Coordinate vector and curve

Question

How can I see that the $j$-th coordinate vector $$\frac{\partial}{\partial x^j}$$ on a manifold is the velocity vector to the $j$-th coordinate curve parametrized by $x^j$. Both, by velocity and direction ?

Answer 1

Because if $p \in M$ and $(U,x(\cdot))$ is a chart on the manifold containing the point $p$, then the tangent vector $\dfrac{\partial}{\partial x_j}\bigg|_p \in T_pM$ is equal to the equivalence class of curves $[t \mapsto x^{-1}(x(p) + t e_j)]$, in other words, in $\Bbb{R}^n$ you consider the straight line $x(p)+ t e_j$ parallel to the $j^{th}$ coordinate axis, and then you "lift it back up" to the get a curve in manifold using $x^{-1}$.