<s>[INST] The projective plane as a smooth surface in a 4-dimensional space 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 ?
 [/INST] Response to question 1
This is a complement to the excellent answer of Federico Fallucca.
The ideal characterizing the image of $f()$ is already generated by
the following four polynomials:
$$
\begin{align}
  & (b - c)^2X^3 + (c - a)^2XY^2 + (a - b)^2XZ^2 + (W - b)(W - c)X - (a - b)(a - c)YZ       & (C_0) \\
  & (c - a)^2Y^3 + (a - b)^2YZ^2 + (b - c)^2YX^2 + (W - c)(W - a)Y - (b - c)(b - a)ZX       & (C_1) \\
  & (a - b)^2Z^3 + (b - c)^2ZX^2 + (c - a)^2ZY^2 + (W - a)(W - b)Z - (c - a)(c - b)XY       & (C_2) \\
  & (b - c)^2(W - a)X^2 + (c - a)^2(W - b)Y^2 + (a - b)^2(W - c)Z^2 + (W - a)(W - b)(W - c) & (C_3)
\end{align}
$$
In particular, this ideal contains the following two polynomials
(naturally derived from the constraints $x^2 + y^2 + z^2 = 1$ and
$ax^2 + by^2 + cz^2 = W$, but superseded by the generators
above):
$$
\begin{align}
  & Y^2Z^2 + Z^2X^2 + X^2Y^2 - XYZ     & (D_0) \\
  & aY^2Z^2 + cX^2Y^2 + bZ^2X^2 - XYZW & (D_1)
\end{align}
$$
Indeed:
$$
\begin{align}
    D_0 &= \frac{(b - c)YZC_0 + (c - a)ZXC_1 + (a - b)XYC_2}{(a - b)(b - c)(c - a)} \\
    D_1 &= \frac{a(b - c)YZC_0 + b(c - a)ZXC_1 + c(a - b)XYC_2}{(a - b)(b - c)(c - a)}
\end{align}
$$
Response to question 2
The real projective hemisphere $S$ (that is, the real unit sphere with
opposite points identified) can be identified with the real projective
plane $P$ by the following map:
$$
\begin{align}
    P & \longrightarrow S \\
    [\bar{x} : \bar{y} : \bar{z}] & \longrightarrow (x, y, z)
  = \frac{1}{\sqrt{\bar{x}^2 + \bar{y}^2 + \bar{z}^2}} \cdot (\bar{x}, \bar{y}, \bar{z})
\end{align}
$$
Thus, for inversing $f()$, it is enough to determine $(x, y, z)$ up to
a non-zero constant scaling factor, that is, to determine
$[x : y : z]$.
Assume $(X, Y, Z, W) = f(x, y, z)$ from now on; in particular:
$$
\begin{align}
  & x^2 + y^2 + z^2 = 1 \\
  & X = y \cdot z \\
  & Y = z \cdot x \\
  & Z = x \cdot y \\
  & W = a \cdot x^2 + b \cdot y^2 + c \cdot z^2
\end{align}
$$
Our goal is to express $[x : y : z]$ as polynomial functions
$g_i(X, Y, Z, W)$, using the constraints above. For any polynomial
$t_i(x, y, z)$, we have:
$$
    [t_i(x, y, z) \cdot x : t_i(x, y, z) \cdot y : t_i(x, y, z) \cdot z]
  = [x : y : z]
$$
unless $t_i(x, y, z) = 0$. The left-hand side is said to be determinate
at $(x, y, z)$ if $t_i(x, y, z)$ does not vanish, and indeterminate
otherwise. We have thus to find polynomials $t_i()$ such that the
left-hand side can be expressed in the $(X, Y, Z, W)$ coordinates,
that is:
$$
    g_i(X, Y, Z, W) = [x : y : z]
$$
Furthermore, we have to find enough of such polynomials $t_i()$ resp.
$g_i()$ such that, for each $(X, Y, Z, W)$ in the image of $f()$, some of
these polynomials is determinate at $(X, Y, Z, W)$, that is,
$g_i(X, Y, Z, W) \neq [0 : 0 : 0]$.
A possibility would be to choose $t_0(x, y, z) = x \cdot y \cdot z$, which
yields:
$$
\begin{align}
    [x : y : z]
 &= [t_0(x, y, z) \cdot x : t_0(x, y, z) \cdot y : t_0(x, y, z) \cdot z] \\
 &= [x \cdot y \cdot z \cdot x : x \cdot y \cdot z \cdot y : x \cdot y \cdot z \cdot z ] \\
 &= [Y \cdot Z : Z \cdot X : X \cdot Y]
\end{align}
$$
and thus:
$$
    g_0(X, Y, Z, W) \overset{\underset{\mathrm{def}}{}}{=}
    [Y \cdot Z : Z \cdot X : X \cdot Y]
$$
which is indeterminate at $(X, Y, Z, W)$ exactly if
$X \cdot Y \cdot Z = 0$, or, equivalently, $x \cdot y \cdot z = 0$.
But more comprehensive polynomials $t_i()$ can be derived using the
following equality:
$$
\begin{align}
    (W - a)
 &= a \cdot x^2 + b \cdot y^2 + c \cdot z^2 - a \\
 &= a \cdot x^2 + b \cdot y^2 + c \cdot z^2 - a \cdot (x^2 + y^2 + z^2) \\
 &= (b - a) \cdot y^2 + (c - a) \cdot z^2
\end{align}
$$
Analogously:
$$
    (W - b) = (c - b) \cdot z^2 + (a - b) \cdot x^2 \\
    (W - c) = (a - c) \cdot x^2 + (b - c) \cdot y^2
$$
from which follows:
$$
\begin{align}
    (W - a) \cdot x^2
 &= (b - a) \cdot x^2 \cdot y^2 + (c - a) \cdot z^2 \cdot x^2 \\
 &= (b - a) \cdot Z^2 + (c - a) \cdot Y^2
\end{align}
$$
and
$$
\begin{align}
    (c - b) \cdot (W - a) \cdot z^2
 &= (W - a) \cdot ((W - b) + (b - a) \cdot x^2) \\
 &= (W - a) \cdot (W - b) + (b - a)^2 \cdot Z^2 + (b - a) \cdot (c - a) \cdot Y^2
\end{align}
$$
Analogously:
$$
\begin{align}
    (W - b) \cdot y^2 &= (c - b) \cdot X^2 + (a - b) \cdot Z^2 \\
    (a - c) \cdot (W - b) \cdot x^2 &= (W - b) \cdot (W - c) + (c - b)^2 \cdot X^2 + (c - b) \cdot (a - b) \cdot Z^2 \\
    (W - c) \cdot z^2 &= (a - c) \cdot Y^2 + (b - c) \cdot X^2 \\
    (b - a) \cdot (W - c) \cdot y^2 &= (W - c) \cdot (W - a) + (a - c)^2 \cdot Y^2 + (a - c) \cdot (b - c) \cdot X^2
\end{align}
$$
Choosing $t_1(x, y, z) = (a - c) \cdot (W - b) \cdot x$ yields:
$$
\begin{align}
    [x : y : z]
 &= [t_1(x, y, z) \cdot x : t_1(x, y, z) \cdot y : t_1(x, y, z) \cdot z] \\
 &= [(a - c) \cdot (W - b) \cdot x^2 : (a - c) \cdot (W - b) \cdot x \cdot y : (a - c) \cdot (W - b) \cdot z \cdot x] \\
 &= [(W - b) \cdot (W - c) + (c - b)^2 \cdot X^2 + (c - b) \cdot (a - b) \cdot Z^2 : (a - c) \cdot (W - b) \cdot Z : (a - c) \cdot (W - b) \cdot Y]
\end{align}
$$
And thus:
$$
    g_1(X, Y, Z, W) \overset{\underset{\mathrm{def}}{}}{=}
    [(W - b) \cdot (W - c) + (c - b)^2 \cdot X^2 + (c - b) \cdot (a - b) \cdot Z^2 : (a - c) \cdot (W - b) \cdot Z : (a - c) \cdot (W - b) \cdot Y]
$$
Choosing $t_2(x, y, z) = (b - a) \cdot (W - c) \cdot y$ resp.
$t_3(x, y, z) = (c - b) \cdot (W - a) \cdot z$ yields, analogously:
$$
    g_2(X, Y, Z, W) \overset{\underset{\mathrm{def}}{}}{=}
    [(b - a) \cdot (W - c) \cdot Z : (W - c) \cdot (W - a) + (a - c)^2 \cdot Y^2 + (a - c) \cdot (b - c) \cdot X^2 : (b - a) \cdot (W - c) \cdot X]
    \\
    g_3(X, Y, Z, W) \overset{\underset{\mathrm{def}}{}}{=}
    [(c - b) \cdot (W - a) \cdot Y : (c - b) \cdot (W - a) \cdot X : (W - a) \cdot (W - b) + (b - a)^2 \cdot Z^2 + (b - a) \cdot (c - a) \cdot Y^2]
$$
The constraints for these last three maps $g_1()$, $g_2()$ and $g_3()$
to all be indeterminate at $f(x, y, z)$ for some $(x, y, z)$ on the
real projective hemisphere are
$t_1(x, y, z) = t_2(x, y, z) = t_3(x, y, z) = 0$, that is,
$
    (a - c) \cdot (W - b) \cdot x
  = (b - a) \cdot (W - c) \cdot y
  = (c - b) \cdot (W - a) \cdot z = 0
$,
that is, by the equalities established above:
$$
    (W - b) \cdot (W - c) + (c - b)^2 \cdot X^2 + (c - b) \cdot (a - b) \cdot Z^2 = 0 \\
    (W - c) \cdot (W - a) + (a - c)^2 \cdot Y^2 + (a - c) \cdot (b - c) \cdot X^2 = 0 \\
    (W - a) \cdot (W - b) + (b - a)^2 \cdot Z^2 + (b - a) \cdot (c - a) \cdot Y^2 = 0
$$
that is, by expressing (X, Y, Z, W) in terms of (x, y, z) and simplifying:
$$
    x \cdot ((c - b) \cdot z^2 + (a - b) \cdot x^2) = 0 \\
    y \cdot ((a - c) \cdot x^2 + (b - c) \cdot y^2) = 0 \\
    z \cdot ((b - a) \cdot y^2 + (c - a) \cdot z^2) = 0
$$
which have no solution on the real projective hemisphere, such that no
point of the hemisphere has an image $f(x, y, z)$ at which all three
maps $g_1()$, $g_2()$ and $g_3()$ are indeterminate.
In other words, the inverse of $f()$ is at least one of the polynomials
$g_1()$, $g_2()$ and $g_3()$, whichever is determinate, depending their
argument (a point $(X, Y, Z, W)$ in the image of $f()$), and there is no
argument at which all three are indeterminate.
Finally, the function $f()$ can be equivalently defined on the real
projective plane:
$$
\begin{align}
    f([x : y : z])
 &= f\left(\frac{x}{\sqrt{x^2 + y^2 + z^2}}, \frac{y}{\sqrt{x^2 + y^2 + z^2}}, \frac{z}{\sqrt{x^2 + y^2 + z^2}}\right) \\
 &= \frac{1}{x^2 + y^2 + z^2} \cdot (y \cdot z, z \cdot x, x \cdot y, a \cdot x^2 + b \cdot y^2 + c \cdot z^2)
\end{align}
$$
and thus the real projective plane and the image of $f()$ are
birationally equivalent.
</s>