connectedness of set of orthogonal matrices with positive determinant Consider the set $A=\{ \bigl(\begin{smallmatrix} x & -y \\ y & x \end{smallmatrix}\bigr) : x,y\in \mathbb{R}: x^2+y^2=1\}$ is this set connected?  I am trying to find a continuous map from $\mathbb{R}$ to $M_2(\mathbb{R})$ whose image is   the given set .
 A: The set $A$ is even a group and is also called the „second special orthogonal group“, denoted by $\operatorname{SO}(2)$. „Special“ refers to the determinant being $1$ and „orthogonal“ refers to both columns being orthogonal to each other. $\operatorname{SO}(2)$ is homeomorphic (even diffeomorphic) to the circle $S^1=\{x,y\in\mathbb R^2|x^2+y^2=1\}\subset\mathbb R^2$ and therefore also called the „circle group“. See its nLab page for more information. This can be shown by the map:
$$f\colon
S^1\rightarrow A=\operatorname{SO}(2),
(x,y)\mapsto\begin{pmatrix}
x & y \\
-y & x
\end{pmatrix}.$$

*

*$f$ is injective since from $f(x,y)=\begin{pmatrix}
x & -y \\
y & x
\end{pmatrix}=\begin{pmatrix}
x' & -y' \\
y' & x'
\end{pmatrix}=f(x',y')$, we can immediatly deduce $x=x'$ and $y=y'$ by just comparing components.


*$f$ is surjective by definition: That of $f$ gives the first part of $A$ (the structure of its elements) and that of $S^1$ gives the second part of $A$ (the condition imposed on those elements).


*$f$ is continuous as it is continous in all four components. We use, that the projection maps $(x,y)\mapsto x$ and $(x,y)\mapsto y$ are continuous (for all four components) as well as that $y\mapsto -y$ and the composition of continuous maps are continuous (for the upper right component).
You can now do the same for the inverse map given by the projection of the matrix onto the first two components. This would imply, that $f$ is a homeomorphism (bijective, continuous with continous inverse map).
(If you are familiar with topology, you can also use the theorem, that a continous and bijective map from a compact space (like $S^1$) to a Hausdorff space (like $\operatorname{SO}(2)$) is a homeomorphism.)
Since $S^1$ is connected (even path-connected, which implies connected), $A=\operatorname{SO}(2)$ is as well.
