I am trying to prove that the the natural map from the special linear group $\operatorname{SL}_2(\mathbb{Z})$ to the special linear group $\operatorname{SL}_2(\mathbb{Z}/n\mathbb{Z})$ is surjective. Clearly, the problem lies in finding an inverse with determinant one. If have following outline of a proof :
Pick a matrix $\big(\begin{smallmatrix}a &b \\c& d\end{smallmatrix}\big)\in \operatorname{SL}_2(\mathbb{Z}/n\mathbb{Z})$. Use the Chinese remainder theorem to find b'=b \text{ mod } n such that $a$ and $b$ are relatively prime... Once I have this I can find an inverse.
The problem is that I cannot figure out how the Chinese remainder theorem could be of use, since there is no reason to assume $a$ and $n$ (capital $N$ in proof) have ggd=1.
If so one could say there is a solution of b'=b \bmod N and b'= 1 \bmod a.
But I fail to see how they do it in general.
Many thanks!