Dual basis for the coordinate curve

Suppose that $M$ is a smooth manifold, $\alpha:I\to \mathbb{R}^n$ a representation of a curve on $M$ and $\phi:M\to \mathbb{R}^n=(x_1,...,x_n)$ a coordinate chart. So, $\gamma(t)=\phi^{-1}\circ\alpha(t)$ is the $\alpha$-th coordinate curve iff $x_\alpha(\gamma(t))=t+c_\alpha$ and $x_i(\gamma(t))=c_i$ where $c_i$'s are suitable constants.

NOW why it holds that for our $\gamma$ with $\dot{x}_\alpha=\frac{d\gamma}{dt}$ that

$\dot{x_\alpha}(\frac{\partial}{\partial x^i})=\delta^\alpha_i$

? I.e. why the vector $\frac{\partial}{\partial x^\alpha}$ is the velocity vector to the $\alpha$-th coordinate curve?

• What is $\dot{x_{\alpha}}$ ? – Sou Nov 26 '17 at 14:22
• @Sou燈馬想: The derivative of the $x_{\alpha}$ coordinate variable. – Faraad Armwood Nov 26 '17 at 14:22
• How can $\dot{x}_{\alpha}$ act on $\partial_i$ ? I think it should act on functions. – Sou Nov 26 '17 at 14:26

As the user in the comments pointed out, we have only defined the action of tangent vectors on functions $f: M \to \mathbb{R}$. Observe how this doesn't fall into the category for evaluation,
$$(x^{\alpha})' \frac{\partial}{\partial x^i} = x^{\alpha}_* \underbrace{\left( \frac{d}{dt} \right) \frac{\partial}{\partial x^i}}_{\textbf{doesn't make sense}}$$