Braid mapping classes lift equivariantly to the Burau cover

Clause (1). For $x\in K$ one has $\omega (h_{\ast }x)=\omega (x)=0$ by [F1], so $h_{\ast }(K)\subseteq K$; hence $h_{\ast }$ descends to an endomorphism ${\bar{h}}_{\ast }$ of ${\pi }_1(X,d)\mathrm{/}K$. Since $\omega$ factors as the isomorphism $\bar{\omega }:{\pi }_1(X,d)\mathrm{/}K\to \mathbb{Z}$ by [F3] and $\omega \circ h_{\ast }=\omega$, one has $\bar{\omega }\circ {\bar{h}}_{\ast }=\bar{\omega }$, so ${\bar{h}}_{\ast }$ is the identity of ${\pi }_1(X,d)\mathrm{/}K$.

The cover is path-connected and locally path-connected. The cover $\tilde{X}$ is connected by [F3] and locally path-connected by [F5], hence path-connected by [F5].

Clause (2). The homeomorphism $h$ fixes $d$, since $d\in \mathrm{\partial }{D}^2$ and $h$ fixes $\mathrm{\partial }{D}^2$ pointwise, so $h\circ p:(\tilde{X},\tilde{d})\to (X,d)$ is based with $(h\circ p{)}_{\ast }{\pi }_1(\tilde{X},\tilde{d})=h_{\ast }(K)\subseteq K=p_{\ast }{\pi }_1(\tilde{X},\tilde{d})$ by [F3] and step 1.1. By the lifting criterion in [F2] there is a unique lift $\tilde{h}:\tilde{X}\to \tilde{X}$ with $p\circ \tilde{h}=h\circ p$ and $\tilde{h}(\tilde{d})=\tilde{d}$. Applying the same construction to $h^{-1}$ yields $g$ with $p\circ g=h^{-1}\circ p$ and $g(\tilde{d})=\tilde{d}$. Then $g\circ \tilde{h}$ and ${\mathrm{id}}_{\tilde{X}}$ are both lifts of $p$ fixing $\tilde{d}$, because $p\circ g\circ \tilde{h}=h^{-1}\circ p\circ \tilde{h}=h^{-1}\circ h\circ p=p$, so $g\circ \tilde{h}={\mathrm{id}}_{\tilde{X}}$ by uniqueness in [F2]; symmetrically $\tilde{h}\circ g={\mathrm{id}}_{\tilde{X}}$. Hence $\tilde{h}$ is a homeomorphism.

Clause (3). By [F4] every deck transformation is ${\tau }_g$ for some $g\in {\pi }_1(X,d)$, with ${\tau }_g(\tilde{d})=\tilde{d}\cdot g$ and ${\tau }_g={\tau }_{g^{\mathrm{\prime }}}$ exactly when $gK=g^{\mathrm{\prime }}K$. Let ${\alpha }_g$ be a loop at $d$ representing $g$ and let $\gamma$ be its lift from $\tilde{d}$, so $\gamma (1)=\tilde{d}\cdot g={\tau }_g(\tilde{d})$; then $\tilde{h}\circ \gamma$ is a lift of $h\circ {\alpha }_g$ starting at $\tilde{h}(\tilde{d})=\tilde{d}$, so $(\tilde{h}\circ \gamma )(1)=\tilde{d}\cdot h_{\ast }g={\tau }_{h_{\ast }g}(\tilde{d})$ by [F4]. Hence $\tilde{h}({\tau }_g\tilde{d})={\tau }_{h_{\ast }g}\tilde{d}$. Since $h_{\ast }$ induces the identity on ${\pi }_1(X,d)\mathrm{/}K$ by step 1.1, the classes $h_{\ast }g$ and $g$ have the same coset modulo $K$, so ${\tau }_{h_{\ast }g}={\tau }_g$ and $\tilde{h}$ fixes every point ${\tau }_g\tilde{d}$ of the fibre $p^{-1}d$. Now let $T$ be any deck transformation: both $\tilde{h}\circ T$ and $T\circ \tilde{h}$ are lifts of $h\circ p$, since $p\circ \tilde{h}\circ T=h\circ p\circ T=h\circ p$ and $p\circ T\circ \tilde{h}=p\circ \tilde{h}=h\circ p$, and they agree at $\tilde{d}$ because $\tilde{h}(T\tilde{d})=T\tilde{d}=T(\tilde{h}\tilde{d})$; by uniqueness in [F2] they are equal.

Clause (4). For two boundary-fixed homeomorphisms $h_1,h_2$, both $\widetilde{h_1{h}_2}$ and ${\tilde{h}}_1\circ {\tilde{h}}_2$ are lifts of $(h_1{h}_2)\circ p$ fixing $\tilde{d}$, so they are equal by uniqueness in [F2]; the inverse statement is step 2.1. If $h_s$ is a homotopy of such homeomorphisms with $h_s(d)=d$ for all $s$, put $H(\tilde{x},s):=h_s(p(\tilde{x}))$ and lift $H$ starting at $\tilde{h}$ by [F6]; then $s\mapsto \tilde{H}(\tilde{d},s)$ lifts the constant path at $d$ from $\tilde{d}$, so it is constant and $\tilde{H}(\tilde{d},1)=\tilde{d}$, and $\tilde{H}(\cdot ,1)$ is a based lift of $h_1$, hence equals ${\tilde{h}}_1$ by uniqueness. For each parameter $s$, uniqueness identifies $\tilde{H}(\cdot ,s)$ with the normalized lift of $h_s$, hence it is a homeomorphism by step 2.1 and deck-equivariant and fibre-fixing by step 3.1. Thus the lifted family is an isotopy of pairs, not equality of its endpoint maps. Homotopic maps induce the same homology maps; for the relative groups the same prism chain homotopy descends to the quotient chain complexes, since every prism of a simplex in the fibre stays in the fibre: each term of The prism operator of a homotopy factors through that simplex times $I$. The chain identity of The singular chain homotopy formula therefore passes to the relative quotient, proving equality of relative homology maps (Homotopic maps induce the same map on singular homology).

Clause (5). The assignment $[h]\mapsto [\tilde{h}]$ into isotopy classes is well defined by step 4.1, because two representatives of a mapping class are isotopic through boundary-fixed homeomorphisms preserving $Q_n$ setwise (Boundary-fixed mapping class group of a punctured disk); it is multiplicative by step 4.1, so composing with the identification $B_n\cong \mathrm{Mod}(D^2,Q_n;\mathrm{\partial }{D}^2)$ of Braid group as boundary-fixed punctured-disk mapping classes, The Artin presentation is complete for geometric braids, in which AC enters, gives a homomorphism into those isotopy classes. The induced homology actions are independent of the representative by the lifted homotopy in step 4.1, and composition of normalized lifts makes them homomorphisms. For homology, each $\tilde{h}$ is a homeomorphism, so by [F7] it induces an automorphism of $H_1(\tilde{X};\mathbb{Z})$; by step 3.1 it commutes with every $(T_{t^k}{)}_{\ast }$, hence with the ring homomorphism ${\mathrm{\Lambda }}_1\to {\mathrm{End}}_{\mathbb{Z}}(M_{\mathrm{r}\mathrm{e}\mathrm{d}})$ determined by $t\mapsto (T_t{)}_{\ast }$ in [F8], and therefore it is a ${\mathrm{\Lambda }}_1$-module automorphism of $M_{\mathrm{r}\mathrm{e}\mathrm{d}}$. Moreover $\tilde{h}$ maps the fibre $p^{-1}d$ to itself by step 3.1, so it is a homeomorphism of pairs $(\tilde{X},p^{-1}d)\to (\tilde{X},p^{-1}d)$ and by [F7] induces an automorphism of $H_1(\tilde{X},p^{-1}d;\mathbb{Z})$ commuting with the deck transformations of the pair; since the deck action determines the ${\mathrm{\Lambda }}_1$-module structure on the relative group by the same universal-property construction as in The reduced Burau homology module, the induced automorphisms are likewise ${\mathrm{\Lambda }}_1$-linear. AC is used only through [F1] and the mapping-class identification cited above; the covering-theoretic and homology steps are choice free.