Proving a function to be an open map

Question

How can we prove that the function $f: S^1 \rightarrow S^1$ defined as $z \mapsto z^2$ is a continuous and open map using topological arguments? Here $S^1$ represents the unit circle in complex plane, centred at the origin.

I could prove that it is a continuous map since the map $F: \Bbb C \rightarrow \Bbb C$ defined as $z \mapsto z^2$ is a continuous map because it is a polynomial function. Moreover, $f$ is just the restriction of $F$ on $S^1$ i.e. $f = F|_{S^1}$ and because restrictions of continuous maps are continuous (via the subspace topology), $f$ is also continuous.

But I could not solve that $f$ is an open map so please help me with that.
Also, please verify whether my proof is correct.

Answer 1

Let $A$ be an open subset of $S^1$. If $A=\emptyset$, then $f(A)=\emptyset$, which is an open set. Otherwise, take $z\in A$ and let $\theta\in\Bbb R$ be such that $e^{i\theta}=z$. Since $A$ is open, there is a $\alpha\in\left(0,\frac\pi2\right)$ such that$$\varphi\in(\theta-\alpha,\theta+\alpha)\implies e^{i\varphi}\in A.$$But then $f(e^{i\varphi})=(e^{i\varphi})^2=e^{2i\varphi}$ and so$$f(A)\supset\{e^{i\varphi}\mid\varphi\in(2\theta-2\alpha,2\theta+2\alpha)\}.$$So, $f(A)$ is an open set.

Answer 2

Your proof of continuity is correct.

The fact that the map is open is an immediate consequence of the following fact from Complex Analysis: Every point on $S^{1}$ has neighborhood on which there is an analytic (hence continuous) branch of square root. The continuity of this square root shows that the $z \to z^{2}$ is (locally, hence globally) open.