Mobius transformation maps $\Bbb R_∞$ onto itself iff we can choose its coefficients to be real

Question

I have seen many solutions which are very intricate and/or long. The solution I had was much shorter which lead me to believe that it was incorrect. It went like this :

$T$ be a Mobius transformation such that $T(\Bbb R_∞)$=$\Bbb R_∞$. $1,0,∞\in \Bbb R_∞$ so their inverse images are also in $\Bbb R_∞$. $(z,z_2,z_3,z_4)$ is the unique Mobius transformation that maps $z_2,z_3,z_4$ to respectively $1,0,∞$. So when we write $T$ as a cross-ratio, the coefficients are all real.

Is this solution really incorrect?

Answer 1

That solution looks correct, at least for one part. The other part --- if the coeffs are real, then the map sends reals to reals --- is trivial.

I think that I'd edit slightly by saying "so their inverse images, $\mathbf{ z_2, z_3, z_4}$ are also in..." so that $z_2, z_3, z_4$ are actually defined. Otherwise it looks good to me.