$GL^+(2, \mathbb R)$ is homeomorphic to $S^1$ × $\mathbb R^3$ Can anyone give me an idea of how to prove that $GL^+(2)$ is homeomorphic to $S^1$ × $\Bbb R^3$ , where 
$$GL^+(2) = \left\{A =  \begin{bmatrix}a&b\\c&d\end{bmatrix}\bigg| \det A > 0\right\}.$$
Any help will be appreciated, thanks a lot!
 A: $GL^+(2)\cong \Bbb R^+ \times SL(2)$ via the isomorphism $A\longmapsto (\det A, \frac{1}{\sqrt{\det A}}A)$. 
Since $SL(2)\cong \Bbb R^2\times S^1$ you have 
$$GL(2)\cong \Bbb R^+\times \Bbb R^2 \times S^1 \cong \Bbb R^3 \times S^1$$
where $\cong$ means homeomorphic.
EDIT: Why $SL(2)\cong \Bbb R^2\times S^1$? This is a non trivial fact known as Iwasawa decomposition; which state that:

Let $K= \left\{\begin{pmatrix}
\cos\theta & -\sin\theta\\
\sin\theta & \cos\theta\
\end{pmatrix}\right\}$; $\quad A=\left\{\begin{pmatrix}
r & 0\\
0 & \frac{1}{r}\
\end{pmatrix}: r>0\right\}$; $\quad N=\left\{ \begin{pmatrix}
1 & x\\
0 & 1\
\end{pmatrix}\right\}$.
Then $SL(2)=KAN$: every matrix $U\in SL(2)$ has a unique decomposition as $U=kan$, where $k\in K$, $a\in A$ $n\in N$.

It's easy to see that $K,A,N$ are one dimensional subspace of $M_2(\Bbb R)$, and that $K\cong S^1$, $A\cong \Bbb R^+\cong \Bbb R$ and $N\cong \Bbb R$. Thus $SL(2)$ is a $3$-dimensional subspace of $M_2(\Bbb R)$ homeomorphic to $\Bbb R^2\times S^1$.
A: We use Gram-Schmidt. $GL^+(2)$ can be seen as the set of linearly independent $v_1,v_2\in\mathbb R^2$ such that $\det(v_1\ v_2)>0$. Let $Orth_2$ be the set of orthogonal pairs $(v_1,v_2)$ with $\det(v_1\ v_2)>0$.
Gram-Schmidt gives the homeomorphism:
$$GL^+(2)\xrightarrow\sim Orth_2\times\mathbb R:(v_1,v_2)\mapsto ((v_1,v_2-\mathrm{proj}_{v_1}v_2),\langle v_1,v_2\rangle).$$
Now, there is also a homomorphism
$$SO(2)\times\mathbb R^2\xrightarrow{\sim}Orth_2:((v_1,v_2),(x,y))\mapsto (e^xv_1,e^yv_2),$$
where $SO(2)$ is the special orthogonal group. It is well-known that $SO(2)=S^1$, so that $GL^+(2)\simeq S^1\times\mathbb R^3$.
The advantage of this approach is that it generalizes readily to any other dimension:
$$GL^+(n)\simeq Orth_n\times\mathbb R^{n(n-1)/2}\simeq SO(n)\times\mathbb R^{n(n+1)/2}.$$
