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In the definition of a smooth (or $C^k$) manifold the charts $\varphi: U \to \mathbb R^n$ are assumed to have the property that for any two of them $\varphi \circ \psi^{-1}$ is smooth ($C^k$).

Does $\varphi \circ \psi^{-1}$ is smooth ($C^k$) imply that both $\varphi$ and $\psi$ are smooth ($C^k$) or is it weaker?

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It wouldn't make sense to say that $\phi$ or $\psi$ is smooth, because $U$ might not even have enough structure to make such a statement meaningful. $U$ might not be a subset of $\mathbb R^m$, for example. Even if $U$ is a subset of $\mathbb R^m$, typically $U$ would not be an open subset of $\mathbb R^m$.

$\phi \circ \psi^{-1}$, on the other hand, maps an open subset of $\mathbb R^n$ into $\mathbb R^n$, so it is perfectly meaningful to say that $\phi \circ \psi^{-1}$ is smooth.

littleO
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    +1. Of course once we have established a smooth manifold structure on $U$, $\phi$ and $\psi$ are smooth by definition - a map from/to a smooth manifold is called smooth if it's smooth when appropriately (pre)composed with charts. – Anthony Carapetis Oct 15 '14 at 09:56
  • Thank you. "typically U would not be an open subset" -- I don't understand. It appears that $U$ is defined to be open? –  Oct 16 '14 at 00:48
  • I just edited my answer to try to be a bit more clear about that. Suppose for example the manifold $M$ is a sphere in $\mathbb R^3$. Then $U$ is an open subset of $M$, in the subspace topology, but $U$ is not an open subset of $\mathbb R^3$. – littleO Oct 16 '14 at 02:31
  • Excellent, thank you. I was confused because I imagined that if you take the northpole of the sphere and an open disk around it (on the sphere) then that disk is also open in $\mathbb R^3$. But that's false. It is neither open nor closed. –  Oct 16 '14 at 04:26