Lee, Introduction to Smooth Manifolds, Change of Coordinates In all versions of John M. Lee's Introduction to Smooth Manifolds, he claims that
$$\left(\psi\circ\varphi^{-1}\right)_*\left.\frac{\partial}{\partial
x^i}\right|_{\varphi\left(p\right)}=\left.\frac{\partial\tilde
x^j}{\partial
x^i}\left(\varphi\left(p\right)\right)\frac{\partial}{\partial\tilde
x^j}\right|_{\psi\left(p\right)},$$
where $\left(U,\varphi\right)=\left(U,\left(x^i\right)\right)$ and $\left(V,\psi\right)=\left(V,\left(\tilde x^i\right)\right)$ are smooth charts of some smooth manifold such that $p\in U\cap V$.
However, this is what I computed:
$$\begin{align}\left(\psi\circ\varphi^{-1}\right)_*\left.\frac{\partial}{\partial x^i}\right|_{\varphi\left(p\right)}&=\psi_*\left(\left(\varphi^{-1}\right)_*\left.\frac{\partial}{\partial x^i}\right|_{\varphi\left(p\right)}\right)\\&=\psi_*\left.\frac{\partial}{\partial x^i}\right|_p\\&=\left.\frac{\partial\tilde x^j}{\partial x^i}\left(\color{red}{p}\right)\frac{\partial}{\partial\tilde x^j}\right|_{\psi\left(p\right)}.\end{align}$$
I highlighted my difference in red. What am I doing wrong?
 A: I think that I discovered the problem: he let
$$\psi\circ\varphi^{-1}\left(x\right)=\left(\tilde x^1\left(x\right),\dots,\tilde x^n\left(x\right)\right)$$
in the middle of his exposition, despite each $\tilde x^i$ already standing for the component functions of $\psi$. In this case, it easily follows that
$$\left(\psi\circ\varphi^{-1}\right)_*\left.\frac{\partial}{\partial
x^i}\right|_{\varphi\left(p\right)}=\left.\frac{\partial\tilde
x^j}{\partial
x^i}\left(\varphi\left(p\right)\right)\frac{\partial}{\partial\tilde
x^j}\right|_{\psi\left(p\right)}.$$
If someone could confirm.
A: Please allow me to introduce another notation. Let $M$ be a smooth manifold with dimension $ \ m \in \mathbb{N}^*$, $U$ and $V$ be open sets, $p \in U \cap V$, $x:U \to \mathbb{R}^m \ $ and $ \ y:V \to \mathbb{R}^m \ $ be charts at $p$ and $ \ i: im(x) \hookrightarrow \mathbb{R}^m \ $ and $ \ j: im(y) \hookrightarrow \mathbb{R}^m \ $ be the inclusion maps. As usual, we denote the coordinate maps $ \ x = (x^1,...,x^m)$, $y= (y^1,...,y^m)$, $i = (i^1,...,i^m) \ $ and $ \ j = (j^1,...,j^m)$. So, for all $ \ \mu \in \{ 1,...,m \}$, the maps $ \ i^{\mu}: im(x) \to \mathbb{R} \ $ and $ \ j^{\mu}: im(y) \to \mathbb{R} \ $ are just restrictions of the projection onto the $\mu$-th coordinate.
Treating tangent vectors as derivations acting on germs, first note that, $\forall h \in C^{\infty} \big( x(p) \big)$, we have
\begin{align*} \displaystyle \left\{ \Big[ d(x^{-1})|_{x(p)} \Big] \left( \left. \frac{\partial}{\partial i^{\mu}} \right|_{x(p)} \right) \right\}(h) & = \left[ \left. \frac{\partial}{\partial i^{\mu}} \right|_{x(p)} \right] (h \circ x^{-1}) = \\ & = [\partial_{\mu} (h \circ x^{-1} \circ i^{-1})] \Big(i \big( x(p) \big) \Big) = \\ & = [\partial_{\mu} (h \circ x^{-1})] \big( x(p) \big) = \\ & = \left( \left. \frac{\partial}{\partial x^{\mu}} \right|_{p} \right) (h) \ \ .
\end{align*}
Then, $\forall \mu \in \{ 1,...,m \}$, we have $$\displaystyle \Big[ d(x^{-1})|_{x(p)} \Big] \left( \left. \frac{\partial}{\partial i^{\mu}} \right|_{x(p)} \right) = \left. \frac{\partial}{\partial x^{\mu}} \right|_{p} \ \ . $$ By the chain rule, $\forall \mu \in \{ 1,...,m \}$, we conclude that
\begin{align*}
\displaystyle \big[ d(y \circ x^{-1})|_{x(p)} \big] \left( \left. \frac{\partial}{\partial i^{\mu}} \right|_{x(p)} \right) & = \Big[ dy|_{x^{-1} (x(p))} \circ d(x^{-1})|_{x(p)} \Big] \left( \left. \frac{\partial}{\partial i^{\mu}} \right|_{x(p)} \right) = \\ & = (dy|_p) \left( \left. \frac{\partial}{\partial x^{\mu}} \right|_{p} \right) = \\ & = \sum_{\nu = 1}^{m} \left[ \left( \left. \frac{\partial}{\partial x^{\mu}} \right|_{p} \right) (j^{\nu} \circ y) \right] \cdot \left. \frac{\partial}{\partial j^{\nu}} \right|_{y(p)} = \\ & = \sum_{\nu = 1}^{m} \left[ \left( \left. \frac{\partial}{\partial x^{\mu}} \right|_{p} \right) (y^{\nu}) \right] \cdot \left. \frac{\partial}{\partial j^{\nu}} \right|_{y(p)} = \\ & = \sum_{\nu = 1}^{m} \frac{\partial y^{\nu}}{\partial x^{\mu}} (p) \cdot \left. \frac{\partial}{\partial j^{\nu}} \right|_{y(p)} \ \ .
\end{align*}
Or we can compute, $\forall \mu \in \{ 1,...,m \}$,
\begin{align*}
\displaystyle \big[ d(y \circ x^{-1})|_{x(p)} \big] \left( \left. \frac{\partial}{\partial i^{\mu}} \right|_{x(p)} \right) & = \sum_{\nu = 1}^{m} \left[ \left( \left. \frac{\partial}{\partial x^{\mu}} \right|_{p} \right) (y^{\nu}) \right] \cdot \left. \frac{\partial}{\partial j^{\nu}} \right|_{y(p)} = \\ & = \sum_{\nu = 1}^{m} [\partial_{\mu} (y^{\nu} \circ x^{-1})] \big( x(p) \big) \cdot \left. \frac{\partial}{\partial j^{\nu}} \right|_{y(p)} = \\ & = \sum_{\nu = 1}^{m} [\partial_{\mu} (y^{\nu} \circ x^{-1} \circ i^{-1})] \Big( i \big( x(p) \big) \Big) \cdot \left. \frac{\partial}{\partial j^{\nu}} \right|_{y(p)} = \\ & = \sum_{\nu = 1}^{m} \left\{ \left[ \left. \frac{\partial}{\partial i^{\mu}} \right|_{x(p)} \right] (y^{\nu} \circ x^{-1}) \right\} \cdot \left. \frac{\partial}{\partial j^{\nu}} \right|_{y(p)} = \\ & = \sum_{\nu = 1}^{m} \left[ \frac{\partial (y^{\nu} \circ x^{-1})}{\partial i^{\mu}} \right] \big( x(p) \big) \cdot \left. \frac{\partial}{\partial j^{\nu}} \right|_{y(p)} \ \ .
\end{align*}
Using that, $\forall \nu \in \{ 1,...,m \}$, $$y^{\nu} \circ x^{-1} = j^{\nu} \circ y \circ x^{-1} = (y \circ x^{-1})^{\nu}$$ is the $\nu$-th coordinate function of the map $y \circ x^{-1}$, we have the desired equality.
