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Can any one give me an example of surjective homomorphism on braid groups on the sphere that is not injective? Such that $B_{n}(S^2)$ is generated by $\sigma_1,\sigma_2, \dots, \sigma_{n-1}$ which are subject to the following relations:

  • $\sigma_{i} \sigma_{j} = \sigma_{j} \sigma_{i}$ if $|i-j| > 1$,

  • $\sigma_{i} \sigma_{i+1} \sigma_{i} = \sigma_{i+1} \sigma_{i} \sigma_{i+1}$ for all $i = 1,2, \dots, n-2$, and

  • $\sigma_1 \sigma_2 \dots \sigma_{n-1}^2 \dots \sigma_2 \sigma_1 = 1$.

Let $\phi: B_{n}(S^2) \rightarrow B_{n}(S^2)$ be a surjective homomorphism. Is there a situation where $\phi$ is not injective?

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A braid group over the sphere is linear (by Bardakov), therefore (by Mal'cev, because it is finitely generated) it is Hopfian, which means that every surjective endomorphism is an isomorphism. Thus the answer to your question is NO.

References:

  1. V. Bardakov, Linear representations of the group of conjugating automorphisms and the braid groups of some manifolds, Siberian Math. J. 46 (2005), 13–23.
  2. V. Bardakov, Linear representations of the braid groups of some manifolds, Acta Appl. Math. 85 (2005), 41–48.
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    $\begingroup$ Linearity is overkill, residual finiteness suffices for a finitely generated group to be Hopfian, and this is a much older result. $\endgroup$
    – Ian Agol
    Commented Apr 3, 2016 at 22:35
  • $\begingroup$ many thunks for Moskovich and for Agol $\endgroup$
    – Nouar Degaichi
    Commented Apr 10, 2016 at 11:39
  • $\begingroup$ can one give me a proof that a finitely generated group is residualy finite? $\endgroup$
    – Nouar Degaichi
    Commented Apr 12, 2016 at 12:54

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