Suppose we have the projective variety $X = \mathbb{V}(y^3 – xz^2) \subseteq \mathbb{P}^2$ and the map $\mathbb{P}^1 \to X$ given by:
$$\phi([s:t]) = [s^3 : st^2 : t^3]$$
First of all: why is this bijective? I can see that $X$ is the projective closure of of the affine variety $\mathbb{V}(y^3 – z^2) = \{(t^2, t^3) : t \in k\}$. So if we restrict to the set where $x\neq 0$ then scale so $x=1$, the points of the image are $[1:t^2:t^3]$. So we can define an inverse on this set by $[1:y:z] \mapsto [1:y/z]$.
But on the rest of $X$, where $x=0$, we have $y=0$ (by the equation defining $X$) and any value of $z$ is in $X$. So now what do we map $[0:0:z]$ to? I want to say $[0:0:z] \mapsto [0:\sqrt[3]{z}]$. But how do we know which cube root to pick to define this map? Or can we recover $t$ some other way?
Once we have the inverse map, I want to prove it is not an isomorphism onto $X$. The inverse map (in terms of set theory) is not a morphism of projective varieties because it is not locally polynomial, in some places it has $y/z$ in its definition. Is this enough? Or maybe could it be possible (a priori) that there are some locally polynomial expressions for this and I need to prove this doesn't exist?
And why are these projective varieties birational? I think we can use $\phi$ as given and the map $[1:y:z] \mapsto [1:y/z]$ for a rational map the other way which is inverse. I mean on the open set where everything is defined (on the set where $x \neq 0$ and I guess $s\neq 0$ for $\psi$).
Best Answer
As mentioned in the comments, we can check $\phi$ doesn't have an inverse, since $X$ has a singularity at $[1:0:0]$.
As for how to show $\phi$ is bijective (and I'm assuming your varieties are only the $k$-points here), it's fairly straightforward to show it directly.
Suppose $[s^3:st^2:t^3]$, if $t=0$, we get $[1:0:0]$, and the only way we can get this point is if $t=0$ respectively.
Then suppose $$[s^3:st^2:t^3]=[s'^3:s't'^2:t'^3],$$ with $t,t'\ne 0$, then $$[s:t]=[st^2:t^3]=[s't'^2:t'^3]=[s':t'].$$
Thus $\phi$ is injective.
Now for surjectivity, suppose $[x:y:z]$ satisfies $y^3-xz^2=0$. If $z=0$, then $y^3=0$, so $y=0$, $x=1$, and $\phi([1:0])=[1:0:0]$.
If $z\ne 0$, then $[y:z]$ is a valid point in $\Bbb{P}^1$. $\phi([y:z])=[y^3:yz^2:z^3]$, and since $y^3=xz^2$, so we have $$\phi([y:z]) = [xz^2:yz^2:z^3] = [x:y:z],$$ as desired.
Finally, as for why $\phi$ is an isomorphism off a proper closed subset, note that when $z\ne 0$, the morphism $[x:y:z]\mapsto [y:z]$ is an inverse to $\phi$.