The configuration braid group $B_n^{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}$ as the fundamental group of an unordered configuration space

Fix $n\in \mathbb{N}$ and the same base configuration $q=(q_1,\dots ,q_n)$ of pairwise distinct points of $\mathrm{int}{D}^2$ that is used in The pure braid group $P{B}_n$ as the fundamental group of an ordered configuration space, with $D^2$ the closed unit disc. Write $p:F_n(D^2)\to C_n(D^2)$ for the quotient of the ordered by the unordered configuration space, so that $p(q)=[q]$ is the orbit of $q$ (Unordered configuration spaces $C_n(X)$). The configuration braid group on $n$ strands is the fundamental group (Based loops and the fundamental group)

$B_n^{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}:={\pi }_1(C_n(D^2),[q]),$

the group of based-loop classes at the orbit $[q]$ in the unordered configuration space of the closed disc, with the first-then-second loop product.

The open-disc model. The inclusion-induced map ${\iota }^C:C_n(\mathrm{int}{D}^2)\to C_n(D^2)$ of The interior-disc and closed-disc configuration spaces are homotopy equivalent induces an isomorphism ${\pi }_1(C_n(\mathrm{int}{D}^2),[q])⟶{\pi }_1(C_n(D^2),[q])$ at the same basepoint (claim 3 of that lemma, The homomorphism on fundamental groups induced by a pointed continuous map), so $B_n^{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}$ may be computed from either disc model exactly as $P{B}_n$ may. Both groups in this definition and in The pure braid group $P{B}_n$ as the fundamental group of an ordered configuration space are taken at the basepoints $[q]$ and $q$ coming from the same tuple $q$, which is what makes the comparison map of the configuration braid short exact sequence, proved in a later item on this page, a map of based fundamental groups.

The basepoint. Since $F_n(D^2)\to C_n(D^2)$ is surjective every basepoint of $C_n(D^2)$ is an orbit, and since $C_n(D^2)$ is path-connected (claim 2 of Ordered configuration spaces cover the unordered ones regularly with deck group $S_n$) the groups at different orbits are isomorphic by conjugation along a path (Conjugating loop classes by a path is an isomorphism of fundamental groups); no particular isomorphism is fixed. For $n=0$ the space $C_0(D^2)$ is a point and $B_0^{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}$ is the one-element group.

The superscript. The decoration $\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}$ records that the group is defined here through configuration spaces, and it is retained until the later geometric identification of $B_n^{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}$ with the braid group given by strand diagrams and with its Artin presentation. No such identification and no presentation is asserted on this page; neither is any identification of $B_n^{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}$ with a group of self-homeomorphisms of the disc.