Suppose that $a,b,c,d$ are distinct. Consider the transform $\mu$ defined as $$ \mu (z) = {(b-d) (c-d) \over (c-b) (z-d)} - {b-d \over c-b} . $$ Simple calculation reveals that $\mu (b) = 1$ $\mu (c) = 0$ and $\mu(d) = \infty$ Furthermore, $\mu (a)$ equals the cross-ratio of $a,b,c,d$

Suppose we have two tetrads with a common cross-ratio $\lambda$ Then, as above, we may construct a transform $\mu_1$ which maps the first tetrad to $(\lambda, 1, 0, \infty)$ and a transform $\mu_2$ which maps the first tetrad to $(\lambda, 1, 0, \infty)$ Then $\mu_2^{-1} \circ \mu_1$ maps the former tetrad to the latter and, by the group property, it is also a Möbius transformation.