Projective varieties of dimension $m$

Question

Let $X \subset \mathbb{CP}^n$ be a projective variety of dimension $m$. Is it true that $X$ can be realized as the image of a holomorphic map $f: \mathbb{CP}^m \to \mathbb{CP}^n$?

To this end, if there is a non-constant map $f: \mathbb{CP}^m \to X$ then it must be a surjection. This follows by the open mapping theorem together with compactness of $\mathbb{CP}^m$ and connectedness of the target. The thing that is unclear to me is whether or not one can always construct a non-constant holomorphic function from $\mathbb{CP}^m$ to an arbitrary projective variety.

Answer 1

One situation where this is impossible is when $X \subset \mathbb{CP}^2$ is a elliptic curve (i.e. a smooth complex projective curve of genus 1, realisable as a cubic hypersurface in $\mathbb{CP}^2$).

If there exists a surjective map $f : \mathbb{CP}^1 \to X$, then by the Riemann-Hurwitz formula, we would have $$ 2g(\mathbb{CP}^1) - 2 = N (2g(X) - 2) + \sum_{P \in \mathbb{CP}^1} (e_P-1),$$ where $N$ is the degree of the covering map $f : \mathbb{CP}^1 \to X$, and for each $P \in \mathbb {CP}^1$, $e_P$ is the ramification index of $f$ at $P$ (which is greater than one at only finitely many $P \in \mathbb{CP}^1$).

Now, $g(\mathbb {CP}^1) = 0$, and $g(X) = 1$, so the left-hand side is negative, while the right-hand side is non-negative. This is a contradiction.

Answer 2

If $X$ is smooth and not the projective space itself, then this is impossible by the following:

Theorem [Lazarsfeld, Thm. 4.1]. Let $X$ be a smooth complex projective variety of dimension $m$. Assume that there is a surjective morphism $\mathbf{P}^m \to X$. Then, $X \cong \mathbf{P}^m$.

There is also a version in positive characteristic, when the morphism is assumed to also be separable; see [Kollár, Cor. V.3.5].

On the other hand, you can have singular $X$ that admit surjective morphisms from $\mathbf{P}^m$ by taking the quotient of $\mathbf{P}^m$ by a finite group.