The fundamental group of the circle

Question

Theorem 1.7. $\pi_1 (\mathbb{S}^1 )$ is an infinite cyclic group generated by the homotopy class of the loop $\omega(s) = (\cos 2\pi s, \sin 2\pi s)$ based at $(1, 0)$. how to prove the following line Note that $[\omega]^n = [\omega_n]$ where $\omega_n(s) = (\cos 2\pi ns, \sin 2\pi ns)$ for $n \in \mathbb{Z}$

Answer 1

The product of two loops is given by

$$f*g = \begin{cases}f(2t) & t\in [0,\frac{1}{2}]\\ g(2t-1) & t\in [\frac{1}{2},1] \end{cases}$$

Analogously, $\omega^n$ is given by

$$\omega^n = \underbrace{\omega* \cdots *\omega}_{n \text{ times}} = \begin{cases}f(nt) & t\in [0,\frac{1}{n}]\\f(nt-1) & t\in [\frac{1}{n},\frac{2}{n}]\\ \vdots & \vdots \\ f(nt - n+1) & t\in [\frac{n-1}{n},1]\end{cases}$$

If $\omega(s) = (\cos 2\pi s, \sin 2\pi s)$, then $\omega^n(s)$ is verified to be $\omega_n(s)$ as defined above.

The last bit is the observation that loop product "commutes" with moving to the equivalence class, i.e., $[f* g] = [f]\cdot [g]$. Putting it all together, this means $$[\omega_n] = [\omega^n] = [\omega]^n$$