Let $M$ be a smooth manifold of dimension $n$ and let $p \in M$. Choose a smooth chart $(U, \phi)$ around $p$ and then we have $\phi : U \to \widehat{U} = \phi[U]$ to be a diffeomorphism with
$\phi(a) = (x^1(a), \dots, x^n(a))$ with $x^i : U \to \mathbb{R}$ being the component functions.
Now we have the coordinate repsentation of $x^j$ to be $\widehat{x^j} = x^j \circ \phi^{-1} : \widehat{U} \to \mathbb{R}$, but does $\widehat{x^j}(a^1, \dots, a^n) = a^j$ for all $(a^1, \dots, a^n) \in \widehat{U}$?
I was told that this is the case, but I'm having some trouble seeing how to show this since $\widehat{x^j}(a^1, \dots, a^n) = x^j(\phi^{-1}(a^1, \dots, a^n))$ and since $x^j$ could be any (smooth) function I don't see how I could show the above.
Best Answer
I believe that definitions might differ in different textbooks, but here's my understanding of the situation. I will try to be consistent with your notations as much as I can.
You have an open set $U\subseteq \mathbb{R}^n$ and a homeomorphism $\phi: U \to \widehat{U}=\phi[U]$ where $p\in\widehat{U} \subseteq M$ is an open set in $M$. Now, we have some standard functions on $\mathbb{R}^n$ that I'm going to call them "coordinate functions" that are defined as follows: $$x^i:U\to \mathbb{R}$$ $$x^i(u^1,\cdots,u^n)=u^i$$
I can define a similar set of coordinate functions on my manifold using $\phi$ and I'm going to call them $\widehat{x^i}$ by defining $\widehat{x^i}=x^i \circ \phi^{-1}$. That's all that there's to it. Nothing more.
Now if $p \in M$, then $M$ has some coordinates like $(a^1,\cdots,a^n)$ which is available after introducing $\phi$. More precisely, $\phi^{-1}(p)=(a^1,\cdots,a^n)$
In our notation, $\widehat{x^j}(p)=x^j\circ\phi^{-1}(p)=x^j(a^1,\cdots,a^n)=a^j$ which is what you wanted to show, I believe.
Your confusion probably stems from the fact that you're assuming that the points of the manifold $M$ are $n$-tuples, ignoring the fact that you need to first introduce a parametrization (i.e. a homeomorphism $\phi$) to refer to the points of $M$ like that. I hope everything is clear now.