proof of converse of Möbius transformation cross-ratio preservation theorem

Suppose that $a,b,c,d$ are distinct. Consider the transform $\mu$ defined as

$$\mu (z)=\frac{(b-d)(c-d)}{(c-b)(z-d)}-\frac{b-d}{c-b}.$$

Simple calculation reveals that $\mu (b)=1$, $\mu (c)=0$, and $\mu (d)=\mathrm{\infty }$.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 mayconstruct a transform ${\mu }_1$ which maps the first tetrad to $(\lambda ,1,0,\mathrm{\infty })$ and atransform ${\mu }_2$ which maps the first tetrad to $(\lambda ,1,0,\mathrm{\infty })$. Then ${\mu }_2^{-1}\circ {\mu }_1$ maps the former tetrad to the latter and, by the group property, it is also aMöbius transformation.