Finite morphism $f:X \to \mathbb{P}_k^n$ is surjective?

Question

Let $X$ be an $n$-dimensional projective $k$-scheme and let $f: X \to \mathbb{P}_k^n$ be a finite morphism.

1. Is $f$ necessarily surjective?

2. If not, then what else do we need to impose such that $f$ is surjective?

What I tried:

1. If $f$ is dominant, then $f(X)$ is dense and closed and hence equal to $\mathbb{P}_k^n$. Now $f$ is dominant if and only if $f(X)$ contains the generic point of $\mathbb{P}_k^n$.
2. Since $f$ is finite, $f(X)$ is closed and thus by the irreducibility of $\mathbb{P}_k^n$ $$ f(X) = \mathbb{P}_k^n \quad \Leftrightarrow \quad \operatorname{dim}f(X) = \operatorname{dim}\mathbb{P}_k^n = n $$ Now if finite morphism do (to some extent) preserve dimensions, then this could be an approach.

Answer 1

Let $X_0\subset X$ be an irreducible component of $X$ of dimension $n$ (you didn't suppose $X$ irreducible!). Its image $Y:=f(X_0)\subset \mathbb P^n_k$ is irreducible and closed ( because $f$ is finite and thus closed).
Since $f\vert X_0:X_0\to Y$ is finite and surjective we have $n=\operatorname {dim}(X_0)=\operatorname {dim} (Y)$.
But now the inclusion of irreducible varieties $Y\subset \mathbb P^n_k$ both of dimension $n$ forces $f(X_0)=Y=\mathbb P^n_k$, so that a fortiori $f(X)=\mathbb P^n_k$, i.e. $f$ is surjective as desired.