Comparing the smooth structure of a manifold $M\subset\mathbb{R}^d$ as a submanifold, and the structure obtained by homeomorphisms $M\to\mathbb{R}^k$.

Motivation Consider $$S^1=\{x \in \mathbb{R}^2: \left|x\right|^2 = 1\}$$ and $$\mathbb{R}/\mathbb{Z}$$ as smooth manifolds, when $$S^1$$ is considered as a smooth submanifold of $$\mathbb{R}^2$$ (here, with the natural smooth sructure of $$\mathbb{R}^2$$) and $$\mathbb{R}/\mathbb{Z}$$ has its smooth structure defined by the atlas obtained from restricting the projection map $$p: \mathbb{R} \to \mathbb{R}/\mathbb{Z}$$ to interval of length half.

Now, suppose we want to show that $$S^1$$ and $$\mathbb{R}/\mathbb{Z}$$ are diffeomorphic. We can take \begin{align} f: \ \mathbb{R}/\mathbb{Z} &\longrightarrow S^1\\ [x] & \longmapsto (\cos(2\pi x), \sin(2\pi x)) \end{align} and show that it is smooth in both directions.

The natural way to show this will be to consider another atlas on $$S^1$$, For example, the one obtained by pairs $$(U_\alpha, \varphi_\alpha)$$ where $$U_\alpha$$ are image of invervals of the form $$(\alpha, \alpha + 2 \pi) \subset \mathbb{R}$$ under the map $$\gamma: t \mapsto (\cos t, \sin t)$$, and $$\varphi_\alpha = (\gamma |_{(\alpha, \alpha + 2 \pi)})^{-1}$$ are inverses of the restriction of $$\gamma$$.

Problem \ Question: The atlas that makes it easy to show the diffeomorphism between the two manifolds is not necessarily equivalent to the structure of $$S^1$$ taken as a submanifold of $$\mathbb{R}^2$$. How can we (easily) know that the smooth structure obtained on $$S^1$$ as a submanifold of $$\mathbb{R}^2$$ is equivalent to the one obtained by the atlas $$\{(U_\alpha,\varphi_\alpha)\}_{\alpha \in \mathbb{R}}$$?

As a submanifold of $$\mathbb{R^2}$$, the smooth structure on $$S^1$$ is defined by diffeomorphisms $$\psi: W \to Q$$ where $$W,Q \subset \mathbb{R}^2$$ and $$\psi(W \cap S^1) \subset \mathbb{R}\times\{0\}$$. Then, the atlas considered on $$S^1$$ is obtained by charts of the form $$(\psi |_{W \cap S^1}, W \cap S^1)$$. In our case, the existence of such diffeomorphisms (of $$\mathbb{R}^2$$) can be established by the theorem about regular zero sets of smooth functions (i.e. $$F(x_1,x_2) = x_1^2+x_2^2 -1$$ and $$S^1 = F^1(\{0\})$$).

In this specific case, it's fairly easy to show by computation that transition maps between some $$(\psi |_{W \cup S^1}, W \cup S^1)$$ and $$(\varphi_\alpha, U_\alpha)$$ is smooth, but it can become a mess in more complicated cases.

Stated more generally: Is there a simple way to show that an atlas obtained from a 'natural' homeomorphisms from open subsets $$\mathbb{R}^k$$ to subsets of some $$k$$-dimensional manifold $$M \subset \mathbb{R}^d$$, establishes on $$M$$ the a smooth structure that is the same as the one it inherits from $$\mathbb{R}^d$$?

• This sounds like a possible application of invariance of domain, but I am not entirely sure. – Prototank Oct 25 '18 at 13:31