The planar exchange character quotient is a flat unitary line bundle

The projection $p(c,\rho ,t)=(c,\rho ,[t])$ is continuous and surjective. For any $[t_0]\in \mathbb{R}\mathrm{/}\mathbb{Z}$, choose an open interval $I=(a,b)$ around a representative of $[t_0]$ with $b-a<1$. Its image is an open arc because its full preimage ${\bigcup }_{n\in \mathbb{Z}}(I+n)$ is open by [F5, F7]; the translates are disjoint by $b-a<1$. On each sheet $I+n$, the quotient map $t\mapsto [t]$ is a homeomorphism onto that arc: it is a continuous bijection and maps open subsets to open subsets by the same saturation check. Taking products with the other two coordinate factors gives evenly covered open neighborhoods in $X$, so [F2] makes $p$ a covering map.

By [F4], a deck transformation $D$ satisfies $pD=p$. Hence $D$ fixes the ${\mathbb{R}}^2$ and $(0,\mathrm{\infty })$ coordinates and has last coordinate $t+k(c,\rho ,t)$ for an integer-valued continuous function $k$, by [F5]. The space $\tilde{X}$ is connected because it is convex: its radius coordinate stays positive on every line segment. Therefore $k$ is constant; thus $D=T_n$ for some $n\in \mathbb{Z}$. Conversely each $T_n$ preserves $p$ and is a homeomorphism, so these are exactly the deck transformations.

The space $\tilde{X}$ is nonempty and convex in ${\mathbb{R}}^4$ because the radius coordinate remains positive along every line segment. For any basepoint $x_0\in \tilde{X}$ and based loop $\alpha$, the straight-line homotopy $H_s(t)=(1-s)\alpha (t)+s{x}_0$ remains in $\tilde{X}$ and fixes both endpoints. Thus every based loop from [F9] is null-homotopic, so [F6] makes $\tilde{X}$ simply connected. By [F3], $p$ is a universal covering map.

The action in [F8] is a group action: composing the $m$ and $n$ actions adds the shifts and multiplies the fiber coordinate by $\chi (m{)}^{-1}\chi (n{)}^{-1}=\chi (m+n{)}^{-1}$. The action is free because $T_n\tilde{x}=\tilde{x}$ implies $n=0$. If $U$ is an evenly covered neighborhood as in step 1.1 and $V$ is one chosen sheet, every orbit over $U$ has exactly one representative in $V\times \mathbb{C}$. The quotient topology therefore identifies ${\mathrm{\Pi }}_{\chi }^{-1}(U)$ homeomorphically with $U\times \mathbb{C}$: the restriction from $V\times \mathbb{C}$ is bijective, and it is open because the saturation of an open subset is the union of its translates under the homeomorphisms in [F8]. The fiber coordinate is the unique representative's complex coordinate. This gives a local trivialization with one-dimensional complex fibers.

On an overlap, two chosen sheets differ locally by a fixed translation $T_n$, because the sheets of a covering are disjoint and indexed by the discrete integers. The corresponding fiber coordinate changes by $z\mapsto \chi (n)z$ or its inverse, according to the direction of the chart change. Since $\mathrm{\mid }\chi (n)\mathrm{\mid }=1$ and $n$ is locally constant, these transition functions are locally constant $U(1)$-valued functions. By [F8], this is a unitary flat complex line bundle.

Let a based loop have a lift from $\tilde{x}$ to $T_n\tilde{x}$; the path-lifting theorem [F10] supplies a unique lift from every prescribed starting point. Keeping $z$ constant along this lift gives endpoint class $[T_n\tilde{x},z]$. The defining equivalence in [F8] gives $[T_n\tilde{x},z]=[\tilde{x},\chi (n)z]$, so monodromy is multiplication by $\chi (n)$. The explicit positive exchange path has lifted angle coordinate $t\mapsto t$ from $0$ to $1$, while the full relative turn has $t\mapsto 2t$ from $0$ to $2$, by the homeomorphism in [F1]. Thus their monodromies are $\chi (1)$ and $\chi (2)$.