TITLE: Relation between the braid group and the mapping class group of the plane
QUESTION [5 upvotes]: According to the following link, page 248, the braid group modulo its center is isomorphic to the mapping class group of the $N$-times punctured plane, i.e. $B_N/Z(B_N)\cong M_N(\mathcal(R)^2)$. Could anybody explain to me, how it is related to the mapping class group of the punctured $2$-sphere, please?

REPLY [11 votes]: The key theorem is the following.

The braid group $B_n$ is isomorphic to the mapping class group
  $$
M_n(D^2) \;=\; \mathrm{MCG}\bigl(D^2 - \{p_1,\ldots,p_n\},\partial D^2\bigr)
$$
  of a disk with $n$ punctures rel the boundary.

Here "rel the boundary" means that all homeomorphisms and isotopies must fix the boundary of the disk pointwise.
The idea of this theorem is that a homeomorphism of the disk "moves" the $n$ punctures, and you can keep track of this motion using a braid.
More precisely, given a homeomorphism $h$ of $D^2 - \{p_1,\ldots,p_n\}$ that fixes $\partial D^2$ pointwise, we can extend $h$ to all of $D^2$ by filling in the punctures.  Now, every homeomorphism of $D^2$ that fixes $\partial D^2$ pointwise is isotopic rel $\partial D^2$  to the identity.  Therefore, there exists some isotopy $h_t\,$ rel $\partial D^2$ for which $h_0$ is the identity map on $D^2$ and $h_1=h$.  Then each puncture $p_i$ follows a path $h_t(p_i)$ under this isotopy, and the union of the graphs
$$
\bigcup_{i=1}^n \{(h_t(p_i),t) \mid t\in[0,1]\}
$$
is a braid embedded in the solid cylinder $D^2 \times [0,1]$.
In more concrete terms, the mapping class group $M_n(D^2)$ is generated by half Dehn twists around pairs $\{p_i,p_{i+1}\}$ of punctures, and these correspond to half-twists of adjacent pairs of strands in the braid group.
In any case, the theorem about a punctured sphere is the following.

Let $M$ be the mapping class group of an $(n+1)$-punctured sphere.  Let $p$ be one of the punctures, and let $M_p$ be the subgroup of $M$ consisting of mapping classes that fix $p$.  Then $M_p$ is isomorphic to $B_n / Z(B_n)$, where $Z(B_n)$ is the center of $B_n$.

Before explaining this theorem, I should mention that the center $Z(B_n)$ of $B_n$ is infinite cyclic, and is generated by a full $360^\circ$ twist of the $n$ strands.  As an element of $M_n(D^2)$, this corresponds to a Dehn twist around a curve that lies close to the boundary of the disk.
Now, there is an obvious homomorphism $M_n(D^2) \to M_p$, obtained by mapping the boundary of the disk to a small circle around the puncture $p$.  Since $M_n(D^2)$ is isomorphic to $B_n$, this gives a homomorphism $B_n \to M_p$.  Now, it should not be too hard to see that:

This homomorphism is onto, and
The center $Z(B_n)$ lies in the kernel of this homomorphism.

The explanation for (2) is that the generator for the center is mapping to a Dehn twist along a small circle surrounding $p$, which is clearly isotopic to the identity.
It should also make sense intuitively that $Z(B_n)$ would be the entire kernel of this homeomorphism, and thus $M_p$ is isomorphic to $B_n / Z(B_n)$.