Surjectivity of $f:\mathbb{S}^{1}\longrightarrow\mathbb{S}^{1}$ via universal covering of $\mathbb{S}^{1}$

Question

Let $f:\mathbb{S}^{1}\longrightarrow\mathbb{S}^{1}$ be given by $f(x,y) = (2xy,x^{2}-y^{2})$. Prove that $f$ is a closed map and $f$ is surjective. Use the universal covering $\phi:\mathbb{R}\longrightarrow\mathbb{S}^{1}, \phi(x)=(\cos(2\pi x),\sin(2\pi x))$ to prove surjectivity.

For the first part, note that $\mathbb{S}^{1}$ is a compact space, so every closed subset is compact. But $f$ is continuous, so it maps compact sets to compact sets, so $f$ is closed. For the second part, I don't know how to proceed. Thanks for any help.

Answer 1

If $(x,y)\in\Bbb S^1$, then $(x,y)=(\sin\alpha,\cos\alpha)$, for some $\alpha\in\Bbb R$. But then$$(x,y)=\left(2\sin\left(\frac\alpha2\right)\cos\left(\frac\alpha2\right),\cos^2\left(\frac\alpha2\right)-\sin^2\left(\frac\alpha2\right)\right),$$and therefore$$(x,y)=f\left(\sin\left(\frac\alpha2\right),\cos\left(\frac\alpha2\right)\right).$$

Answer 2

The universal covering is helpful, but you can avoid to use it.

Identifying $\mathbb R^2$ with $\mathbb C$ via $(x,y) = x + iy$, we see that $S^1 = \{ z \in \mathbb C \mid \lvert z \rvert = 1\}$. Let $\tau : S^1 \to S^1, \tau(x+iy) = y + ix$ (which is the same as $\tau(z) = i \overline z$) be the map flipping real and imaginary part. It is a homeomorphism, thus $f$ is surjective if and only $\bar f = \tau \circ f$ is surjective.

Clearly $\bar f(z) = z^2$. Since each $w \in \mathbb C$ has a square root $z_w$ (if $w \ne 0$, then it has in fact exactly two square root differing by the factor $-1$) and $\lvert z_w \rvert = \sqrt{\lvert w \rvert}$, we see each $w \in S^1$ has the form $\bar f(z_w) = z_w^2$. Thus $\bar f $ is surjective.