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There cannot be a surjective homomorphism $B_2 \to B_n$ for any $n > 2$ because $B_2$ is commutative and $B_n$ is not. It seems plausible that if $m < n$, there cannot be a surjective homomorphism $B_m \to B_n$.

If $m>n$, there are surjective maps $B_m \to B_n$ obtained by forgetting a certain number of strands, but these are not homomorphisms: e.g., if $f: B_3 \to B_2$ is the map forgetting the third strand, then $f(\sigma_1 \sigma_2 \sigma_1) = \sigma_1$, but $f(\sigma_1) \cdot f(\sigma_2) \cdot f(\sigma_1) = \sigma_1^2$. Here, $\sigma_i$ is the braid swapping the $i$-th and the $(i+1)$-st strand using a single positive crossing.

How do these observations generalize? I.e., under what conditions on $m,n \geq 2$ does there exist a surjective homomorphism $B_m \to B_n$? My suspicion is that this requires $m = n$ or, similar to the case of symmetric groups, $(m,n) = (4,3)$, is this true?

I originally posted this on MathSE but since this seems to be sort of hard I thought it would be appropriate to post it on MO. I hope this sort of cross-post is acceptable.

Edit: As pointed out to me in the comments by @YCor, in the case $4 \neq m > n > 2$ there are indeed no surjective homomorphisms $B_m \to B_n$ by Theorem 3.1 in Lin (integral homomorphisms have cyclic image). The seemingly easier case $m < n$ is what remains unanswered.

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    $\begingroup$ Isn't the rank of the braid group $B_m$ given by $\dim H_1(\mathrm{Conf}_m(\mathbb{R}^2); \mathbb{Q}) = \binom{m}{2}$? That settles the question for the existence of surjective morphisms $B_m \to B_n$ for $m < n$ (not the other way around obviously). Edit: well, no, of course, that's the pure braid group. (For which the "forgetting strands" maps are surjective homomorphisms, BTW.) I just leave this here in case someone thinks the same mistake. $\endgroup$
    – Najib Idrissi
    Jul 12, 2020 at 11:50
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    $\begingroup$ You probably get information and restrictions on homomorphisms $B_m\to B_n$, $m>n$ from arxiv.org/abs/math/0404528 (because for $n\ge 3$ and large $m$ $B_m$ should not surject onto $S_n$). See also arxiv.org/abs/1910.07177 $\endgroup$
    – YCor
    Jul 12, 2020 at 12:27
  • $\begingroup$ @YCor Thanks for pointing out these sources. Lin's Theorem 3.1 b) states that for $4 \neq m > n$, homomorphisms $B_m \rightarrow B_n$ are integral (i.e., their image is torsion-free cyclic), so this actually settles my question for $4 \neq m > n > 2$, without any discussion of the cardinalities of finite quotients as in Chudnovsky et al. Do you agree? $\endgroup$
    – Levi Ryffel
    Jul 12, 2020 at 12:51
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    $\begingroup$ If there is a surjection $B_m \rightarrow B_n$ then there is a surjection $B_m\rightarrow S_n$. When $m < n$ this already seems unlikely to me. $\endgroup$
    – Mark Wildon
    Jul 12, 2020 at 20:12
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    $\begingroup$ You might also want to look at this paper of Chen, Kordek and Margalit. I think it’s the state of the art: arxiv.org/abs/1910.00712v1 . $\endgroup$
    – HJRW
    Jul 15, 2020 at 6:25

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