the morphism from $SL(2,\mathbb{Z})$ to $SL(2,\mathbb{R})$ For every morphism $\rho: SL(2,\mathbb{Z}) \to GL(2,\mathbb{R})$, then $Im(\rho)\subset SL(2,\mathbb{R})$? Thanks. 
 A: It is known that $SL(2,\mathbb{Z})$ is an amalgamated product $\mathbb{Z}_4 \underset{\mathbb{Z}_2}{\ast} \mathbb{Z}_6$; more precisely, we have the presentation $$SL(2,\mathbb{Z})= \langle x,y \mid x^4=y^6=1, x^2=y^3 \rangle.$$ Therefore, the map $x \mapsto \left( \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right)$, $y \mapsto \mathrm{Id}$ extends to a morphism $SL(2,\mathbb{Z}) \to GL(2,\mathbb{R})$ whose image is not contained into $SL(2,\mathbb{R})$.
NB: It is worth noticing that, from the presentation above, any morphism $SL(2,\mathbb{Z}) \to GL(2,\mathbb{R})$ sends $y$ to a matrix of determinant $1$ and $x$ to a matrix of determinant $\pm 1$. 
However, there is no matrix $M \in GL(2,\mathbb{R})$ satisfying $M^4=\mathrm{Id}$, $M^2 \neq \mathrm{Id}$ and $\det(M)=-1$. 
Indeed, such a matrix would be diagonalizable over $\mathbb{C}$, and its two eigenvalues $\lambda_1,\lambda_2$ would have to satisfy $(\lambda_1,\lambda_2) \in \{ (\pm i, \mp i), (\pm 1, \pm 1), (\pm 1, \mp 1)\}$; but $\lambda_1 \lambda_2= \det(M)=-1$ implies $(\lambda_1, \lambda_2)= (\pm 1, \mp 1)$, hence $M^2= \mathrm{Id}$.
Therefore, if the morphism is supposed one-to-one, its image is necessarily contained into $SL(2,\mathbb{R})$. 
A: One can actually write down quite explicitly a nontrivial surjective character $SL_2(\mathbf Z) \to \mu_{12}$, the group of $12$-th roots of unity; this can easily be used to construct counter-examples to what you ask.
This character comes from the theory of modular forms: let $\mathbf H$ be the upper-half plane with its usual action of $ SL_2(\mathbf Z)$, and $Y=SL_2(\mathbf Z)\backslash\mathbf H$, considered as a complex-analytic space. There exists a coherent sheaf $\omega$ on $Y$ which is a line bundle away from the points corresponding to the orbits of $i, \rho$, and whose $12$th power is the trivial line bundle on $Y$. (This is the "Hodge bundle", which comes from the family of elliptic curves living over $Y$.) By descent, the data of $\omega$ on $Y$ determines a nontrivial cohomology class in $H^1(SL_2(\mathbf Z), \mathcal O^\times_{\mathbf H-S})$, where $S$ consists of the union of the orbits of $i$ and $\rho$ in $\mathbf H$. Moreover this class is killed by the twelvth-power map on $\mathcal O^\times_{\mathbf H-S}$. By the Kummer sequence, this class lies in the image of $$H^1(SL_2(\mathbf Z), \mu_{12})= \text{Hom}(SL_2(\mathbf Z), \mu_{12})$$ in $H^1(SL_2(\mathbf Z), \mathcal O^\times_{\mathbf H-S})$. Therefore, there is an element of order $12$ in $\text{Hom}(SL_2(\mathbf Z), \mu_{12})$.
