Push vector field from Euclidean space to tangent space of manifold via charts If $\phi:S^2\to\mathbb{R}^2$, $(x,y,z)\mapsto (\frac{x}{1-z},\frac{y}{1-z})$ (hence $\phi^{-1}:(u,v)\mapsto (\frac{2u}{u^2+v^2+1},\frac{2v}{u^2+v^2+1},\frac{u^2+v^2-1}{u^2+v^2+1})$),  and I were to consider the vector field $\frac{\partial}{\partial_u}$ on $\mathbb{R}^2$, im unsure as to how to push this into $(x,y,z)$ coordinates. Do I calculate the derivative matrix of $\phi^{-1}$ and multiply it by the matrix $
 \begin{pmatrix}
\frac{\partial}{\partial_u}\\
0 \end{pmatrix} $
and then just use $\phi$ to change all coordinates back from $(u,v)$ to $(x,y,z)$? But then how do I change from $\frac{\partial}{\partial_u}$ to $\frac{\partial}{\partial_k}$ for $k=x,y,z$?
 A: You have a map $\phi: S^2 \subset \mathbb{R}^3 \to \mathbb{R}^2$. Let $u,v$ be the coordinates function on $S^2$ and $x,y$ that on $\mathbb{R}^2$. You have $\phi^{-1}: \mathbb{R}^2 \to S^2 \subset \mathbb{R}^3$ and you wish to sent a vector field on $\mathbb{R}^2$ to one on $S^2$. Then you want to use the push-forward $\phi^{-1}_*:=\phi^{-1}_{*,\phi(p)}: T_{\phi(p)} \mathbb{R}^2 \to T_pS^2$ given by:
$$\left(\phi^{-1}_*X_{\phi(p)}\right) (f) = X_{\phi(p)} (f \circ \phi^{-1})$$
where $f$ is a germ at $p \in U \subset S^2$ i.e $\phi^{-1}_*: X_{\phi(p)} \in T_{\phi(p)}\mathbb{R}^2\mapsto \phi^{-1}_*X_{\phi(p)} \in T_p S^2$. Recall first that the bases for $T_{\phi(p)}S^2$ and $T_p\mathbb{R}^2$ are given by:
$$\left\{\frac{\partial}{\partial u}\Bigr|_{\phi(p)},\frac{\partial}{\partial v}\Bigr|_{\phi(p)}\right\} \ \ \ \left\{\frac{\partial}{\partial x}\Bigr|_{p},\frac{\partial}{\partial y}\Bigr|_{p}\right\}$$
Thus if you wish to push forward tangent vectors you have to compute:
$$\phi^{-1}_* \left(\frac{\partial}{\partial u}\Bigr|_p\right) = a\frac{\partial}{\partial x}\Bigr|_{p}+ b \frac{\partial}{\partial y}\Bigr|_{p}$$
$$\phi^{-1}_* \left(\frac{\partial}{\partial v}\Bigr|_p\right)=c\frac{\partial}{\partial x}\Bigr|_{p}+ d \frac{\partial}{\partial y}\Bigr|_{p}$$
The above follows from the fact that $\phi^{-1}_*$ is linear i.e basis vectors map to basis vectors. You can recover the coefficients of the matrix:
$$A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$
by evaluating both sides of the equations with $x,y$. Remember $x,y$ are coordinate functions i.e $x \circ \phi^{-1}$ gives the first coordinate of the function $\phi^{-1}$. The given matrix above is the transition matrix between the two bases i.e $A^{-1}$ can take you the other way.
