Calculating normalization of the projective cusp $V(x^2z-y^3)$ I've understood the normalization of the affine cusp $V(x^2-y^3)\subset\mathbb{A}^2$ is just $\phi:\mathbb{A}^1\to \operatorname{Spec}(\mathbb{C}[x,y]/(x^2-y^3))$ coming from the algebra map
$$
\phi^*:\mathbb{C}[x,y]/(x^2-y^3) \to \mathbb{C}[t] \\
x \mapsto t^3 \\
y\mapsto t^2
$$
How does this extend to the projective curve $V=V(x^2z-y^3)$? Do we get a similar map $\phi:\mathbb{P}^1\to\operatorname{Proj}(\mathbb{C}[x,y,z]/(x^2z-y^3))$ coming from the map
$$
\phi^*:\mathbb{C}[x,y,z]/(x^2z-y^3) \to \mathbb{C}[t] \\
x \mapsto t^3 \\
y\mapsto t^2 \\
z \mapsto 1
$$
by just extending the affine algebra map? What happens for the point $(1:0:0)$ with $z=0$?
 A: Since $\DeclareMathOperator{\Proj}{Proj} \DeclareMathOperator{\Spec}{Spec} \newcommand{\C}{\mathbb{C}} \mathbb{P}^1 = \Proj(\C[t_0, t_1])$, then $\C[t_0, t_1]$ is the homogeneous coordinate ring of $\mathbb{P}^1$, so the desired homomorphism of graded rings is
\begin{align*}
\phi^*:\mathbb{C}[x,y,z]/(x^2z-y^3) &\to \mathbb{C}[t_0, t_1] \\
x &\mapsto t_0^3 \\
y &\mapsto t_0^2 t_1 \\
z &\mapsto t_1^3 \, .
\end{align*}
From this, we see that the point $(t_0 : t_1) = (1:0)$ maps to the point $(x:y:z) = (1:0:0)$.
Now $\newcommand{\P}{\mathbb{P}} \newcommand{\A}{\mathbb{A}} \P^1$ is covered by two copies of $\A^1$, which map homeomorphically onto the subsets of $\P^1$ where $t_0 \neq 0$ and $t_1 \neq 0$, respectively. These two maps $\psi_0, \psi_1: \mathbb{A}^1 \hookrightarrow \P^1$ correspond to the ring homomorphisms
\begin{align*}
\psi_0^*: \mathbb{C}[t_0, t_1] &\to \mathbb{C}[t]\\
t_0 &\mapsto 1\\
t_1 &\mapsto t
\end{align*}
and
\begin{align*}
\psi_1^* : \mathbb{C}[t_0, t_1] &\to \mathbb{C}[t]\\
t_0 &\mapsto t\\
t_1 &\mapsto 1 \, .
\end{align*}
The map you wrote in the question is the composition $\psi_1^* \circ \phi^* = (\phi \circ \psi_1)^*$, corresponding to the map
\begin{align*}
\mathbb{A}^1 &\overset{\psi_1}{\hookrightarrow} \P^1 \overset{\phi}{\to} \mathbb{V}(x^2 z - y^3)\\
t &\mapsto [t : 1] \mapsto [t^3 : t^2 : 1] \, .
\end{align*}
You noted that the point $(1:0:0)$ does not lie in the image of this map, which is because the point $(1:0)$ which maps to it under $\phi$ is not contained in the image of $\psi_1$, but rather in the image of $\psi_0$.
As a note, not every map of graded rings induces a map of $\Proj$s. In general, given a map $\varphi: S_\bullet \to R_\bullet$ of graded rings, there is an induced morphism of schemes
$$
\varphi^*: \Proj(R_\bullet) \setminus \mathbb{V}(\varphi(S_+)) \to \Proj(S_\bullet)
$$
where $S_+$ is the irrelevant ideal. (See $\S6.4$ of Vakil's The Rising Sea or Tag 01MX of the Stacks Project.) Basically the problem is that, if $\varphi$ has a nontrivial kernel, then elements in this kernel will get mapped to "$(0: 0 : \cdots : 0)$", which is not a point in projective space. Fortunately in the example above,
\begin{align*}
\mathbb{V}(\phi^*(S_+)) &= \mathbb{V}(\phi^*(x,y,z)) = \mathbb{V}(t_0^3, t_0^2 t_1, t_1^3) = \varnothing
\end{align*}
so we obtain a map defined on all of $\P^1$.
