prove that $S^1$ is smooth submanifold of $\mathbb{R}^2$ using the definition with diffeomorphism I'm tring to prove that the unit circle $$S^1=\{(x_1,x_2)\in\mathbb{R}^2\text{ such that }x_1^2+x_2^2=1\}$$ is an embedded submanifold of $\mathbb{R}^2$ using the following Characterization:
A nonempty subset $M \subset\mathbb{R}^n$ is an m-manifold iff:
For every $p\in M$, there are two open sets $O,W\subset\mathbb{R}^n$ with $0_n\in O$ and $p ∈ M ∩ W$,
and a smooth diffeomorphism $ϕ: O → W$, such that $ϕ(0_n) = p$ and
$$ϕ(O ∩ (\mathbb{R}^m × {0_{n−m}})) = M ∩ W$$.
My attempt:
lets fix $a\in S^1$, there exists a unique $\theta_0\in[0,2\pi)$ such that $a=(cos(\theta_0),sin(\theta_0))$
and let $\phi$ be the diffeomorphism
\begin{array}{cccc}
\phi : & (0,\infty)\times(\theta_0-\pi,\theta_0+\pi) & \longrightarrow & \mathbb{R}^2\backslash D_{\theta_0+\pi}\\
~~ & (r,\theta) & \mapsto &(rcos(\theta),rsin(\theta))  
\end{array}
where $D_{\theta_0+\pi}$ is the half line at the origin with polar angle $\theta_0$
From here i dont know how to proceed. Should i modify this map or use as it is? and who are the two open sets of the charactrization?
 A: The way you constructed your $\phi$ is almost good. Intuitively, the characterization you presented requires $\phi$ to be a diffeomorphism between 2 open sets $W$ - the ambient space around $\mathbb S^1$, and $O$ - a deformed version of $W$ such that $\mathbb S^1 \cap W$ is "squished and flattened" into $\mathbb R \times \left \{ 0 \right \}$ (by $\phi^{-1}$). You could think of $\phi|_{O\cap (\mathbb R \times \left \{ 0 \right \})}$ as a parameterization of the circle using $\theta$.
Your $\phi$ is a good candidate besides the fact there needs to be some alignment so that $0$ is mapped to $a$: substituting $\theta+\theta_0$ into the trig functions instead of $\theta$, and using a radius $1+r$ instead of $r$. This way we get $W=\mathbb R^2 \setminus D_{\theta_0+\pi}$ as you said but
$$O=(-1,\infty)\times(-\pi,\pi)$$
From the naturality of polar coordinates it's pretty easy to convince $\phi$ is bijective, two points can't have the same radius and angle, and every point has a radius and an angle. $\phi$ is also smooth due to its components being elementary functions. Hence, it is a diffeomorphism.
$\phi(O\cap (\mathbb R \times \left \{ 0 \right \}))\subseteq \mathbb S^1 \cap W$ since any point of the form $(0,\theta)$ is mapped to the unit circle at an angle which isn't $\theta_0+\pi$, and $\mathbb S^1 \cap W \subseteq\phi(O\cap (\mathbb R \times \left \{ 0 \right \})) $ using the reverse argument.
