Let $w, \alpha_n: I=[0,1] \rightarrow \mathbb{S}^1$, $w(s)=e^{2\pi i s}$, $\alpha_n(s)=e^{2 \pi i n s}$. Let $[u] \in \pi_1(\mathbb{S}^1,1)$ be the path homotopy class of the path $u$, an element of the fundamental group of the circle based at $1$. With the usual group multiplication I want to understand why $[w]^n=[\alpha_n]$ for $n \in \mathbb{Z}$.
I'm reading Lee's "Introduction to topological manifolds" and he proves this statement with the induction:
- $\alpha_1= w$
- $\alpha_{-1}=\bar{w}$ (where the bar means the operation of reversing the path)
- $\alpha_{n-1} \cdot w$ is a reparametrization of $\alpha_n$ (therefore it is path homotopic to $\alpha_n$ and so the inductive step follows; the dot stands for path composition)
I get the single ingredients, by I don't see how to build a proof by induction from them, especially on $\mathbb{Z}$.
Thank you for the time!
1) $n >0$. We apply induction. The base case is $[w]^1 = [w] = [\alpha_1]$. The inductive step $(n-1) \mapsto n$ is this: $[w]^n = [w]^{n-1} [w] = [\alpha_{n-1}][w] = [\alpha_{n-1} \cdot w] = [\alpha_n]$.
2) $n < 0$. We apply what we know for $n > 0$. Write $m = -n > 0$. Then $[w]^n = ([w]^m)^{-1} = [\alpha_m]^{-1} = [\overline{\alpha_m}]$. But obviously $\overline{\alpha_m} = \alpha_n$ and we are done.
Note that the case $n=0$ is trivial. $\alpha_0$ is the constant path which represents the neutral element $e \in \pi_1(S^1,1)$ and $[w]^0 = e$ by definition.