Math4202 Topology II (Lecture 17)

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 $\hat{F}(s,1)=\hat{g}(s)$.

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

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

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.

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