I am trying to understand the notes here: http://unapologetic.wordpress.com/2011/04/13/cotangent-vectors-differentials-and-the-cotangent-bundle/.
Specifically, this sentence: If we have local coordinates $(U,x)$ at $p$, then each coordinate function $x^i$ is a smooth function...
Question: what is the smooth function $x^i$?
My understanding is that if we have a manifold $M$, a point $p \in M$, and a chart or "local coordinate" $(U,x)$ at $p$, then a smooth function $f$ is a map from $M$ to $\mathbb{R}^n$.
Am I right that $x^i: M \to \mathbb{R}^n$ too, but it outputs the $i$-th standard basis vector? i.e. $x^i(p) = e_i$?
I think if $x: M \mapsto \mathbb R^n$ such that $x(p)= q$, then $x^i(p) = q^i$. The coordinate function $x^i$ returns the $i$th coordinate of the vector $q$.