My question is different from this question since my question asks to clarify the proof, while the other question asks to check if the proof is correct.
Let $z_1, z_2, z_3\in\mathbb{C}_\infty$ be distinct points. Then $\alpha(z):=[z; w_1, w_2, w_3]=[\mu(z); \mu(w_1), \mu(w_2), \mu(w_3)]=\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}$ for all $z\in\mathbb{C}_\infty$ (where $\mu$ is a Möbius map). This means that Möbius maps preserve cross-ratios.
Here's the proof I have in the notes:
$\mu^{-1}$ is a Möbius map, and:
$$\alpha\circ \mu^{-1}(\mu(w_1))=\alpha(w_1)=0 $$ $$\alpha\circ \mu^{-1}(\mu(w_2))=\alpha(w_2)=1 $$ $$\alpha\circ \mu^{-1}(\mu(w_3))=\alpha(w_1)=\infty $$
Thus $\alpha \circ \mu^{-1}(z)=[z;\mu(w_1), \mu(w_2), \mu(w_3)], \forall z\in \mathbb{C}_\infty$.
What I don't understand is how this proves that Möbius maps preserve cross-ratios. I would assume that we would have to take $\alpha(\mu(w))$ to show the preservation.
Moreover, if we show that some map is equal to $\alpha$ at just three points, how does this prove that two cross-ratios are equal? Perhaps I don't understand something about the meaning of cross-ratios.