Let $\mathbb P^n=\mathbb {CP^n}$. A covering by curves of genus $g$ is a set of complete curves $\{C_\lambda\}$ such that
(i) every $C_\lambda$ is a complete curve of genus $g$ on $\mathbb P^n$.
(ii) there exists a zariski open subset $U\subset \mathbb P^n$, such that for every closed point $x\in U$ there exists an unique curve $C_\lambda$ containing $x$.
For example, fibers of the linear projection $$\mathbb P^n \dashrightarrow \mathbb P^{n-1}$$ define a covering by curves of genus $0$.
My question is:
For which genera $g$ there exists a covering by curves for $\mathbb P^n$?
My attempt:
A general fiber of the composition $$\mathbb P^n \rightarrow \mathbb P^n\dashrightarrow \mathbb P^{n-1}$$ (where the second map is a linear projection) is of genus $1+\frac{1}{2}d^{n-1}(nd-d-n-1)$, and it defines a covering by curves of such genus. I don't know if these are the only possibilities?
Pick your favourite nonsingular complete curve $C$ in $\mathbb P^n$ and project it to a curve $C'$ in $\mathbb A^n$, given as the set of common zeros of $n-1$ polynomials $f_1, \ldots, f_{n-1} \in \mathbb C[X_1, \ldots, X_n]$.
For every $a = (a_1, \ldots, a_{n-1}) \in \mathbb C^{n-1}$, you can define a curve $C'_a := \{p \in \mathbb A^n : f_i(p) = a_i \text{ for } i=1, \ldots, n-1\}$.
Observe that for every $p \in \mathbb A^n$ there is exactly one $a = (f_1(p), \ldots, f_{n-1}(p))$ such that $p \in C'_a$.
Let $A \subset \mathbb C^{n-1}$ be the set of those $a$ such that $C_a$ has the same genus as $C = C_0$, where $C_a$ is the projectivization of $C'_a$. Note that $A$ has nonempty Zariski interior in $\mathbb{C}^{n-1}$.
I now claim that $\{C_a\}_{a\in A}$ is a covering by curves of $\mathbb P^n$, of the same genus as $C$. I leave the remaining details to you, if you don't mind.