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I'm having trouble with the proof of the following remark from page 59 of Tu's book on Manifolds. The part I'm worried about is where he gets that $\phi\circ \psi^{-1}$ is $C^\infty$. Is he allowed to make that assumption? I thought we could only say this if the two charts were $C^\infty$-compatible.

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Other relevant definitions from his book:

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justin
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1 Answers1

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A smooth manifold is a $C^{\infty}$ manifold by definition and the transition functions of a $C^{\infty}$ manifold are supposed to be $C^{\infty}$ by definition.

Tsemo Aristide
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  • He defines a smooth/$C^\infty$ manifold as being one with a maximal atlas. Does it follow from the presence of a maximal atlas that all of the transition functions are $C^\infty$ compatible? I guess this is probably a corollary. – justin Feb 23 '16 at 01:40
  • Not from the maximal atlas. Generally, $M$ is a $C^k$ manifold if and only if there exists an atlas $(U_i,\phi_i)$ such that $\phi_i\circ {\phi_j}^{-1}$ is a differentiable map of class $C^k$, in particular, if $k=+\infty$, $M$ is a smooth manifold. – Tsemo Aristide Feb 23 '16 at 01:43
  • Okay. I added in pictures of his definitions to my question. – justin Feb 23 '16 at 01:50
  • Pairwise $C^{\infty}$-compatible atlas means that $\phi_{\alpha}\circ {\phi_{\beta}}^{-1}$ is a $C^{\infty}$ map. – Tsemo Aristide Feb 23 '16 at 01:53
  • Okay. What I was worried about was what if $(U,\phi)$ and $(V,\psi)$ are not compatible with the maximal $C^\infty$ atlas? I mean all I know is that the charts in the atlas are pairwise $C^\infty$ compatible, but might they not be compatible with other charts on $M$? – justin Feb 23 '16 at 01:59
  • By definition a chart is compatible with all the charts – Tsemo Aristide Feb 23 '16 at 02:01
  • The relevant discussion is here: https://math.stackexchange.com/a/3660188/97129 . – rainman May 26 '20 at 02:56