Let $\text{GL}_2(\mathbb{R})$ be the set of real $2\times 2$ invertible matrices (where the operation is matrix multiplication). It has $\text{SL}_2(\mathbb{R})=\{ A \in \text{GL}_2(\mathbb{R}| \det A = 1)$ as a normal subgroup. Similarly, let $\text{gl}_2(\mathbb{R})$ be the set of real $2\times 2$ matrices (where the operation is matrix addition). It has $\text{sl}_2(\mathbb{R})=\{ A \in \text{gl}_2(\mathbb{R}| \text{tr} A = 0)$ as a normal subgroup. It is standard to show that $\text{GL}_2(\mathbb{R})/\text{SL}_2(\mathbb{R})\cong\mathbb{R}^{\times}$ and $\text{gl}_2(\mathbb{R})/\text{sl}_2(\mathbb{R})\cong\mathbb{R}^+$.
Recently I saw that the surjective homomorphism $\text{gl}_2(\mathbb{R})\to\text{sl}_2(\mathbb{R})$ given by
$$\begin{pmatrix} a & b \\ c & d \end{pmatrix}\mapsto
\begin{pmatrix} a – d & b \\ c & d – a \end{pmatrix}$$ has a kernel isomorphic to $\mathbb{R}^+$, allowing us to write $\text{gl}_2(\mathbb{R})/\mathbb{R}^+\cong\text{sl}_2(\mathbb{R})$.
All the above begs the question: is there a surjective homomorphism $\text{GL}_2(\mathbb{R})\to\text{SL}_2(\mathbb{R})$ with kernel isomorphic to $\mathbb{R}^{\times}$?
Note: the map $\begin{pmatrix} a & b \\ c & d \end{pmatrix}\mapsto \frac{1}{ad-bc}\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ does not work.
Best Answer
$\newcommand\SL{\text{SL}_2(\mathbb{R})} \newcommand\SLpm{\text{SL}^{\pm}_2(\mathbb{R})} \newcommand\ZS{\{\pm 1\}} \newcommand\ZG{{\mathcal Z}} \newcommand\GL{\text{GL}_2(\mathbb{R})} \newcommand\PSL{\text{PSL}_2(\mathbb{R})} \newcommand\Rx{\mathbb{R}^{\times}} \newcommand\Rp{\mathbb{R}^{+}}$
The answer is no. The answer is no even if one ignores the requirement that the kernel be isomorphic to $\Rx$. In fact we have:
Theorem. There is no surjective homomorphism (continuous or otherwise) from $\GL$ to $\SL$.
In order to prove it, let us begin with an auxiliary result, certainly well known to specialists.
Lemma. Any normal subgroup $H\trianglelefteq \GL$ either contains $\SL$ or is contained in the center of $\GL$, which we denote by $\ZG$ (and which is isomorphic to $\Rx$).
Proof. Clearly $H\cap \SL$ is a normal subgroup of $\SL$. On the other hand we know that $\SL$ has exactly one nontrivial normal subgroup, namely $\ZS $. Thus, either $H$ contains $\SL$, or $$ H\cap \SL\subseteq \ZS. $$ In the latter case we will prove that $H\subseteq \ZG$. To see this, pick $h$ in $H$, and observe that for every $g$ in $\SL$, one has that $$ hgh^{-1}g^{-1}\in H\cap \SL\subseteq \ZS, $$ by virtue of both $H$ and $\SL$ being normal. This means that the map $g\mapsto hgh^{-1}g^{-1}$ maps $\SL$ into $\ZS$, but since the former is connected, the range of this map must be $\{1\}$, meaning that $h$ commutes with $\SL$. Observing further that $\GL$ is generated by $\SL\cup \ZG$, we see that $h$ commutes with $\GL$, whence $h\in \ZG$, as required. QED
Proof (of the Theorem). Suppose, by contradiction, that there exists a surjective homomorphism $$ \varphi :\GL\to \SL, $$ and let $H$ be the kernel of $\varphi $, hence a normal subgroup of $\GL$. Considering the two alternatives of the Lemma, it is clear that $H$ doesn't contain $\SL$, or otherwise the quotient $\GL/H$ would be commutative. Hence $H\subseteq \ZG$.
Observe that $\ZG/H \simeq \varphi (\ZG)$ is contained in the center of $\SL$, namely $\ZS$. This implies that the index of $H$ in $\ZG$ is at most 2.
Should $[\ZG:H]=1$, we would necessarily have that $H=\ZG$, in which case $$ \SL\simeq \GL/\ZG \simeq \SL/\ZS=\PSL, $$ which is a contradiction because $\PSL$ has a trivial center while $\SL$ doesn't.
We are then left with the alternative according to which $[\ZG:H]=2$. On the other hand, it is easy to see that the only subgroup of index $2$ of $\ZG\simeq\Rx$ is $\Rp$, hence $H=\Rp$. Consequently $$ \SL\simeq \GL/\Rp \simeq \SLpm= \{g\in \GL: \text{det}(g)=\pm 1\}. $$ This is again impossible since $\SLpm$ admits a nontrivial homomorphism into a commutative group (namely the determinant), while $\SL$ doesn't. QED