Isomorphism between a mapping class group and the fundamental group of a moduli space 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.)
 A: I think you mean to say that $X$ is the space parametrizing ordered $(d-3)$-tuples of pairwise distinct points in $\mathbf P^1$ with $\{0,1,\infty\}$ removed. Then $X$ is isomorphic to $\mathcal M_{0,d}$, which is the space of $d$ distinct ordered points on $\mathbf P^1$ modulo the action of $\mathrm{PGL}_2(\mathbf C)$. (Note that $\mathrm{PGL}_2(\mathbf C)$ is the automorphism group of the Riemann sphere.  $\mathrm{PSL}_2(\mathbf R)$ is the automorphism group of the upper half plane.)
In general, the orbifold fundamental group of $\mathcal M_{g,d}$ is isomorphic to the mapping class group of a genus $g$ surface with $d$ ordered punctures. When $g=0$, the orbifold $\mathcal M_{0,d}$ is actually a manifold, and it does not matter in what sense one takes the fundamental group. I don't know any way of identifying $\pi_1(\mathcal M_{g,d})$ with the mapping class group that does not involve the action of the mapping class group on Teichmüller space. (Which is not to say that there couldn't be one.) In general it seems to me the most direct way possible to compute a fundamental group, to explicitly write down a universal cover and a group of deck transformations!
