Picture below is from the Do Carmo's Riemannian Geometry. In my view, the $x$ and $y$ are differentiable homeomorphism. Thus, $x^{-1}\circ y$ and $y^{-1}\circ x$ are differentiable homeomorphism. Why the proof of this book is so complex ? Whether I misunderstand it ?
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1You cannot immediately say ${\bf x}^{-1}\circ {\bf y}$ and its inverse are differentiable--you need to apply the Inverse Function Theorem to conclude this (which is what do Carmo does). – Glare Dec 09 '20 at 04:03
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@Glare Sorry, I am not understand you well. In my view, since $x$ and $y$ are differentiable homeomorphism. Thus, $x^{-1}$ existence and differentiable, so $x^{-1}\circ y$ is differentiable. Why it is wrong ? – Enhao Lan Dec 09 '20 at 05:05
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1You can say $x^{-1}$ exists since $x$ is a homeomorphism, but again, you cannot say $x^{-1}$ is differentiable without the inverse function theorem. A counterexample to what you describe could be the function $x: t\mapsto t^3$, which is a differentiable homeomorphism but does not have differentiable inverse, as $x^{-1}: s\mapsto s^{1/3}$ is not differentiable at $s=0$. – Glare Dec 09 '20 at 05:08
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1@Glare Thanks, I understand. For example, $y=x^3$ is differentiable, but $x=y^{1/3}$ is not differentiable at $y=0$. – Enhao Lan Dec 09 '20 at 05:10
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1@Glare Haha, I think a same example with you . – Enhao Lan Dec 09 '20 at 05:11