Is any map preserving cross ratios a Mobius map?

Question

If a map $f:\mathbb{C}_\infty \to \mathbb{C}_\infty$ preserves all cross ratios, is it necessarily a Mobius map?

Answer 1

Yes, any function preserving all cross-ratios is necessarily a Möbius map and the proof for this is outlined below, assuming the result that Möbius maps preserve cross-ratios.

Definition For any $4$ distinct points $z_1,z_2,z_3,z_4\in \mathbb{C}^{\infty}$, their cross-ratio is defined as$$ [z_1,z_2,z_3,z_4]=\frac{(z_1-z_3)(z_2-z_4)}{(z_1-z_2)(z_3-z_4)} $$ such that if $z_i$ is $\infty$ for some $i\in \{1,2,3,4\}$ then $[z_1,z_2,z_3,z_4]$ is the limit of the ratio as $z_i\to \infty$.

First one can prove that any function $f$ which preserves all cross-ratios must be injective. Say $f(z_1)=f(z_2)$ and $z_1,z_2,z_3,z_4$ are distinct points in $\mathbb{C}^{\infty}$. Now the cross-ratio $[f(z_1),f(z_3),f(z_2),f(z_4)]$ is undefined inspite of $[z_1,z_3,z_2,z_4]$ being a valid non-zero cross-ratio. In fact even if one defines $[a,b,a,c]=0$ then $[f(z_1),f(z_3),f(z_2),f(z_4)]=0$, so the relation $[z_1,z_3,z_2,z_4]=[f(z_1),f(z_3),f(z_2),f(z_4)]$ does not hold.

Theorem If $(z_1,z_2,z_3,z_4)$ and ($w_1,w_2,w_3,w_4$) are tuples of distinct points in $\mathbb{C}^{\infty}$ then there is a Möbius map $g$ satisfying $$g(z_i)=w_i\quad \forall\quad i\in \{1,2,3,4\}$$ (Hint: Find a Möbius map sending $z_1, z_2,z_4$ to $0,1,\infty$ respectively and notice $[0,1,x,\infty]=x$. Use the fact that cross-ratios are invariant under Möbius maps.)

Now if $z_1,z_2,z_3,z_4$ are distinct points in $\mathbb{C}^{\infty}$ and $f$ is any map preserving cross-ratios, then $[z_1,z_2,z_3,z_4]=[f(z_1),f(z_2),f(z_3),f(z_4)]$. Using the above theorem, there is a Möbius map $g$ satisfying $$g(z_i)=f(z_i)\quad \forall\quad i\in \{1,2,3,4\}$$ We wish to prove $f=g$. Let $w$ be a point in $\mathbb{C}^{\infty}$, distinct from $z_1,z_2$ and $z_3$. Then $$[f(z_1),f(z_2),f(z_3),f(w)]=[z_1,z_2,z_3,w]=[g(z_1),g(z_2),g(z_3),g(w)]$$ Letting $f(z_1)=a, f(z_2)=b$ and $f(z_3)=c$, we get $$[a,b,c,f(w)]=[a,b,c,g(w)]$$ $$\frac{(a-c)(b-f(w))}{(a-b)(c-f(w))}=\frac{(a-c)(b-g(w))}{(a-b)(c-g(w))}$$ From the injectivity of $g$ and $f$, $a,b,c$ are distinct, and $f(w)$ and $g(w)$ are distinct from each of them. So, $$(b-f(w))(c-g(w))=(b-g(w))(c-f(w))$$ $$c\,(f(w)-g(w))=b\,(f(w)-g(w))$$ $$f(w)-g(w)=0$$ As the choice of $w$ was arbitrary, we have proved $f$ is the Möbius map $g$.