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Currently I am reading Tu's book 'An introduction to Manifolds', and I have some problem about the Proposition 6.10 in the book.

I have already known that $f(x): x \mapsto x^3$ is not diffeomorphism since its inverse mapping is not differentiable when $x=0$. The Proposition 6.10 in the book said that

If $\left(U, \phi\right)$ is a chart on a manifold $M$ of dimension n, then the coordinate map $\phi : U → \phi\left(u\right) ⊂ R^n$ is a diffeomorphism

If I take $U$ as $R$ and $\phi(x)=x^3$ to make a chart $(R,x^3)$ on manifold $R$, then according to this proposition $\phi(x)=x^3$ is diffeomorphism. What is my mistake here?

Tanblade
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    $\mathbb{R},x^3$ is not a chart – julio_es_sui_glace Jun 26 '23 at 14:10
  • @SomeCallMeTim try to show that it is differentiable rather than asking why it isn't (it is not) – julio_es_sui_glace Jun 26 '23 at 14:11
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    Also what is your definition of chart? – julio_es_sui_glace Jun 26 '23 at 14:12
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    @JulesBesson Not so fast. $R$ is not $\Bbb R$ with the standard differentiable structure. – Ted Shifrin Jun 26 '23 at 15:12
  • Ok then I understand his mistake better but he did not define it properly – julio_es_sui_glace Jun 26 '23 at 15:13
  • $R$ does not have the same differential structure as $\mathbb{R}$ that is why it is a diffeomorphism (between two different manifolds) but not an homeomorphism! By setting such an $R$ you basically force $x^3$ to become diffeomorphic. – julio_es_sui_glace Jun 26 '23 at 15:15
  • @JulesBesson. Thanks for your comments. I am still confused why $(R, x^3)$ is not a chart. The Definition 5.1 in the book said that if there is a homeomorphism $\phi$ from U onto an open subset of $R^n$ then we can call $(U, \phi)$ a chart. The $x^3$ is homeomorphism, why $(U, X^3)$ is not a chart? – Tanblade Jun 26 '23 at 17:59
  • It is a chart but not for $\mathbb{R}$, if you set it to be a chart (which is what you do when you define formally a manifold), you create a second manifold $R$ but it has not the same structure as $\mathbb{R}$. What I meant to say is that $(\mathbb{R},x^3)$ is not a chart, but $(R,x^3)$ is – julio_es_sui_glace Jun 26 '23 at 18:34
  • check this https://math.stackexchange.com/questions/680202/finding-a-diffeomorphism-between-two-smooth-structures-of-bbb-r?rq=1 – julio_es_sui_glace Jun 26 '23 at 18:37
  • @JulesBesson When I consider $(R, x^3)$ as a chart, the mapping $x^3$ is actually defined on another manifold $R_1$ rather than the original manifold $R$. Do I understand correctly? And is there a method to avoid such mistake? – Tanblade Jun 26 '23 at 19:12
  • Tis is a classic case of snake biting his own tail. Always verify that you did not assumed the answer before you proved it, but it is hard to tell, just be rigorous – julio_es_sui_glace Jun 26 '23 at 19:38

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