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For each $z \in \mathbb{Z}$, let $\omega_z:[0,1] \to S^1$ given by $\omega_{z}(t)=(\cos(2\pi z t), \sin(2 \pi z t))$. How I can prove that $\omega_{a+b}$ and $\omega_a \ast \omega_b$ are homotopic, for each $a,b \in \mathbb{Z}$?

In this context, $f\ast g$ denote the "concatenation". I like hints!!

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You can verify it directly by the definition of concatenation of loops. Consider the homotopy $H$ bwtween $g(t)=(a+b)t$ and $f=\begin{cases}2at&0\le t\le\frac{1}{2}\\a+b(2t-1)&\frac{1}{2}\le t\le1\end{cases}$. Let $\zeta(x)=(cos(2\pi x), sin(2\pi x))$. Then $\zeta\circ H$ is the homotopy between $\zeta\circ g$ and $\zeta\circ f$, and $\zeta\circ g$, $\zeta\circ f$ are exactly $\omega_{a+b}$ and $\omega_a*\omega_b$.

However, if you have learnt the fundamental group of $S^1$, you can see it in this way:

Since $\pi_1(S^1)\cong \mathbb{Z}$, let $p$ be the isomorphism between $\pi_1(S^1)$ and $ \mathbb{Z}$. Because $p([\omega_{a+b}])=a+b, p([\omega_a*\omega_b])=p([\omega_a]*[\omega_b])=p([\omega_a])+p([\omega_b])=a+b$, $[\omega_{a+b}]$ and $[\omega_a*\omega_b]$ must be the same in $\pi_1(S^1)$. Hence $\omega_{a+b}$ is homotopic to $\omega_a*\omega_b$.

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  • $\begingroup$ Can you give me more details about how I can verify it directly by the definition? $\endgroup$
    – pmk
    Commented Jul 21, 2021 at 15:01
  • $\begingroup$ I have updated the answer. $\endgroup$
    – wyhorgyh
    Commented Jul 21, 2021 at 16:34

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