Use of intermediate value theorem to show that $p = (\cos 2\pi x, \sin 2 \pi x)$ is a covering map for $S^1$.

Question

I am not completely following the argument given in Munkres' topology to show that $p = (\cos 2\pi x, \sin 2 \pi x)$ is a covering map for $S^1$ is a covering map for $S^1$. Here is the outline of the proof:

Consider the set $U \subseteq S^1$ consisting of points of $S^1$ having positive first coordinate. The set $p^{-1}(U)$ is the union of $$ V_n = (n - \frac{1}{4}, n + \frac{1}{4}) $$ The map $p$ is injective when restricted to $\overline V_n$, because $\sin 2\pi x$ is strictly monotonic on such an interval. Furthermore, p carries $\overline V_n$ surjectively onto $\overline U$, and $V_n$ to $U$, by the intermediate value theorem.

I am not sure how the intermediate value theorem (IVT) applies. Do I need to identify each point on the circle with its angle and then use the IVT?

Answer 1

As Munkres writes, the fact that $p$ is a covering map comes from elementary properties of the sine and cosine functions. Which properties do we need?

1. $\sin$ is strictly monotonically increasing on each interval $[2 \pi n - \frac \pi 2 , 2 \pi n + \frac \pi 2]$.
   This implies that $\sin 2 \pi t$ is strictly monotonically increasing, in particular injective, on each interval $\overline V_n = [n - \frac 1 4, n + \frac 1 4]$. Hence $p$ is injective on $\overline V_n$.

2. $\cos$ is nonnegative on each interval $[2 \pi n - \frac \pi 2 , 2 \pi n + \frac \pi 2]$.
   This implies that $\cos 2 \pi t$ is nonnegative on each interval $\overline V_n$.

3. I think it is also well-known that $\sin$ maps each interval $[2 \pi n - \frac \pi 2 , 2 \pi n + \frac \pi 2]$ bijectively onto $[-1,1]$. Anyway, this can be proved by the IVT. Let $y \in [-1,1]$. Since $\sin (2 \pi n - \frac \pi 2) = \sin (-\frac \pi 2) = -1$ and $\sin (2 \pi n + \frac \pi 2) = \sin (\frac \pi 2) = 1$, we find a (unique) number $t \in [2 \pi n - \frac \pi 2 , 2 \pi n + \frac \pi 2]$ such that $\sin t = y$.
   I guess this is what Munkres means.

Let $(x,y) \in \overline U$. This means $x^2 + y^2 = 1$ and $x \ge 0$. In particular $y \in [-1, 1]$. By 3. we find $t \in \overline V_n$ such that $\sin 2 \pi t = y$. Since $x \ge 0$, we get $$x = \sqrt{1 - y^2} = \sqrt{1 - \sin^2 2\pi t} = \sqrt{\cos^2 2\pi t}. $$ By 2. we see that $\sqrt{\cos^2 2\pi t} = \cos 2\pi t$, thus $$(x,y) = p(t) .$$