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Is there an epimorphism $f\colon \mathrm{GL}(2,\mathbb{Z})\to \mathrm{GL}(2,\mathbb{Z})$ which is not injective? Here, $\mathrm{GL}(2,\mathbb{Z})$ is the group of invertible $2\times 2$ matrices with integer entries.

Added. The answer is no: this group is Hopfian. I know a proof in which it is shown that this group is finitely generated (easy) and residually finite(tricky). I suspect that there exists another more "elementary" proof if we use isomorphisms of this group.

I'd really like to see such a proof and so I would be thankfull if someone wrote it (if there exists of course).

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    Please try to use mark-up. It makes things easier to read.2011-01-11
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    Why is it tricky to show that $\operatorname{GL}(2,\mathbb Z)$ is residually finite? You automagically have lots of surjections $\operatorname{GL}(2,\mathbb Z)\twoheadrightarrow \operatorname{GL}(2,\mathbb Z_p)$ for all primes $p$, for example...2011-01-11
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    I mean it's the tricky part of this proof.2011-01-11
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    I think Mariano's remark means you don't need to know $GL(2,\mathbb{Z})$ is finitely generated.2011-01-12
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    Steve - you do need to know finite generation if you want to deduce that it's non-Hopfian.2011-01-12
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    Regarding other possible proofs: there's another proof that free groups are Hopfian, using Nielsen transformations. It's given in Lyndon & Schupp. Now, $GL_2(\mathbb{Z})$ is virtually free. It may be possible to use these two facts to come up with a different proof, though I don't know how.2011-01-13
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    @Henry Wilton: Right, sorry, I was thinking of residually finite.2011-01-13

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