The shapes of GL(2,R), SL(2,R) and {det < 0}

Idea. In the coordinates $(z,w)$ the three sets are $\{|z|>|w|\}$, $\{|z|^2=1+|w|^2\}$, $\{|w|>|z|\}$. In each of them one of $z,w$ never vanishes, so polar coordinates split off a circle.

1. $\{|z|>|w|\}$. Here $z\ne0$. The map
$$\Phi:S^1\times\{(\rho,w)\in\mathbb{R}\times\mathbb{C}:\rho>|w|\}\to\{|z|>|w|\},\qquad(e^{i\alpha},\rho,w)\mapsto(\rho e^{i\alpha},w)$$
is continuous, bijective and has the continuous inverse $(z,w)\mapsto(z/|z|,|z|,w)$. So it is a homeomorphism. The second factor $K=\{(\rho,w):\rho>|w|\}$ is homeomorphic to $\mathbb{R}^3$: the map $(\rho,w)\mapsto(\ln\rho,\,w/\rho)$ is a homeomorphism from $K$ onto $\mathbb{R}\times D$ with $D=\{u\in\mathbb{C}:|u|<1\}$ (inverse $(s,u)\mapsto(e^s,e^su)$), and $D\cong\mathbb{C}\cong\mathbb{R}^2$ through $u\mapsto u/(1-|u|)$ (inverse $v\mapsto v/(1+|v|)$). Therefore $GL_2^+\cong S^1\times\mathbb{R}^3$.

2. $SL_2$. $\det A=1$ means $|z|^2=1+|w|^2$, so $|z|=\sqrt{1+|w|^2}>0$. The map
$$S^1\times\mathbb{C}\to SL_2(\mathbb{R}),\qquad(e^{i\alpha},w)\mapsto\bigl(\sqrt{1+|w|^2}\,e^{i\alpha},\,w\bigr)$$
is continuous and bijective, with continuous inverse $(z,w)\mapsto(z/|z|,w)$. So $SL_2(\mathbb{R})\cong S^1\times\mathbb{C}=S^1\times\mathbb{R}^2$.

3. $\{\det<0\}=\{|w|>|z|\}$. Now $w\ne0$ and the same proof as in 1 with the roles of $z$ and $w$ exchanged gives
$$(e^{i\beta},\rho,z)\mapsto(z,\rho e^{i\beta}),\qquad S^1\times\{(\rho,z):\rho>|z|\}\to\{|w|>|z|\},$$
a homeomorphism onto the set, and the factor is again $\cong\mathbb{R}^3$.

Check. The circle factor is the obvious one: in $GL_2^+$ the circle $\{\rho e^{i\alpha},0\}$ is $\rho SO(2)$ (the matrices $\rho R(\alpha)$), and $SO(2)\subset GL_2^+$ is a deformation retract (polar decomposition).

Remark. $GL_2(\mathbb{R})$ has two components, $GL_2^+$ and $\{\det<0\}$, each homotopy equivalent to a circle, and they are exchanged by $A\mapsto\mathrm{diag}(1,-1)A$. In particular $\pi_1(SL_2(\mathbb{R}))\cong\mathbb{Z}$. By contrast, $SO(2)$ itself corresponds to the circle $\{|z|=1,w=0\}$.

Lesson. Choosing coordinates in which a defining condition becomes $|z|\gtrless|w|$ or $|z|^2=1+|w|^2$ makes the topology of matrix groups visible.