Möbius transformations preserve circlines and conjugate their reflections

Writing $C=C(a,b,c)$, the defining points $a,b,c$ map to $M(a),M(b),M(c)\in C(M(a),M(b),M(c))$. If $z\mathrm{\notin }\{a,b,c\}$, then [L2] says $z\in C$ exactly when $[a,b;c,z]\in \mathbb{R}$. By [L1] this is equivalent to $[M(a),M(b);M(c),M(z)]\in \mathbb{R}$, which is exactly the condition $M(z)\in C(M(a),M(b),M(c))$. Thus $M(C)=C(M(a),M(b),M(c))$ is again a circline.

If $C=C(a,b,c)$, then [L3] gives a Möbius map $N$ with $N(a)=\mathrm{\infty }$, $N(b)=0$, and $N(c)=1$. Step 1.1 makes $N(C)$ a circline. Its defining triple $\mathrm{\infty },0,1$ already lies in $C(\mathrm{\infty },0,1)$, and for every $w\mathrm{\notin }\{\mathrm{\infty },0,1\}$ one has $[\mathrm{\infty },0;1,w]=w$, so [L2] makes $w\in C(\mathrm{\infty },0,1)$ exactly when $w\in \mathbb{R}$. Hence $N(C)=C(\mathrm{\infty },0,1)=\widehat{\mathbb{R}}$, so every circline admits a Möbius normalization to the standard real circline.

Let $N$ and $N^{\mathrm{\prime }}$ be two normalizing maps for $C$, and put $H:=N^{\mathrm{\prime }}\circ N^{-1}$. By [L4], $H$ is Möbius and preserves $\widehat{\mathbb{R}}$. The map $\tilde{H}:={\sigma }_{\widehat{\mathbb{R}}}\circ H\circ {\sigma }_{\widehat{\mathbb{R}}}$ is Möbius by conjugating the coefficients of a fractional-linear formula. Moreover, $H$ and $\tilde{H}$ agree at $0$, $1$, and $\mathrm{\infty }$, because these points and their $H$-images lie in $\widehat{\mathbb{R}}$. By [L3], $\tilde{H}=H$, equivalently $H\circ {\sigma }_{\widehat{\mathbb{R}}}={\sigma }_{\widehat{\mathbb{R}}}\circ H$.

Since $N^{\mathrm{\prime }}=H\circ N$, step 3.1 gives $(N^{\mathrm{\prime }}{)}^{-1}\circ {\sigma }_{\widehat{\mathbb{R}}}\circ N^{\mathrm{\prime }}=N^{-1}\circ H^{-1}\circ {\sigma }_{\widehat{\mathbb{R}}}\circ H\circ N=N^{-1}\circ {\sigma }_{\widehat{\mathbb{R}}}\circ N\mathrm{.}$ Thus the reflection ${\sigma }_C$ is independent of the normalizing map.

Choose a normalizing map $N$ for $C$. Since step 1.1 makes $M(C)$ a circline, $N\circ M^{-1}$ normalizes $M(C)$ to $\widehat{\mathbb{R}}$. Using the well-defined reflection from step 4.1 gives ${\sigma }_{M(C)}=(N\circ M^{-1}{)}^{-1}\circ {\sigma }_{\widehat{\mathbb{R}}}\circ (N\circ M^{-1})=M\circ {\sigma }_C\circ M^{-1}\mathrm{.}$