Question about $\mathbb{P^1}$ - without loss of generality doubt

Question

I want to show that:

Given three distinct points $P_1,P_2,P_3 \in \mathbb{P}^1$ and three distinct points $Q_1,Q_2,Q_3 \in \mathbb{P}^1$, there is a unique isomorphism $f: \mathbb{P}^1 \rightarrow \mathbb{P}^1$ such that $f(P_i)=Q_i$ for $i \in \{1,2,3\}$.

My question is:

Why do we can suppose without loss of generality that $P_1=0, P_2=1$ and $P_3=\infty$?

If someone can carefully explain this I'd be grateful.

By the way, I think I can solve the rest of the problem, no need to answer the whole problem.

Thanks!

Answer 1

Suppose we are guaranteed to have such an isomorphism $\phi_Q$, given any three $Q_i$; that is, $\phi_Q$ maps the triple $(0,1, \infty)$ to $(Q_1, Q_2, Q_3)$.

Then we would have such a $\phi_R$, which maps $(0, 1, \infty)$ to $(R_1, R_2, R_3)$, for any $R_i$'s we want.

What would happen to $(Q_1, Q_2, Q_3)$, using the isomorphism $\phi_R \circ \phi_Q^{-1}$?