The geometric half twist acts on the lifted-edge basis by the Burau block

Set-up. By the hypotheses and [F1] the homeomorphism $h_i$ represents ${\sigma }_i$, fixes $d$, and induces the displayed meridian substitution; by [F2] the based lift ${\tilde{h}}_i$ exists, fixes $p^{-1}d$ pointwise, and acts on $U$ by a ${\mathrm{\Lambda }}_1$-module automorphism. Choose the actual lifted meridian path $a_j^{(k)}$ in $\tilde{X}$ from $v_k$ to $v_{k+1}$. Under the spine identification its relative class is ${ϵ}_j^{(k)}$ by [F4]. Since ${\tilde{h}}_i$ fixes the fibre, its image is another path in $\tilde{X}$ with these endpoints. Thus the following calculation applies ${\tilde{h}}_i$ to paths in its domain and transports their classes to the spine, without applying a map of $\tilde{X}$ directly to the quotient $\mathrm{\Sigma }$.

The action on the edge classes. Fix $k$ and $j$. By [F2] the path ${\tilde{h}}_i\circ a_j^{(k)}$ is the lift of $h_i\circ x_j$ from $v_k$. If $j\mathrm{\notin }\{i,i+1\}$, then $h_i\circ x_j$ is homotopic to $x_j$ relative to $d$ by [F1], so by [F3] the lift ${\tilde{h}}_i\circ a_j^{(k)}$ is homotopic rel endpoints to $a_j^{(k)}$ and $({\tilde{h}}_i{)}_{\ast }({ϵ}_j^{(k)})={ϵ}_j^{(k)}$. If $j=i+1$, then $h_i\circ x_{i+1}$ is homotopic rel $d$ to $x_i$, so ${\tilde{h}}_i\circ a_{i+1}^{(k)}$ is homotopic rel endpoints to the lift of $x_i$ from $v_k$, which is $a_i^{(k)}$; hence $({\tilde{h}}_i{)}_{\ast }({ϵ}_{i+1}^{(k)})={ϵ}_i^{(k)}$. If $j=i$, then $h_i\circ x_i$ is homotopic rel $d$ to the loop $x_i{x}_{i+1}{x}_i^{-1}$; its lift from $v_k$ is the concatenation $a_i^{(k)}\ast a_{i+1}^{(k+1)}\ast (a_i^{(k+1)}{)}^{-1}$: the first factor lifts $x_i$ from $v_k$ to $v_{k+1}$, the second lifts $x_{i+1}$ from $v_{k+1}$ to $v_{k+2}$, and the third lifts $x_i^{-1}$ from $v_{k+2}$ back to $v_{k+1}$. For two composable paths $a,b$ between fibre points, put $P=a\ast b$ and map ${\mathrm{\Delta }}^2$ to the path by $P(u_1\mathrm{/}2+u_2)$; its boundary is $b-P+a$, so $[P]=[a]+[b]$ in relative homology. The map $a(u_1)$ has boundary $a^{-1}-c+a$, where $c$ is constant at its initial vertex and lies in the fibre, so $[a^{-1}]=-[a]$ (The singular boundary operator, The standard topological simplex and its affine face maps). Thus $({\tilde{h}}_i{)}_{\ast }({ϵ}_i^{(k)})={ϵ}_i^{(k)}+{ϵ}_{i+1}^{(k+1)}-{ϵ}_i^{(k+1)}$.

The block in the design basis. In the basis $e_j={ϵ}_j^{(j-1)}$ of [F4], step 1.2 gives $({\tilde{h}}_i{)}_{\ast }(e_i)=({\tilde{h}}_i{)}_{\ast }({ϵ}_i^{(i-1)})={ϵ}_i^{(i-1)}+{ϵ}_{i+1}^{(i)}-{ϵ}_i^{(i)}=(1-t)e_i+e_{i+1}$, because ${ϵ}_i^{(i)}=t{ϵ}_i^{(i-1)}=t{e}_i$ and ${ϵ}_{i+1}^{(i)}=e_{i+1}$; likewise $({\tilde{h}}_i{)}_{\ast }(e_{i+1})=({\tilde{h}}_i{)}_{\ast }({ϵ}_{i+1}^{(i)})={ϵ}_i^{(i)}=t{e}_i$, and $({\tilde{h}}_i{)}_{\ast }(e_j)=e_j$ for $j\mathrm{\notin }\{i,i+1\}$. These are exactly the columns of the Burau block $B_i$ of The unreduced Burau matrices in the column convention, so $({\tilde{h}}_i{)}_{\ast }$ acts as $B_i$; this is the assertion.

Conclusion and the use of AC. The topological action of the Artin generator ${\sigma }_i$ on the frozen relative lifted-edge basis equals the matrix ${\rho }_n^{\mathrm{m}\mathrm{a}\mathrm{t}}({\sigma }_i)=B_i$ of the matrix representation, on the nose and not merely up to conjugacy. AC enters exactly through the published mapping-class identification and the meridian action used in [F1]; the covering-theoretic and relative-homology steps are choice free.