Projective Plane Embedding Ambiguity I'm working in a problem in do Carmo: let $F: \mathbb{R}^3 \to \mathbb{R}^4$ be given by 
$$
F(x,y,z) \;\; =\;\; (x^2 - y^2, xy, xz, yz).
$$
Let $\varphi: \mathbb{S}^2 \to \mathbb{R}^3$ be the restriction $\varphi = F|_{\mathbb{S}^2}$.  Observing that $\varphi(p) = \varphi(-p)$ for all $p \in \mathbb{S}^2$ we can unambiguously define the mapping $\tilde{\varphi}: \mathbb{RP}^2 \to \mathbb{R}^4$ given by $\tilde{\varphi}[p] = \varphi(p)$.  Ultimately we want to show that $\tilde{\varphi}$ is an immersion, and subsequently an embedding of the projective plane into $\mathbb{R}^4$.
My Problem: It's not hard to see, considering $\mathbb{RP}^2 \approx \mathbb{S}^2/\mathbb{Z}_2$, that the map $\tilde{\varphi}$ is well-defined, but when I try to take the differential, wouldn't we expect $d\tilde\varphi_p$ to be well-defined as well?  This is to say, shouldn't we expect $d\tilde{\varphi}_p = d\tilde{\varphi}_{-p}$, or at least the columns of each Jacobian to span the same 2-dimensional affine subspace of $\mathbb{R}^4$?  
Example 1: First consider the case that we restrict $F$ to the upper and lower open hemispheres.  We can rewrite these restrictions as
\begin{eqnarray*}
F^{z+}(x,y) & = & \left (x^2 - y^2, xy, x\sqrt{1-x^2-y^2}, y\sqrt{1-x^2-y^2} \right ) \\
F^{z-}(x,y) & = & \left ( x^2 - y^2, xy, -x\sqrt{1-x^2-y^2}, -y\sqrt{1-x^2-y^2} \right )
\end{eqnarray*}
where we can see that $F^{z+}(x,y) = F^{z-}(-x,-y)$.  This yields two Jacobians which are incompatible at antipodal points:
\begin{eqnarray*}
dF^{z+}_{(x,y)} & =& \left [ \begin{array}{cc}
2x & -2y \\
y & x \\
\frac{1- 2x^2 - y^2}{\sqrt{1-x^2-y^2}} & -\frac{xy}{\sqrt{1-x^2-y^2}} \\
-\frac{xy}{\sqrt{1-x^2-y^2}} & \frac{1-x^2-2y^2}{\sqrt{1-x^2-y^2}}
\end{array} \right ] \\
dF^{z-}_{(x,y)} & = & \left [ \begin{array}{cc}
2x & -2y \\
y & x \\
-\frac{(1-2x^2-y^2)}{\sqrt{1-x^2-y^2}} & \frac{xy}{\sqrt{1-x^2-y^2}} \\
\frac{xy}{\sqrt{1-x^2-y^2}} & -\frac{(1-x^2-2y^2)}{\sqrt{1-x^2-y^2}} \\
\end{array} \right ].
\end{eqnarray*}
Wouldn't we expect that if $F(p) = F(-p)$ for all $p \in \mathbb{R}^3$ that at least $dF^{z+}_{(x,y)} = dF^{z-}_{(-x,-y)}$, or at least have their columns span the same affine subspace?  
Example 2: Another approach I've considered for this problem is using the stereographic projection $\pi^{-1}:\mathbb{R}^2 \to \mathbb{S}^2/\{N\}$ given by 
$$
\pi^{-1}(u,v) \;\; =\;\; \left ( \frac{2u}{u^2+v^2+1}, \frac{2v}{u^2+v^2+1}, \frac{u^2+v^2-1}{u^2+v^2+1} \right )
$$
and writing $\varphi|_{\mathbb{S}^2/\{N\}} = F \circ \pi^{-1}$. The problem that I see with this latter approach is that antipodal coordinates in the plane don't map to antipodal coordinates in $\mathbb{S}^2/\{N\}$.  
Overall my main question is how we appropriately compute the Jacobian $d\tilde{\varphi}_p$, and how are we guaranteed that the Jacobian is well-defined on antipodal points of $\mathbb{S}^2$?  I assume that this condition is necessary in order to show that $\tilde{\varphi}$ is an immersion.  Am I mistaken?
 A: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Proj}{\mathbf{P}}$Define $\varphi:S^{2} \to \Reals^{4}$ by $\varphi = F \circ i$, with $i:S^{2} \to \Reals^{3}$ the inclusion, so that
$$
\varphi:(x, y, z) \in S^{2} \mapsto (x^{2} - y^{2}, xy, xz, yz) \in \Reals^{4}.
$$
To take the differential, you've covered two open hemispheres of $S^{2}$ by graph charts,
$$
\sigma^{\pm}(x, y) = \left(x, y, \pm\sqrt{1 - x^{2} - y^{2}}\right).
$$
Note, however, that $\sigma^{+}(x, y)$ and $\sigma^{-}(x, y)$ are not antipodal on the sphere; instead, $\sigma^{+}(x, y)$ and $\sigma^{-}(-x, -y)$ are antipodal. Composing with $F$, you get
$$
F^{z^{+}}(x, y)
  = (F \circ \sigma^{+})(x, y) = (x^{2} - y^{2}, xy, xz, yz)
  = (F \circ \sigma^{-})(-x, -y)
  = F^{z^{-}}(-x, -y),
$$
upon which the Jacobian calculation comes out as expected. (The respective columns are negatives of each other, as you'd expect since each map $F \circ \sigma^{\pm}$ is the other pre-composed with the map $(x, y) \mapsto (-x, -y)$ of $\Reals^{2}$.)
As a separate matter, the fact that $\varphi:S^{2} \to \Reals^{4}$ factors through the antipodal map, and therefore induces a smooth map $\tilde{\varphi}:\Reals\Proj^{2} \to \Reals^{4}$, automatically guarantees the Jacobian of $\tilde{\varphi}$ is well-defined: The two local lifts (map a small piece of $\Reals\Proj^{2}$ to either of two small pieces of $S^{2}$, then push forward by $F$) are identical, and therefore must have the same Jacobian with respect to local coordinates on the projective plane.
