Induce Lie algebra isomorphism

Question

In the picture below, how to show the Lemma 2.2.6 ? According to 2.2.9, I can deal RHS of 2.2.12, but LHS of 2.2.12 I don't know how to deal.

Answer 1

(1) $$\psi(p)=p',\ \psi(U)=U'$$ $$(\psi_\ast X) f (p') =df\ \psi_\ast X_p =d(f\circ \psi)\ X_p\ \ast$$

Define $$ F_{X,f} : U'\subset N\rightarrow \mathbb{R},\ F_{X,f}(p'):=(\psi_\ast X) f (p') $$

and $$ G_{X,f} : U\subset M\rightarrow \mathbb{R},\ G_{X,f}(p)=d(f\circ \psi)\ X_p $$

Note that $\ast$ implies that $$ F_{X,f}\circ \psi =G_{X,f} $$

(2) Note that $F_{X,f}$ is a function on $U'$ Hence $$ F_{Y,F_{X,f}} \circ \psi =G_{Y,F_{X,f}} $$

Here $$ F_{Y,F_{X,f}}=\psi_\ast Y\ F_{X,f} (p')= \psi_\ast Y\ \bigg( \psi_\ast X f\bigg) (p') $$

and \begin{align*} G_{Y,F_{X,f}}&=d(F_{X,f}\circ \psi)\ Y_p\\&= dG_{X,f}\ Y_p\\&= \frac{\partial }{\partial t}\bigg|_{t=0} G_{X,f}(C_Y(t,p)) \\&= \frac{\partial }{\partial t} d(f\circ \psi)\ X_{C_Y(t,p)}\\&= \frac{\partial }{\partial t}\frac{\partial }{\partial s}\bigg|_{s=0} (f\circ \psi )\circ C_X(s,C_Y(t,p) ) \end{align*}

where $C_W$ is flow of vector field $W$ on $M$ So we complete the proof by (3)

(3) If $h$ is a function on $M$ then \begin{align*} Xh(p)&=\frac{\partial }{\partial t} h\circ C_X(t,p) \\&\\ YXh&=Y_p (Xh)\\&= \frac{\partial }{\partial t} (Xh)\circ C_Y(t,p)\\&= \frac{\partial }{\partial t} (Xh) \bigg(C_Y(t,p) \bigg)\\&= \frac{\partial }{\partial t}\bigg(\frac{\partial }{\partial s} h\circ C_X (s, C_Y(t,p) ) \bigg) \end{align*}