Conditions for a well-defined map on a variety $A\subset \mathbb{P}^2$

Question

Let $A\subset \mathbb{P}^2$ be the set defined by the equation $X^3=Y^2Z$. I have the map $f:\mathbb{P}^1\rightarrow V$, $[S,T]\rightarrow [S^2T, S^3, T^3]$. I want to find another rational map $g:V\rightarrow \mathbb{P}^1$ (not necessarily a morphism) such that $f\circ g$ and $g\circ f$ are the identity wherever they are defined. My candidate function is $g([X,Y])=[X^2,YZ]$. It does the job, but also arises a problem. If $Y=0$, then also $X=0$ (because in $V$ we have $X^3=YZ$) and we can assume $Z=1$. So then, $g([0,0,1])=[0,0]$, which is not a point in $\mathbb{P}^1$. Why does this happen if my map is well-defined? Also, how can I avoid it?

Answer 1

If such a map exists, you have an isomorphism from $V$ to $\mathbb{P}^1$. However, as $V$ is the projectivization of the cusp, it has a singularity at $[0,0,1]$, while $\mathbb{P}^1$ is regular. So there is no such map.