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 $\tilde{f}(1)=\bar{e_0}$ .

To produce a class in $\pi_1(B,b_0)$ that $\phi$ sends to $\bar{e_0}$ , project the path down: set $f=p\circ\tilde f:I\to B$ . Then $f(0)=p(\tilde f(0))=p(e_0)=b_0$ and $f(1)=p(\tilde f(1))=p(\bar{e_0})=b_0$ (since $\bar{e_0}\in p^{-1}(b_0)$ ), so $f$ is a loop at $b_0$ .

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

Finally, by definition of the lifting correspondence, $\tilde f$ is the unique lift of $f$ starting at $e_0$ , so $\phi([f])=\tilde f(1)=\bar{e_0}$ . Since $\bar{e_0}\in p^{-1}(b_0)$ was arbitrary, $\phi$ is surjective. $\square$

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 ~~$f,g\in \pi_1(E,b_0)$~~ $[f],[g]\in \pi_1(B,b_0)$ ( $\phi$ is defined on loops in $B$ , not $E$ ). So $\tilde{f},\tilde{g}:I\to E$ are path homotopic.

Why path homotopic: by definition $\tilde f(0)=\tilde g(0)=e_0$ , and the assumption $\phi([f])=\phi([g])$ gives $\tilde f(1)=\tilde g(1)=:\bar{e_0}$ . Since $E$ is simply connected, any two paths in $E$ with the same endpoints 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

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

So $F$ is a path homotopy from $f$ to $g$ in $B$ : the $t=0$ and $t=1$ slices are constant at $b_0$ , and the $s=0,1$ slices are $f$ and $g$ respectively.

Therefore $[f]=[g]$ , which shows that $\phi$ is injective. Combined with surjectivity from the previous theorem, $\phi$ is a bijection. $\square$

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 $\phi([f]*[g])=\phi(\tilde{\tilde{g}})=m+n$ .