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This may seem tautological but if you have a smooth manifold $M$ of dimension $n$ by definition any chart $x:M \supset U\to \mathbb{R}^n$ is also smooth, right? The reasoning is, the differentiable structure of $M$ is determined by the collection of charts.

Bob
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  • what is your question? – user300 Nov 02 '16 at 06:11
  • Is a chart belonging to the atlas of a smooth manifold always smooth? – Bob Nov 02 '16 at 06:12
  • if the chart maps are smooth, the manifold is said to be smooth. there are of course other types of manifold, for example: analytic manifold (in that case maps are needed to be analytic) and like that! – user300 Nov 02 '16 at 06:14

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Yes; all the charts in your smooth atlas are smooth. By definition, a function $f: M \to \mathbb{R}^k$ on a smooth manifold $M$ is smooth if for every point $x \in M$ there is a chart $\varphi: U \to \mathbb{R}^n$, where $x \in U$, such that $f \circ \varphi^{-1}: \varphi(U) \to \mathbb{R}^k$ is smooth. Since $\varphi \circ \varphi^{-1}: \varphi(U) \to \varphi(U)$ is the identity, $\varphi$ is smooth. See Chapter 2 in John Lee's text on Smooth Manifolds for more details.

ಠ_ಠ
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