Immersion $(x, y, z)\mapsto (xy, xz, yz)$ from sphere I need to prove that the map $f:\mathbb{S}^2\to\mathbb{R}^3$ given by
$$f(x,y,z) = (xy,xz,yz)$$
is an immersion except for a finite number of points.
Any tips on how to get started? I've tried using local charts, but it gets too messy:
Denoting the stereographic projection from the north pole by $\varphi$, I found that
$$ f\circ \varphi^{−1}(x,y)=\left( \frac{4xy}{(1+x^2+y^2)^2},\frac{2x(x^2+y^2−1)}{(1+x^2+y^2)^2},\frac{2y(x^2+y^2−1)}{(1+x^2+y^2)^2} \right),$$
so I basically need to show that this is an immersion, which is possible, just very hard. I feel that there must be some other way.
 A: Since your map $f$ is a restriction of a global map $F \colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$, it is easier to work with $dF$ and then restrict $dF|_p$ to $T_p(S^2)$ to get $df|_p$. Working with $F$, we have
$$ dF|_{(x,y,z)} = \begin{bmatrix} y & x & 0 \\ z & 0 & x \\ 0 & z & y \end{bmatrix}. $$
The determinant of $dF|_{(x,y,z)}$ is $-2xyz$ so whenever $xyz \neq 0$, the map $dF|_{(x,y,z)}$ is an isomorphism. In particular, whenever $(x,y,z) \in S^2$ and $xyz \neq 0$ we have
$$ \dim df|_{(x,y,z)}(T_p(S^2)) = \dim dF|_{(x,y,z)}(T_p(S^2)) = \dim T_p(S^2) = 2 $$
so $f$ is an immersion at such points. This leaves you three cases:

*

*Assume $(x,y,z) \in S^2$ and $x = 0$. Then we are looking at points of the form $(0,y,z)$ with $y^2 + z^2 = 1$ and the differential of $f$ at such points is the restriction of $dF$ to $T_pS^2$. The differential $dF|_{(0,y,z)}$ is equal to
$$ \begin{pmatrix} y & 0 & 0 \\ z & 0 & 0 \\ 0 & z & y \end{pmatrix}. $$
It has rank two (since $y$ and $z$ are not $0$) and kernel
$$ \operatorname{span} \begin{pmatrix} 0 \\ -y \\ z \end{pmatrix}. $$
Note that $(0,-z,y)^T$ belongs to $T_{(0,y,z)}S^2$ iff $\left< (0,y,z), (0,-y,z) \right> = z^2 - y^2 = 0$. This means that $z = \pm y$ so $z,y \in \{ \pm \frac{1}{\sqrt{2}} \}$. At such points the map $f$ fails to be an immersion while at all other points $df|_{(0,y,z)} = \left( dF|_{(0,y,z)} \right)|_{T_{(0,y,z)}S^2}$ has no kernel so $f$ is an immersion.

*The other cases where $y = 0$ or $z = 0$ are completely analogous.

In the end, you get that the $f$ is an immersion away from the 12 points
$$\left \{ \left( 0,\pm\frac{1}{\sqrt{2}},\pm\frac{1}{\sqrt{2}} \right) \right \} \cup \left \{ \left(\pm\frac{1}{\sqrt{2}}, 0, \pm\frac{1}{\sqrt{2}} \right) \right \} \cup \left \{ \left( \pm\frac{1}{\sqrt{2}},\pm\frac{1}{\sqrt{2}}, 0 \right) \right \}. $$
