For the smooth map $F: \mathbb{R} P^2 \to \mathbb{R}^4$ prove $dF_p$ is injective $\forall p \in \mathbb{R} P^2.$ Can I please have help proving the problem below? How can I extend my proof for $dF_p.$ My proof is below. Also, could another approach to this be through contradiction? Thank you.
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For the smooth map $F: \R P^2 \to \R^4$ prove that $dF_p$ is injective for all $p$ in $\R P^2.$
$\textit{Proof.}$ Observe $f([x_1:y_1:z_1]) = \frac{1}{x_1^2 + y_1^2 + z_1^2}(x_1^2-y_1^2,x_1y_1,x_1z_1,y_1z_1).$ First identify $\R P^2$ with the quotient $S^2/\{\pm{1}\}$ of $S^2$ which can be obtained by identifying various antipodal points.
Now the map, which is given lefts to the map. $$F:\R P^2 \to \R^4, F(x_1,y_1,z_1) = (x_1^2-y_1^2,x_1y_1,x_1z_1,y_1z_1)$$ and now we will show, $F(x_1,y_1,z_1) = F(x_1', y_1', z_1')$ then $(x_1,y_1,z_1) = \pm (x_1',y_1',z_1').$ Now $(a,b,c,d) = F(x_1,y_1,z_1) = F(x_1',y_1',z_1').$ These act as image points. Initially, assume $a \ne 0,$ then $z_1^2 = \frac{c_1^2 + d_1^2}{a} = (z_1')^2$ hence $z_1 = \pm z_1'.$
When $z_1 \ne 0,$ then $x_1 = \frac{c}{z_1} = \pm x_1'$ and $y = \frac{d}{z} = \pm y'$ hence it follows the claim.
Now when $z_1, z_1' = 0$ then $x_1^2 = (a+1)/2 = (x_1')^2$ hence $x_1 = \pm x_1'.$ When $x_1 \ne 0,$ then $y_1 = \frac{b}{x_1} = \pm y_1'$ and also proves the claim.
When $x_1, x_1' = 0,$ we get $z_1^2 = 1 = (z_1')^2$ and also follows claim.
Conversely, assuming $a=0,$ then $x_1^2 = y_1^2$ and $(x_1')^2 = (y_1')^2.$ When $b=0,$ then $x_1,y_1,x_1',y_1' = 0$ and $z_1' = 1 = (z_1')^2$ hence claim is verified. When $b> 0$ then $x_1 = y_1 = \pm \sqrt{b}$ and $z_1 = \pm c\sqrt{b}.$ This similarly applies to $(x',y',z')$ thus $(x_1,y_1,z_1) = \pm (x_1',y_1',z_1').$ As well as when $b<0,$ then $x_1 = -y_1 = \pm \sqrt{-b}$ and $z_1 = \pm c / \sqrt{-b}.$
 A: Consider the mapping
$$ G :\mathbb R^3\to \mathbb R^4, \qquad G(x, y, z) = (x^2-y^2,xy,xz,yz),$$
which has Jacobian
$$\begin{pmatrix} 2x & -2y & 0 \\y & x & 0 \\z & 0 &x \\0 & z & y \end{pmatrix}$$
For each $(x, y, z)\in \mathbb R^3$, the Jacobian matrix is the linear mapping $dG_{(x, y, z)} : \mathbb R^3 \to \mathbb R^4$. Now for $p = [x:y:z]$, the tangent plane $T_p \mathbb{RP}^2$ identified as
$$ \{ v\in \mathbb R^3: v\cdot (x, y, z) = 0\}$$
and $dF_p$ is just the restriction of $dG_p$ to $T_p\mathbb{RP}^2$. We need to show that $dG_p$ is injectie when restricted to the tangent plane.
Note that when $x$ is non zero, the minor
$$\begin{pmatrix} 2x & -2y & 0 \\y & x & 0 \\z & 0 & x \end{pmatrix}$$
has full rank. Similar you can minors that is invertible when $y$ is nonzero. Thus $dG_p$ is injective whenever $x, y$ are nonzero. Since $dF_p$ is the restriction, $dF_p$ is also injective in this case.
When $x, y$ are both zero, then the matrix is
$$\tag{1} \begin{pmatrix} 0 & 0 & 0 \\0 & 0 & 0 \\z & 0 &0 \\0 & z & 0 \end{pmatrix}$$
If you restrict to the sphere, there's only two possibilities $z = \pm 1$, and in both case the tangent plane are
$$ T_{[0,0,1]} \mathbb{RP}^2 = \operatorname{span}\{ (1,0,0), (0,1,0)\}$$
and it is clear that $dF_{[0,0,1]}$ is also injective (say, using the last two rows in (1). Thus $dF_p$ is injective for all $p$.
