This question was originally posted on math.SE by myself nearly a year ago. I've been thinking again about the problem after it recently received a little attention, but little progress was made in finding a solution. So, I feel it's sufficiently difficult to post on mathoverflow.
It's an easy exercise to show that if $B_n$ is Artin's classical braid group on $n$ strings, then $B_n$ can be embedded in $B_{n+1}$ (and in a canonical way). A similar statement can be proved for the pure braid group $P_n$. This property is useful for proving various properties of the classical braid groups. One therefore asks if a similar property holds for braid groups on other surfaces such as the sphere.
Let $P\mathcal{S}_n$ be the pure $n$-string braid group on the sphere $S^2$. Fox's definition of this group is the fundamental group of the configuration space $F_{n}S^2=\prod_n S^2\setminus\{(x_1,\ldots,x_n)|\exists i\neq j, x_i=x_j\}$ and then the full braid group $\mathcal{S}_n$ is defined to be the fundamental group of the configuration space $B_nS^2=F_nS^2/\Sigma_n$ where $\Sigma_n$ is the action of the symmetric group by permuting coordinates of the elements of $F_nS^2$.
It was proven by Fadell and Van Buskirk that the braid group $\mathcal{S}_n$ has presentation given by the braid generators $\sigma_1,\ldots,\sigma_{n-1}$ and relations$$\begin{eqnarray}\sigma_i\sigma_j\sigma_i&=&\sigma_j\sigma_i\sigma_j&(\mbox{ for }|i-j|=1)\\\sigma_i\sigma_j&=&\sigma_j\sigma_i&(\mbox{ for }|i-j|>1)\\\gamma&=&1&\end{eqnarray}$$where $\gamma=(\sigma_1\sigma_1\ldots\sigma_{n-1})(\sigma_{n-1}\ldots\sigma_2\sigma_1)$.
With that framework now built up, my question is, can $\mathcal{S}_n$ be embedded in to $\mathcal{S}_{n+1}$ for $n\geq 3$ (and similarly for their pure counter parts)? The naive 'add a string on the end' map will not work because, for instance, the braid $\gamma$ becomes non-trivial when a string is added on the end.
I would think that the answer is no because of the dependence of the relation $\gamma=1$ on $n$, but a proof eludes me.