Exercise 2.3 (Smoothness does not depend on the choice of charts I)

Question

Let $M$ be a smooth manifold with or without boundary, and suppose $f: M \rightarrow \mathbb{R}^k$ is a smooth function. Show that $f \circ \varphi^{-1}: \varphi(U) \rightarrow \mathbb{R}^k$ is smooth for every smooth chart $(U, \varphi)$ for $M$.

Elu Thingol · Accepted Answer

\begin{proof}
Let $f:M 	o \mathbb{R}^k$ be a smooth function and let $(U,\varphi)$ be a smooth chart. Choose some arbitrary $p \in U$. Since $f$ is smooth, we can find a chart $(V_p,\psi)$ containing $p$ such that
\begin{equation} \label{f_is_smooth}
    f \circ \psi^{-1}: \psi(V_p) 	o \mathbb{R}^k
\end{equation}
is smooth. Also, since $(U,\varphi)$ and $(V_p,\psi)$ are smoothly compatible, the transition function
\begin{equation} \label{transition_is_smooth}
    \psi \circ \varphi^{-1}: \varphi(U \cap V_p) 	o \psi(U \cap V_p)
\end{equation}
is smooth as well. Placing both ef{f_is_smooth} and ef{transition_is_smooth} to the common open domain $U \cap V_p$, we see that
\begin{equation*}
    (f \circ \psi^{-1}) \circ (\psi \circ \varphi^{-1}) = f \circ \varphi^{-1}: \varphi(U \cap V_p) 	o \mathbb{R}^k
\end{equation*}
must be smooth. 
ewline

Since our choice of $p\in U$ was arbitrary, we can find such neighbourhood $V_p$ for any point $p$ in $U$; thus, $f \circ \varphi^{-1}$ must be smooth on the whole domain $\bigcup_{p \in U} \varphi(U \cap V_p) = \varphi(U)$. In other words, $f$ is smooth for $(U,\varphi)$.
\end{proof}

Exercise 2.3 (Smoothness does not depend on the choice of charts I)

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Exercise 2.3 (Smoothness does not depend on the choice of charts I)

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