The two-point unordered cover of the plane and the monodromy of a half turn

The transposition acts by $(c,w)\mapsto (c,-w)$. Let $x=(x_1,x_2)\in F_2(\mathbb{C})$. By [F2] one has $(\tau \cdot x{)}_1=x_{{\tau }^{-1}(0)+1}=x_2$ and $(\tau \cdot x{)}_2=x_{{\tau }^{-1}(1)+1}=x_1$, so $\tau \cdot x=(x_2,x_1)$. Hence $c(\tau \cdot x)=\frac{x_2+x_1}{2}=\frac{x_1+x_2}{2}=c(x),w(\tau \cdot x)=x_1-x_2=-(x_2-x_1)=-w(x),$ using the field laws of [F5]. This is claim 1.

$\bar{\mathrm{\Phi }}$ is continuous. The composite $\bar{\mathrm{\Phi }}\circ p:F_2(\mathbb{C})\to \mathbb{C}\times Q$ has components $p_1(x)=c(x)$ and $p_2(x)=\rho (w(x))$: the first is continuous by [F1], and the second is the composite of the continuous $w$ of [F1] with the quotient map $\rho$, which is continuous by [F4]. Hence $\bar{\mathrm{\Phi }}\circ p$ is continuous by the product criterion [F7], and therefore $\bar{\mathrm{\Phi }}$ is continuous by the characteristic property of the quotient map $p$ [F3] (equivalently, [F4] applied to the final topology of $p$).

The half turn $\gamma$ is a continuous path in ${\mathbb{C}}^{\times }$ from $1$ to $-1$. On the closed interval $[0,\frac{1}{2}]$ the map $\gamma (t)=(1-2t)+2ti$ is built from the continuous inclusion $t\mapsto t$ of $[0,\frac{1}{2}]$ into $\mathbb{C}$ (the restriction of the identity, [F7]) by the continuous operations of multiplication by the constants $\mp 2$ and $i$ and of addition, so it is continuous by [F5] and [F7]; the same holds on $[\frac{1}{2},1]$ for $\gamma (t)=-(2t-1)+(2-2t)i$. At $t=\frac{1}{2}$ both formulas give $i$, and $[0,\frac{1}{2}]\cup [\frac{1}{2},1]=I$ is a finite closed cover, so $\gamma$ is continuous by [F7]. For $0\le t\le \frac{1}{2}$ one has $\gamma (t)=u(t)$ with $u(t)=(1-2t)+2ti$, whose real and imaginary parts are $1-2t$ and $2t$, so $u(t)\overline{u(t)}=(1-2t{)}^2+(2t{)}^2$ by [F6]; this is a sum of two squares of real numbers vanishing only if $1-2t=0$ and $2t=0$ simultaneously, which is impossible, so $\gamma (t)\mathrm{\ne }0$. For $\frac{1}{2}\le t\le 1$ the same computation with the real and imaginary parts $-(2t-1)$ and $2-2t$ gives $\mathrm{\mid }\gamma (t){\mathrm{\mid }}^2=(2t-1{)}^2+(2-2t{)}^2\mathrm{\ne }0$. Finally $\gamma (0)=1$ and $\gamma (1)=-1$.

$\bar{\mathrm{\Phi }}$ and $\bar{\mathrm{\Psi }}$ are well defined and mutually inverse. If $x$ and $\tau \cdot x$ are representatives of the same orbit, then by step 1.1 their coordinates are $(c(x),w(x))$ and $(c(x),-w(x))$, and $\rho (-w(x))=\{w(x),-w(x)\}=\rho (w(x))$; since $p$ is the orbit map [F3], $\bar{\mathrm{\Phi }}$ is well defined. Likewise, if $w^{\mathrm{\prime }}=\pm w$ then by step 1.1 and [F1] $\mathrm{\Psi }(c,-w)=\tau \cdot \mathrm{\Psi }(c,w)$, so $\mathrm{\Psi }(c,w)$ and $\mathrm{\Psi }(c,w^{\mathrm{\prime }})$ lie in the same orbit and $\bar{\mathrm{\Psi }}$ is well defined. Moreover $\bar{\mathrm{\Psi }}(\bar{\mathrm{\Phi }}([x]))=[\mathrm{\Psi }(c(x),w(x))]=[x]$ and $\bar{\mathrm{\Phi }}(\bar{\mathrm{\Psi }}(c,\rho (w)))=\bar{\mathrm{\Phi }}([\mathrm{\Psi }(c,w)])=(c,\rho (w))$ by [F1] and the definition of $\bar{\mathrm{\Phi }}$. So $\bar{\mathrm{\Phi }}$ is a bijection with inverse $\bar{\mathrm{\Psi }}$.

$\bar{\mathrm{\Psi }}$ is continuous. The quotient map $\rho$ is open: for an open $O\subseteq {\mathbb{C}}^{\times }$, its saturation ${\rho }^{-1}(\rho (O))=O\cup (-O)$ is open, since multiplication by $-1$ is a homeomorphism by [F5]; hence $\rho (O)$ is open by [F4]. It follows that $Q_0:={\mathrm{id}}_{\mathbb{C}}\times \rho :\mathbb{C}\times {\mathbb{C}}^{\times }\to \mathbb{C}\times Q$ is an open continuous surjection: on each basic open box it has the open image $U\times \rho (O)$, and every open set is a union of such boxes by [F7]. An open continuous surjection is a quotient map. The continuous map $p\circ \mathrm{\Psi }:\mathbb{C}\times {\mathbb{C}}^{\times }\to C_2(\mathbb{C})$ is constant on the fibres of $Q_0$ by step 2.1. Therefore it factors continuously through $Q_0$ by the quotient characteristic property [F4], and its factor is exactly $\bar{\mathrm{\Psi }}$.

Claim 2. Steps 1.2, 2.1 and 3.1 exhibit $\bar{\mathrm{\Phi }}$ as a continuous bijection with continuous inverse $\bar{\mathrm{\Psi }}$, that is, a homeomorphism; hence $C_2(\mathbb{C})\cong \mathbb{C}\times ({\mathbb{C}}^{\times }\mathrm{/}\{\pm 1\})$.

$\alpha$ is a based loop at $[q]$ and $\tilde{\alpha }$ is a lift. The map $t\mapsto w_0\gamma (t)$ is continuous as a product of continuous complex-valued maps [F5, F7] and takes values in ${\mathbb{C}}^{\times }$: $\mathrm{\mid }{w}_0\gamma (t)\mathrm{\mid }=\mathrm{\mid }{w}_0\mathrm{\mid }\mathrm{\mid }\gamma (t)\mathrm{\mid }\mathrm{\ne }0$ by [F6] and step 1.3. Hence $t\mapsto (c_0,\rho (w_0\gamma (t)))$ is continuous into $\mathbb{C}\times Q$ by the product criterion [F7], and composing with the continuous $\bar{\mathrm{\Psi }}$ of step 3.1 gives that $\alpha$ is continuous. Since $\rho (w_0)=\{w_0,-w_0\}=\rho (-w_0)$, one has $\alpha (0)=\bar{\mathrm{\Psi }}(c_0,\rho (w_0))=\bar{\mathrm{\Psi }}(c_0,\rho (-w_0))=\alpha (1)$, and $\alpha (0)={\bar{\mathrm{\Phi }}}^{-1}((c_0,\rho (w_0)))=[q]$ because $\bar{\mathrm{\Phi }}([q])=(c_0,\rho (w_0))$ by [F1] and step 2.1. So $\alpha$ is a based loop at $[q]$. The path $\tilde{\alpha }(t):=\mathrm{\Psi }(c_0,w_0\gamma (t))$ is continuous into $F_2(\mathbb{C})$ by [F1], starts at $\mathrm{\Psi }(c_0,w_0)=q$, and satisfies $p\circ \tilde{\alpha }=\alpha$ because $p(\mathrm{\Psi }(c,w))=\bar{\mathrm{\Psi }}(c,\rho (w))$ for all $w\in {\mathbb{C}}^{\times }$ by the definition of $\bar{\mathrm{\Psi }}$ in step 2.1.

The endpoint of the lift is $\tau \cdot q$, so the monodromy is nontrivial. By [F8] $p$ is a covering map, so [F9] gives a unique lift of the path $\alpha$ starting at $q$; by step 4.2 the path $\tilde{\alpha }$ is such a lift, hence it is that unique lift. Its endpoint is $\tilde{\alpha }(1)=\mathrm{\Psi }(c_0,w_0\gamma (1))=\mathrm{\Psi }(c_0,-w_0)$ by step 1.3, and by [F1] $\mathrm{\Psi }(c_0,-w_0)=(c_0+\frac{w_0}{2},c_0-\frac{w_0}{2}),q=\mathrm{\Psi }(c_0,w_0)=(c_0-\frac{w_0}{2},c_0+\frac{w_0}{2}),$ so $\tilde{\alpha }(1)=(q_2,q_1)=\tau \cdot q$ by step 1.1; equivalently $q\cdot [\alpha ]=\tau \cdot q$ in the sense of [F9], and the permutation ${\sigma }_{\alpha }$ with $\tilde{\alpha }(1)={\sigma }_{\alpha }\cdot q$ is $\tau$. Since the action is free and $\tau \mathrm{\ne }\mathrm{id}$, one has $\tau \cdot q\mathrm{\ne }q$ by [F2], so the monodromy is nontrivial: the half turn of the difference coordinate does not lift to a loop in $F_2(\mathbb{C})$. The deck transformation $\tau$ carries the lift starting at $q$ to the lift starting at $\tau \cdot q$ and carries its endpoint $\tau \cdot q$ to $q$. Thus the monodromy transposes both points of the fibre and fixes neither. A trivial two-sheeted covering has identity monodromy around every loop, so this covering is not trivial.

Conclusion. Claim 1 is step 1.1, claim 2 is step 4.1, and claim 3 is steps 1.3, 4.2 and 5.1: the transposition acts on centre and difference coordinates by $(c,w)\mapsto (c,-w)$, the unordered space of two points is homeomorphic to $\mathbb{C}\times ({\mathbb{C}}^{\times }\mathrm{/}\{\pm 1\})$ through $\bar{\mathrm{\Phi }}$, and the half turn of the difference coordinate is a based loop at $[q]$ with nontrivial endpoint monodromy $\tau$. No choice principle was used.