Using quotient to show that a map $\mathbb{RP}^2\rightarrow\mathbb{R}^3$ is an immersion at all but six points We want to show that $\phi:\mathbb{RP}^2\rightarrow \mathbb{R}^3$ given by $\phi([x,y,z])=\frac{(yz,xz,xy)}{x^2+y^2+z^2}$ is an immersion at all but 6 points. Recall that a map $f$ is an immersion at $x$ if the matrix $Df|_x$ is injective.
We know that $\mathbb{RP}^2$ can be given by the quotient map $q:\mathbb{R}^3-\{0\}\rightarrow\mathbb{RP}^2$ by identifying one dimensional subspaces of $\mathbb{R}^3$. I believe that we have a natural map $(\phi\circ q)(x,y,z)=\frac{(yz,xz,xy)}{x^2+y^2+z^2}$. I think that the points where $\phi\circ q$ fails to be an immersion correspond to the points where $\phi$ fails to be an immersion. Then I wrote down the Jacobian of $D(\phi\circ q)|_{(x,y,z)}$ in hopes of obtaining $6$ one-dimensional subspaces where the matrix was non-injective, that is, has zero determinant. However, setting the determinant to be zero, I obtained the equation $xyz(-x^2+y^2+z^2)(x^2-y^2+z^2)(x^2+y^2-z^2)=0$, which in fact gives us three two-dimensional subspaces and $3$ cone-shapes in $\mathbb{R}^3$. This does not seem to correspond to $6$ points in $\mathbb{RP}^2$.
What am I misunderstanding here?
This was a "simple" question in a qualifying topology/geometry exam, so I do not believe the solution is too involved.
 A: Let us write $\xi = (x,y,z)$.
$\phi$ is an immersion at $[\xi]$ if the rank of $D\phi \mid_{[\xi]}$ is  $2$. Solving $\det D(\phi \circ q) \mid_{\xi} = 0$ gives you points at which $D(\phi \circ q) \mid_{\xi}$ has rank $< 3$. This does of course not exclude that the rank of $D\phi \mid_{[\xi]}$ is  $2$. In fact, we have
$$D(\phi \circ q) \mid_\xi = D\phi \mid_{[\xi]} \circ Dq \mid_\xi \tag{1}$$
which shows that $D(\phi \circ q) \mid_{\xi}$ will always have rank $\le 2$ (note that $D\phi \mid_{[\xi]}$ has rank $\le 2$) so that the determinant of the Jacobian of $D(\phi \circ q) \mid_{\xi}$ is always $0$.
Let $n(\xi) = x^2+y^2+z^2 = \lVert \xi \rVert^2$. Then the Jacobian of $D(\phi \circ q) \mid_{\xi}$ is
$$J(\xi) = \begin{pmatrix}
\frac{-2xyz}{n(\xi)^2} & \frac{z(n(\xi)-2y^2)}{n(\xi)^2} & \frac{y(n(\xi)-2z^2)}{n(\xi)^2}\\
\frac{z(n(\xi)-2x^2)}{n(\xi)^2} & \frac{-2xyz}{n(\xi)^2} & \frac{x(n(\xi)-2z^2)}{n(\xi)^2} \\
\frac{y(n(\xi)-2x^2)}{n(\xi)^2} & \frac{x(n(\xi)-2y^2)}{n(\xi)^2} & \frac{-2xyz}{n(\xi)^2}
\end{pmatrix}$$
It is easy to see that $\sqrt{n(\xi)} J(\xi) = J(\xi/\sqrt{n(\xi)}) = J(\xi/\lVert \xi \rVert)$. Thus $J(\xi)$ and $J(\xi/\lVert \xi \rVert)$ have the same rank. It therefore suffices to determine the rank for $\xi $ with $\lVert \xi \rVert = 1$. In that case
$$J(\xi) = \begin{pmatrix}
-2xyz & z(1-2y^2) & y(1-2z^2)\\
z(1-2x^2) & -2xyz & x(1-2z^2) \\
y(1-2x^2) & x(1-2y^2) & -2xyz
\end{pmatrix}$$
Case 1. $x,y, z \ne 0$.
Assume that $J(\xi)$ has rank $< 2$. Then all determinants of $2 \times 2$-submatrices must vanish. In particular
$$0 = (-2xyz)^2 - z(1-2x^2)z(1-2y^2) = -(1 -2(x^2 + y^2))z^2 = -(1 -2(1-z^2))z^2 = -(2z^2-1)z^2$$
which implies $2z^2-1 = 0$, i.e. $z^2 = 1/2$. Similarly $x^2 = 1/2$ and $y^2 = 1/2$. But then $1 = x^2 +y^2 + z^2 = 3/2$ which is a contradiction. Therefore $J(\xi)$ has rank $2$.
Case 2. Exactly one of $x,y,z$ is $0$.
Let us first consider $z = 0$. Then
$$J(\xi) = \begin{pmatrix}
0 & 0 & y\\
0  & 0 & x \\
y(1-2x^2) & x(1-2y^2) & 0
\end{pmatrix}$$
This matrix has rank $2$ unless $y(1-2x^2) = x(1-2y^2) = 0$. This means $x^2 = y^2 = 1/2$. Thus we get four points $\xi = (\pm \sqrt{1/2},\pm \sqrt{1/2},0)$ in which $J(\xi)$ has rank $1$.
The same is true for the four points $\xi = (\pm \sqrt{1/2},0,\pm \sqrt{1/2})$ and the four points $\xi = (0,\pm \sqrt{1/2},\pm \sqrt{1/2})$.
Case 3. Exactly two of $x,y,z$ are $0$.
Consider $y = z = 0$. Then
$$J(\xi) = \begin{pmatrix}
0 & 0 & 0\\
0  & 0 & x \\
0 & x & 0
\end{pmatrix}$$
which has rank $2$. The same is true if $x = z = 0$ and $x = y = 0$.
We conclude from $(1)$ that $D\phi \mid_{[\xi]}$ must have rank $2$ with the possible exception of six points $[\xi]$, the equivalence classes of the $12$ points $\xi$ detected in case 2. In fact the rank of $D\phi \mid_{[\xi]}$ is $1$ at these points. This follows from the fact $Dq \mid_{\xi}$ has rank $2$ in all points. This is fairly obvious, but let us prove it. Again it suffices to consider $\lVert \xi \rVert = 1$. With the possible exception of the $12$ points of case 2 this follows from $(1)$ because we know that $D(\phi \circ q) \mid_\xi$ has rank $2$ for all other points. We can now argue by symmetry that also in the above $12$ points $\xi_i$ the rank must be $2$. Formally, we can chose a rotation $R$ on $\mathbb R^3$ such that $R(\xi_i)$ is a non-exceptional point. Since $q(R(-\xi)) = q(-R(\xi)) = q(R(\xi))$, we see that $q \circ R$ induces a map $R' : \mathbb{RP}^2 \to \mathbb{RP}^2$ such that $q \circ R = R' \circ q$. Now $R$ and $R'$ are diffeomorphisms and it follows that the rank of $D(\phi \circ q) \mid_{\xi_i}$ agrees with the rank of $D(\phi \circ q) \mid_{R(\xi_i)}$.
