Prove that $f$ determines a continuous map $\tilde{f}: \mathbb{R}P^2 \to \mathbb{R}^4$

Question

Let $\mathbb{S}^2$ denote the unit sphere in $\mathbb{R}^3$. Let $f:\mathbb{S}^2 \to \mathbb{R}^4$ be defined by $f(x,y,z)=(x^2-y^2,xy,yz,zx)$.

Prove that $f$ determines a continuous map $\tilde{f}: \mathbb{R}P^2 \to \mathbb{R}^4$ where $\mathbb{R}P^2$ is the real projective plane. $\tilde{f}$ is a homeomorphism onto a topological subspace of $\mathbb{R}^4$.

I know $\mathbb{R}P^2$ is homeomorphic to the quotient space $\mathbb{S}^2/R$ where $R$ identifies each pair of antipodal points in $\mathbb{S}^2$. Then I have a quotient map $p:\mathbb{S}^2\to \mathbb{R}P^2$. Then $\tilde{f}=fp^{-1}$?

Answer 1

Check that $f(x,y,z)=f(-x,-y,-z)$, and use the universal property of the quotient.

In particular, any map $f:X \to Y$ where $x \sim y \implies f(x)=f(y)$, then $f$ induces a unique continuous map $\tilde f: (X/\sim )\to Y$, where the quotient map followed by $\tilde f$ is the same as $f$.

You should verify that $\tilde f$ is injective (when is $f(x)=f(y)$?)

To see that it is an open mapping, note that $\tilde f$ is a continuous bijection from a compact space to a Hausdorff space so it admits continuous inverse.