Continuous Bijection From $S^1$ to $[0,2\pi)$ I came up with the following (I believe) continuous bijection $\theta: S^1  \rightarrow [0,2\pi)$:
$\begin{align} \theta(x,y) & = 
\begin{cases} 
      \arctan(\frac{y}{x}) & x > 0, 0 \leq y < 1 \\
      \frac{\pi}{2} & x = 0, y = 1 \\
      \arctan(\frac{-x}{y}) + \frac{\pi}{2} & x < 0, 0 < y < 1 \\
      \pi & x = -1, y = 0 \\
      \arctan(\frac{-y}{-x}) + \pi & x < 0, -1 < y < 0 \\
      \frac{3\pi}{2} & x = 0, y = -1 \\
      \arctan(\frac{x}{-y}) + \frac{3\pi}{2} & x > 0, -1 < y < 0 \\
\end{cases}
\end{align}$
This looks ugly though. Is there a more natural way to write a continuous bijection from $S^1$ to $[0,2\pi)$?

you cannot find such a bijection since the image of a compact set by a continuous function is compact, $S^1$ is compact and $[0,2\pi)$ is not compact.
