# Differentiable structures on Unit circle same

Let $$\mathbb{S}^1$$ have the manifold structure induced by it being a regular submanifold of $$\mathbb{R}^2$$. Let $$\{$$ $$U_i$$,$$\phi_i$$ $$\}_{i=1}^4$$ be the differentiable structure on $$\mathbb{S}^1$$ where $$U_i$$ are the open semicircles and each $$\phi_i$$ projects $$U_i$$ to an axis.

I would like to show that these smooth structures determine the same smooth structure. To do so, I must show that their union is also a smooth atlas.

Let $$\{$$ $$(V_{\alpha}\cap \mathbb{S}^1,\psi_{\alpha}) \}_{\alpha}$$ be the smooth structure given by the adapted charts. On $$V_{\alpha}\cap \mathbb{S}^1 \cap U_i$$ , we have

$$\phi_{i}(\psi_{\alpha}^{-1}(\psi(p))=\phi_{i}(\psi_{\alpha}(x^1(p))=\phi_{i}(p)=p=\psi_{\alpha}^{-1}(\psi_{\alpha}(p))$$

Moreover, $$\psi_{\alpha}({\phi^{-1}(\phi(p))}=\psi_{\alpha}(p)$$

Similarly for other charts.

Is this correct? How do I fix this?

Let us see which differentiable structure is induced on $$S^1$$ as a submanifold of $$\mathbb R^2$$. Define $$\phi : (-1,1) \times \mathbb R \to (-1,1) \times \mathbb R, \phi(x,y) = (x, y - \sqrt{1-x^2}).$$ This map is a diffeomorphism (with inverse $$\phi^{-1}(x,y) = (x, y + \sqrt{1-x^2})$$) such that
1. $$\phi((-1,1) \times (0,\infty)) = (-1,1) \times [0,\infty) \cup$$ open lower half disk of $$D^2$$ = $$W_1$$.
2. $$\phi(U_1) = (-1,1) \times \{0\}$$, where $$U_1$$ is the open upper half circle of $$S^1$$.
Thus $$\phi : (-1,1) \times (0,\infty) \to W$$ is a chart on $$\mathbb R^2$$ such that $$\phi(S^1 \cap (-1,1) \times (0,\infty)) = W \cap (\mathbb R \times \{0\})$$.
This gives us the chart $$\phi_1 : U_1 \stackrel{\phi}{\to} W \cap (\mathbb R \times \{0\}) \stackrel{proj}{\to} (-1,1)$$ on $$S^1$$ as a submanifold of $$\mathbb R^2$$. You see that this is nothing else than the map in your question. For the other $$U_i$$ you can apply the same construction.