Consider the map $v \colon \mathbb{S}^2 \to \mathbb{R}^6$ defined by
$$ v(x, y, z) = (x^2, y^2, z^2, \sqrt{2}xy, \sqrt{2}yz, \sqrt{2}zx) $$
Furthermore, let $ M := v(\mathbb{S}^2) $ and $P^2\mathbb{R}$ denote the projective plane.

I want to show that the map $v$ defines a homeomorphism $V$ from $P^2\mathbb{R}$ onto $M$.

Well, I know that $P^2\mathbb{R} \cong {\mathbb{S}^2}/{\sim}$ where $(x, y, z) \sim (-x, -y, -z)$. If I somehow manage to prove that $M \cong {\mathbb{S}^2}/{\sim}$, it'd be neat.