The pre-image of a curve under a surjective morphism

Question

The following lemma and proof come from the paper Toward a numerical theory of ampleness, Chapter I, Section 4, Lemma 1 by Kleiman:

Let $f: V'\rightarrow V$ be a surjective morphism from a proper irreducible variety $V'$to a integral curve $V$. Then there is a integral curve $X'$ on $V'$ such that $f(X')=V$.

Proof. By Chow's lemma we can assume that $V'$ is projective. Let $r'=\dim V'$ and let $X'$ be the section of $V'$ by a suitably general linear space of codimension $r'-1$. Then $X'$ is a variety of dimension 1. Let $F'$ be the generic fibre of $f$. We have $\dim X'\cap F'=(r'-1)-(r'-1)=0$. Therefore $f(X')=V$.

My questions about the proof are:

1. Where is the projectivity hypothesis used in the proof?

2. What is "suitably general linear space" exactly? And what is a section of it?

3. Why $\dim X'\cap F'=(r'-1)-(r'-1)$?

4. Why does the final conclusion follows from the equation in question 3?

Thanks in advance.

Answer 1

This is a 'cutting by hyperplanes' argument, which requires some kind of quasi-projectivity. I'm going to reorder it a bit but the idea is the same.

Assuming $V'$ is projective is tantamount to assuming there is a closed embedding $V' \subset \mathbb{P}^n$ for some $n$.

Claim: For a general hyperplane $H$ in $|\mathcal{O}_{\mathbb{P}^n}(1)|$, the restriction $H \cap V' \to V$ is surjective, and $H \cap V'$ is integral of dimension $r' - 1$.

Once we show this, we can repeat this process until we arrive at a linear space $L$, which is general in a reasonable way, so that $X' = L \cap V'$ is an integral curve and the map $X' \to V$ is surjective. The fibers are used because of the following lemma.

Lemma: Consider a morphism $X \to Y$ with $X$ projective and $Y$ irreducible, of dimension $m \geq m'$ respectively. A general fiber $F$ satisfies $\dim F = m - m'$ if and only if map is surjective.

Proof: Since $X$ is a projective variety, its image in $Y$ is closed, so if it is not surjective, then $X$ surjects onto a strict closed subset. Hence in this case a general fiber is empty, so its dimension will certainly be less than $m - m'$. This proves one direction.

Conversely, the dimension formula and this lemma from the stacks project imply that a general fiber of a surjective map in the above setup will have said dimension, thus proving the lemma.

Now we can prove the claim.

Proof of claim: The second part of the assertion follows from Bertini's theorem so we focus on the first. That is, by the lemma, we must show that for general hyperplanes $H \in |\mathcal{O}_{\mathbb{P}^n(1)}|$ and points $x$ in $V$, the fiber over $P$ of $H \cap V' \to V$ has dimension $r' - 2$. To this end, consider the following subvariety $$\Lambda \subset |\mathcal{O}(1)| \times V' = \mathbb{P}^n_{a_0, \dots, a_n} \times V'$$ by $\sum a_i x_i = 0$, where $x_i$ are the coordinates of the projective space $V'$ is embedded in. Note that $\Lambda$ has dimension $n + r' - 1$.

$\Lambda$ the incidence variety colloquially writted as the union $\bigcup_{H \in |\mathcal{O}(1)|} \{H\} \times (H \cap V')$ to emphasize that its fibers over $H$ are precisely the corresponding hyperplane sections on $V'$.

Now let $f: V' \to V$ denote the morphism we started with, and let $\Lambda \to |\mathcal{O}(1)| \times V$ be the map $(H, x) \mapsto (H, f(x))$. As we've seen, a general fiber of this map will have dimension $(n + r' - 1) - (n + 1) = r' - 2$. This proves the result.