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For $(U,x_1,...x_n),(V,y_1,...y_n)$, we may assume the charts are $(U,\phi),(V,\psi)$ respectively. Then $x_i:=r_i\circ\phi$, $y_i:=r_i\circ\psi$. In this sense, $x_i$ is not related to $y_1,...,y_n$ at all. However, in the https://people.maths.ox.ac.uk/hitchin/hitchinnotes/Differentiable_manifolds/Chapter_3.pdf p.60 the last but one line, it uses $y_i(x_1,...x_n)$. I don't understand why we can see $y_i$ as a function with variables $x_1,...x_n$.

Moreover, on some textbooks, a differential $\omega$ on $U\cap V$, has local expression $fdx_1\wedge...\wedge dx_n$ associated with $(U,x_1,...x_n)$ and has local expression $gdy_1\wedge...\wedge dy_n$ associated with $(V,y_1,...y_n)$, then they claim $\omega=fdx_1\wedge...\wedge dx_n=gdy_1\wedge...\wedge dy_n$ on $U\cap V$. However, they are in different charts, it doesn't make sense to say they are equal, right?

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  • $\begingroup$ I presume $r_i$ is the $i$th coordinate function on $\Bbb R^n$? Note that writing $y=y(x)$ on the intersection of the charts is just looking at $\psi\circ\phi^{-1}$ on the appropriate domain. By definition of a smooth manifold, that will be a smooth function. $\endgroup$ – Ted Shifrin Jan 24 '17 at 20:11
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This is a convention...if you have a chart $(U,\phi=(x_1,x_2,\ldots, x_n))$, where $x_i:U\longrightarrow \mathbb{R}$, then it's very usefull consider a function $f:U\longrightarrow \mathbb{R}$ as a function of variables $x_i$, but really this is not a composition between $f$ and $\phi$. In this same spirit you define $\partial f /\partial x_i:=\partial (f\circ\phi^{-1})/\partial r_i$. Now (in your notattions) if $(U,\phi),(V,\psi)$ are two chart on a point $p\in U\cap V$ then 2 basis of $T_pX$ are given by $\lbrace\partial /\partial x_i|p \rbrace $ and $\lbrace\partial /\partial y_i|p \rbrace$ then you have (do some simple compuations) $\partial /\partial x_i|p=\sum_j (\partial y_j/\partial x_i) \partial /\partial y_j|p$. Now you can use the fact that $$df=\sum_i \partial f/\partial x_i dx_i$$ and the proprerty of the wedge product.

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