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Suppose $A,\,B$ are (possibly noncommutative) rings, and $GL_n(-)$ is the group of invertible $n\times n$ matrices over a given ring. Suppose $f:A\to B$ is surjective, does it necessarily follow that $f_\ast:GL_n(A)\to GL_n(B)$ is surjective for $n>1$? If not, why not?

I know that this need not be true for $n=1$, but is true for the subgroup $E_n(-)$ generated by elementary matrices. If it isn't true in general, are there any conditions upon the rings or homomorphism that ensures this is true?

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    $\begingroup$ A quick consideration: if $A\to B$ is surjective then ${Mat}_{n\times n}(A)\to{Mat}_{n\times n}(B)$ is surjective too, and $GL_n(A)$ is the same as $GL_1(\operatorname{Mat}_{n\times n}(A))$. So if you believe that it need not be true for $n=1$, then... :) $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jan 2, 2017 at 20:19
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    $\begingroup$ A necessary condition is that $A^*\rightarrow B^*$ is surjective (consider diagonal matrices). $\endgroup$
    – abx
    Commented Jan 2, 2017 at 20:53
  • $\begingroup$ @abx Why is a diagonal matrix the image of a diagonal matrix? Consider $R=\mathbb{C}[x, y, a, b]/(xb+ya-1)$ and then quotient out $(x-a, y-b)$. The matrix $\begin{pmatrix}x&0\\0&y\end{pmatrix}$ is the image of $\begin{pmatrix}(x+a)/2&(x-a)/2\\(y-b)/2&(y+b)/2\end{pmatrix}$, but, I believe, not the image of a diagonal matrix. This is not a counter-example to the claim, just a note that the proof is not obvious. $\endgroup$
    – Andrei Smolensky
    Commented Jan 4, 2017 at 9:24
  • $\begingroup$ On the other hand, it must be surjective "on the determinants". $\endgroup$
    – Andrei Smolensky
    Commented Jan 4, 2017 at 13:58
  • $\begingroup$ @Andrei It suffices to have it for scalar diagonal matrices. Still not obvious, though $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jan 6, 2017 at 14:31

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