We work over the complex numbers.
Let $\mathbb{A}^1\times \mathbb{A}^1\to \mathbb{A}^1$ be the morphism given by $(x,y)\mapsto x-y$.
Does this extend to a morphism $\mathbb{P}^1\times \mathbb{P}^1\to \mathbb{P}^1$?
I have the feeling it does not because it contracts the diagonal (which is ample on $\mathbb{P}^1\times \mathbb{P}^1$), but I can't think of an "easier" argument.
I’m not entirely sure this works, but here’s an idea: the “minus” operation is defined on $\mathbb{P}^1$ (it’s basically $y \longmapsto y,x \longmapsto -x$ in homogeneous coordinates) so it’s obviously equivalent to consider the addition.
Let $\mu: \mathbb{P}^1 \times \mathbb{P}^1 \rightarrow \mathbb{P}^1$ be an extension of the addition, it is proper. Now, it’s easy to see that $\mu(\{\infty\} \times \mathbb{A}^1)=\{\infty\}$, and same the other way around. By the rigidity lemma, it follows that $\mu$ is constant: contradiction.
