For some fixed integer $d \geq 3$, let $M(0, d)$ be the mapping class group of self-homeomorphisms of the Riemann sphere which fix each of the $d$ points $0, 1, ... , d-2, \infty$. Let $X$ be the configuration space of ordered $(d-3)$-element subsets of the Riemann sphere minus $\{0, 1, \infty\}$. We may identify $X$ with the moduli space $\mathcal{M}_{0, d}$ of all $d$-element subsets of the Riemann sphere modulo automorphisms in $\mathrm{PSL}_{2}(\mathbb{C})$. Then, as has been pointed out to me under another question that I asked here, $M(0, d)$ should be isomorphic to the fundamental group of $X$. The closest thing I have seen to a proof of this involves viewing $\mathcal{M}_{0, d}$ as an orbifold which is covered by a contractible Teichmuller space acted on by $M(0, d)$.
Is there a way to show directly that, more precisely, Birman's map $\pi_{1}(X, (0, 1, ... , d-2, \infty)) \to M(0, d)$ is an isomorphism? Birman's map is defined by lifting a loop on $X$ to a path on $M(0, 0)$ and taking its endpoint in $M(0, d) \subset M(0, 0)$.
EDIT: One way of attacking the problem is to consider the long exact sequence of fundamental groups induced by the fibration $C(0, d) \to C(0, 3) \to \mathcal{M}_{0, d}$ (where $C(0, d)$ is the group of self-homeomorphisms fixing $0, 1, ... , d-2, \infty$) as follows:
$$... \to \pi_{1}C(0, 3) \to \pi_{1}\mathcal{M}_{0, 3} \to \pi_{0}C(0, d) \to \pi_{0}C(0, 3) \to \pi_{0}\mathcal{M}_{0, d} \to 1$$
Note that $\pi_{1}\mathcal{M}_{0, 3} \to \pi_{0}C(0, d) = M(0, d)$ is Birman's map. It is well known that $\pi_{0}C(0, 3) = M(0, 3)$ and $\pi_{0}\mathcal{M}_{0, 3}$ are both trivial, so Birman's map is a surjection. So the question becomes whether $\pi_{1}C(0, 3)$ is trivial as well. (I believe it suffices to show that $\pi_{1}C(0, 0)$ is trivial.)