First off, we have $\mathbb{Z}_4^+=\{0,1,2,3\}$, and $\mathbb{Z}_{10}^\times = \{1,3,7,9\}$.
Now, I'm going to tell you that $f:\mathbb{Z}_4^+ \to \mathbb{Z}_{10}^\times$ is a homomorphism for which $f(1) = 3$. What you have to figure out is: what are the values of $f(2),f(3),$ and $f(4)$, and why is this an isomorphism?
At this point, you should have the following:
$$
\begin{align}
f(1) &= 3\\
f(2) &= f(1+1) = f(1)f(1) = f(1)^2 = 9\\
f(3) &= \cdots = f(1)^3=7\\
f(4) &= \cdots = f(1)^4 = 1
\end{align}
$$
In other words, we've defined $f$ by $f(n) = 3^n \pmod{10}$ for $n = 0,1,2,3$. Now, to prove $f$ is a homomorphism, it suffices to state that
$$
f(n+m) = 3^{n+m}=3^n\cdot3^m \pmod{10} = f(n)f(m)
$$
$\newcommand\SL{\text{SL}_2(\mathbb{R})}
\newcommand\SLpm{\text{SL}^{\pm}_2(\mathbb{R})}
\newcommand\ZS{\{\pm 1\}}
\newcommand\ZG{{\mathcal Z}}
\newcommand\GL{\text{GL}_2(\mathbb{R})}
\newcommand\PSL{\text{PSL}_2(\mathbb{R})}
\newcommand\Rx{\mathbb{R}^{\times}}
\newcommand\Rp{\mathbb{R}^{+}}$
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
Best Answer
Consider $f:G\rightarrow \mathbb{R}^*\times\mathbb{R}^*$ defined by $f\left(\matrix{a&b\cr 0&c}\right)=(a,c)$ show that $f$ is an epimorphism and its kernel is $N$.
$\mathbb{R}^*\times\mathbb{R}^*$ is the multiplicative group.