Definition. Let $M$ a smooth manifold of dimension $n$. A funciton $f\colon M\to \mathbb{R}$ is said to be smooth at a point $p$ in $M$ if there is a chart $(U,\varphi)$ about $p$ in $M$ such that $f\circ\varphi^{-1}$ is smooth in $\varphi(p)$.
The definition of the smothness of a function $f$ at a point is independet of the chart $(U,\varphi)$, for if $f\circ \varphi^{-1}$ is smooth at $\varphi(p)$ and $(V,\psi)$ is any other chart about $p$ in $M$, then on $\psi(U\cap V)$, $$f\circ\psi^{-1}=(f\circ \varphi^{-1})\circ (\varphi\circ\psi^{-1}).$$
It is clear that $\varphi\circ\psi^{-1}$ is smooth in $\psi(p)$.
Question. Why also $\big(f\circ\varphi^{-1}\big)$ is smooth in $\psi(p)?$
Reference: An Introduction to Manifolds by Loring Tu (Second Edition, page no. 59, Remark 6.2).
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Best Answer
We have
$\varphi : U \to U' \subset \mathbb R^n, f \circ \varphi^{-1} : U' \to \mathbb R$
$\psi : V \to V' \subset \mathbb R^n, f \circ \psi^{-1} : V' \to \mathbb R$
The transition function between $\varphi$ and $\psi$ is then
$\varphi \circ \psi^{-1} : W' = \psi(U \cap V) \to \varphi(U \cap V) = W''.$
$W'$ is an open neigborhood of $\psi(p)$ and $W''$ an open neigborhood of $\varphi(p)$. We get
$(f \circ \psi^{-1}) \mid_{W'} = (f \circ \varphi^{-1}) \mid_{W''} \circ \phantom{.} (\varphi \circ \psi^{-1}).$
Hence if $f \circ \varphi^{-1}$ is smooth in $\varphi(p)$, then so is $(f \circ \varphi^{-1}) \mid_{W''}$. Therefore $(f \circ \psi^{-1}) \mid_{W'}$ is smooth in $\psi(p)$ and the same is true for $f \circ \psi^{-1}$.
This is what has to be shown. The question whether $\big(f\circ\varphi^{-1}\big)$ is smooth in $\psi(p)$ does not make sense. In general $\psi(p)$ is not even contained in $U'$.