Suppose a function $f:\mathbb {RP}^1\to \mathbb {RP}^1$ satisfy: $$ \left[f(a),f(b);f(c),f(d)\right]=\left[a,b;c,d\right] $$ for all $a,b,c,d \in \mathbb {RP}^1$.
What can the function be in general? Möbius transformations are certainly one type, but are there any other? Suppose the function is linear, or differentiable, I can prove that there are none. But can we do this without these assumptions?
Möbius transformations are the only such functions. Indeed, just note that $[0,\infty;x,1]=x$ for all $x\in\mathbb{RP}^1$, and so we must have $$x=[0,\infty;x,1]=[f(0),f(\infty);f(x),f(1)].$$ This implies $a=f(0)$, $b=f(\infty)$, and $c=f(1)$ are all distinct (otherwise the cross-ratio on the right would be the same for all $x$). But given three distinct points $a,b,c\in\mathbb{RP}^1$, the function $g(x)=[a,b;x,c]$ is a Möbius transformation $\mathbb{RP}^1\to\mathbb{RP}^1$. The equation above then says that $g(f(x))=x$ for all $x$ so $f$ must be the inverse of $g$, which is also a Möbius transformation.
