$f:(0,\pi)\to\{{(x,y)}\space\space\vert \space \space x^2 +y^2 =1, y>0\}$, such that $f(\theta)=(\cos(\theta),\sin(\theta)).$
This $f$ maps an argument $\theta$ ( counter-clockwise rotation ) to a vector of $\Bbb{R^2}$, which lies on the "partial" top half of the unit circle. Continuity of $f$ is clear after we look at it componentwise, and the issue I have is the existence and continuity of the inverse function $f^{-1}$. This is my current idea $$f^{-1}:(x\in(-1,1), y\in(0,1)]^T\to(0,\pi),$$ $$f^{-1}((x,y)^T)=\Big(\cot^{-1}\Big(\frac{x}{y}\Big)\Big).$$
All of this is assuming $f$ is a homeomorphism, which I think it is. How can I find the inverse mapping?
Without invoking the cotangent you can observe that cos is bijective on the domain $(0, \pi)$. So you can use as inverse $(x,y)\mapsto arccos(x)$. This is a continous map, so the proposed $f$ is an homeomorphism.