An exchange closes only after forgetting labels

Track the ordered pair. The midpoint of $q_1$ and $q_2$ is $0$, so the published half-twist formula in [L4] gives the ordered path ${\tilde{\sigma }}_1(t)=(\rho (t),-\rho (t))\mathrm{.}$ Its coordinates lie in $D^{\circ }$ because $\parallel \rho (t)\parallel \le h<1$, and they are distinct because their difference is $2\rho (t)\mathrm{\ne }0$. Thus this is a path in $F_2(D^{\circ })$, with endpoints ${\tilde{\sigma }}_1(0)=(q_1,q_2)=Q,{\tilde{\sigma }}_1(1)=(q_2,q_1)\mathrm{.}$

Since $q_1\mathrm{\ne }{q}_2$, the terminal ordered tuple $(q_2,q_1)$ is not $Q$. By [L7], ${\tilde{\sigma }}_1$ is a path in the ordered configuration space but is not a based loop at $Q$.

Forget the labels. By [L2], the endpoint tuples have the same orbit: $[(q_1,q_2)]=[(q_2,q_1)]=[Q]\mathrm{.}$ By [L3], the quotient projection is continuous, so $\alpha (t):=[{\tilde{\sigma }}_1(t)]$ is a continuous path in $C_2(D^{\circ })$ with $\alpha (0)=\alpha (1)=[Q]$. By [L5, L6], this is the slice path $S({\sigma }_1)$ of the positive geometric half twist, so it is the unordered based loop promised by the counterexample.

The ordered coordinate path fails to close at $Q$, while its unordered image closes at $[Q]$. Hence forgetting labels can close a nonlooping ordered coordinate path, refuting the claim in the statement.