For $(a, b, c)$ pairwise distinct: $$ \begin{align} f: & S^2 & \mapsto & \quad \mathbb{R}^4 \\ & (x, y, z) & \mapsto & \quad (X, Y, Z, W) = (y \cdot z, x \cdot z, x \cdot y, a \cdot x^2 + b \cdot y^2 + c \cdot z^2) \end{align} $$ is a smooth double covering map of the real unit sphere $S^2$ to a surface in the real 4-dimensional space $\mathbb{R}^4$, and opposite points have the same image, such that it can also be interpreted as an injective embedding of the real projective plane. In particular, the image of $f()$ is a surface, that is, each point has a non-degenerate tangential plane.
The image of $f()$ can be defined in the $(X, Y, Z, W)$ coordinates as follows.
If $X \cdot Y \cdot Z \neq 0$: $$ (Y \cdot Z)^2 + (X \cdot Z)^2 + (X \cdot Y)^2 = X \cdot Y \cdot Z \\ a \cdot (Y \cdot Z)^2 + b \cdot (X \cdot Z)^2 + c \cdot (X \cdot Y)^2 = X \cdot Y \cdot Z \cdot W $$ If $X = Y = 0$ and $Z \neq 0$: $$ (a - b)^2 \cdot (1 - 4 \cdot Z^2) = (2 \cdot W - (a + b))^2 $$ If $X = Z = 0$ and $Y \neq 0$: $$ (a - c)^2 \cdot (1 - 4 \cdot Z^2) = (2 \cdot W - (a + c))^2 $$ If $Y = Z = 0$ and $X \neq 0$: $$ (b - c)^2 \cdot (1 - 4 \cdot Z^2) = (2 \cdot W - (b + c))^2 $$ If $X = Y = Z = 0$: $$ (W - a) \cdot (W - b) \cdot (W - c) = 0 $$
Question 1:
Is the image of $f()$ an algebraic set in the $(X, Y, Z, W)$ coordinates ? Can it be defined by a single ideal, without case analysis ?
If yes, what is the ideal in question, explicitly ?
If no, how to prove it ? Furthermore, the image of f() consists then of five algebraic pieces fitting together smoothly, without being, as a whole, algebraic; I find it surprising, what happens where the pieces meet ?
Question 2:
If $X \cdot Y \cdot Z \neq 0$, or, equivalently, $x \cdot y \cdot z \neq 0$, then $$ [x : y : z] = [Y \cdot Z : X \cdot Z : X \cdot Y] $$ Is there a rational formula for each of the other pieces ?
If yes, what are the respective formulæ, explicitly ?
If no, how to prove it ?