Question: Is every finite simple group a quotient of a braid group?
Context: The braid group on two strands $ B_2 $ is isomorphic to $ \mathbb{Z} $ and so the infinite family of abelian finite simple groups (cyclic of prime order) are all quotients of $ B_2 $.
The braid group on three strands $ B_3 $ is a universal central extension of $ PSL(2, \mathbb{Z}) $, $ B_3 \cong \mathbb{Z} : PSL(2,\mathbb{Z}) $. So the infinite family of finite simple groups $ PSL(2,p)\cong PSp(2,p) $ are all quotients of $ B_3 $.
This pattern continues with higher $ B_n,n \geq 3 $. There is a quotient of $ B_n $ which is a (finite index) subgroup of $ Sp(2m,\mathbb{Z}) $ for some $ m $ ($ m $ is the rank of the homology of a certain space, details given in this paper https://link.springer.com/article/10.1007/BF02566275.) And moreover if we take this subgroup mod $ p $ then we get all of $ Sp(2m,p) $. Thus we have that $ Sp(2m,p) $ for all primes $ p $ are quotients of that $ B_n $. Assuming that every integer is the $ m $ corresponding to some $ n $ then we have that every finite simple group $ PSp(2m,p) $ is a quotient of some Braid group $ B_n $.
Since $ PSp(2m,p) \cong PSL(2,p) $ then $ PSL(2,5) $ and $ PSL(2,7) $ are indeed quotients of $ B_3 $ and all cyclic groups are quotients of $ B_2 $. The next group to check would be $ A_6 $.
So one way to answer this question would just be to show that $ A_6 $ is not a quotient of any braid group $ B_n $.
Best Answer
The alternating group $A_6$ is not a quotient of any braid group $B_n$. This follows from direct calculation. Since $$B_n = \langle x_1, \dots, x_{n-1} \mid [x_i,x_j] = 1 ~ \text{for} ~ |i-j|>1, x_ix_{i+1}x_i = x_{i+1}x_ix_{i+1} ~\text{for}~ 1 \le i \le n-2\rangle,$$ we can enumerate homomorphisms $B_n \to A_6$ by counting $x_1,\dots,x_{n-1} \in A_6$ satisfying the braid relations. It turns out that $$|\mathrm{Hom}(B_n, A_6)| = \begin{cases}3960 &:n = 3, \\ 5400 &: n=4, \\ 360 &: n = 5. \end{cases}$$ The image of a homomorphism $B_n \to A_6$ ($n \le 5$) never has order greater than $60$. Now observe that there are exactly $|G| = 360$ homomorphisms $B_n \to A_6$ factoring as $B_n \to \mathbb Z \to A_6$. The calculation for $n=5$ demonstrates that every homomorphism $B_5 \to A_6$ has this form. Since $B_{n+1}$ is generated by two copies of $B_n$, it follows by induction that every homomorphism $B_n \to A_6$ for $n \ge 5$ factors through $\mathbb Z$ and hence generates a subgroup of order at most $5$.