Is $\text{GL}_2(\mathbb{R})/\mathbb{R}^{\times}$ isomorphic to $\text{SL}_2(\mathbb{R})$? 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.
 A: $\newcommand\SL{\text{SL}_2(\mathbb{R})}
\newcommand\SLpm{\text{SL}^{\pm}_2(\mathbb{R})}
\newcommand\ZS{\{\pm 1\}}
\newcommand\ZG{{\mathcal Z}}
\newcommand\GL{\text{GL}_2(\mathbb{R})}
\newcommand\PSL{\text{PSL}_2(\mathbb{R})}
\newcommand\Rx{\mathbb{R}^{\times}}
\newcommand\Rp{\mathbb{R}^{+}}$
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
