The unreduced module fits an exact sequence with the reduced module

The homology of the fibre. Let $A=p^{-1}d$. Since $A$ is discrete, every singular simplex ${\mathrm{\Delta }}^n\to A$ is constant by [F1]; hence $C_n(A;\mathbb{Z})$ is free on $A$ and, by [F2], the boundary of the constant simplex at $a$ is $\frac{1+(-1{)}^n}{2}[a]$ in degree $n\ge 1$. Therefore ${\mathrm{\partial }}_1=0$ on $C_1(A)$ and ${\mathrm{\partial }}_2$ is the identity on $C_2(A)$, so $H_1(A)=0$ and $H_0(A)={⨁}_{a\in A}\mathbb{Z}[a]$ is free on the fibre.

The ${\mathrm{\Lambda }}_1$-identifications. The deck action is free and transitive on $A=\mathbb{Z}$-torsor $\{T_{t^k}\tilde{d}\}$, so $H_0(A)$ is the free ${\mathrm{\Lambda }}_1$-module of rank one on the class of $\tilde{d}$, with $t\cdot [T_{t^k}\tilde{d}]=[T_{t^{k+1}}\tilde{d}]$; we identify it with ${\mathrm{\Lambda }}_1$, $[T_{t^k}\tilde{d}]\leftrightarrow t^k$. The spine is path-connected: its vertices $v_k$ are joined by finite strings of edges of type $1$, and every point of an edge is joined to an endpoint. The homotopy equivalences of [F5] therefore make $\tilde{X}$ path-connected, so $H_0(\tilde{X})=\mathbb{Z}$ by [F2], and the map $H_0(A)\to H_0(\tilde{X})$ sends every point class to the single generator; under the identification this is $\sum a_k{t}^k\mapsto \sum a_k=\epsilon$, which is ${\mathrm{\Lambda }}_1$-linear with kernel $(t-1)$ by [F6].

The connecting map on the basis. Use the deck-equivariant identification $U\cong H_1(\mathrm{\Sigma },{\mathrm{\Sigma }}^0)$ of [F5], under which $H_0(A)\cong H_0({\mathrm{\Sigma }}^0)$ identifies $[v_k]=[T_{t^k}{v}_0]$ with $t^k$; naturality of the pair sequence [F3] identifies the connecting maps. For the relative class of the oriented edge $e_i^{(k)}$ from $v_k$ to $v_{k+1}$, the boundary is $[v_{k+1}]-[v_k]$, so ${\mathrm{\partial }}_{\ast }({ϵ}_i^{(k)})=[v_{k+1}]-[v_k]=t^{k+1}-t^k=t^k(t-1)$ in ${\mathrm{\Lambda }}_1$; for the design basis $e_i={ϵ}_i^{(i-1)}$ this gives ${\mathrm{\partial }}_{\ast }(e_i)=t^{i-1}(t-1)$. Hence for $x={\sum }_i{x}_i{e}_i$ one has ${\mathrm{\partial }}_{\ast }(x)={\sum }_i{x}_i{t}^{i-1}(t-1)=(t-1){\sum }_i{t}^{i-1}{x}_i=(t-1)\sigma (x)$, so ${\mathrm{\partial }}_{\ast }=(t-1)\sigma$. Since ${\mathrm{\Lambda }}_1$ is a domain and $t-1\mathrm{\ne }0$ by [F6], $\ker {\mathrm{\partial }}_{\ast }=\ker \sigma$.

The exact sequence. The long exact sequence [F3] of the pair reads $H_1(A)\to H_1(\tilde{X})\to H_1(\tilde{X},A)\stackrel{{\mathrm{\partial }}_{\ast }}{\to }{H}_0(A)\stackrel{\epsilon }{\to }{H}_0(\tilde{X})\to H_0(\tilde{X},A)\to 0$. By step 1.1 the first term vanishes, so the first map is injective with image $\ker {\mathrm{\partial }}_{\ast }$; by step 1.2 the map $\epsilon$ is surjective, so $H_0(\tilde{X},A)=0$ by exactness and the displayed segment is the asserted four-term sequence of ${\mathrm{\Lambda }}_1$-modules. All maps are ${\mathrm{\Lambda }}_1$-linear by [F4], so the sequence is a sequence of ${\mathrm{\Lambda }}_1$-modules; the annihilation of $H_0(\tilde{X},A)$ and the identification of the last map with $\epsilon$ are step 1.2.

Consequences and non-splitting. By exactness in step 2.1 the image of $M_{\mathrm{r}\mathrm{e}\mathrm{d}}$ in $U$ is exactly $\ker {\mathrm{\partial }}_{\ast }$, which step 1.3 identifies with $\ker \sigma$; the invariant vector $v=(1,\dots ,1{)}^T$ of The invariant vector and the invariant covectors of the unreduced Burau satisfies ${\mathrm{\partial }}_{\ast }(v)=(t-1)\sigma (v)=(t-1)(1+t+\cdots +t^{n-1})$, a product of two nonzero elements of the domain ${\mathrm{\Lambda }}_1$ by [F6] and The invariant vector and the invariant covectors of the unreduced Burau(c), hence nonzero; so $v\mathrm{\notin }\ker {\mathrm{\partial }}_{\ast }$. Nothing in the argument produces a ${\mathrm{\Lambda }}_1$-linear splitting of the sequence, and none is asserted; the integral structure is exactly the displayed four terms. AC enters only through the $B_n$-equivariance clause, via the braid lifts of [F4]; the exact sequence, the fibre computation, the connecting map and the non-splitting observation are choice free.