The projective plane as a smooth surface in a 4-dimensional space

Question

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 ?

Answer 1

$\bf{Question \ 1:}$

I think you should be more clear:

First of all you say that your map is an embedding of $S^2$ in $\mathbb R^4$. An embedding is injective, and as you said just one second later, your map sends opposites points to the same point.

Moreover, it is not so clear why your map $S^2/\mathbb Z_2\to \mathbb R^4$ should be injective and which is exactly the role of your coefficients $a,b,c$. We have to prove that if two points $(x_0',y_0', z_0')$, $(x_0,y_0,z_0)\in S^2$ have the same image, then they are equal or just the opposite to each other.

I have thought to this trick, although I don't know if there is a more easy proof. Remembering that our points are on $S^2$, let us compute

$$(x_0'+y_0'+z_0')^2=1+2x_0'y_0'+2x_0'z_0'+2y_0'z_0'=1+2x_0y_0+2x_0z_0+2y_0z_0=(x_0+y_0+z_0)^2 \implies x_0'+y_0'+z_0'=(-1)^\alpha(x_0+y_0+z_0).$$

However, we can do the same also for others combinations:

$$(x_0'+y_0'-z_0')^2=\dots=(x_0+y_0-z_0)^2 \implies x_0'+y_0'-z_0'=(-1)^\beta(x_0+y_0-z_0),$$

$$(x_0'-y_0'-z_0')^2=\dots=(x_0-y_0-z_0)^2 \implies x_0'+-y_0'-z_0'=(-1)^\gamma(x_0-y_0-z_0).$$

To sumarize, we have got the following linear system

$$ \begin{pmatrix}1 & 1& 1\\ 1 &1& -1\\ 1 &-1&-1 \end{pmatrix}\begin{pmatrix} x_0' \\ y_0' \\ z_0'\end{pmatrix}=\begin{pmatrix}(-1)^\alpha & 0& 0\\ 0 &(-1)^\beta& 0\\ 0 &0&(-1)^\gamma \end{pmatrix}\begin{pmatrix}1 & 1& 1\\ 1 &1& -1\\ 1 &-1&-1 \end{pmatrix}\begin{pmatrix} x_0 \\ y_0 \\ z_0\end{pmatrix}.$$

However, the left matrix is invertible and its inverse is

$$\begin{pmatrix}1 & 1& 1\\ 1 &1& -1\\ 1 &-1&-1 \end{pmatrix}^{-1}=\frac{1}{2}\begin{pmatrix}1 & 0& 1\\ 0 &1& -1\\ 1&-1&0 \end{pmatrix}$$

Therefore, for each fixed $\alpha,\beta, \gamma$ we have got a unique solution

$$ \begin{pmatrix} x_0' \\ y_0' \\ z_0'\end{pmatrix}=\frac{1}{2}\begin{pmatrix}1 & 0& 1\\ 0 &1& -1\\ 1&-1&0 \end{pmatrix}\begin{pmatrix}(-1)^\alpha & 0& 0\\ 0 &(-1)^\beta& 0\\ 0 &0&(-1)^\gamma \end{pmatrix}\begin{pmatrix}1 & 1& 1\\ 1 &1& -1\\ 1 &-1&-1 \end{pmatrix}\begin{pmatrix} x_0 \\ y_0 \\ z_0\end{pmatrix}=\frac{1}{2}\begin{pmatrix}1 & 0& 1\\ 0 &1& -1\\ 1&-1&0 \end{pmatrix}\begin{pmatrix}(-1)^\alpha & (-1)^\alpha& (-1)^\alpha\\ (-1)^\beta &(-1)^\beta& (-1)^{\beta+1}\\ (-1)^\gamma &(-1)^{\gamma+1}&(-1)^{\gamma+1} \end{pmatrix}\begin{pmatrix} x_0 \\ y_0 \\ z_0\end{pmatrix}=\frac{1}{2}\begin{pmatrix}(-1)^\alpha+(-1)^\gamma & (-1)^\alpha+(-1)^{\gamma+1}& (-1)^\alpha+(-1)^{\gamma+1}\\ (-1)^\beta+(-1)^{\gamma+1} &(-1)^\beta+(-1)^\gamma& (-1)^{\beta+1}+(-1)^\gamma\\ (-1)^\alpha+(-1)^{\beta+1} &(-1)^\alpha+(-1)^{\beta+1}&(-1)^\alpha+(-1)^{\beta}\end{pmatrix}\begin{pmatrix} x_0 \\ y_0 \\ z_0\end{pmatrix}.$$

Observe that if $(\alpha,\beta,\gamma)=(0,0,0)$, then the solution is trivial, and is $(x_0,y_0,z_0)$. Instead, if $(\alpha,\beta,\gamma)=(1,1,1)$, then the solution is trivial too, and it is the opposite point $(-x_0,-y_0,-z_0)$.

It remains to discuss the others $6$ possible solutions:

$(\alpha,\beta,\gamma)=(1,0,0)$:

$$\begin{pmatrix} x_0' \\ y_0' \\ z_0'\end{pmatrix}=\frac{1}{2}\begin{pmatrix}(-1)^\alpha+(-1)^\gamma & (-1)^\alpha+(-1)^{\gamma+1}& (-1)^\alpha+(-1)^{\gamma+1}\\ (-1)^\beta+(-1)^{\gamma+1} &(-1)^\beta+(-1)^\gamma& (-1)^{\beta+1}+(-1)^\gamma\\ (-1)^\alpha+(-1)^{\beta+1} &(-1)^\alpha+(-1)^{\beta+1}&(-1)^\alpha+(-1)^{\beta}\end{pmatrix}\begin{pmatrix} x_0 \\ y_0 \\ z_0\end{pmatrix}=\begin{pmatrix}0 & -1& -1\\ 0&1& 0\\ -1 &-1&0\end{pmatrix}\begin{pmatrix} x_0 \\ y_0 \\ z_0\end{pmatrix}=\begin{pmatrix}-y_0-z_0 \\ y_0 \\ -x_0-y_0\end{pmatrix}$$

Now we have impose that this solution belongs to $S^2$:

$$(-y_0-z_0)^2+y_0^2+(-x_0-y_0)^2=1+2y_0^2+2y_0z_0+2x_0y_0=1 \implies y_0(y_0+z_0+x_0)=0 \implies y_0=0 \quad \text{or} \quad z_0=-x_0-y_0$$

If $z_0=-x_0-y_0$, then we would have $\begin{pmatrix} x_0' \\ y_0' \\ z_0'\end{pmatrix}=\begin{pmatrix}-y_0-z_0 \\ y_0 \\ -x_0-y_0\end{pmatrix}=\begin{pmatrix} x_0\\ y_0 \\ z_0\end{pmatrix}$, and we are done.

Instead, if $y_0=0$, then we would have $\begin{pmatrix} x_0' \\ y_0' \\ z_0'\end{pmatrix}=\begin{pmatrix}-y_0-z_0 \\ y_0 \\ -x_0-y_0\end{pmatrix}=\begin{pmatrix} -z_0\\0 \\ -x_0\end{pmatrix}$. However, we observe that we never used until now our coefficients $a,b,c$. We remember that we have also the relation

$$a(x_0')^2+b(y_0')^2+c(z_0')^2=ax_0^2+by_0^2+cz_0^2 \implies az_0^2+cx_0^2=ax_0^2+cz_0^2 \implies (a-c)(z_0^2-x_0^2)=0 \implies z_0=\pm x_0$$

because $a\neq c$, by assumption. If $z_0=x_0$,then we have obtained that $\begin{pmatrix} x_0' \\ y_0' \\ z_0'\end{pmatrix}=\begin{pmatrix} -x_0\\0 \\ -x_0\end{pmatrix}$, so it is the opposite point of $\begin{pmatrix} x_0 \\ y_0 \\ z_0\end{pmatrix}=\begin{pmatrix} x_0\\0 \\ x_0\end{pmatrix}$; instead if $z_0=-x_0$, then they are the same point $\begin{pmatrix} x_0' \\ y_0' \\ z_0'\end{pmatrix}=\begin{pmatrix} x_0\\0 \\ -x_0\end{pmatrix}=\begin{pmatrix} x_0 \\ y_0 \\ z_0\end{pmatrix}$.

We have concluded this case. You can do the same for the remain $5$ systems. Clearly, you will use there that $a\neq b$,$a \neq c$ and $b\neq c$.

Thus, we can conclude that $S^2/\mathbb Z_2\to \mathbb R^4$ is injective.

Now let us try to get the image of this interesting map:

Let us call by $X,Y,Z, W$ the variables of $\mathbb R^4$. We observe that

$$\begin{cases} aYZ=X(ax^2) \\ bXZ=Y(by^2) \\ cXY=Z(cz^2) \end{cases}\implies \begin{cases} aY^2Z^2=XYZ(ax^2) \\ bX^2Z^2=XYZ(by^2) \\ cX^2Y^2=XYZ(cz^2) \end{cases} \implies aY^2Z^2+ bX^2Z^2+cX^2Y^2=XYZ(ax^2+by^2+cz^2)=XYZW$$

Therefore one of the equations is

$$aY^2Z^2+bX^2Z^2+cX^2Y^2=XYZW.$$

We can use the same approach to deduce the remain equations:

$$Y^2Z^2+X^2Z^2+X^2Y^2=XYZ,$$

$$(W-a)(W-b)(W-c)+(c-b)^2X^2(W-a)+(c-a)^2Y^2(W-b)+(a-b)^2Z^2(W-c)=0$$

$$X[(b-c)^2X^2+(W-b)(W-c)]+[(c-a)^2Y^2X-(c-a)(b-a)YZ+(b-a)^2Z^2X]=0$$ $$Y[(a-c)^2Y^2+(W-a)(W-c)]+[(c-b)^2X^2Y-(a-b)(c-b)XZ+(a-b)^2Z^2Y]=0$$ $$Z[(a-b)^2Z^2+(W-a)(W-b)]+[(b-c)^2X^2Z-(a-c)(b-c)XY+(a-c)^2Y^2Z]=0$$

Are they enough? Let us take a point $(X, Y, Z, W)$ satisfying these equations and prove that it belongs to the image of our map:

First suppose that $XYZ\neq 0$. Since

$XYZ=Y^2Z^2+X^2Z^2+X^2Y^2>0$, then $\frac{YZ}{X}, \frac{XZ}{Y}, \frac{XY}{Z}>0$.

Observe that if $X>0$, then only two options can occur, i.e. $Y>0$ and $Z>0$, or $Y<0$ and $Z<0$. Instead if $X<0$, then the others two options are $Y>0$, $Z<0$, or $Y<0$ and $Z>0$. This suggests to define

$x:=\sqrt{\frac{YZ}{X}}, y:=sign(XY)\sqrt{\frac{XZ}{Y}}$, and $z:=sign(XZ)\sqrt{\frac{XY}{Z}}$. By construction, we have $(x,y,z)\in S^2$.

Then

$$\begin{cases} y\cdot z=sign(XY) \cdot sign(XZ)\cdot \vert X\vert=X; \\ x \cdot z=sign(XZ)\cdot \vert Y\vert=Y;\\ x\cdot y=sign(XY)\cdot \vert Z\vert=Z;\\ ax^2+by^2+cz^2=W \end{cases}.$$

Thus our point $(X,Y,Z,W)$ belongs to our image.

Now let us suppose that $XYZ=0$. By symmetry of the problem, let us study what happens just when $Z=0$. In this case we would have

$$ 0=XYZ=Y^2Z^2+X^2Z^2+X^2Y^2=X^2Y^2.$$

Therefore $X=0$, or $Y=0$. Suppose $Y=0$, then the third equation tells us

$$(W-a)(W-b)(W-c)=-(c-b)^2X^2(W-a)$$.

If $W\neq a$, then we have $X^2=-\frac{(W-b)(W-c)}{(c-b)^2}$. In particular, $W-b$ and $W-c$ have to be opposite sign. Let us define

$$x:=0, \quad y:=sign(X)\sqrt{\vert \frac{W-c}{b-c}\vert }, \quad z:=\sqrt{\vert -\frac{W-b}{b-c}\vert }.$$

Then

$$\begin{cases} y\cdot z=sign(X)\sqrt{\vert \frac{W-c}{b-c}\vert \cdot \vert -\frac{W-b}{b-c}\vert}=X; \\ x \cdot z=0=Y;\\ x\cdot y=0=Z;\\ ax^2+by^2+cz^2=b\vert \frac{W-c}{b-c}\vert+c\vert -\frac{W-b}{b-c}\vert=W \end{cases}.$$

Therefore, it remains only to study the points of the form $(X, 0,0, a)$, $(0,Y,0,b)$, and $(0,0,Z,c)$.

Here we have to use one of our remain three equations, depending by which points are we considering. For instance, let us consider the point $(X, 0,0, a)$. Then the fourth equation tells us

$$X[(b-c)^2X^2+(a-b)(a-c)]=0$$

If $X\neq 0$, then we would have $X^2=-\frac{(a-b)(a-c)}{(b-c)^2}$. In this case our point belongs to the image, since the point of coordinates

$$x:=0, \quad y:=sign(X)\sqrt{\vert \frac{a-c}{b-c}\vert }, \quad z:=\sqrt{\vert -\frac{a-b}{b-c}\vert }$$

maps to $(X, 0,0, a)$.

It remain to study the point $(0,0,0,a)$. Here it is simply: it belongs to our image since $(x,y,z)=(1,0,0)$ maps exactly to $(0,0,0,a)$.

We have proved that the image of our map is an algebraic surface given by the six equations

$$p_1(X,Y,Z,W):=aY^2Z^2+bX^2Z^2+cX^2Y^2-XYZW=0,$$ $$p_2(X,Y,Z,W):=Y^2Z^2+X^2Z^2+X^2Y^2-XYZ=0,$$ $$p_3(X,Y,Z,W):=(W-a)(W-b)(W-c)+(c-b)^2X^2(W-a)+(c-a)^2Y^2(W-b)+(a-b)^2Z^2(W-c)=0,$$ $$p_4(X,Y,Z,W):=X[(b-c)^2X^2+(W-b)(W-c)]+[(c-a)^2Y^2X-(c-a)(b-a)YZ+(b-a)^2Z^2X]=0,$$ $$p_5(X,Y,Z,W):=Y[(a-c)^2Y^2+(W-a)(W-c)]+[(c-b)^2X^2Y-(a-b)(c-b)XZ+(a-b)^2Z^2Y]=0,$$ $$p_6(X,Y,Z,W):=Z[(a-b)^2Z^2+(W-a)(W-b)]+[(b-c)^2X^2Z-(a-c)(b-c)XY+(a-c)^2Y^2Z]=0.$$

As every algebraic varieties on the field of real numbers, our surface is described as the zero locus of just a single equation, the sum of the squares of our $6$ polynomial:

$$\mathbb P^2(\mathbb R)\cong Z(p_1^2+\dots +p_6^2)\subseteq \mathbb R^4.$$

$\bf{Question \ 2:}$

I am sorry, I don't understand exactly what you want. What I can say is that if you consider the projection of you map to the first three coordinates, so basically if you consider the map sending $(x,y,x)\to (y\cdot z, x\cdot z, x\cdot y)\in \mathbb R^3$, then your map is an isomorphism only for $xyz\neq 0$ (I suggest you to follow my proof above of the injectivity of the map and see where are the obstructions, that should be when $x$ or $y$ or $z$ are equal to zero)

So, outside $xyz=0$, our map in $\mathbb R^3$ is an isomorphism to the image $XYZ\neq 0$, and it is defined by the equation

$$ Y^2Z^2+X^2Z^2+X^2Y^2-XYZ=0.$$

This surface is the so-called $\textbf{Roman Surface}$: https://en.wikipedia.org/wiki/Roman_surface

Answer 2

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.