Open sets on the unit circle $S^1$ While doing some Analysis 3 exercises, I came across the problem of showing that a function $f:S^1\to S^1$ is open, cts and surjective. Here I am thinking about $S^1$ as a metric subspace of the reals. I thought about using the function $(x,y)\mapsto(x^2-y^2,2xy)$, i.e. just squaring the corresponding complex number, but I'm not sure how to prove that this is open as I'm having trouble understanding the notion of open sets in this metric.
I've seen some solutions using sets like $I:=\{e^{it} | t∈(a,b)\}$, but I need help/an explanation as to why this is open and how open sets work in $S^1$ in general.
Any answer would be greatly appreciated, thank you.
 A: 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.

*

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


*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

*

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


*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.
A: 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 $S^1(a,b) = \phi((a,b)) = \{ e^{it} \mid t\in (a,b) \}$.

*

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


*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 $\phi((a,b)) = S^1 \setminus \phi(K)$.


*$\phi$ 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 $S^1$.


*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.


*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
$$f(S^1(a,b)) = f(\{ e^{it} \mid t\in (a,b) \}) = \{ e^{2it} \mid t\in (a,b) \} = \{ e^{is} \mid s \in (2a,2b) \} = S^1(2a,2b) .$$
