Consider the torus
$S^1 \times S^1 := [(x_1,x_2,x_3,x_4) \in E^4 \space | \space x_1^2+x_2^2=1,\space x_3^2+x_4^2=1]$.
And the sphere
$S^2 := [(x_1,x_2,x_3) \in E^4 \space | \space x_1^2+x_2^2+x_3^2=1]$.
Let $f: S^1 \times S^1 \to S^2$ be the smooth map such that
$f(z,w) = \left( \space[\bar{z},\bar{w}]\sigma_1 \left[ \begin{array}{c} z \\ w \end{array} \right], [\bar{z},\bar{w}]\sigma_2 \left[ \begin{array}{c} z \\ w \end{array} \right], [\bar{z},\bar{w}]\sigma_3 \left[ \begin{array}{c} z \\ w \end{array} \right]\right)$
Here $z=x_1+ix_2, w=x_3+ix_4$, and $\sigma_1,\sigma_2$ and $\sigma_3$ are the Pauli matrices:
$\sigma_1 = \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right], \sigma_2 = \left[ \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right], \sigma_3 = \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right]$
Question: Compute $T_pf$ for any point p in the torus. Is $T_pf$ always a linear equivalence?
I just know that if given that $f(p)=q$ ,
$T_pf: T_p S^1 \times S^1 \to T_qS^2, \quad (p,\underline{u}) \mapsto (q, f_\underline{u}(p))$.
But following that I have no idea. Can anyone help?
