I'm studying Differential Geometry using the book "Diferential Geometry of Curves and Surfaces - Manfredo P. do Carmo", and he defines covariant derivative as:
Let $S \subset \mathbb{R}^3$ a surface, $p \in S$ and $\omega:S \rightarrow \bigcup\limits_{p \in S} T_p S$ a field, such that
$$\omega(p) \in T_pS, \hspace{0.1cm} \mbox{for all $p$ $\in S$}. $$
Consider a parametrization $\sigma : U\subset \mathbb{R}^2 \rightarrow V \cap S$, with $\sigma(q) = p$, and a curve $\alpha(t) := \sigma(u(t),v(t))$, satisfying $$\alpha(0) = p $$ $$\alpha'(0) = u'(0) \frac{\partial\sigma}{\partial u}(q) +v'(0) \frac{\partial\sigma}{\partial u} (q) = y \in T_pS. $$
Using the symbols above, we can write the $\omega$ field as follows
$$\omega(\sigma(u,v)) = a(u,v) \frac{\partial\sigma}{\partial u} (u,v) + b(u,v)\frac{\partial\sigma}{\partial v}(u,v), $$ where $b, a: U \rightarrow \mathbb{R}$ $\in \mathcal{C}^{\infty} (U,\mathbb{R}) $
So, we define the covariant derivative of $\omega$ in the diretion $y$ ($D_y \omega (p)$) as:
$$D_y \omega (p) = \pi_{T_pS}\circ \left(\left.\frac{d \omega(\alpha(t))}{dt}\right\rvert_{t=0} \right)$$ $=\left((a\circ\sigma^{-1} \circ \alpha)'(0) + \Gamma_{11}^1(q) a(q) u'(0) + \Gamma_{12}^1 (q) a(q) v'(0) + \Gamma_{12}^{1}(q) b(q) u'(0) + \Gamma_{22}^1(q) b(q) v'(0) \right)\sigma_u(q) + + \left((b\circ\sigma^{-1} \circ \alpha)'(0) + \Gamma_{11}^2(q) a(q) u'(0) + \Gamma_{12}^2(q) a(q) v'(0) + \Gamma_{12}^{2}(q) b(q) u'(0) + \Gamma_{22}^2(q) b(q) v'(0) \right)\sigma_v(q). $
where $\pi_{T_pS}$ is the projection on $T_pS$ and $\Gamma_{ij}^{k}$ are the Christoffel symbols.
Writing the covariant derivative in this form is clear that this definition does not depend of the curve $\alpha$ chosen. But why this derivative does not depend on the parametrization $\sigma$? Can someone help me?