First I need some notation (it's all standard I think). For a manifold $M$, let $F_nM = F_{0,n}M$ be the space of $n$-tuples of distinct points on $M$ ; let $B_nM = B_{0,n}M = F_nM / \Sigma_n$. When $M= \mathbb{R}^2$ the fundamental group of $B_nM$ is the braid group $B_n$, and that of $F_nM$ is the pure braid group $P_n$.

Further, $F_{m,n} M$ is defined to be $F_n N$ where $N$ is $M$ with $m$ points removed. Likewise for $B_{m,n} M$.

Now this question is about $\pi_1 F_n S^2$. The fundamental group of $B_n S^2$(the "braid group of the sphere") is usually presented as a quotient of $B_n$, adding just one relation to Artin's usual ones ; so the fundamental group of $F_n S^2$ is a quotient of $P_n$.

However, a classical result asserts that there is a fibration

$$ F_{m+r, n-r}M \to F_{m,n}M \to F_{m,r}M$$

for all $m,n$ and $r\le n$. Taking $M=S^2$, $m=0$ and $r=1$ this becomes

$$ F_{n-1} \mathbb{R}^2 \to F_n S^2 \to S^2 $$

And the long exact sequence of homotopy groups gives in particular

$$ \mathbb{Z} \to P_{n-1} \to \pi_1 F_n S^2 \to 0$$

using that $\pi_1(S^2) = 0$ and $\pi_2(S^2) = \mathbb{Z}$.

This gives a presentation of $\pi_1 F_n S^2$ as a quotient of $P_{n-1}$ rather than $P_n$, adding just one relator (with a very simple proof indeed). The group $P_n$ can be generated by $n(n-1)/2$ generators and no fewer, and so $P_{n-1}$ can be generated by $(n-1)(n-2)/2$ generators, giving a much smaller set of generators for $\pi_1 F_n S^2$.

So my questions are: (EDITED)

(1) Did I get something wrong in the above argument?

(2) Does someone know what the image of $\mathbb{Z}$ in $P_{n-1}$ is?

(3) Has this presentation been studied algebraically? Is it easier to work with the group $\pi_1 F_n S^2$ presented as a quotient of $P_{n-1}$ than with the presentation as a quotient of $P_n$ ? Any reference to a work in this direction?

The answer by Ryan Budney below covers (1) and (2), I think. Any help with (3) appreciated.

Thanks !

Pierre