What is the flaw in the answer for proving any rational function on $P{P}^{1}$ is constant?

Let$\varphi :P{P}^{1}\to {\mathrm{\forall}}^{1}$ be a rational function. Since $\mathrm{\forall}}^{1}\subset P{P}^{1$ we can think $\varphi :P{P}^{1}\to P{P}^{1}$ as rational function. As every rational map from $P{P}^{1}\to P{P}^{n}$ is regular, so $\varphi$ is regular. We also know that any regular function on $P{P}^{1}$ is constant, so $\varphi$ is constant.

Let