Continuous Bijection From $S^1$ to $[0,2\pi)$

Question

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 $[0,2\pi)$?

Answer 1

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.

Answer 2

A continuous bijection from a compact space to a Hausdorff space is a homeomorphism and so no such function can exist. These spaces are clearly not homeomorphic by the cut-point properties of the spaces - $S^1$ has no cut-points, $[0,2\pi)$ has every point except one ($0$) being a cut-point.