Ordered and Unordered Configuration Spaces — Examples

$\mathrm{\Phi }$ is well defined. Let $(z_1,z_2)\in F_2(\mathbb{C})$, so $z_1\mathrm{\ne }{z}_2$. If $z_2-z_1=0$ then adding $z_1$ gives $z_2=z_1$ by [F2], a contradiction; hence $z_2-z_1\mathrm{\ne }0$ and $\mathrm{\Phi }(z_1,z_2)\in \mathbb{C}\times {\mathbb{C}}^{\times }$.

$\mathrm{\Psi }$ is well defined. Let $(c,w)\in \mathbb{C}\times {\mathbb{C}}^{\times }$, so $w\mathrm{\ne }0$. The two coordinates of $\mathrm{\Psi }(c,w)$ differ by $(c+\frac{w}{2})-(c-\frac{w}{2})=w\mathrm{\ne }0$, hence are distinct, and $\mathrm{\Psi }(c,w)\in F_2(\mathbb{C})$.

Linear combinations on the plane are continuous. Let $\alpha ,\beta \in \mathbb{C}$ and consider $L:{\mathbb{C}}^2\to \mathbb{C}$, $L(z_1,z_2):=\alpha z_1+\beta z_2$. Let $x=(x_1,x_2)\in {\mathbb{C}}^2$ and $\epsilon >0$, and put $\delta :=\epsilon \mathrm{/}(1+\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid })>0$, a positive real. If $\mathrm{\mid }{z}_k-x_k\mathrm{\mid }<\delta$ for $k=1,2$, that is, if $z$ lies in the basic open box $B(x_1,\delta )\times B(x_2,\delta )$ of [L4], then by [L3] $\mathrm{\mid }L(z)-L(x)\mathrm{\mid }\le \mathrm{\mid }\alpha \mathrm{\mid }\mathrm{\mid }{z}_1-x_1\mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid }\mathrm{\mid }{z}_2-x_2\mathrm{\mid }<(\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid })\delta \le \epsilon ,$ with strict inequality $\mathrm{\mid }L(z)-L(x)\mathrm{\mid }<\epsilon$ in every case: if $\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid }=0$ then $L(z)-L(x)=0<\epsilon$, and otherwise $(\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid })\delta =\epsilon (\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid })\mathrm{/}(1+\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid })<\epsilon$. Hence the preimage of the ball $B(L(x),\epsilon )$ contains the box $B(x_1,\delta )\times B(x_2,\delta )$ around $x$, and since balls form a basis of the topology of $\mathbb{C}$ and boxes a basis of the topology of ${\mathbb{C}}^2$ [L4], $L$ is continuous by [L5].

$\mathrm{\Phi }$ and $\mathrm{\Psi }$ are mutually inverse, so $\mathrm{\Phi }$ is a bijection. Let $(z_1,z_2)\in F_2(\mathbb{C})$ and put $c:=\frac{z_1+z_2}{2}$, $w:=z_2-z_1$. By the field laws of [F2] and $2\cdot \frac{1}{2}=1$, $c-\frac{w}{2}=\frac{z_1+z_2-(z_2-z_1)}{2}=\frac{2z_1}{2}=z_1,c+\frac{w}{2}=\frac{z_1+z_2+(z_2-z_1)}{2}=\frac{2z_2}{2}=z_2,$ so $\mathrm{\Psi }(\mathrm{\Phi }(z_1,z_2))=(z_1,z_2)$. Conversely, for $(c,w)\in \mathbb{C}\times {\mathbb{C}}^{\times }$ the first coordinate of $\mathrm{\Phi }(\mathrm{\Psi }(c,w))$ is $\frac{(c-\frac{w}{2})+(c+\frac{w}{2})}{2}=c$ and the second is $(c+\frac{w}{2})-(c-\frac{w}{2})=w$, so $\mathrm{\Phi }(\mathrm{\Psi }(c,w))=(c,w)$. Thus $\mathrm{\Phi }$ has the two-sided inverse $\mathrm{\Psi }$ and is a bijection.

$\mathrm{\Phi }$ is continuous. The two components of $\mathrm{\Phi }$ are the restrictions to the subspace $F_2(\mathbb{C})\subseteq {\mathbb{C}}^2$ of the continuous maps $L_{\frac{1}{2},\frac{1}{2}}(z_1,z_2)=\frac{z_1+z_2}{2}$ and $L_{-1,1}(z_1,z_2)=z_2-z_1$ of step 1.3, hence are continuous by [F1]. The second of them takes all its values in the subspace ${\mathbb{C}}^{\times }\subseteq \mathbb{C}$ by step 1.1, so it is a continuous map $F_2(\mathbb{C})\to {\mathbb{C}}^{\times }$ by the subspace criterion of [F1]; therefore $\mathrm{\Phi }$, whose target is the product $\mathbb{C}\times {\mathbb{C}}^{\times }$, is continuous by the product criterion of [L5].

$\mathrm{\Psi }$ is continuous. The target $F_2(\mathbb{C})$ is a subspace of ${\mathbb{C}}^2$, so by the subspace criterion of [F1] it suffices to show that the composite $H:\mathbb{C}\times {\mathbb{C}}^{\times }\to {\mathbb{C}}^2$, $H(c,w)=(c-\frac{w}{2},c+\frac{w}{2})$, is continuous. By the product criterion of [L5] it suffices that the two components $L_{1,-\frac{1}{2}}(c,w)=c-\frac{w}{2}$ and $L_{1,\frac{1}{2}}(c,w)=c+\frac{w}{2}$ be continuous as maps into $\mathbb{C}$. The domain $\mathbb{C}\times {\mathbb{C}}^{\times }$ is a subspace of ${\mathbb{C}}^2$: by [L4] its basic open sets are the boxes $U\times (V\cap {\mathbb{C}}^{\times })$ with $U,V$ open in $\mathbb{C}$, and these are exactly the traces $(U\times V)\cap (\mathbb{C}\times {\mathbb{C}}^{\times })$ of the boxes $U\times V$ on ${\mathbb{C}}^2$, so the two topologies coincide. Hence continuity of the components follows from the continuity of $L_{1,\mp \frac{1}{2}}$ on ${\mathbb{C}}^2$ in step 1.3 together with the restriction clause of [F1].

Conclusion. By step 2.1 and step 1.2 the map $\mathrm{\Phi }$ is a bijection $F_2(\mathbb{C})\to \mathbb{C}\times {\mathbb{C}}^{\times }$ with inverse $\mathrm{\Psi }$; by steps 2.2 and 2.3 both $\mathrm{\Phi }$ and $\mathrm{\Psi }$ are continuous. Hence $\mathrm{\Phi }$ is a homeomorphism and $F_2(\mathbb{C})\cong \mathbb{C}\times {\mathbb{C}}^{\times }$, as claimed. No choice principle was used.

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.

The formula defines a left action on the whole product. For $\sigma \in S_n$ and $y\in X^n$ the tuple $\sigma \cdot y$ with $(\sigma \cdot y{)}_i=y_{{\sigma }^{-1}(i-1)+1}$ is well defined in $X^n$, because ${\sigma }^{-1}(i-1)+1$ is a label for every $i$ by [F2]; and the formal verification of $\mathrm{id}\cdot y=y$ and $(\sigma \tau )\cdot y=\sigma \cdot (\tau \cdot y)$ for $\sigma ,\tau \in S_n$ is the reindexing computation of [F2], which never uses the distinctness of coordinates, so it applies to all of $X^n$. Hence [F3] and [F4] apply to this action.

The collision witness. Fix an element $x\in X$, which exists by [F4], and put $a:=(x,x,\dots ,x)\in X^n$, the tuple with $a_i=x$ for every label $i$; its coordinates collide, and $a\mathrm{\notin }{F}_n(X)$ when $n\ge 2$ because $a_1=a_2$. So $a$ is a point of $X^n$ to which the freeness conclusion of [F2] does not apply.

The transposition fixes $a$. By step 1.1 the tuple $\tau \cdot a\in X^n$ is defined, and for every label $i$ one has $(\tau \cdot a{)}_i=a_{{\tau }^{-1}(i-1)+1}=x=a_i$, because all coordinates of $a$ equal $x$; hence $\tau \cdot a=a$ by [F5].

Freeness fails. The transposition $\tau$ is not the identity by [F3], yet it fixes the point $a$ of $X^n$ by step 2.1. Therefore the action of $S_n$ on $X^n$ is not free, in the sense of the definition in [F4].

Conclusion. The claim stated in the refuted statement is false for every $n\ge 2$ and every nonempty $X$, with the explicit witness $a=(x,\dots ,x)$ fixed by the transposition $\tau =(01)$. The contrast with [F2] is exactly the removal of the collision diagonals: freeness of the coordinate permutation action is a property of $F_n(X)$, not of the full product $X^n$.