Pure geometric braids and ordered configuration loops

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 ${\iota }^C\circ S(\beta )$.

The inverse-loop map preserves the endpoint permutation. Put $a:={\iota }_{\ast }^C[S(\beta )]$. By [L15] and the inverse identity in the fundamental group [L9], ${\pi }_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}(a^{-1})={\pi }_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}(a{)}^{-1}$: indeed ${\pi }_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}(a){\pi }_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}(a^{-1})={\pi }_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}(a{a}^{-1})=1$. By [L4], $\mathrm{\Phi }([\beta ])=a^{-1}$, and step 1.2 gives ${\pi }_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}(a)={\pi }_{\mathrm{g}\mathrm{e}\mathrm{o}}([\beta ]{)}^{-1}$. Thus for every $[\beta ]\in G_n$, ${\pi }_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}(\mathrm{\Phi }([\beta ]))={\pi }_{\mathrm{g}\mathrm{e}\mathrm{o}}([\beta ])\mathrm{.}$ Consequently $\mathrm{\Phi }([\beta ])$ lies in $\ker {\pi }_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}$ if and only if $[\beta ]$ lies in $G_n^{\mathrm{p}\mathrm{u}\mathrm{r}\mathrm{e}}$. Since $\mathrm{\Phi }$ is an isomorphism by [L4], its restriction is an isomorphism $G_n^{\mathrm{p}\mathrm{u}\mathrm{r}\mathrm{e}}\to \ker {\pi }_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}$.

The short exact sequence identifies the ordered group with the kernel. By [L7], $p_{\ast }:P{B}_n\to B_n^{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}$ is injective and has image $\ker {\pi }_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}$. Regard its codomain as this image. Then it is surjective onto the kernel by the definition of image [L11], hence bijective by [L12]. It is a group homomorphism by [L7], so it is an isomorphism by [L12]. Composing its inverse with the restriction in step 2.1 gives an isomorphism $\mathrm{\Psi }:G_n^{\mathrm{p}\mathrm{u}\mathrm{r}\mathrm{e}}\to P{B}_n$.

Compute the ordered representative. If $[\beta ]$ is pure, [L1] gives $z_{\beta }(0)=Q=z_{\beta }(1)$, so [L17] makes $z_{\beta }$ a based loop in $F_n(\mathrm{int}{D}^2)$ at $Q$. Using [L2], [L5], [L8], and [L14], $p_{\ast }({\iota }_{\ast }^F[z_{\beta }])=[p\circ {\iota }^F\circ z_{\beta }]=[{\iota }^C\circ p_{\mathrm{int}{D}^2}\circ z_{\beta }]={\iota }_{\ast }^C[S(\beta )]\mathrm{.}$ Since $p_{\ast }$ is a homomorphism by [L7], it carries inverses to inverses: $p_{\ast }(x)p_{\ast }(x^{-1})=p_{\ast }(x{x}^{-1})=1$ for each $x$ by [L9]. Thus $p_{\ast }(({\iota }_{\ast }^F[z_{\beta }]{)}^{-1})=({\iota }_{\ast }^C[S(\beta )]{)}^{-1}=\mathrm{\Phi }([\beta ])\mathrm{.}$ The preimage under $p_{\ast }$ is unique by its injectivity [L7]; hence the isomorphism in step 3.1 is exactly $\mathrm{\Psi }([\beta ])=({\iota }_{\ast }^F[z_{\beta }]{)}^{-1}$. This also proves the formula is independent of the representative braid.

Check the product order explicitly. Let $[\gamma ],[\beta ]$ be pure and put $a:={\iota }_{\ast }^C[S(\gamma )]$ and $b:={\iota }_{\ast }^C[S(\beta )]$. By [L1, L10] and the homomorphism property in [L8], $\mathrm{\Phi }([\gamma ][\beta ])=(ba{)}^{-1}\mathrm{.}$ In the group $B_n^{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}$, $a^{-1}{b}^{-1}$ is a two-sided inverse of $ba$: $(ba)(a^{-1}{b}^{-1})=b(a{a}^{-1})b^{-1}=1$ and $(a^{-1}{b}^{-1})(ba)=a^{-1}(b^{-1}b)a=1$. Thus $(ba{)}^{-1}=a^{-1}{b}^{-1}=\mathrm{\Phi }([\gamma ])\mathrm{\Phi }([\beta ])$. The composite $\mathrm{\Psi }=p_{\ast }^{-1}\circ \mathrm{\Phi }$ on pure classes is therefore multiplicative; this shows directly that inversion of the reversed slicing product gives the ordered configuration product in geometric stacking order.

Empty and one-strand cases. For $n=0$, [L13] gives the unique empty braid and trivial $P{B}_0$ and $S_0$; exactness [L7] then makes $B_0^{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}$ trivial, so each group, kernel, and displayed map is the unique one-element group map. For $n=1$, [L19] says every geometric braid is pure and $P{B}_1$ and $S_1$ are trivial, so [L7] gives $B_1^{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}=\mathrm{im}{p}_{\ast }=\ker {\pi }_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}$, also trivial. The isomorphism [L4] then makes $G_1^{\mathrm{p}\mathrm{u}\mathrm{r}\mathrm{e}}$ trivial, and the formula in step 4.1 is the unique isomorphism. The zero-strand case is the empty case, and no additional zero-valued parameter is present. [L4, L7, L13, L19, step 2.1, step 3.1, step 4.1] $\mathrm{\square }$