi_1(B,b_1) and \hat{g}:\pi_1(B,b_0)\to \pi_1(B,b_1) are path homotopic.

Proof

Since f,g are path homotopic, then there exists F:I\times I\to B such that

\hat{F} is a homotopy between \hat{f} and \hat{g}, where \hat{F}(s,0)=\hat{f}(s) and \hat{F}(s,1)=\hat{g}(s).

Definition of lifting correspondence

Let p: E\to B be a covering map, and p^{-1}(b_0)\subseteq E be the fiber of b_0.

Let [f]\in \pi_1(B,b_0), then define \phi:\pi_1(E,b_0)\to p^{-1}(b_0) as follows:

\phi([f])=\tilde{f}(1), and \tilde{f}(0)=e_0. Note that p(\tilde{f}(1))=p(f(1))=b_0.

Example

Let E=\mathbb{R} and B=S^1. Then p^{-1}(b_0)=\mathbb{Z}.

Theorem for surjective lifting correspondence

Let \phi:\pi_1(E,b_0)\to p^{-1}(b_0) be a lifting correspondence. If E is path connected, then \phi is surjective.

Proof

Consider p^{-1}(b_0)=\{e_0,e_0',e_0'',\cdots\}, take \bar{e_0}\in p^{-1}(b_0), E is path connected.

Since E is path connected, then \exists \tilde{f}:I\to E such that \tilde{f}(0)=e_0 and \tilde{f}(1)=\bar{e_0}.

Therefore [f]\in \pi_1(B,b_0).

Theorem for bijective lifting correspondence

Let \phi:\pi_1(E,b_0)\to p^{-1}(b_0) be a lifting correspondence.

If E is simply connected, then \phi is a bijection.

Proof

By previous theorem, it is sufficient to show that \phi is one-to-one (i.e., \phi is injective).

Suppose \phi([f])=\phi([g]), then f,g\in \pi_1(E,b_0). So \tilde{f},\tilde{g}:I\to E are path homotopic.

So \exists \tilde{F}:I\times I\to E such that

• \tilde{F}(s,0)=e_0
• \tilde{F}(s,1)=\bar{e_0}
• \tilde{F}(0,t)=\tilde{f}(t)
• \tilde{F}(1,t)=\tilde{g}(t)

Define F=p\circ \tilde{F}:I\times I\to B, then

• F(s,0)=p(e_0)=b_0
• F(s,1)=p(\bar{e_0})=b_0
• F(0,t)=f(t)
• F(1,t)=g(t)

Therefore [f]=[g], which shows that \phi is a bijection.

Theorem for fundamental group for circle

Let E=\mathbb{R} and B=S^1. Then \phi:\pi_1(E,b_0)\to \pi_1(B,b_0)\simeq \mathbb{Z}. is a isomorphism.

(fundamental group for circle is \mathbb{Z})

Proof

Since \mathbb{R} is simply connected, then \phi is a bijection.

It is suffice to show that \phi satisfies the definition of homomorphism. \phi([f]*[g])=\phi([f])+\phi([g]).

Suppose f,g\in \pi_1(S^1,b_0), then \exists \tilde{f},\tilde{g}:S^1\to \mathbb{R} such that \phi([f])=n, \phi([g])=m, then \tilde{f}:S^1\to \mathbb{R} and \tilde{g}:S^1\to \mathbb{R} such that

• \tilde{f}(0)=0
• \tilde{f}(1)=n
• \tilde{g}(0)=0
• \tilde{g}(1)=m

Take \tilde{\tilde{g}}(x)=\tilde{g}(x)+n, then \phi([f]*[g])=\phi(\tilde{\tilde{g}})=m+n.