Showing that a function from the unit sphere to $SO(2)$ is continuous. Consider the usual unit sphere $S^1$ and the group $SO(2).$ There are several equivalent definitions for this two, the ones I am using follows below.
$$ S^1 = \{ x = (x_1,x_2) \in \mathbb R^2\colon x_1 = \cos \theta \, \wedge x_2 = \sin \theta, \, \theta \in [0,2\pi]\}.$$
$$ SO(2)  = \Bigg\{ A \in Mat(2,\mathbb R)\colon A = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}, \, \theta \in [0,2\pi] \Bigg\}.$$
Now, consider the map $f\colon S^1 \to SO(2)$ defined by
$$ f(\cos\theta , \sin \theta) = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}.$$
I am trying to prove that $f$ is a homemorphism between $S^1$ and $SO(2)$ (consequently, both spaces must be homeomorphic). I was able to prove bijectivity and was also able to find the inverse of $f$, $g\colon SO(2) \to S^1$ which is defined by
$$ g\Bigg( \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}\Bigg) = (\cos \theta,\sin \theta).$$
I am having trouble with proving the continuity of $f$ and $g$, thought. I believe this can be done reffering to "coordinate functions" somehow, but I am not really familiar with those.
Any help is apreciatted.
 A: Choose homeomorphisms $\varphi:(0,\pi)\cup(\pi,2\pi)\rightarrow U$ defined by $\varphi(\theta)= (\cos \theta,\sin \theta)$, where the coordinate neighborhood $U=S^1-\{(\pm1,0)\}$, and $\phi:(0,\pi)\cup(\pi,2\pi)\rightarrow V$ defined by $\phi(\theta)=\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}$, where the neoghtborhood $V=\text{SO}(2)-\{\begin{pmatrix} \pm1 & 0 \\ 0 & \pm1 \end{pmatrix}\}$. Then $\phi^{-1}\circ f\circ \varphi(\theta)=\theta$ and $\varphi^{-1}\circ g\circ \phi(\theta)=\theta$ are continuous and thus $f$ and $g$ are continuous on $U$ and $V$, reapectively. Similrly, choose homeomorphisms $\varphi:(-\frac{\pi}{2},\frac{\pi}{2})\cup(\frac{\pi}{2},\frac{3\pi}{2})\rightarrow U'$ defined by $\varphi(\theta)= (\cos \theta,\sin \theta)$, where the coordinate neighborhood $U'=S^1-\{(0,\pm1)\}$, and $\phi:(-\frac{\pi}{2},\frac{\pi}{2})\cup(\frac{\pi}{2},\frac{3\pi}{2})\rightarrow V'$ defined by $\phi(\theta)=\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}$, where the neoghtborhood $V'=\text{SO}(2)-\{\begin{pmatrix} 0 & \pm1 \\ \mp1 & 0 \end{pmatrix}\}$. Then $\phi^{-1}\circ f\circ \varphi(\theta)=\theta$ and $\varphi^{-1}\circ g\circ \phi(\theta)=\theta$ are continuous and thus $f$ and $g$ are continuous on $U'$ and $V'$, respectively. Since $U\cup U'=S^1$ and $V\cup V'=\text{SO}(2)$, these $f$ and $g$ are countinuous.
A: Note, that $S^1$ and $\operatorname{SO}(2)$ are both compact and Hausdorff. This means, that just from $f$ and $g$ being bijective and continuous, you can already conclude, that they are homeomorphisms (which also means you only need one of the maps to prove both spaces are homeomorphic) using the following two theorems:

*

*Continuous maps from compact spaces to Hausdorff spaces are closed (and proper). See here.

*A bijective continuous map, which is open/closed (they imply each other, see here), is a homeomorphism.

Note, that both $S^1$ and $\operatorname{SO}(2)$ are metric spaces, so you can also use the $\varepsilon$-$\delta$-criterium to prove continuity (with which you can reduce it on the continuity of $\sin$ and $\cos$), instead of proving, that preimages of open sets are open.
