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I've been reading the text Introduction to Manifold by Loring W. Tu and according to his definition,

A smooth or $ C^{\infty}$ is a topological manifold $M$ together with a maximal atlas and a topological manifold is a Hausdorff, second countable and locally Euclidean space.

Now, consider the graph of the function, $y = |x| ~~, -1 < x < 1$ in $\mathbb{R}^2$. The graph of this function is a smooth manifold since being a subset of $\mathbb{R}^2$, it is second countable and Hausdorff, and it also has an atlas consisting of a single chart, $$ \phi(x,y) = x ~~\text{with}~~\phi^{-1}(x) = (x, |x|) ,~~ -1 < x < 1 .$$

Later on in the book, he proved that the coordinate map, $\phi$ is a diffeomorphism which means that its inverse is $C^{\infty}$, but from the example above, $ \phi^{-1}(x)$ is not differentiable at $ x = 0$, how can this be though?? I know I did something wrong to get to this conclusion but I don't know what I did wrong. Thank you very much.

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    $\begingroup$ No, with the atlas it inherits sitting in $\Bbb R^2$ it is not a smooth manifold. $\phi^{-1}$ is obviously not even differentiable, let alone smooth. The differentiable structure Tu wants to give it is a different one. Can you figure that out? $\endgroup$ – Ted Shifrin Dec 25 '16 at 21:58
  • $\begingroup$ @TedShifrin Thank you for the reply. I know that the graph is a topological manifold, is that right? To show that a topological manifold is a smooth manifold according to Tu, we need to show that it has an atlas. So by what you said, I would guess that this topological manifold does not have any atlas since it would fail the compatibility of charts, is that what you meant? $\endgroup$ – Khoa ta Dec 25 '16 at 22:21
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    $\begingroup$ You have not yet given the right atlas. :) In fairness, this is a confusing point for all of us when we begin with this material. $\endgroup$ – Ted Shifrin Dec 25 '16 at 22:26
  • $\begingroup$ @TedShifrin Thank you for the reply. Can you clarify what you meant more? I'm confused now. $\endgroup$ – Khoa ta Dec 25 '16 at 22:34
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    $\begingroup$ We've already agreed that as this curve sits in $\Bbb R^2$ it is not a smooth submanifold but only a topological submanifold. But you said it was homeomorphic to $\Bbb R$. How can you use this latter fact? $\endgroup$ – Ted Shifrin Dec 25 '16 at 22:38

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