$M= \{ A \in Mat_{2 \times 2}{\mathbb{R}}| \det(A)=1 \}$ is homeomorphic to $S^{1} \times \mathbb{R}^{2}$ Let's consider a group $M$ (under multiplication) of all matrices $A$ of size $2 \times 2$ over $\mathbb{R}$ so that $\det(A)=1$. How to show that the group is homeomorphic to the $S^{1} \times \mathbb{R^{2}}$?

Topology on $M$ is induced by the norm $||A||=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}}$, where $A = \begin{pmatrix}
x_{1} &amp;&amp; x_{2}\\
x_{3} &amp;&amp; x_{4}\\
\end{pmatrix}$
so it's the same as considering matrix as a point in $\mathbb{R^{4}}$. According to $Q = S^{1} \times \mathbb{R}^{2}$,, the topology on $Q$ is induced by the standart one from $\mathbb{R}^{4}$. 

The common idea is to start with considering the $x_{1} x_{4} - x_{2} x_{3}=1$, but i can not find some rigorous ways how to conclude that it's precisely homeomorphic to $Q$.

Any help would be much appreciated.
The group $M$, which I am going to rename $G$, acts transitively on $\mathbb{R}^2 \setminus \{(0, 0)\}$ which is clearly homeomorphic to $\mathbb{R} \times S^1$. The stabilizer of the vector 
$$
\begin{pmatrix} 1 \\ 0 \end{pmatrix}
$$
is the set of matrices of the form
$$
\begin{pmatrix} 1 &amp; * \\ 0 &amp; 1 \end{pmatrix}
$$
which is clearly homeomorphic to $\mathbb{R}$. 

If you've seen group actions, that's a complete answer. 

EDIT: I translated this into a language not using group actions in my previous answer, but made quite a mistake (the equations I wrote always admit a trivial solution).
This is not true !
Let $x_1=a+b$, $x_2=u+v$, $x_3=u-v$ and $x_4=a-b$. Then, $a^2-b^2-u^2+v^2=1$. Let $b=r\cos t$, $u=r\sin t$, then $a^2+v^2=1+r^2$. Let $a=\sqrt{1+r^2}\cos s$ and $v=\sqrt{1+r^2}\sin s$. Therefore, $\mathrm{SL}(2,\mathbb{R})$ can be parametrized by $t,s\in\mathbb{S}^1$ and $r\in\mathbb{R}^+$, and is diffeomorphic to $\mathbb{S}^1\times\mathbb{S}^1\times\mathbb{R}^+$.