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Let $X$ be a closed, orientable surface of genus at least 2, and let $\phi: \pi_1(X) \to \pi_1(X)$ be a surjective homomorphism. Is $\phi$ necessarily injective?

flag	asked Apr 28 2010 at 3:42 Lucas Culler 497●4●10

Andy Putman · Accepted Answer · 2010-04-28 04:52:26Z UTC

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Yes. Surface groups are Hopfian. More generally, all residually finite groups are Hopfian -- see Theorem IV.4.10 in Lyndon and Schupp's book "Combinatorial Group Theory".

edited Apr 28 2010 at 4:52

answered Apr 28 2010 at 3:49

Andy Putman
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I knew I could count on you. Should I just keep trying to prove it myself or is there a reference? – Lucas Culler Apr 28 2010 at 3:54

I gave a reference for a more general statement -- there's probably a more direct proof for surface groups, but the more general statement is extremely useful. For the residual finiteness of surface groups, see Hempel's beautiful 1-page paper "Residual finiteness of surface groups". – Andy Putman Apr 28 2010 at 3:57

And I should also point out that the proof in Lyndon and Schupp I referred to above is only one paragraph long and mostly self-contained. – Andy Putman Apr 28 2010 at 3:58

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Not to nitpick, but all finitely generated residually finite groups are Hopfian. – Steve D Apr 28 2010 at 4:31

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The proof is easy enough to write in a comment. Suppose g is in the kernel of a surjection $\phi:G\to G$ and let $q: G\to Q$ be a finite quotient with $q(g)\neq 1$. Then it's easy to see that $q\circ\phi^n$ are all distinct maps $G\to Q$. But there are only finitely many maps from a finitely generated group to a fixed finite group. – HW Apr 28 2010 at 17:19

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Self-homomorphisms of surface groups

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