A homeomorphism $T$ from extended complex plane to itself preserving cross ratio is a Mobius map. Cross ratio preserving means $(Ta,Tb,Tb,Td)=(a,b,c,d)$ where $(a,b,c,d)=\dfrac{(a-b)(c-d)}{(a-d)(c-b)}$.
 If we assume $T$ fixes infinity, can we prove $T$ is affine?
 A: My answer needs two facts about Möbius transforms:


*

*They preserve cross ratios (this can be checked by a direct calculation).

*A Möbius transform can map three arbitrary points to any other three arbitrary points (this is not completely obvious, but at least very intuitive since Möbius transforms have 3 independent complex degrees of freedom).


You also need to know that for any fixed pairwise distinct $u, v, w \in \mathbb{C} \cup \{\infty\}$, the function $f : \mathbb{C} \cup \{\infty\} \to \mathbb{C} \cup \{\infty\}, f(x) = (u, v; w, x)$ is bijective (again an easy calculation). This means that the images of three distinct points $a, b, c$ under your map $T$ already completely determine it. 
However, there also is a Möbius $M$ transform that maps $a, b, c$ to $T(a), T(b), T(c)$ and, by virtue of being a Möbius transform preserves cross-ratios. Therefore $M = T$.
We can express 
$$M(z) = \frac{\alpha z + \beta}{\gamma z + \delta}$$
Suppose $M$ fixes $\infty$, i.e. $M(\infty) = \frac{\alpha \cdot \infty + \beta}{\gamma \cdot \infty + \delta} = \infty$. This is the case iff $\gamma = 0$. In this case $M(z) = \frac\alpha\delta z + \frac\beta\delta$ is affine.
