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I'm trying to prove the statement:

Given a smooth atlas $\{(U_{\alpha},\phi_{\alpha})\}$, if $U$ is an open subset of $U_a$ for some $(U_a,\phi_a)$ in the atlas, then $U$ is smooth compatible with the charts in the atlas.

My proof goes like: Let $(V,\psi)$ be a chart in the atlas. Suppose $U\cap V$ is nonempty. By assumption, $U_a$ and $V$ are compatible. Therefore, the maps \begin{align*} \phi_a \circ\psi^{-1}:\psi(U_a\cap V) \longrightarrow \phi_a(U_a\cap V),\\ \psi \circ\phi_a^{-1}:\phi_a(U_a\cap V) \longrightarrow \psi_a(U_a\cap V). \end{align*} are smooth. In particular, \begin{align*} \phi_a \circ\psi^{-1}:\psi(U\cap V) \longrightarrow \phi_a(U\cap V),\\ \psi \circ\phi_a^{-1}:\phi_a(U\cap V) \longrightarrow \psi_a(U\cap V). \end{align*} are smooth. Since $U$ is open, $\phi_a$ being a homeomorphism on $U_a$ is also homeomorphisms from $U$ onto some open subset of Euclidean space. Therefore, $(U,\phi_a)$ and $(V,\psi)$ are smooth compatible.

I'm not sure if the arguments after "In particular" are valid. Can someone check if this is true? Thank you.

123
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    I know what it means for a chart to be smoothly compatible with charts from an atlas, but you're asking about a subset. In place of the second occurrence of $U$, I believe you mean $(U, \varphi_{\alpha}|_U)$. – Michael Albanese May 07 '22 at 11:14
  • @MichaelAlbanese yes, exactly. I meant $(U,\phi_a|_U)$. – 123 May 07 '22 at 11:40

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