# What is the flaw in the answer for proving any rational function on PP^1 is constant? Le

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.
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Zernerqcw
A regular function on a variety X is, by definition, a regular map $X\to {\mathrm{\forall }}^{1}$. Just because every regular function on $P{P}^{1}$ is constant, that doesn't mean that every regular map to a variety other than ${\mathrm{\forall }}^{1}$ is constant. In particular, $\varphi$, being a regular map $P{P}^{1}\to P{P}^{1}$, need not be constant.