I am curious about a step on the proof that shows Lie derivative of a vector field is equivalent Lie bracket.
Following comes from Nakahara. We define integral curves by vector field X and Y as following:
$\frac{d\sigma^{\mu}(s,x)}{ds} = X^{\mu}(\sigma(s,x))$ and $\frac{d\tau^{\mu}(t,x)}{dt} = Y^{\mu}(\tau(t,x))$.
Then, we define Lie derivative, $\mathcal{L}_X Y = lim_{\epsilon\to 0}\frac{1}{\epsilon}[(\sigma_{-\epsilon})_{*}Y|_{\sigma_\epsilon(x)} - Y|_x]$.
We know $Y|_{\sigma_\epsilon(x)} \approx [Y^\mu(x)+\epsilon X^\nu(x)\partial_\nu Y^\mu(x)]e_\mu |_{x+\epsilon X}$.
Now, I am not understanding how he is pushing this over with $(\sigma_{-\epsilon})_{*}$. He defines the $\nu$ component of the pushovered vector as $[Y^\mu(x)+\epsilon X^\lambda(x)\partial_\lambda Y^\mu(x)]\partial_\mu[x^\nu - \epsilon X^\nu(x)]e_\nu|_x$.
But I think it should be $[Y^\mu(x)+\epsilon X^\lambda(x)\partial_\lambda Y^\mu(x)]\frac{\partial x^\nu}{\partial (x^\mu + \epsilon X^\mu)}e_\nu|_x$ since $\sigma_\epsilon$ takes $x^\mu \to \approx x^\mu + \epsilon X^\mu$.
(And since pushover is defined by $W^{\alpha} = V^\mu \frac{\partial y^\alpha(x)}{\partial x^\mu}$ when $V = V^\mu \frac{\partial}{\partial x^{\mu}}, f_*V = W^\alpha \frac{\partial}{\partial y^{\alpha}}$ and $f: M \to N$ such that x is the coordinate of M and y is the coordinate of N.)
What am I missing here? Or Am I being completely stupid and not seeing somehow that $\frac{\partial x^\nu}{\partial (x^\mu + \epsilon X^\mu)} = \partial_\mu[x^\nu - \epsilon X^\nu(x)]$?
Thank you.