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  • $\begingroup$ Thank you. See my remark to the original question -- it seems you are proving it. I note that Richard's proof also works for it. $\endgroup$ Commented Mar 15, 2010 at 16:24
  • $\begingroup$ ahh I see. should read the questions more carefully . . . $\endgroup$ Commented Mar 15, 2010 at 16:29
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    $\begingroup$ This is a bit weaker than Petya's original question since there the entries in the lifted matrix are constrained to be 0, 1 or -1. For the proof of the surjectivity of $SL_n(\mathbf{Z})\to SL_n(\mathbf{Z}/N\mathbf{Z})$ my favoured argument goes roughly as follows. Show that $SL_n(\mathbf{Z}/N\mathbf{Z})$ is generated by elementary matrices of the form $I + E_{ij}$ where $E_{ij}$ is a matrix unit with $i\ne j$. As these matrices all lie in the image, the map is surjective. I haven't seen this argument in the textbooks :-) $\endgroup$ Commented Mar 15, 2010 at 18:12
  • $\begingroup$ Nice argument, Robin! $\endgroup$ Commented Mar 15, 2010 at 18:27