# Why are these loops homotopic? [closed]

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 $$\omega_a*\omega_b$$.