How to show that the Lie derivative $L_{Y}Z$ is equivalent to the Lie bracket $[Y, Z]$? How can you show that, if $Y,Z \in \Gamma(TM)$ and $Y$ is complete, then: $$L_{Y}Z=\frac{\text{d}}{\text{d}t}\phi^{-1}_{t*}(Z)\bigg\rvert_{t=0}\equiv[Y, Z]$$
Where $\phi_{t}$ is in the one parameter group generated by $Y$. I know it should be something related to using charts to get it "down" to $\mathbb{R}^n$ but I don't see how to do that with $\phi^{-1}_{t*}$.
 A: We assume to be on a smooth manifold $M$. Let $X,Y$ be two vector fields on $M$ and $\varphi_t$ be the one-parameter group generated by $X$. Our claim is
$$[X,Y]_p = \lim_{t \to 0^+} \frac{(\varphi_{-t})_{*|\varphi_t(p)}Y_{\varphi_t(p)}-Y_p}{t},$$
so essentially you want to show that the vectors on both sides are equal. Two vectors at $p$ coincide when the directional derivatives of each smooth function defined around $p$ with respect to both of them are equal. So take a function $f$ defined around $p$ and calculate 
$$((\varphi_{-t})_{*|\varphi_t(p)}Y_{\varphi_t(p)}-Y_p)f,$$
where here $t$ is small enough that $\varphi_t(p)$ sits in the same chart as $p$ - you can assume this because you will take the limit $t \to 0^+$. You have
$$
((\varphi_{-t})_{*|\varphi_t(p)}Y_{\varphi_t(p)}-Y_p)f = Y_{\varphi_{t}(p)}(f\circ \varphi_{-t})-Y_pf,
$$
so one first step is to write $f \circ \varphi_{-t}$ in a simpler way. Consider the function $F(t,\cdot):= f\circ \varphi_{-t}-f$. You can see that $F(0,p)=0$ and a result from analysis tells you that there exists a function $g=g(t,q)$ such that $f \circ \varphi_{-t} - f = tg$. Such a function is defined as
$$g(t,p) = \int_0^1 \frac{\partial F}{\partial t}(st,p)\,ds.$$
Also, 
$$g = \frac{f\circ \varphi_{-t} - f}{t}$$
so when $t \to 0^+$ you get $g(0,p) = -X_pf$. All of this is good because now $f\circ \varphi_{-t} = f+tg$ and then
\begin{align}
Y_{\varphi_{t}(p)}(f\circ \varphi_{-t}) -Y_pf&= Y_{\varphi_{t}(p)}(f+tg) -Y_pf\\
& = Y_{\varphi_t(p)}f-Y_pf+tY_{\varphi_t(p)}g.
\end{align}
Divide by $t$:
\begin{align}
\frac{Y_{\varphi_{t}(p)}(f\circ \varphi_{-t})-Y_pf}{t}&=\frac{Y_{\varphi_t(p)}f-Y_pf+tY_{\varphi_t(p)}g}{t} \\
& = \frac{Y_{\varphi_t(p)}f-Y_pf}{t}+Y_{\varphi_t(p)}g.
\end{align}
Now as $t \to 0^+$ the first term goes to $X_p(Yf)$, whereas the second one goes to $-Y_p(Xf)$, hence your limit is
$$X_p(Yf)-Y_p(Xf) = [X,Y]_pf$$
which shows the claim.
A: HINT: Apply $\dfrac d{dt}\Big|_{t=0} \phi_{-t*}Z$ to a smooth function $f$ and use $Yf = \dfrac d{dt}\Big|_{t=0} f\circ\phi_t$. The tricky part will be differentiating $Z_{\phi_t(p)}(f\circ\phi_{-t})$; perhaps the easier way to do this is to think about differentiating $Z_{\phi_s(p)}(f\circ\phi_{-t})$ and then applying the chain rule. 
Comment: No need to do anything in charts.
