Path-class independence and equivariance in the topological construction

Use the established assignment. The first two assertions are exactly the independence and equivariance proved in [F1]. The reverse directed arrow is defined using $X_{C^{\mathrm{\prime }},C}$, by the same assignment with the chambers interchanged.

Track an argument around the separating hyperplane. Represent $p(C,C^{\mathrm{\prime }})$ and $p(C^{\mathrm{\prime }},C)$ by their prescribed two-piece paths and concatenate them to a loop $\mathrm{\Gamma }$ based at $a_C$. Apply the continuous nonvanishing map $z\mathrm{/}\mathrm{\mid }z\mathrm{\mid }$ from [F2] to obtain the unit-circle loop $\lambda (t)=z(\mathrm{\Gamma }(t))\mathrm{/}\mathrm{\mid }z(\mathrm{\Gamma }(t))\mathrm{\mid }$. On the slit plane containing the image of $X_C$, use the continuous argument $A$ in $(-\pi \mathrm{/}2,3\pi \mathrm{/}2)$; on the slit plane containing the image of $X_{C^{\mathrm{\prime }}}$, use the continuous argument $B$ in $(-3\pi \mathrm{/}2,\pi \mathrm{/}2)$. At the first intermediate point, whose real part is positive, $A=B$. At the second, whose real part is negative, $A=B+2\pi$. Along the four successive path segments use $A,B,B,A-2\pi$, respectively. These branches agree at the joins and give a continuous real argument $\theta$ for $\lambda$ satisfying $\theta (1)-\theta (0)=-2\pi$. The conclusion is unaffected by the particular representatives, since the stated restrictions hold throughout each path segment.

Normalize the basepoint and compute degree. Set $\mu (t)=\lambda (t)\mathrm{/}\lambda (0)$; multiplication by the fixed unit complex number $\lambda (0{)}^{-1}$ is a homeomorphism of the unit circle, so $\mu$ is a loop based at $1$. By [F5], $\beta =h^{-1}\circ \mu$ is a loop in $\mathbb{R}\mathrm{/}\mathbb{Z}$ based at $[0]$. Define $\tilde{\beta }(t)=(\theta (t)-\theta (0))\mathrm{/}(2\pi )$. This is continuous, starts at $0$, and satisfies $h(q(\tilde{\beta }(t)))=\mu (t)$ by the sine and cosine formula for $h$ and the definition of $\theta$. Injectivity of $h$ therefore gives $q\circ \tilde{\beta }=\beta$. Thus $\tilde{\beta }$ is precisely the lift in [F6], and $\mathrm{deg}(\beta )=\tilde{\beta }(1)=-1$.

Distinguish the classes. By [F4] and step 2.1, $\beta$ is not endpoint-fixed homotopic to the constant loop at $[0]$. If $\mathrm{\Gamma }$ were endpoint-fixed homotopic to the constant loop at $a_C$, composing that homotopy with the continuous map $u\mapsto h^{-1}((z(u)\mathrm{/}\mathrm{\mid }z(u)\mathrm{\mid })\mathrm{/}\lambda (0))$ would give such a homotopy of $\beta$, because this map sends $a_C$ to $[0]$. This is a contradiction. If $p(C^{\mathrm{\prime }},C)$ were $p(C,C^{\mathrm{\prime }}{)}^{-1}$, their concatenation would represent the identity at $a_C$ by [F4], which would give exactly the prohibited homotopy of $\mathrm{\Gamma }$. Hence the reverse arrow is not the inverse path class.