# Find a tangent mapping $TF\colon\,TS^{3}\to TS^{2}$

I need to find a tangent mapping $TF\colon\,TS^{3}\to TS^{2}$, where $$F(z,w)=(z\overline{w}+w\overline{z},iw\overline{z}-iz\overline{w}, z\overline{z}-w\overline{w})$$ and $S^{3}=\{|z|^2+|w|^2=1\}\subset \mathbb{C}^{2}$.

This is what I did. Firstly, I introduced a mapping $$f(x_{1}, x_{2}, x_{3}, x_{4})=F(z, w).$$ After some computations I obtained: $$f(x_{1}, x_{2}, x_{3}, x_{4})=(2x_1x_3+2x_2x_4, 2x_1x_3-2x_1x_4,x_1^2+x_2^2-x_3^2-x_4^2).$$ It is easy to find both $TS^3$ and $TS^2$. We have $$T_{x_{0}}S^3=\{(y_1, y_2, y_3, y_4)\,:\,2x_{1}^{0}y_1+2x_{2}^{0}y_2+2x_{3}^{0}y_3+2x_{4}^{0}y_4=0\}.$$ Analogously $$T_{x_{0}}S^2=\{(y_1, y_2, y_3)\,:\,2x_{1}^{0}y_1+2x_{2}^{0}y_2+2x_{3}^{0}y_3=0\}.$$ I know that the mapping I'm looking for is of the form: $$T_{x_{0}}f=i_{\psi}\circ(\psi\circ f \circ \varphi^{-1})'(\varphi(x_{0}))\circ i^{-1}_{\varphi},$$ where $\varphi$ and $\psi$ are some maps from relevant atlases. The most natural choice is to pick stereographic projections. Each atlas consists of two maps becouse of the poles. To simplify, let's pick $$\varphi (x_1, x_2, x_3, x_4)=\delta_{3}(x_1, x_2, x_3, x_4)=\frac{1}{1-x_4}(x_1, x_2, x_3)$$ defined on $S^{3}\setminus \{(0,0,0,1)\}$ and $$\psi (x_1, x_2, x_3)=\delta_2 (x_1, x_2, x_3)=\frac{1}{1-x_3}(x_1, x_2)$$defined on $S^{2}\setminus \{(0,0,1)\}$. I computed $\varphi^{-1}$ as well so I can handle with $(\psi\circ f \circ \varphi^{-1})'(\varphi(x_{0}))$. As far as the remaining mappings are concerned I only know that $i_{\varphi}:\mathbb{R}^{3}\to TS^{3}$ and $i_{\psi}:\mathbb{R}^{2}\to TS^{2}$. My problem is: what are the formulas for $i_{\varphi}, i_{\psi}?$ I cannot figure them out.

Hint: There is no need to go into charts. You can simply view your map $F$ as a map $\mathbb R^4\to\mathbb R^4$ and verify that the restriction of $F$ to $S^3\subset \mathbb R^4$ has values in $S^2\subset\mathbb R^3$. This implies that for each point $(z,w)\in S^3$ and each tangent vector $v\in T_{(z,w)}S^3$, the derivative $Df(z,w)(v)$ will automatically lies in $T_{F(z,w)}S^2$. So you can compute the tangent map as the restriction of the usual derivative.
Probably things get even easier, if you view $F$ as the map $\mathbb C^2\to\mathbb C\times\mathbb R$ given by $F(z,w)=(2z\bar w,|z|^2-|w|^2)$ without going to real coordinates. In this form you can almost read off the derivative (just let me know if you want me to write it out),