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There's a naive exercise that I'm having trouble to finish.

Suppose $M$ is a smooth manifold and $f:$ $M$ $\rightarrow$ $\mathbb{R}^{k}$ 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 $M$.

Well, by definition of smooth function, for every point $p$ in M there is a smooth chart $(U,\varphi)$ such that the function above is smooth. My idea is to take, for every $p$ $\in$ $M$, such a chart. Sounds like a good start, but I couldn't find an argument to finish the problem, even because there are charts $(U',\psi)$ such that $p$ $\in$ $U$ $\subset$ $U'$ that I haven't considered in the beginning. If $U'$ $\subset$ $U$, we just have to restrict the domain, but in the first case I couldn't go on. What am I missing?

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Some books do this differently. Is the definition of a smooth function $f:M\to \mathbb{R}^k$: for any point $p$ there exists a smooth chart $(U, \phi)$ such that $f\circ \phi^{-1}$ is smooth? If so, then there is a typo right under the block text because it says "smooth chart". –  Matt Sep 18 '12 at 13:56
    
Hint: Use the fact that the transition maps $\psi\circ\phi^{-1}$ are diffeomorphisms –  Jason DeVito Sep 18 '12 at 13:59
    
Thanks, Matt. I corrected it. –  Br09 Sep 18 '12 at 14:10

1 Answer 1

up vote 3 down vote accepted

Let ($U,\phi)$ be a smooth chart and let $x\in \phi(U)\subseteq \mathbb R^m$. We have to show that $f\circ\phi^{-1}$ is smooth at $x$. Let $p=\phi^{-1}(x)$. Since $f\colon M\to \mathbb R^k$ is smooth, there exists a smooth map $(V,\psi)$, $p\in V$, such that $f\circ\psi^{-1}$ is smooth. Smooth charts are compatible, that is the map $\psi\circ\phi^{-1}\colon\phi(U\cap V)\to\psi(U\cap V)$ is smooth. Then on $\phi(U\cap V)$ we have $f\circ\phi^{-1}=(f\circ\psi^{-1})\circ(\psi\circ\phi^{-1})$ is ths composition of smooth maps, and since $p\in U\cap V$, we see that $f\circ\phi^{-1}$ is smooth at $x=\phi(p)$.

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