Open sets on the unit circle $S^1$

I shall use a little piece of complex analysis:

Each non-constant holomorphic map $\phi : U \to \mathbb C$ defined on an open $U \subset \mathbb C$ is an open map.

Hence the map $F : \mathbb C \to \mathbb C, F(z) = z^2$, is an open map. Its restriction to $S^1$ yields your map $f$.

Let $V \subset S^1$ be open. There exists an open $V' \subset \mathbb C$ such that $V' \cap S^1 = V$. Thus $W = F(V') \cap S^1$ is open in $S^1$. We claim $W = f(V)$ which will prove that $f$ is an open map.

1. $f(V) = F(V) \subset F(V')$. Since trivially $f(V) \subset S^1$, we get $f(V) \subset W$.

2. For each $w \in W = F(V') \cap S^1$ there exists $z \in V'$ such that $F(z) = w$. We have $\lvert z \rvert^2 = \lvert z^2 \rvert = \lvert w \rvert = 1$, thus $\lvert z \rvert = 1$ and therefore $z \in V' \cap S^1 = V$. We have $f(z) = F(z) = w$. Thus $W \subset f(V)$.

We can also use the above result about holomorphic maps to prove that the map $$\phi : \mathbb R \to S^1, \phi(t) = e^{it},$$ (which is a surjection) is an open map. In fact $f(z) = e^z$ is a non-constant holomorphic map, thus an open map. If $W \subset \mathbb R$ is open, then $W' = \mathbb R \times W$ is an open subset of $\mathbb R^2 = \mathbb C$, hence $f(W')$ is open in $\mathbb C$. We have $$f(W') = \{ e^xe^{iy} \mid x \in \mathbb R, y \in W \} .$$ But $e^xe^{iy} \in S^1$ iff $e^x = 1$, thus $$f(W') \cap S^1 = \{ e^{iy} \mid y \in W \} = \phi(W) .$$ This shows in particular that all sets $$S^1(a,b) = \{ e^{iy} \mid y \in (a,b) \} = \phi((a,b))$$ are open in $S^1$. Moreover, they from a basis for the topology on $S^1$. In fact, let $U \subset S^1$ be open and $z_0 \in U$. There is $t_0 \in \mathbb R$ such that $\phi(t_0) = z_0$. Since $\phi^{-1}(U)$ is open in $\mathbb R$ and contains $t_0$, we find $r > 0$ such that $(t_0-r, t_0+r) \subset \phi^{-1}(U)$. This shows $z_0 \in \phi((t_0-r, t_0+r)) \subset U$.

Let us finally note that

1. If $b -a > 2\pi$, then $S^1(a,b) = S^1$.

2. If $b -a \le 2\pi$, then the resriction $\phi_{a,b} : (a,b) \to S^1(a,b)$ is a homeomorphism. To see that, note that $\phi_{a,b}$ is a bijection which is open.

Here is another approach.

Let us first understand the topology on $S^1$.

The continuous map $$\phi : \mathbb R \to S^1, \phi(t) = e^{it} =\cos t + i\sin t,$$ has the property $$\phi(s) = \phi(t) \text{ iff } s - t = 2\pi k \text{ for some } k \in \mathbb Z .\tag{*} $$ We have $\phi([0,2\pi]) = S^1$. Thus also $\phi([a,2\pi+a]) = S^1$ for each $a \in \mathbb R$. Consider an open interval $(a,b)$ and let $S^1(a,b) = \phi((a,b)) = \{ e^{it} \mid t\in (a,b) \}$.

1. If $b - a > 2\pi$, then $S^1(a,b) = S^1$ (which is trivially open in $S^1$).

2. If $b - a \le 2\pi$, then $S^1(a,b)$ is open in $S^1$: The set $K = [a,2\pi +a] \setminus (a,b)$ is compact, hence $\phi(K) \subset S^1$ is compact, thus closed in $S^1$. Therefore $S^1 \setminus \phi(K)$ is open in $S^1$. We have $S^1 = \phi([a,2\pi +a]) = \phi(K \cup (a,b)) = \phi(K) \cup \phi((a,b))$. But $K$ and $(a,b)$ are disjoint, thus $s \in K$ and $t \in (a,b)$ cannot have the same image under $\phi$ (note that by (*) the only two distinct points in $[a,2\pi +a]$ having the same image under $\phi$ are $a$ and $2\pi + a$). We conclude that $\phi(K)$ and $ \phi((a,b))$ are disjoint, hence $\phi((a,b)) = S^1 \setminus \phi(K)$.

3. $\phi$ is an open map: Each open $U \subset \mathbb R$ can be written as $U = \bigcup_{t \in U}(t-r(t),t+r(t))$ with suitable $r(t) > 0$. Thus $\phi(U) = \bigcup_{t \in U}\phi((t-r(t),t+r(t)))$ is open in $S^1$.

4. If $b - a \le 2\pi$, then the restriction $\phi_{a,b} : (a,b) \to \phi((a,b)) = S^1(a,b)$ of $\phi$ is a homeomorphism: By (*) it is a bijection, by 3. it is an open map.

5. The sets $S^1(a,b)$ form a basis for the topology on $S^1$: Let $V \subset S^1$ be open and $z_0 \in V$. There is $t_0 \in \mathbb R$ such that $\phi(t_0) = z_0$. Then $\phi^{-1}(V)$ is an open subset of $\mathbb R$ containing $t_0$. There exists $r > 0$ such that $(t_0 -r t_0 +r) \subset \phi^{-1}(V)$. Hence $z_0 \in \phi((t_0 -r t_0 +r)) \subset V$.

Let us now show that $f$ is an open map.

It suffices to show that the images of the basis elements $S^1(a,b)$ are open in $S^1$. But $$f(S^1(a,b)) = f(\{ e^{it} \mid t\in (a,b) \}) = \{ e^{2it} \mid t\in (a,b) \} = \{ e^{is} \mid s \in (2a,2b) \} = S^1(2a,2b) .$$