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Here is a part of a lecture note:

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I need some help to solve the exercise.

I want to show that if $\psi\circ f\circ\phi^{-1}$ is differentiable and $\alpha, \psi$ and $\beta,\phi$ are in the same charts, i.e, if $\psi^{-1}\circ\alpha$ and $\beta^{-1}\circ \phi$ are differentiable, then $\alpha\circ f\circ\beta^{-1}$ is differentiable. As $\phi^{-1}\circ \phi$ and $\psi^{-1}\circ\psi$ are differentiable, so is $\psi^{-1}\circ\psi\circ f \circ \phi^{-1}\circ \phi=f$. How do I insert $\alpha$ and $\beta$? Can I say that $\psi\circ\psi^{-1}\circ\psi^{-1}\circ\alpha$ and similarly $\beta^{-1}\circ\psi\circ\psi\circ\psi^{-1}$ are differentiable which means that their composition with $f$ is differentiable?

Xena
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    You are wrong when you say that $f$ is differentiable. This is what you have just defined, it makes no sense to use it again. Instead, write $$\alpha f \beta^{-1}= \alpha \psi^{-1}(\psi f \phi^{-1})\phi \beta^{-1}.$$ – Giuseppe Negro Sep 09 '13 at 16:55
  • How do we know that $\alpha\psi^{-1}$ is differentiable? – Xena Sep 09 '13 at 17:04
  • This is by assumption. Try reviewing the definition of "atlas". – Giuseppe Negro Sep 09 '13 at 17:13
  • In my own lecture notes it says that for two homeomorphisms $\phi_1$ and $\phi_2$ being in the same chart means that $\phi_1^{-1}\phi_2$ is differentiable, is this wrong? – Xena Sep 09 '13 at 17:32
  • Of course not. Note that the roles of $\phi_1$ and $\phi_2$ can be reversed, and so $\phi_2^{-1}\phi_1$ must be differentiable too. – Giuseppe Negro Sep 09 '13 at 17:35
  • Yes, but you put the inverse on the right, like $\alpha\psi^{-1}$, how is this possible? – Xena Sep 09 '13 at 17:42
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    Right, that was not the point. If $\phi_j\colon M\to \mathbb{R}^n$ then the correct assumption is that $\phi_1\phi_2^{-1}$ is differentiable. Indeed, it is this one which maps (an open set of) $\mathbb{R}^n$ onto (an open set of) $\mathbb{R}^n$: $$\phi_1\phi_2^{-1}\colon \mathbb{R}^n \to M \to \mathbb{R}^n.$$ If you put the inverse on the left you have $$\phi_1^{-1}\phi_2\colon M \to \mathbb{R}^n\to M, $$ and you do not have a "built-in" notion of differentiability on $M$, so a priori it makes no sense saying that $\phi_1^{-1}\phi_2$ is differentiable. – Giuseppe Negro Sep 09 '13 at 18:26
  • Makes sense, thank you very much! – Xena Sep 09 '13 at 18:42

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