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?
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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".
answered Apr 28, 2010 at 3:49
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$\begingroup$ I knew I could count on you. Should I just keep trying to prove it myself or is there a reference? $\endgroup$ Commented Apr 28, 2010 at 3:54
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$\begingroup$ 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". $\endgroup$ Commented Apr 28, 2010 at 3:57
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$\begingroup$ 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. $\endgroup$ Commented Apr 28, 2010 at 3:58
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4$\begingroup$ Not to nitpick, but all finitely generated residually finite groups are Hopfian. $\endgroup$– Steve DCommented Apr 28, 2010 at 4:31
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5$\begingroup$ 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. $\endgroup$– HJRWCommented Apr 28, 2010 at 17:19