Define $\omega_n: I \to S^1$ by $\omega_n(t)=\exp (2\pi int)$. Show that the concatenation $\omega_p * \omega_q$ is homotopic, rel endpoints, to $\omega_{p+q}$. (Consider first the special case of $p=2, q=1$. Then do the general case.)
The problem is that I don't really know what I have to do here. Aren't the concatenation $\omega_p * \omega_q$ the same function as $\omega_{p+q}$? I don't really understand why it helps to consider the special case $p=2, q=1$, the product of the paths seems simply to be $\omega_3(t)$. What am I getting wrong here? I would greatly appreciate any help.
They trace out the same path, but with a different parametrization. For instance, $\omega_2*\omega_1$ goes around the circle once as $t$ goes from $0$ to $1/4$, once again as $t$ goes from $1/4$ to $1/2$, and finally a third time as $t$ goes from $1/2$ to $1$. On the other hand, $\omega_3$ goes around once as $t$ goes from $0$ to $1/3$, again as $t$ goes from $1/3$ to $2/3$, and a third time as $t$ goes from $2/3$ to $1$.
You can get a homotopy between them by just linearly interpolating between the parametrizations. For instance, you could take the homotopy $F:I\times I\to S^1$ such that for any fixed $s$, $F(t,s)$ goes around the circle once (at constant speed) between $t=0$ and $t=s/4+(1-s)/3$, between $t=s/4+(1-s)/3$ and $s/2+2(1-s)/3$, and between $t=s/2+2(1-s)/3$ and $t=1$. I'll let you think about how to write this out more formally, show that it is continuous, and generalize to arbitrary $p$ and $q$.