The cyclic cover retracts onto the lifted flower and has a deck-equivariant spine model

Restriction of the cover to the closed subspaces $T$ and $W$. Let $q:E\to B$ be a covering and $A\subseteq B$ any subspace. For $a\in A$ choose an evenly covered open $V\ni a$, so $q^{-1}(V)={⨆}_j{V}_j$ with $q\mathrm{\mid }{V}_j:V_j\to V$ a homeomorphism. Then $(q{\mathrm{\mid }}_{q^{-1}A}{)}^{-1}(V\cap A)={⨆}_j(V_j\cap q^{-1}A)$, the pieces are open in $q^{-1}A$, and $q$ maps each piece homeomorphically onto $V\cap A$; hence $V\cap A$ is evenly covered and $q{\mathrm{\mid }}_{q^{-1}A}$ is a covering. This applies to $A=T$ and $A=W$, which are closed and not open, and supplies the conclusion of [F3] beyond its published open case.

Clause (1): lifting the deformation retraction. Lift the homotopy $F(\tilde{x},u):=H(p(\tilde{x}),u)$ through $p$ with initial lift ${\mathrm{id}}_{\tilde{X}}$, obtaining by [F2] a unique $\tilde{H}:\tilde{X}\times I\to \tilde{X}$ with $p\circ \tilde{H}=F$ and $\tilde{H}(\tilde{x},0)=\tilde{x}$. For every deck transformation $T$, the map $(\tilde{x},u)\mapsto T\tilde{H}(T^{-1}\tilde{x},u)$ is a lift of $F$ with the same value $\tilde{x}$ at $u=0$, hence equals $\tilde{H}$ by uniqueness of homotopy lifts; thus ${\tilde{H}}_u\circ T=T\circ {\tilde{H}}_u$ for all $u$, and $\tilde{H}$ is deck-equivariant. If $\tilde{x}\in p^{-1}(W)$ then $p(\tilde{H}(\tilde{x},u))=H(p(\tilde{x}),u)=p(\tilde{x})$, so $u\mapsto \tilde{H}(\tilde{x},u)$ is a path in the discrete fibre $p^{-1}(p(\tilde{x}))$, hence constant with value $\tilde{x}$; in particular $\tilde{H}$ fixes $p^{-1}d$ pointwise. Also $p(\tilde{H}(\tilde{x},1))=r(p(\tilde{x}))\in W$, so ${\tilde{H}}_1(\tilde{X})\subseteq p^{-1}(W)$. Therefore $\tilde{H}$ is a deck-equivariant homotopy of pairs from the identity to a retraction onto $(p^{-1}(W),p^{-1}d)$ fixing $p^{-1}(W)$ pointwise: a deformation retraction, which is clause (1).

The lifted tether trees. By 1.1 the restriction $p^{-1}(T)\to T$ is a covering. Put $v_k:=T_{t^k}(\tilde{d})$ and let $T_k$ be the connected component of $p^{-1}(T)$ containing $v_k$; the deck action permutes the components, so $T_{t^k}(T_0)=T_k$. For each component the projection $p\mathrm{\mid }{T}_k:T_k\to T$ is a homeomorphism: since $T$ is simply connected and path-connected, the lifting criterion [F2] lifts ${\mathrm{id}}_T$ to a map $s:T\to T_k$ (the subgroup condition being vacuous), and then $s\circ p\mathrm{\mid }{T}_k$ and ${\mathrm{id}}_{T_k}$ are two lifts of $p\mathrm{\mid }{T}_k$ agreeing at $v_k$, so they are equal by uniqueness of lifts [F2]. Hence $p$ identifies $T_k$ with $T$, the fibrewise preimage of $d$ is exactly $\{v_k:k\in \mathbb{Z}\}$, the points ${\sigma }_i^{(k)}:=(p\mathrm{\mid }{T}_k{)}^{-1}(p_i)$ are the unique points of $T_k$ over $p_i$, and the lift of $t_i$ is an edge $T_{i,k}$ of $T_k$ joining $v_k$ to ${\sigma }_i^{(k)}$.

The lifted circles and the graph $p^{-1}(W)$. Fix $i$ and parametrize $C_i$ by $c_i:[0,1]\to C_i$ periodically with $c_i(0)=p_i$, positively oriented. Since $\mathbb{R}$ is simply connected, the map $\mathbb{R}\to C_i$, $u\mapsto c_i(u-\lfloor u\rfloor )$, lifts through $p$ to a map $\epsilon :\mathbb{R}\to \tilde{X}$ with $\epsilon (0)={\sigma }_i^{(0)}$; its image is connected and contains $\epsilon (k)$ for every $k\in \mathbb{Z}$. The points $\epsilon (k)$ lie over $p_i$, and $\epsilon (k+1)$ is the endpoint of the lift of one full circle traversal from $\epsilon (k)$, i.e. $\epsilon (k)$ acted on by the monodromy of the class of $c_i$; that class has total winding $1$ because $\omega$ is conjugation invariant and $\omega (x_i)=1$, so by the defining property of $t$ in The Burau infinite cyclic cover the monodromy is $T_t$ and $\epsilon (k)={\sigma }_i^{(k)}$ by induction. Every component of the $1$-manifold $p^{-1}(C_i)$ contains some point of $p^{-1}(p_i)$ (follow a lifted circle arc back to $p_i$), and $p^{-1}(C_i)\cap p^{-1}(T)=p^{-1}(p_i)=\{{\sigma }_i^{(k)}\}$, so the connected set $\epsilon (\mathbb{R})={\bigcup }_k{E}_{i,k}$ equals all of $p^{-1}(C_i)$, with $E_{i,k}$ the lifted arc of $c_i$ from ${\sigma }_i^{(k)}$ to ${\sigma }_i^{(k+1)}$. Hence $p^{-1}(W)={⨆}_k{T}_k\cup {\bigcup }_{i,k}{E}_{i,k}$ is the locally finite graph with vertices $v_k,{\sigma }_i^{(k)}$ and edges $T_{i,k},E_{i,k}$, a one-dimensional CW complex carrying the weak topology, and the deck action sends $T_{i,k}\mapsto T_{i,k+1}$ and $E_{i,k}\mapsto E_{i,k+1}$.

Clause (3): the cellular chain complex. The CW complex $\mathrm{\Sigma }$ has zero-cells $\{v_k\}$ and one-cells $\{e_i^{(k)}\}$ and no higher cells. With integral coefficients, $C_1(\mathrm{\Sigma };\mathbb{Z})={⨁}_{i,k}\mathbb{Z}{e}_i^{(k)}$ and $C_0(\mathrm{\Sigma };\mathbb{Z})={⨁}_k\mathbb{Z}{v}_k$ by [F5], and the cellular boundary is $d_1(e_i^{(k)})=v_{k+1}-v_k$: the relative class of the oriented edge is sent by the connecting morphism to the class of its boundary $[\mathrm{\partial }{e}_i^{(k)}]=[v_{k+1}]-[v_k]$ under the conventions of [F5]. The deck action makes these chain groups ${\mathrm{\Lambda }}_1$-modules with $t\cdot e_i^{(k)}=e_i^{(k+1)}$ and $t\cdot v_k=v_{k+1}$; writing $e_i:=e_i^{(0)}$ and $v:=v_0$, they are free modules $C_1(\mathrm{\Sigma })={⨁}_{i=1}^n{\mathrm{\Lambda }}_1{e}_i$ and $C_0(\mathrm{\Sigma })={\mathrm{\Lambda }}_{1}v$, and ${\mathrm{\Lambda }}_1$-linearity of $d_1$ gives ${\mathrm{\partial }}_1{e}_i=(t-1)v$. For the pair $(\mathrm{\Sigma },{\mathrm{\Sigma }}^0)$ one has $C_1(\mathrm{\Sigma },{\mathrm{\Sigma }}^0)=C_1(\mathrm{\Sigma })$ and $C_0(\mathrm{\Sigma },{\mathrm{\Sigma }}^0)=0$, so $C_1(\mathrm{\Sigma },{\mathrm{\Sigma }}^0)={⨁}_{i=1}^n{\mathrm{\Lambda }}_1{ϵ}_i$ with ${ϵ}_i$ the relative class of $e_i$ and $H_1(\mathrm{\Sigma },{\mathrm{\Sigma }}^0)=C_1(\mathrm{\Sigma },{\mathrm{\Sigma }}^0)$ because there are no $2$-cells.

Clause (2): the collapse onto the spine. Let $\mathrm{\Sigma }$ be the quotient of $p^{-1}(W)$ obtained by collapsing each tree $T_k$ to the vertex $v_k$; its cells are the vertices $v_k$ and the images $e_i^{(k)}$ of the arcs $E_{i,k}$, each an edge $v_k\to v_{k+1}$, so $\mathrm{\Sigma }$ is the lifted spine with ${\mathrm{\Sigma }}^0=\{v_k\}=p^{-1}d$ and with deck translation $t\cdot v_k=v_{k+1}$, $t\cdot e_i^{(k)}=e_i^{(k+1)}$. Let $q:p^{-1}(W)\to \mathrm{\Sigma }$ be the quotient map. Parametrize each tether edge $T_{i,k}$ by $u\in [0,1]$ from $v_k$ to ${\sigma }_i^{(k)}$, each circle edge $E_{i,k}$ by $v\in [0,1]$ from ${\sigma }_i^{(k)}$ to ${\sigma }_i^{(k+1)}$, and let $L_{i,k}:[0,3]\to p^{-1}(W)$ be the concatenation of $T_{i,k}$, $E_{i,k}$ and the reverse of $T_{i,k+1}$. Define $j:\mathrm{\Sigma }\to p^{-1}(W)$ by $j(v_k)=v_k$ and by mapping $e_i^{(k)}$ onto the arc $L_{i,k}$ increasingly; then $j$ is continuous. On the unit parameter $v$ of every spine edge, $q\circ j$ has parameter $\eta (v)=\max (0,\min (1,3v-1))$: the two tether thirds collapse. The interpolation $(1-u)\eta (v)+uv$ defines a deck-equivariant homotopy $qj\simeq {\mathrm{id}}_{\mathrm{\Sigma }}$ relative to the vertices. Define ${\tilde{H}}_u$ on $p^{-1}(W)$ by ${\tilde{H}}_u(T_{i,k}(v)):=T_{i,k}((1-u)v)$ and ${\tilde{H}}_u(E_{i,k}(v)):=L_{i,k}({\psi }_u(v))$ with ${\psi }_u(v):=(1-u)(1+v)+3uv$. The values at ${\sigma }_i^{(k)}$ agree from the two incident circle edges and the tether, at ${\sigma }_i^{(k+1)}$ likewise, and at $v_k$ all definitions give $v_k$; on the locally finite closed-cell cover this defines a continuous homotopy fixing every root $v_k$, with ${\tilde{H}}_0=\mathrm{id}$ and ${\tilde{H}}_1=j\circ q$. Hence $q$ is a homotopy equivalence of pairs with homotopy inverse $j$: $q\circ j\simeq {\mathrm{id}}_{\mathrm{\Sigma }}$ relative to ${\mathrm{\Sigma }}^0$ and $j\circ q\simeq \mathrm{id}$ through the homotopy $\tilde{H}$, which fixes ${\mathrm{\Sigma }}^0=p^{-1}d$ and is deck-equivariant because every formula is stated in the canonical cell parameters and $T_{t^k}\circ L_{i,l}=L_{i,l+k}$.

The homology isomorphisms are ${\mathrm{\Lambda }}_1$-linear. By step 1.2 the inclusion-induced map $H_1(p^{-1}(W))\to H_1(\tilde{X})$ is an isomorphism, natural for the deck actions, and it maps $H_1(p^{-1}d)$ identically; by step 2.1 the homotopy equivalence $q$ induces isomorphisms $H_n(p^{-1}(W))\to H_n(\mathrm{\Sigma })$ and, by the five lemma applied to the commuting map of pair long exact sequences established in [F6], an isomorphism $H_1(p^{-1}(W),p^{-1}d)\to H_1(\mathrm{\Sigma },{\mathrm{\Sigma }}^0)$, while $q{\mathrm{\mid }}_{p^{-1}d}$ is the identity onto ${\mathrm{\Sigma }}^0$. Since all these maps commute with the deck actions, they are isomorphisms of ${\mathrm{\Lambda }}_1$-modules for the structures induced by those actions, and the absolute case gives $H_1(\tilde{X})\cong H_1(\mathrm{\Sigma })$ while the relative case gives $H_1(\tilde{X},p^{-1}d)\cong H_1(\mathrm{\Sigma },{\mathrm{\Sigma }}^0)$; this is clause (2).

The kernel and the ranks. Write an element of $C_1(\mathrm{\Sigma })$ as ${\sum }_{i=1}^n{a}_i{e}_i$ with $a_i\in {\mathrm{\Lambda }}_1$. Since $d_1$ is ${\mathrm{\Lambda }}_1$-linear and ${\mathrm{\partial }}_1{e}_i=(t-1)v$, one has ${\mathrm{\partial }}_1\left({\sum }_i{a}_i{e}_i\right)=\left({\sum }_i{a}_i\right)(t-1)v$; as $C_0(\mathrm{\Sigma })={\mathrm{\Lambda }}_{1}v$ is free of rank one and ${\mathrm{\Lambda }}_1$ is a domain with $t-1\mathrm{\ne }0$ by [F7], this vanishes exactly when ${\sum }_i{a}_i=0$. Hence $H_1(\mathrm{\Sigma })=\ker {\mathrm{\partial }}_1={⨁}_{i=1}^{n-1}{\mathrm{\Lambda }}_1(e_i-e_n)$, free of rank $n-1$, and $H_1(\mathrm{\Sigma },{\mathrm{\Sigma }}^0)={⨁}_{i=1}^n{\mathrm{\Lambda }}_1{ϵ}_i$ is free of rank $n$; this is clause (3).

Conclusion. Clause (1) is step 1.2, clause (2) is step 3.1, and clause (3) is step 4.1; all constructions used one fixed contraction data set and explicit formulas, so no choice principle is used.