I just finished correcting my answer on visualizing braid groups as fundamental groups of configuration spaces, and in the process became interested in the other pictorial definition of the braid group $B_n$, namely as the mapping class group of the $n$-punctured closed unit disk.
Some definitions. If $X$ is a topological space, let $F_n(X)$ be the subspace of $X^n$ consisting of tuples with distinct coordinates, on which the symmetric group $S_n$ acts by permuting coordinates, and then define the quotient $SF_n(X):=F_n(X)/S_n$. The braid group of $X$ is defined to be the fundamental group $B_n(X):=\pi_1(SF_n(X))$. Note $B_n=B_n(\Bbb C)$ is the usual braid group.
The automorphism group ${\rm Aut}(X)$ is the group of homeomorphisms $X\to X$ which fix its boundary $\partial X$ pointwise. Denote by ${\rm Aut}_0(X)$ those automorphisms which are isotopic to the identity map. The mapping class group is ${\rm Mod}(X):={\rm Aut}(X)/{\rm Aut}_0(X)$.
Denote by $\overline{{\Bbb D}^2}$ the closed unit disk. Then $B_n(\Bbb C)\cong {\rm Mod}(\overline{\Bbb D^2}-\{x_1,\cdots,x_n\})$ for any choice of $n$ distinct points. It seems obvious to me that $B_n(\Bbb C)=B_n(\overline{\Bbb D^2})$, which inspires the question:
- For what kinds of surfaces/manifolds does $B_n({\cal S})={\rm Mod}({\cal S}-\{x_1,\cdots,x_n\})$?
And this leads to the larger question:
- What is the general relationship between braid groups and mapping class groups?
Sorry if these facts are well-known somewhere. (I am pretty new to homotopy theory and algebraic topology in general too, so I might be a bit slow. It's possible I am biting off more than I am supposed to be chewing.) Here is my likely invalid argument:
Claim. ${\rm Mod}({\cal S}-\{x_1,\cdots,x_n\})=B_n({\cal S})$ for "nice" spaces $\cal S$.
"Proof". Assume ${\rm Aut}(S)$ acts transitively on $n$-subsets of ${\cal S}$. (Intuitively it feels like this should follow automatically from $\cal S$ being homogeneous, i.e. ${\rm Aut}({\cal S})$ only acting transitively on $\cal S$ itself, by cordoning off a nbhd around any $n$ points, but I haven't tried proving it.) We should be able to identify ${\rm Aut}({\cal S}-\{x_1,\cdots,x_n\})$ with ${\rm Stab}_{{\rm Aut}({\cal S})}(\{x_1,\cdots,x_n\})$ the setwise (not pointwise) stabilizer. By the orbit-stabilizer theorem, we have
$${\rm Aut}(S)/{\rm Stab}_{{\rm Aut}(\cal S)}(\{x_1,\cdots,x_n\})\cong SF_n({\cal S}) $$
So to get $B_n({\cal S})$ we apply $\pi_1$ to the left side. Here I invoke a lemma:
Lemma. If $G$ is connected, simply connected, $H$ a subgroup, and $H^\circ$ the connected component of the identity in $H$, then $\pi_1(G/H)\cong H/H^\circ$ via $[\gamma]\mapsto\gamma(1)H^\circ$.
If we apply with $G={\rm Aut}({\cal S})$ and $H={\rm Stab}_{{\rm Aut}({\cal S})}(\{x_1,\cdots,x_n\})$ then $H/H^\circ$ should be the identified group ${\rm Aut}({\cal S}-\{x_1,\cdots,x_n\})$ modulo isotopy, i.e. ${\rm Mod}({\cal S}-\{x_1,\cdots,x_n\})$, no?
My argument must have issues, because the claim doesn't work for ${\cal S}=\Bbb C$: as I understand it, we instead have that $\Bbb C\cong\Bbb S^2-\{\rm pt\}$ and so ${\rm Mod}({\Bbb C}-\{x_1,\cdots,x_n\})$ is a subgroup of $B_{n+1}(\Bbb S^2)$ with elements having a marked string always trivial, and this seems like a different group.
Let $\mathcal S$ be a compact surface, possibly with boundary. Let $\text{Homeo}^+(\mathcal S)$ refer to the group of orientation-preserving homeomorphisms that fix the boundary pointwise with the compact-open topology. Throwing an $n$ in there means we add $n$ marked points in the interior, which I prefer over deleting points. (I'd be worried about the validity of any particular version of this result if we delete points instead of marking them.) I don't know the correct topology on $\text{Homeo}^+$ if you use noncompact surfaces.
Then the correct general statement of your dream is that $$\text{Homeo}^+(\mathcal S,n) \to \text{Homeo}^+(\mathcal S) \to SF_n(\mathcal S)$$ is a fiber bundle. The tools you need to analyze this are the homotopy long exact sequence, the Earle-Eels theorem (you can find a statement and proof in Appendix B here, and that $\text{Homeo}^+(D^2)$ is contractible.
Some special cases:
When $\mathcal S = D^2$, you immediately obtain that $B_n = \text{MCG}(\mathcal S,n)$.
Ignore the issues with topologizing $\text{Homeo}^+(\mathbb R^2)$. The trick with your proof is that this is not actually simply connected! $\text{Homeo}^+(\mathbb R^2)$ should be the same as $\text{Homeo}^+(S^2,1)$ (send every homeomorphism to its compactification). This is homotopy equivalent to $SO(2) = S^1$.
When $\mathcal S = \Sigma_{g,k}$, a genus $g$ surface with $k$ boundary components, and either $g \geq 2$ or $k \geq 1$, applying Earle-Eels you obtain the exact sequence $$1 \to B_n(\mathcal S) \to \text{MCG}(\mathcal S,n) \to \text{MCG}(\mathcal S) \to 1.$$
When $\mathcal S = S^2$, using Smale's theorem that $\text{Homeo}^+(S^2) \simeq SO(3)$, and using Earle-Eels you can prove that the components of $\text{Homeo}^+(S^2,n)$ are contractible, so you get a short exact sequence $$1 \to \mathbb Z/2 \to B_n(S^2) \to \text{MCG}(S^2,n) \to 1.$$
When $\mathcal S = T^2$, $\text{Homeo}^+(T^2) \simeq SL_2(\mathbb Z) \times T^2$. If you can get some control on $\text{Homeo}^+(T^2,n)$ you should get something interesting here, but a couple mindless attempts didn't work. (Idea: work with Diff instead so that at each marked point you can 'pull apart' the marked points and obtain diffeomorphisms of $T^2$ minus some open discs without control on the way the diffeomorphisms behave on the boundary. This space might be assailable with EE.)
The fiber sequence above generalizes perfectly well to higher-dimensional manifolds. You might enjoy playing with it for 3-manifolds, when $\text{Homeo}^+(M)$ is known in many examples. $S^2 \times S^1$ might be fun.
You would probably enjoy Farb's primer on mapping class groups. The fiber sequence above just comes from modifying his proof of Birman's exact sequence to having $n$ points instead of one.