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?
Edit: I also posted this on MO since this may be harder than it looks. In the comments there, a solution for the case $m > n$ has turned up in the literature (Theorem 3.1 in Lin). Any thoughts on the case $m < n$ are very much appreciated.
