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How to prove that if $a,b,c,d$ and $a',b',c',d'$ are 2 quadruples of distinct points in extended complex plane, and if the cross ratios of these quadruples are equal then there exists Möbius transformation $M$ such that $M(a)=a', M(b)=b',M(c)=c',M(d)=d'$

I am reading a paper and there is already proof that if there exists then the cross ratio is preserved but the remaining is left as an exercise.Any hint would be appreciated.

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One definition of cross ratio is particularly apt here: if you begin with $(z_1,z_2,z_3,z_4)$ and send it by a linear fractional transformation to $(0,1,\infty, u),$ then the cross ratio of $(z_1,z_2,z_3,z_4)$ is defined to be $u.$

After that your exercise is easy.

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Lemma. The group of Möbius transformations $\mathrm{PGL}_2(\Bbb C)$ acts sharply 3-transitively on the Riemann sphere and has three complex dimensions as a manifold. That is, for any pair of 3-tuples of points from the extended complex plane, there is one and only one transformation mapping one point to the other point. I will leave this lemma as a different exercise - it is only linear algebra.

$\bullet$ Let $(z_1,z_2,z_3,z_4)$ and $(w_1,w_2,w_3,w_4)$ be two tuples with the same cross-ratio $r\in\Bbb C$. Since linear equations have unique solutions, $v=w_4$ is the only solution to $[w_1,w_2,w_3;v]=r$.

So let $A=\bigl(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\bigr)\in\mathrm{PGL}_2(\Bbb C)$ be the unique Möbius transformation that sends $(z_1,z_2,z_3)$ to the point $(w_1,w_2,w_3)$. Applying $A$ to $(z_1,z_2,z_3,z_4)$ and taking the cross-ratio, we have

$r=[z_1,z_2,z_3,z_4]=[Az_1,Az_2,Az_3,Az_4]=[w_1,w_2,w_3,Az_4]$

hence $Az_4=w_4$ (by $\bullet$) and $A(z_1,z_2,z_3,z_4)=(w_1,w_2,w_3,w_4)$ as desired. Note this shows that Möbius transformations act sharply transitively on $4$-tuples with a given cross-ratio.