A unique Möbius transformation carries any ordered triple of distinct sphere points to any other

Define $N(z):=\{\begin{array}{ll}{\displaystyle \frac{(z-b)(c-a)}{(z-a)(c-b)}},& a,b,c\in \mathbb{C},\\ {\displaystyle \frac{z-b}{c-b}},& a=\mathrm{\infty },\\ {\displaystyle \frac{c-a}{z-a}},& b=\mathrm{\infty },\\ {\displaystyle \frac{z-b}{z-a}},& c=\mathrm{\infty },\end{array}$ and define $N^{\mathrm{\prime }}$ by the same formula with $(a,b,c)$ replaced by $(a^{\mathrm{\prime }},b^{\mathrm{\prime }},c^{\mathrm{\prime }})$. In each case the displayed formula is Möbius and direct substitution gives $N(a)=\mathrm{\infty }$, $N(b)=0$, $N(c)=1$ and $N^{\mathrm{\prime }}(a^{\mathrm{\prime }})=\mathrm{\infty }$, $N^{\mathrm{\prime }}(b^{\mathrm{\prime }})=0$, $N^{\mathrm{\prime }}(c^{\mathrm{\prime }})=1$.

The composition $(N^{\mathrm{\prime }}{)}^{-1}\circ N$ is Möbius by [L1] and carries $(a,b,c)$ to $(a^{\mathrm{\prime }},b^{\mathrm{\prime }},c^{\mathrm{\prime }})$, so the required map exists.

If another Möbius map had the same three values, then composing with $N$ and $N^{\mathrm{\prime }}$ would produce a Möbius map fixing $0$, $1$, and $\mathrm{\infty }$; writing it as $(\alpha z+\beta )\mathrm{/}(\gamma z+\delta )$ forces $\beta =\gamma =0$ and then $\alpha =\delta$, so it is the identity. Thus the map is unique.