understanding the degree of a function from the circle to the circle

Question

What is the most intuitive way of seeing that $$S^1\to S^1,\ z\mapsto z^k$$

has degree $k\ ({\rm mod}\ 2)$ from a topological point of view (i.e. not appealing to algebraic/analytic facts)?

Answer 1

You can make a video of it, that would look something like this.

First, think of the function as an equation $w = z^k$ with dependent variable $z$ and independent variable $w$.

On the left hand side of the video, show the unit circle in the $z$-plane. On the right hand side, show the unit circle in the $w$-plane.

Initially, the base point $z=1 \, (= 1+0i)$ is shown as a glowing dot on in the $z$-circle, and the corresponding point $w = 1^k = 1$ is shown as a glowing dot in the $w$-circle.

Now the video shows $z$ moving at a slow angular velocity in the counterclockwise direction around the left hand circle. Its image $w$ moves counterclockwise around the right circle, but you will see that $w$ is moving exactly $k$ times faster than $z$, i.e. the angular velocity of $w$ is $k$ times the angular velocity of $z$.

So, for instance, as the glowing dot representing $z$ moves through $1/k$ of its circle from the point $1+0i$ to the point $\cos(2 \pi / k) + i \, \sin(2 \pi /k) = e^{2 \pi i / k}$, the glowing dot representing $w$ will move around the entire circumference of its circle from $1$ to $$(e^{2 \pi i / k})^k = e^{2 \pi i} = \cos(2 \pi) + i \sin(2 \pi) = 1 + 0i = 1 $$ And as $z$ makes one full revolution around its circle, $w$ will make $k$ full revolutions around its circle.