The planar two-point configuration space retracts to the quotient circle

For an ordered pair put $c=(x+y)\mathrm{/}2$ and $z=(x-y)\mathrm{/}2$. This is an invertible linear coordinate change, with inverse $(c,z)\mapsto (c+z,c-z)$; the collision-free condition is $z\mathrm{\ne }0$. In polar coordinates write $z=ru$ with $r=\mathrm{\parallel }z\mathrm{\parallel }>0$ and $u\in S^1$. These formulas and their inverses are continuous, so they give a homeomorphism $F_2({\mathbb{R}}^2)\cong {\mathbb{R}}^2\times (0,\mathrm{\infty })\times S^1$. Transposition fixes $c,r$ and sends $u$ to $-u$. If $A:=S^1\mathrm{/}(u\sim -u)$ has quotient topology, the map $(c,r,u)\mapsto (c,r,[u])$ is an open quotient map: the circle quotient is open because the saturation of an open set is its union with its antipodal image, and products of open rectangles show the same for the product map. Its fibers are exactly the transposition orbits. Thus it induces a homeomorphism $C_2({\mathbb{R}}^2)\cong X:={\mathbb{R}}^2\times (0,\mathrm{\infty })\times A\mathrm{.}$

Let $\eta :\mathbb{R}\mathrm{/}\mathbb{Z}\to S^1$ be the homeomorphism $\eta ([t])=(\cos 2\pi t,\sin 2\pi t)$ from [F5]. The antipodal map on $S^1$ corresponds under $\eta$ to $[t]\mapsto [t+1\mathrm{/}2]$. The continuous map $D_2:\mathbb{R}\mathrm{/}\mathbb{Z}\to \mathbb{R}\mathrm{/}\mathbb{Z}$, $D_2([t]):=[2t]$, is well-defined and continuous by [F6] applied to $t\mapsto [2t]$ on $\mathbb{R}$; its fibres are exactly the pairs $[t],[t+1\mathrm{/}2]$. Hence $\psi :A\to \mathbb{R}\mathrm{/}\mathbb{Z}$, $\psi ([\eta ([t])]):=[2t]$, is well-defined and continuous by the quotient topology. Its inverse is $[s]\mapsto [\eta ([s\mathrm{/}2])]$; changing $s$ by an integer changes $s\mathrm{/}2$ by a half-integer, which is the same class in $A$. This inverse is continuous by [F6] applied to the continuous map $s\mapsto [\eta ([s\mathrm{/}2])]$ on $\mathbb{R}$; the two formulas compose to the identity, so $\psi$ is a homeomorphism. Under the homeomorphism of step 1.1, $q_0$ is $(0,1,[(1,0)])$, which $\psi$ sends to $[0]$.

On $X$ define $r(c,\rho ,a):=\psi (a)$ and $i([t]):=(0,1,{\psi }^{-1}([t]))$. These maps are continuous, preserve the specified basepoints, and $r\circ i$ is the identity. The maps $H_s(c,\rho ,a):=((1-s)c,(1-s)\rho +s,a),0\le s\le 1,$ are a continuous based homotopy from ${\mathrm{id}}_X$ to $i\circ r$: the radius is positive since $(1-s)\rho +s>0$, and the basepoint $(0,1,[(1,0)])$ is fixed for every $s$.

By [F3], $(r\circ i{)}_{\ast }=r_{\ast }{i}_{\ast }=\mathrm{id}$ and the based homotopy in step 3.1 gives $(i\circ r{)}_{\ast }=i_{\ast }{r}_{\ast }=\mathrm{id}$ on the based-loop classes of [F2]. Thus $r_{\ast }$ and $i_{\ast }$ are inverse isomorphisms. The pointed homeomorphism of step 1.1 transports this isomorphism back to $C_2({\mathbb{R}}^2)$; applying [F4] gives ${\pi }_1(C_2({\mathbb{R}}^2),q_0)\cong \mathbb{Z}$.

In the coordinates of step 1.1, $\gamma$ keeps $c=0$ and $r=1$ and has $u(t)=(\cos \pi t,\sin \pi t)$. Its image under $r$ is $[t]={\omega }_1(t)$, which is the positive generator by [F4]. Replacing $\pi t$ by $2\pi t$ describes a full turn of the relative direction; its image is $[2t]={\omega }_2(t)$, so its class is twice $[\gamma ]$. This identifies exchange and winding orientation without using a diagram.