# Showing that this map descends to the quotient in an injective way

Let $f : \mathbb{S}^3 \to \mathbb{S}^2$ be the map

$$f(z_1,z_2) = (2z_1 \overline{z_2}, \vert z_1 \vert^2 - \vert z_2 \vert^2),$$

where we regard $\mathbb{S}^3 \subset \mathbb{C}^2$ and $\mathbb{S}^2 \subset \mathbb{C} \times \mathbb{R}$. In the 3-sphere we have the following equivalence relation: $(z_1,z_2) \sim (w_1,w_2)$ iff there exists $\lambda \in \mathbb{S}^1$ such that $z_i = \lambda w_i$, $i=1,2$. I want to show that $z \sim w$ iff $f(z) = f(w)$, so that $f$ induces an injective continuous function $\overline{f} : \mathbb{C}\mathbb{P}^1 \to \mathbb{S}^2$. I already showed that if $z \sim w$, then $f(z) = f(w)$ (this is straightforward). How do I prove the converse?

• Perhaps you should also mention that we regard $S^2 \subset \mathbb{C} \times \mathbb{R}$. – Paul Frost Aug 27 '18 at 7:29

We have $(z_1,z_2) \in S^3$ if and only if $\lvert z_1 \rvert^2 + \lvert z_2 \rvert^2 = 1$. Hence $f(z_1,z_2) = f(w_1,w_2)$ implies $2 \lvert z_1 \rvert^2 - 1 = \lvert z_1 \rvert^2 - \lvert z_2 \rvert^2 = \lvert w_1 \rvert^2 - \lvert w_2 \rvert^2 = 2 \lvert w_1 \rvert^2 - 1$ and therefore $\lvert z_1 \rvert = \lvert w_1 \rvert$. We conclude that $z_1 = \lambda w_1$ where $\lambda = z_1/w_1 \in S^1$. But now $2\lambda w_1 \overline{z_2} = 2 z_1 \overline{z_2} = 2 w_1 \overline{w_2}$, hence $\lambda \overline{z_2} = \overline{w_2}$ and therefore $z_2 = \lambda \overline{\lambda } z_2 = \lambda \overline{\lambda \overline{z_2}} = \lambda \overline{\overline{w_2}} = \lambda w_2$.