Braids lift to the LKB cover and act Lambda-linearly

The homeomorphism $\sigma$ preserves $D\setminus P$, commutes with the interchange of coordinates and fixes $d_1,d_2\in \mathrm{\partial }D$; it therefore induces a homeomorphism, again denoted $\sigma$, of $C$ with $\sigma (c_0)=c_0$. Its induced automorphism of ${\pi }_1(C,c_0)$ is ${\sigma }_{\ast }$ The homomorphism on fundamental groups induced by a pointed continuous map.

The assignment depends only on the isotopy class of $\sigma$ relative to $\mathrm{\partial }D\cup P$. Once the based lifts are constructed below, an isotopy ${\sigma }_t$ from $\sigma$ to a homeomorphism ${\sigma }^{\mathrm{\prime }}$, relative to $\mathrm{\partial }D\cup P$, induces a homotopy $H(s,t)$ of basepoint-fixing maps of $C$; lift $H$ starting from $\tilde{\sigma }$ by [F1]. The lifted homotopy $\tilde{H}$ satisfies $\tilde{H}(-,t)({\tilde{c}}_0)={\tilde{c}}_0$ for every $t$: the path $t\mapsto \tilde{H}(-,t)({\tilde{c}}_0)$ lifts the constant path at $c_0$ and starts at ${\tilde{c}}_0$, so it is constant by uniqueness of path lifting [F1]. Therefore $\tilde{H}(-,1)$ is the unique lift of ${\sigma }^{\mathrm{\prime }}$ fixing ${\tilde{c}}_0$, and it is homotopic to $\tilde{\sigma }$ relative to the basepoint section; the induced maps on $H_2(\tilde{C})$ agree.

One has $\mathrm{\Phi }\circ {\sigma }_{\ast }=\mathrm{\Phi }$. For the $a$-component, the tracks form an integral one-cycle $Z_{\alpha }={\alpha }_1+{\alpha }_2$ in $D\setminus P$: its boundary is zero because the terminal unordered pair equals the initial pair. Winding is additive on this cycle, and [F3] gives $a(\alpha )={\sum }_j\mathrm{wind}(Z_{\alpha },p_j)$. A boundary-fixed disk homeomorphism is orientation preserving by [F2]. It sends a positively oriented small meridian about $p$ to a positively oriented Jordan meridian about $\sigma (p)$; hence it permutes the puncture winding coordinates of every one-cycle. Equivalently $\mathrm{wind}({\sigma }_{\ast }{Z}_{\alpha },p_j)=\mathrm{wind}(Z_{\alpha },{\sigma }^{-1}(p_j))$. Summing gives $a(\sigma \circ \alpha )=a(\alpha )$. For the $b$-component, $b$ is the exponent of the image of $\alpha$ under the map ${\pi }_1(C,c_0)\to {\pi }_1(C_2(D),c_0)$ induced by forgetting the punctures, where $C_2(D)$ is the unordered two-point configuration space of the unpunctured disk. By [F2] choose an isotopy ${\sigma }_t$ from $\mathrm{id}$ to $\sigma$ relative to $\mathrm{\partial }D$; forgetting the punctures turns it into a based homotopy from the identity of $C_2(D)$ to the homeomorphism induced by $\sigma$. Hence $\sigma$ induces the identity on ${\pi }_1(C_2(D),c_0)$ and therefore preserves the exponent $b$. Thus $\mathrm{\Phi }({\sigma }_{\ast }\alpha )=q^{a(\sigma \circ \alpha )}{t}^{b(\sigma \circ \alpha )}=q^{a(\alpha )}{t}^{b(\alpha )}=\mathrm{\Phi }(\alpha )$ for every $[\alpha ]$.

By step 2.1 the automorphism ${\sigma }_{\ast }$ of ${\pi }_1(C,c_0)$ preserves $\ker \mathrm{\Phi }={\pi }_1(\tilde{C},{\tilde{c}}_0)$. The covering-space lifting criterion [F1] applied to $\sigma \circ p:(\tilde{C},{\tilde{c}}_0)\to (C,c_0)$ therefore produces a unique lift $\tilde{\sigma }:(\tilde{C},{\tilde{c}}_0)\to (\tilde{C},{\tilde{c}}_0)$ with $p\circ \tilde{\sigma }=\sigma \circ p$.

The lift commutes with every deck transformation. Let $T$ be a deck transformation and let $\gamma$ be a loop in $C$ at $c_0$ representing the class corresponding to $T$ under $\mathrm{Deck}(\tilde{C}\mathrm{/}C)\cong {\pi }_1(C,c_0)\mathrm{/}\ker \mathrm{\Phi }$. Then $\tilde{\sigma }T{\tilde{\sigma }}^{-1}$ is again a deck transformation, and the class it corresponds to is $[\sigma \circ \gamma ]$, whose image under $\mathrm{\Phi }$ equals $\mathrm{\Phi }([\gamma ])$ by step 2.1. As the deck group is ${\mathbb{Z}}^2$ and the correspondence is through $\mathrm{\Phi }$, the two deck transformations $\tilde{\sigma }T{\tilde{\sigma }}^{-1}$ and $T$ coincide. Hence $\tilde{\sigma }T=T\tilde{\sigma }$ for every $T$, and in particular $\tilde{\sigma }$ commutes with the generators $q$ and $t$ of the deck group.

Consequently ${\tilde{\sigma }}_{\ast }:H_2(\tilde{C};\mathbb{Z})\to H_2(\tilde{C};\mathbb{Z})$ is $\mathrm{\Lambda }$-linear: it commutes with the deck automorphisms $q_{\ast },t_{\ast }$ that define the module structure. It is invertible because ${\sigma }^{-1}$ is again a boundary-fixing homeomorphism of $(D,P)$ satisfying $\mathrm{\Phi }\circ ({\sigma }^{-1}{)}_{\ast }=\mathrm{\Phi }$, so by step 3.1 it has a lift fixing ${\tilde{c}}_0$; the composite of the two lifts in either order is a lift of the identity fixing ${\tilde{c}}_0$, hence equals the identity of $\tilde{C}$ by uniqueness in [F1]. Thus ${\tilde{\sigma }}_{\ast }\in {\mathrm{Aut}}_{\mathrm{\Lambda }}(H_2(\tilde{C};\mathbb{Z}))$.

The assignment is multiplicative. If $\tau$ is another such homeomorphism, then $\tilde{\sigma }\circ \tilde{\tau }$ lifts $\sigma \circ \tau$ and fixes ${\tilde{c}}_0$, so by uniqueness $\widetilde{\sigma \circ \tau }=\tilde{\sigma }\circ \tilde{\tau }$; on homology $(\widetilde{\sigma \circ \tau }{)}_{\ast }={\tilde{\sigma }}_{\ast }\circ {\tilde{\tau }}_{\ast }$. Together with step 1.2 this makes $[\sigma ]\mapsto {\tilde{\sigma }}_{\ast }$ a homomorphism from the boundary-fixed mapping class group to ${\mathrm{Aut}}_{\mathrm{\Lambda }}(H_2(\tilde{C};\mathbb{Z}))$.

Finally, the classical braid group is identified with the boundary-fixed mapping class group of the punctured disk by Braid group as boundary-fixed punctured-disk mapping classes, which is where the Axiom of Choice is used; under this identification the homomorphism of step 6.1 is the claimed map $[\sigma ]\mapsto {\tilde{\sigma }}_{\ast }$ on $B_n$.