Braids as Fundamental Groups of Configuration Spaces

The quotient is a covering on the open disk. By [L3], $\mathrm{int}{D}^2$ is Hausdorff; applying [L4] with $X=\mathrm{int}{D}^2$ gives an evenly covered neighbourhood at every unordered configuration, and continuity and surjectivity from [L2] show that $p^{\circ }$ is a covering. For $n=0$ both spaces are points, so this conclusion still holds.

Lift the motion. Since $\alpha$ is based at $[Q]$ by [L7] and $p^{\circ }(Q)=[Q]$, [L5] gives a unique continuous lift $\tilde{\alpha }:I\to F_n(\mathrm{int}{D}^2)$ with $\tilde{\alpha }(0)=Q$ and $p^{\circ }\circ \tilde{\alpha }=\alpha$; write its coordinates as $z_1,\dots ,z_n$, taking the empty tuple when $n=0$.

Check the strand conditions. By [L1], the coordinate projections of $\tilde{\alpha }$ are continuous; because every lifted tuple lies in $F_n(\mathrm{int}{D}^2)$, the coordinates remain interior and pairwise distinct, and they start at $Q$. At the top, $p^{\circ }(\tilde{\alpha }(1))=\alpha (1)=[Q]$, so [L2] places the terminal tuple in the coordinate-permutation orbit of $Q$, giving endpoint set exactly $\{q_1,\dots ,q_n\}$; these conditions are vacuous for $n=0$.

The lift is the required braid and traces the original motion. By [L6], $\beta =(z_1,\dots ,z_n)$ is a geometric braid based at $Q$: its strands are the graphs $t\mapsto (z_j(t),t)$, so height is their parameter and distinct strands cannot meet. Its unordered slice is $[(z_1(t),\dots ,z_n(t))]=p^{\circ }(\tilde{\alpha }(t))=\alpha (t)$ for every $t$, by [L5]; this also holds for the unique empty braid when $n=0$.

The unique lift has all the endpoint, collision, continuity and slicing properties required in the statement, establishing the claim.

Specify the initial lift. By [L1], $p^{\circ }$ is a covering. The constant map ${\tilde{H}}_0:I_s\to F_n(\mathrm{int}{D}^2)$, ${\tilde{H}}_0(s)=Q$, is continuous and lifts the edge $H(s,0)=[Q]$ because $p^{\circ }(Q)=[Q]$.

Lift the full square. Apply [L3] to $G=H$ with $Y=I_s$, height coordinate $t$, and initial lift from step 1.1. There is a unique jointly continuous lift $\tilde{H}:I_s\times I_t\to F_n(\mathrm{int}{D}^2)$ with $p^{\circ }\circ \tilde{H}=H$ and $\tilde{H}(s,0)=Q$.

Each lifted height path is the braid trace. Fix $s\in I$. By step 2.1, $t\mapsto \tilde{H}(s,t)$ lifts the based loop $H(s,-)$ from $Q$. Uniqueness in [L1] identifies it with the lift whose coordinate graphs form the geometric braid traced by that loop. Its terminal tuple projects to $H(s,1)=[Q]$, so its endpoint set is $Q$. This includes $n=1$, where there are no pairwise-collision conditions.

Identify the two boundary braids. By step 2.1, $\tilde{H}(0,-)$ lifts $\alpha$ from $Q$ and $\tilde{H}(1,-)$ lifts $\beta$ from $Q$. Uniqueness in [L1] identifies these restrictions with the traces of the two boundary loops, so they are the exact braids at the ends of the claimed isotopy.

The lifted coordinates give a braid isotopy. Define $Z_j(s,t)$ to be the $j$th coordinate of $\tilde{H}(s,t)$ under the fixed real-complex identification. The lift in step 2.1 is jointly continuous into $F_n(\mathrm{int}{D}^2)\subseteq (\mathrm{int}{D}^2{)}^n$; the coordinate projections are continuous by [L4, L5]. Thus each $Z_j$ is jointly continuous. By step 3.1, every slice is a geometric braid with bottom $Q$ and top endpoint set $Q$. It therefore meets the braid-isotopy conditions [L6]. By [L7], joint continuity also keeps each labelled top endpoint fixed during the isotopy. For $n=0$, both configuration spaces are points and this is the unique empty braid isotopy.

The jointly continuous lifted family has the prescribed boundary braids from step 3.2 and satisfies every braid-isotopy condition by step 4.1.

The ordered tuple varies continuously. Apply the fixed real-complex identification from [L6] to each coordinate $z_j$. By [L2], each resulting coordinate map $I\to \mathrm{int}{D}^2$ is continuous. The product topology on $(\mathrm{int}{D}^2{)}^n$ is the topology for which continuity into the product is checked coordinatewise [L8], so $z:I\to (\mathrm{int}{D}^2{)}^n,z(t)=(z_1(t),\dots ,z_n(t))$ is continuous. Pairwise distinctness in [L3] places its image in $F_n(\mathrm{int}{D}^2)$; because that space has the subspace topology, the same map, with this restricted codomain, is continuous ([L1]). For $n=0$ this is the constant map to the one-point empty tuple.

Pass to the unordered quotient. By [L7], $p_n$ is continuous, so $S(\beta )=p_n\circ z:I\to C_n(\mathrm{int}{D}^2)$ is continuous. At the bottom, $z(0)=Q$ by [L4], so $S(\beta )(0)=[Q]$. At the top, the set of the coordinates of $z(1)$ is $Q$ by [L5], hence $z(1)$ is a coordinate permutation of $Q$ and $p_n(z(1))=[Q]$ by the orbit description in [L7]. Thus $S(\beta )(1)=[Q]$, as required for a based loop ([L4, L5, L7]).

Identify each height slice. The graph of the $j$th coordinate is $t\mapsto (z_j(t),t)$ by [L9], so its intersection with height $t$ returns the point $z_j(t)$. Taking all labels, their unordered orbit is exactly $p_n(z(t))=S(\beta )(t)$. The labels are retained in $z$ and then forgotten precisely by the quotient, with no reparametrization of height. For $n=0$ the slice and loop are the unique empty configuration.

The sliced path is continuous, has both endpoint values $[Q]$, and at each height equals the unordered slice of the given level-preserving braid.

Trace is well-defined on path classes. Define $T([\alpha ])$ to be the braid-isotopy class of the braid traced by the unique lift of $\alpha$ from $Q$, which exists by [L1]. If $\alpha$ and ${\alpha }^{\mathrm{\prime }}$ represent the same based path-homotopy class by [L5, L6], then [L2] makes their traced braids braid-isotopic. Thus $T$ is independent of the representative.

Slice is well-defined on braid-isotopy classes. Suppose $Z$ is a braid isotopy from $\beta$ to ${\beta }^{\mathrm{\prime }}$. Apply the real-complex identification [L10] to its coordinates. By joint continuity in [L4] and the product criterion [L7], the tuple map $z:I_s\times I_t\to (\mathrm{int}{D}^2{)}^n,z(s,t)=(Z_1(s,t),\dots ,Z_n(s,t))$ is continuous. Each slice is collision-free by [L11, L12], so its image lies in $F_n(\mathrm{int}{D}^2)$; the subspace topology [L8] makes the restricted map continuous. Composing with the continuous orbit projection [L15] gives $H(s,t):=p_n(z(s,t))\in C_n(\mathrm{int}{D}^2)\mathrm{.}$ The fixed bottom tuple gives $H(s,0)=[Q]$ for every $s$, and the setwise top condition gives $H(s,1)=[Q]$ for every $s$, by [L12] and the orbit description [L9]. At $s=0$ and $s=1$, $H$ is respectively the slice of $\beta$ and of ${\beta }^{\mathrm{\prime }}$ by [L3, L13]. The switch $(t,u)\mapsto (u,t)$ is continuous by the product-topology criterion [L7], so $K(t,u):=H(u,t)$ is jointly continuous and is a path homotopy relative to its endpoints by [L6]. Therefore the slice classes agree, and slicing descends to braid-isotopy classes.

Slice after trace is the original loop. For any $[\alpha ]$, the trace lemma says that the unordered slice of the traced braid equals $\alpha (t)$ at every $t$ [L1]. Thus $S(T([\alpha ]))=[\alpha ]$ in the based path-homotopy class set.

Trace after slice is the original braid. By [L11], the ordered coordinate path $t\mapsto (z_1(t),\dots ,z_n(t))$ is a continuous path in $F_n(\mathrm{int}{D}^2)$ starting at $Q$, using [L7, L8, L10]. Its projection is $S(\beta )$ by [L3]. It is therefore an ordered lift of $S(\beta )$ from $Q$; uniqueness in [L1] makes the traced braid exactly $\beta$, so $T(S([\beta ]))=[\beta ]$ as a braid-isotopy class.

The two well-defined assignments satisfy both inverse identities by steps 1.3 and 1.4. Consequently tracing and slicing induce mutually inverse bijections on the stated classes. For $n=0$ both configuration spaces and class sets are singletons; for $n=1$ the configuration spaces identify with the disk and there are no collision conditions, so the same constructions apply [L14].

The lower half is the first slice loop. For $0\le t\le \frac{1}{2}$, [L1] gives $S(\gamma \star \beta )(t)=[(z_1(2t),\dots ,z_n(2t))]=S(\beta )(2t),$ which is the first half of the concatenation $S(\beta )\ast S(\gamma )$ by [L3].

The upper half is the second slice loop. For $\frac{1}{2}\le t\le 1$, [L1] gives the ordered tuple $(w_{\pi (\beta )(1)}(2t-1),\dots ,w_{\pi (\beta )(n)}(2t-1))$. Since $\pi (\beta )$ is a permutation, this is a reordering of the coordinates of $\gamma$ at height $2t-1$; [L4] therefore gives $S(\gamma \star \beta )(t)=S(\gamma )(2t-1),$ the second half of $S(\beta )\ast S(\gamma )$.

The piecewise paths agree at the seam. At $t=\frac{1}{2}$, the first half has value $S(\beta )(1)=[Q]$ and the second has value $S(\gamma )(0)=[Q]$ by [L2]. The stacked slice is also this same orbit by [L1]. Thus the two formulas establish the pointwise identity of based loops $S(\gamma \star \beta )=S(\beta )\ast S(\gamma )$ on all of $I$, including the shared endpoint of the two closed halves.

Taking path-homotopy classes of this equality and using the product convention [L3] gives $[S(\gamma \star \beta )]=[S(\beta )][S(\gamma )]$ in the group ${\pi }_1(C_n(\mathrm{int}{D}^2),[Q])$ of [L5]. For $n=0$ all loops are the unique empty loop and the identity holds; for $n=1$, $\pi (\beta )$ is the identity and the same two-half calculation applies without collision conditions.

Fix the common basepoint. By [L1], the definition of $B_n^{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}$ is instantiated at $q=Q$, so the open-to-closed inclusion is based at the same orbit $[Q]$ on both sides. By [L5] and [L7], ${\iota }_{\ast }^C$ is a group isomorphism from the open-disk fundamental group onto $B_n^{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}$.

Raw slicing is a bijection. By [L3], the class map $S:[\beta ]\mapsto [S(\beta )]$ is well defined and has tracing as its inverse. It is therefore bijective, including the zero- and one-strand cases.

Raw slicing reverses stacking. For $[\gamma ],[\beta ]\in G_n$, [L2] defines their product by $[\gamma \star \beta ]$, and [L4] gives $S([\gamma ][\beta ])=[S(\gamma \star \beta )]=[S(\beta )][S(\gamma )]\mathrm{.}$ This is the anti-homomorphism identity for the first-then-second loop product of [L8]. Since $S$ is bijective by step 1.2, it is an anti-isomorphism.

The inverse-loop map is multiplicative. Write $a:={\iota }_{\ast }^C[S(\gamma )],b:={\iota }_{\ast }^C[S(\beta )]\mathrm{.}$ By step 2.1 and the homomorphism property [L7], $\mathrm{\Phi }([\gamma ][\beta ])=(ba{)}^{-1}\mathrm{.}$ In the group $B_n^{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}$, the element $a^{-1}{b}^{-1}$ is a two-sided inverse of $ba$: associativity gives $(ba)(a^{-1}{b}^{-1})=b(a{a}^{-1})b^{-1}=1$ and $(a^{-1}{b}^{-1})(ba)=a^{-1}(b^{-1}b)a=1$. Uniqueness of inverses therefore gives $(ba{)}^{-1}=a^{-1}{b}^{-1}$. Hence $\mathrm{\Phi }([\gamma ][\beta ])=({\iota }_{\ast }^C[S(\gamma )]{)}^{-1}({\iota }_{\ast }^C[S(\beta )]{)}^{-1}=\mathrm{\Phi }([\gamma ])\mathrm{\Phi }([\beta ]),$ so $\mathrm{\Phi }$ is a group homomorphism by [L10].

The map is bijective and hence an isomorphism. The formula for $\mathrm{\Phi }$ is the composite of the bijection $S$ from step 1.2, the isomorphism ${\iota }_{\ast }^C$ from step 1.1, and inversion in $B_n^{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}$. Inversion is a bijection because applying it twice returns the original element. Thus $\mathrm{\Phi }$ is bijective by [L11]; together with step 3.1 it is a group isomorphism by [L12]. This also proves the stated anti-isomorphism property of raw slicing.

Boundary cases. When $n=0$, [L13] gives the unique empty braid and [L14] gives singleton configuration spaces and one based motion; by [L3] and [L5], the class sets and inclusion map are also singletons. When $n=1$, [L13]–[L14] give no collision condition; the same tracing/slicing bijection, based inclusion isomorphism, and product calculation above apply. These cases require no additional choice or basepoint path. [L3, L5, L9, L13, L14, step 1.1, step 1.2, step 2.1, step 3.1, step 4.1] $\mathrm{\square }$

Fix the shared basepoint and the pure subgroup. In every parameterized configuration group take $q=Q$, so $P{B}_n$, $B_n^{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}$, both inclusion-induced maps, and $p_{\ast }$ are based at $Q$ or $[Q]$ as appropriate ([L4, L5, L6, L7]). By [L1], ${\pi }_{\mathrm{g}\mathrm{e}\mathrm{o}}$ is a homomorphism and its identity fiber is exactly the set of pure geometric classes. By [L11] this set $G_n^{\mathrm{p}\mathrm{u}\mathrm{r}\mathrm{e}}=\ker {\pi }_{\mathrm{g}\mathrm{e}\mathrm{o}}$ is a subgroup of $G_n$.

The coordinate motion computes covering monodromy. For any braid $\beta$, [L2] gives $p_{\mathrm{int}{D}^2}\circ z_{\beta }=S(\beta )$, and [L17] makes $z_{\beta }$ a path in the ordered configuration space. By the commutative square in [L14], ${\iota }^F\circ z_{\beta }$ is a lift, starting at $Q$, of ${\iota }^C\circ S(\beta )$ to $F_n(D^2)$. It is the unique such lift by [L3]. Its terminal coordinate satisfies $z_j(1)=q_{{\pi }_{\mathrm{g}\mathrm{e}\mathrm{o}}(\beta )(j)}$ by [L18], so the label record $e$ of [L16] is ${\pi }_{\mathrm{g}\mathrm{e}\mathrm{o}}(\beta )$. By [L20] the lift endpoint is $\sigma \cdot Q$ for endpoint monodromy $\sigma$, and [L16] gives $e={\sigma }^{-1}$. The same coordinate tuple has label record $e={\pi }_{\mathrm{g}\mathrm{e}\mathrm{o}}(\beta )$ by [L18], while [L15] identifies ${\pi }_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}=\sigma$. It follows that ${\pi }_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}({\iota }_{\ast }^C[S(\beta )])={\pi }_{\mathrm{g}\mathrm{e}\mathrm{o}}([\beta ]{)}^{-1}\mathrm{.}$ The use of the closed-disc lift here is valid because the open coordinate path is also a path in $F_n(D^2)$ and the quotient square in [L14] identifies its projection with ${\iota }^C\circ S(\beta )$.