Given the quadric $Q=V(XT-YZ)\subset\mathbb{P}^3$ and the lines $L_{X,Y}=V(X,Y)\subset Q$ and $L_{Z,T}=V(Z,T)\subset Q$, we have the morphism $\Phi: Q\rightarrow\mathbb{P}^{1}$ given by:
$\Phi(X,Y,Z,T)=\begin{cases} (Z,T),\;\text{ in } Q\setminus L_{Z,T}\\ (X,Y),\;\text{ in } Q\setminus L_{X,Y}\\ \end{cases}$,
which is clearly well defined since both images coincide in $(Q\setminus L_{Z,T})\cap (Q\setminus L_{X,Y})$.
My question is the following:
I want to prove that the morphism $\Phi$ cannot be written as $\Phi(X,Y,Z,T)=(P_{1}(X,Y,Z,T),\;P_{2}(X,Y,Z,T))$ for all $(X,Y,Z,T)\in\mathbb{P}^3$, where $P_{1}$ and $P_{2}$ are homogeneous polynomials of the same degree.
My attempt has been the following:
My idea was to prove that such polynomials cannot exists because they would share some common zeros and then the point $[0:0:0:0]$ will be in the image of the morphism, which cannot be possible.
To do so, I use the relation given by the quadric $Q$, i.e. $XT-YZ$, to express $P_{1}$ and $P_{2}$ as polynomials with the form $P_{1}(X,Y,Z,T)=F_{1}(X,Y,T)+G_{1}(X,Z,T)$ and $P_{2}(X,Y,Z,T)=F_{2}(X,Y,T)+G_{2}(X,Z,T)$, where $F$ and $G$ are homogeneous polynomials of the same degree as $P_{1}$ and $P_{2}$.
Then, I have tried to see how the polynomials $P_{1}$ and $P_{2}$ behave when applied on points from $L_{X,Y}$ and $L_{Z,T}$ (especially in the points $[1:0:0:0],\;[0:1:0:0],\;[0:0:1:0],\;[0:0:0:1]$) and imposing the conditions given by the morphism $\Phi$.
However, I have not been capable to prove that $P_{1}$ and $P_{2}$ share any common zero in $Q$. I would be very grateful if someone could help me to figure out how to show that both polinomyals $P_{1}$ and $P_{2}$ vanish at some common point in $Q$.
Thanks!