Degree $2$ birational map of $\mathbb{P}^1$

Question

I know that there are many examples of birational degree $2$ projective maps.

One example based on the $2$-dimensional Lyness map is $$ \varphi([x,y,z]) = [x y, yz + z^2, x z] $$ with inverse map $$ \psi([x,y,z]) = [x z+z^2, x y, y z]. $$ The composite maps satisfy $$ \varphi(\psi([x,y,z])) = y z(x+z)[x, y, z], \qquad \psi(\varphi([x,y,z])) = x z(y+z)[x, y, z]. $$

I am failing to find nontrivial examples of one dimension less. By nontrivial, here I mean it is not a degree $1$ map or projectively equivalent to such a map. The setup is $$ \varphi([x,y]) = [ax^2+bxy+cy^2, dx^2+exy+fy^2] $$ with inverse map $$ \psi([x,y]) = [a'x^2+b'xy+c'y^2, d'x^2+e'xy+f'y^2] $$ given constants $\,a,b,c,d,e,f,a',b',c',d',e',f'\,$ such that the composite maps satisfy $$ \varphi(\psi([x,y])) = p(x,y)[x, y], \qquad \psi(\varphi([x,y])) = q(x,y)[x, y] $$ for some homogeneous cubic polynomials $\,p(x,y),q(x,y).\,$

Is there a simple example, or else is there a simple reason that nontrivial examples do not exist?

Answer 1

Since any automorphisms of $\mathbb{P^1}$ are linear transformations, we only need to extend such a rational map $\varphi$ to a morphism $\bar{\varphi} \colon \mathbb{P^1} \to \mathbb{P^1}$. There are several ways to explain the claim.

The easiest approach is to express $\varphi$ by using the usual coordinate $z = x/y$. Then $\varphi$ turns out to be a rational function of $z$ and hence it defines a morphism from $\mathbb{P^1}$ to $\mathbb{P^1}$.

Another way is to use Riemann's removable singularity theorem. In this case the compactness of the codomain $\mathbb{P^1}$ assures the continuous extension of $\varphi$ to the entire domain, and it is actually holomorphic by Riemann's theorem.

This behavior is a special case of the categorical equivalence between the category of

1. smooth projective curves and dominant morphisms;
2. quasi-projective curves and dominant rational maps;
3. function fields of dimension 1 over k, and k-homomorphisms;
4. compact Riemann surfaces and surjective holomorphic maps (if the base field is $\mathbb{C}$).

The equivalence between 1. and 2. provides a completion of the rational map $\varphi$ (which is dominant since the inverse map $\psi$ exists) to a morphism. The essential point of the argument above is the compactness (= projectiveness) of $\mathbb{P^1}$. In general, a dominant rational map between smooth projective curves can be completed to a (dominant) morphism.