Parameterizations of the circle $S^1$ and open arcs in $S^1$ The circle $S^1 = \{ z \in \mathbb C \mid \lvert z \rvert = 1\}$ inherits its standard topology from the plane $\mathbb  C$ with the Euclidean topology.
The purpose of this question is to give a standard reference in this forum which covers some basic facts about parameterizations of $S^1$ and open arcs in $S^1$.
A base for the topology on $S^1$ is given by the sets $S(z,\epsilon) = B(z,\epsilon) \cap S^1$ with $z \in S^1$ and $\epsilon > 0$. Here $B(z,\epsilon) = \{ w \in \mathbb C  \mid \lvert w - z \rvert < \epsilon \}$ is the open disk with center $z$ and radius $\epsilon$. Note that $S(z,\epsilon) = S^1$ for $\epsilon > 2$ and $S(z,2) = S^1 \setminus \{-z\}$.
The sets $S(z,\epsilon)$ with $\epsilon \le 2$ are often called open arcs on $S^1$. I think formally an open arc on $S^1$ should be defined as an open subset of $S^1$ which is homeomorphic to an open interval $(a,b) \subset \mathbb R$. So do both definitions agree?
To address this question let us consider any parameterization of $S^1$ which we understand as a continuous surjection
$$\varphi : [0,1] \to S^1$$
such that

*

*$\varphi(0) = \varphi(1)$.


*$ \varphi$ is "essentially injective" which means that $\varphi(s) = \varphi(t)$ if and only $s = t$ or $\{s,t\} =\{0,1\}$.
The standard example is $\omega(t) = e^{2 \pi i t} = \cos(2\pi t) + i \sin(2\pi t)$.
Given a parameterization $\varphi$ of $S^1$, we can extend it to a periodic map
$$\phi : \mathbb R \to S^1$$
with period $1$. It wraps $\mathbb R$ infinitely often around $S^1$ and is called a periodic parameterization of $S^1$.
The above standard parameterization gives us the well-known
$$\Omega : \mathbb R \to S^1, \Omega(t) = e^{2 \pi i t} = \cos(2\pi t) + i \sin(2\pi t)$$
which is the standard universal covering of the circle.
Here are some tasks:

*

*Show that each periodic parameterization $\phi$ is an open map. In particular, $\phi$ maps $(0,1)$ homeomorphically onto $S^1 \setminus \{z\}$ where $z = \phi(0) =\phi(1)$.


*Show that for each periodic parameterization $\phi$ there exists a unique homeomorphism $h : \mathbb R \to \mathbb R$ such that $\phi = \Omega \circ h$. Note that either $h(t + 1) = h(t)+1$ for all $t$ or $h(t + 1) = h(t)-1$ for all $t$.


*Show that each open arc has the form $\Omega((a,b))$ for some open interval $(a,b)$ with $b -a \le 1$.


*Show that each $\Omega((a,b))$ with $b - a \le 1$ is an open arc.
 A: Periodicity implies that $\phi([a,a+1]) = S^1$ for all $a \in \mathbb R$. In fact, let $k$ be the unique integer such that $a-1 \le k < a$. Then $$\phi([a,a+1]) = \phi([a,k+1] \cup [k+1,a+1]) =  \phi([a,k+1]) \cup \phi([k+1,a+1]) \\ = \phi([a-k,1]) \cup \phi([0,a-k]) =  \phi([a-k,1] \cup [0,a-k])  = \phi([0,1] = S^1 .$$
Let us next show that if $\phi(s)= \phi(t)$ iff $s -t \in \mathbb Z$. Clearly if $s = t + k$ with $k \in \mathbb Z$, then periodicity implies $\phi(s)= \phi(t)$. For the converse let $m$ be the unique integer such that $m \le s < m+1$. With $s' = s - m$ let $n$ be the unique integers such that $s' + n \le t < s' + n +1$. Setting $t' = t - n$ we get $\phi(s') = \phi(s) = \phi(t) = \phi(t')$ with  $0 \le s' < 1$ and $s' \le t' < s' +1$. We shall show $s' = t'$ which proves $s - t \in \mathbb Z$. If $s' = 0$, this follows from the definition of a parameterization. The same is true if $0 < s'$ and $t' \le 1$. The case $0 < s'$ and $1 < t'$ cannot occur. Otherwise $\phi(s') = \phi(t'-1)$ with $0 < t'- 1 < s' < 1$ which implies $s' = t' - 1$, contradicting  $t' < s' + 1$.
Task 1.
It suffices to show that $\phi((a,b))$ is open if $(a,b)$ has length $\le 1$ (i.e. $b - a \le 1$). This is due to the fact that these intervals form a base for the topology of $\mathbb R$.
Let $C = [a,a+1] \setminus (a,b)$. This is a compact set, thus $\phi(C)$ is compact and $S^1 \setminus \phi(C)$ is open in $S^1$.  We have $\phi((a,b)) \cup \phi(C) = \phi((a,b) \cup C) = \phi([a,a+1]) = S^1$. Moreover $\phi((a,b)) \cap \phi(C) = \emptyset$: Otherwise there exist $s \in (a,b)$ and $t \in C$ such that $\phi(s) = \phi(t)$ which implies $s=t$ or $\{s,t\} =\{a,a+1\}$, both of which are impossible. Therefore $\phi((a,b)) = S^1 \setminus \phi(C)$ which finishes the proof.
Here is an alternative proof. The map $p = \phi \mid_{([a,a+1]} : [a,a+1] \to S^1$ is closed, thus a quotient map. Let $z = \phi(a) = \phi(a+1)$. Then $U = S^1 \setminus \{z\}$ is open in $S^1$ so that  $p' \mid_{p^{-1}(U)} : p^{-1}(U) \stackrel{p}{\to} U$ is a quotient map. But $p^{-1}(U) = (a,a+1)$ and $p' :(a,a+1) \to S^1 \setminus \{z\}$ is a bijection. Since bijective quotient maps are homeomorphisms, we see that $p'$ is an open map, hence $p'((a,b)) =\phi((a,b))$ is open in $S^1 \setminus \{z\}$ and therefore open in $S^1$.
Task 2.
We have $\phi(0) = \phi(1) = z$ and $z = \Omega(\tau)$ for some $\tau$. Define
$$h'' : (\tau,\tau+1) \stackrel{\Omega}{\to} S^1 \setminus \{z\} \stackrel{\phi^{-1}}{\to}(0,1) .$$
This is a homeomorphism. It can either be strictly increasing or strictly decreasing.
If it strictly increasing, it extends to a homeomorphism $h' : [\tau,\tau+1] \to [0,1]$ such that $h'(\tau) = 0, h'(\tau+1) = 1$. Extend $h'$ to a map $h :\mathbb R \to \mathbb R$ such that $h(t + 1) = h(t)+1$ for all $t$. This $h$ is clearly a homeomorphism and by construction we have $\phi = \Omega \circ h$.
The case that $h''$ is strictly decreasing can be treated similarly.
Task 3.
Let $S(z,\epsilon)$ be an open arc. We first consider the special case $z=1$.
We have $\Omega((-1/2, 1/2)) =  S^1 \setminus \{-1\}$. For $t \in ( -1/2, 1/2)$ we have $\Omega(t) \in S(1,\epsilon)$ iff  $f(t) = \lvert \Omega(t)- 1 \rvert < \epsilon$. But
$$f(t) = \sqrt{(\cos(2\pi t) -1)^2 + \sin^2 (2\pi t)} = \sqrt 2 \sqrt{1 -\cos(2 \pi t)} ,$$
thus $f(t) < \epsilon$ iff $\cos(2 \pi t) > 1 - \epsilon^2/2$. Let $c \in (0,1/2]$ be the unique point such that $\cos(2 \pi c) =1 - \epsilon^2/2$ (recall $0 < \epsilon \le 2$). Thus
$\Omega(t) \in S(1,\epsilon)$ for $t \in (-c,c)$ and $\Omega(t) \notin S(1,\epsilon)$ for $t \in [-1/2,-c] \cup [c,1/2]$. We conclude that $\Omega((-c,c)) = S(1,\epsilon)$.
For an arbitrary $z \in S^1$ the map $\mu : \mathbb C \to \mathbb C, \mu(w) = w/z$, is a homeomorphism which preserves the absolute value. Geometrically it is a rotation. We have $\mu(S(z,\epsilon)) = S(1,\epsilon)$. Thus $\mu(S(z,\epsilon)) = \Omega((-c,c))$ for some $c$. Since $z= \Omega(\tau)$, we get $$\Omega((\tau-c,\tau+c)) = \{\ \Omega(\tau + t) \mid t \in (-c,c) \} = \{ \Omega(\tau)\Omega(t) \mid t \in (-c,c) \} = \{ z \Omega(t) \mid t \in (-c,c) \} = \{ \mu^{-1}(\Omega(t)) \mid t \in (-c,c) \} = \mu^{-1}(\Omega((-c,c))) = S(z,\epsilon) .$$
Task 4.
We have seen above that for $c \in (0,1/2]$ we have $\Omega((-c,c)) = S(1,\epsilon)$ with $\epsilon = \sqrt 2 \sqrt{1 - \cos(2 \pi c)}$ . For an arbitrary $(a,b)$ with $b - a \le 1$ let $c = (b-a)/2$. With $z = \Omega(a + c)$ we get
$$\Omega((a,b)) = \{ \Omega(a + c + t) \mid t \in (-c,c) \} = \{ \Omega(a + c) \Omega(t) \mid t \in (-c,c) \} = \{ z \Omega(t) \mid t \in (-c,c) \} = S(z,\epsilon) .$$
