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I need to show that $S^1$ is smooth manifold with the Following structure

$$ U=\{(\cos \alpha, \sin \alpha): \alpha \in(0,2 \pi)\}, \quad V=\{(\cos \alpha, \sin \alpha): \alpha \in(-\pi, \pi)\} . $$ and $$ \begin{aligned} &\varphi: U \rightarrow \mathbb{R}, \quad \varphi(\cos \alpha, \sin \alpha)=\alpha, \quad \alpha \in(0,2 \pi) \\ &\psi: V \rightarrow \mathbb{R}, \quad \psi(\cos \alpha, \sin \alpha)=\alpha, \quad \alpha \in(-\pi, \pi) \end{aligned} $$

I've managed to show that the maps :

$$f:(0;2\pi)\rightarrow S^1 -\{(1;0)\}; \ \alpha\mapsto(\cos(\alpha);\sin(\alpha))$$ and

$$g:(-\pi;\pi)\rightarrow S^1 -\{(-1;0)\}; \ \alpha\mapsto(\cos(\alpha);\sin(\alpha))$$

are bijection; and it is clear that they are continuous (the coordinates functions are continuous) . But how do we show the inverse is also continous??

any hint would be appreciated.

Donnie Darko
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    $S^1 \subset \mathbb{R}^2$ has the subspace topology, so open sets are given by $U \cap S^1$ where $U$ is an open set of $\mathbb{R}^2$. So given an open interval $(a,b)$ can you express $f(a,b) \in S^1$ in this form, hence showing that $(a,b)$ open $\implies f(a,b) = (f^{-1})^{-1}(a,b)$ open i.e. that $f^{-1}$ continuous – Dylan Jun 13 '22 at 11:08
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    The map $\phi : \mathbb R \to S^1, \phi(t) = e^{it} = \cos t + i \sin t$, is open which proves that $f$ and $g$ are open maps. See my answer to https://math.stackexchange.com/q/3873485. – Paul Frost Jun 13 '22 at 12:13
  • thank you all. for @Dylan : is the open set in $\mathbb{R}^2$ is gonna be what is above the line passing through the two points $(cos(a),sin(a))$ and $(cos(b),sin(b))$?? – Donnie Darko Jun 13 '22 at 13:17
  • @PaulFrost Im looking into it , thak your sir. – Donnie Darko Jun 13 '22 at 13:18
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    @DonnieDarko Sorry, I didn't see this notification. Well above is not the word I would use, but the open set in $S^1$ will be the arc of the circle between $(\cos(a),\sin(a))$ and $(\cos(b),\sin(b))$. The open sets are what you would expect, visually they look like arced open intervals of the real line, if that helps. – Dylan Jun 14 '22 at 14:21
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    Although $S^1$ is a smooth manifold, the question has nothing to do with smooth structures. It is a general topology question. I suggest to edit the tags. – Kritiker der Elche Jun 15 '22 at 08:02

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