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Let $:f:X\to Y$ be a projective surjective morphism between two normal varieties over $\mathbb{C}$. Assume that $f$ has only $1$-dimensional fibers. Let $D$ be a multi-section of $f$, i.e., $D$ is a prime Weil divisor on $X$ such that $f|_D: D\to Y$ is a generically finite surjective morphism. Let $D\to Y''\to Y $ be the Stein factorization of $f|_D$ and $Y'\to Y''$ the normalization morphism. Now we do a fiber product between $f:X\to $Y and $\phi:Y'\to Y$, i.e., $X\times_Y Y'$. Let $X'$ be the normalization of the main component of $X\times_Y Y'$ dominating $Y'$. Let $f':X'\to Y'$ and $\phi':X'\to X$ be the corresponding projections.

Now in this scenario, I have seen in several papers so far now (without any explanation) that $\phi'^{-1}D$ splits into irreducible components, at least one of which becomes a section for $f':X'\to Y'$ over a dense open subset of $Y'$.

My question is why is this true? How would one prove it formally? I actually need to make all the components of $\phi'^{-1}D$ into sections over some dense open set, so I believe I will have to perform this kind base changes few more times, then how I do know this process stops after finitely many steps?

** I assumed $D$ is only a prime Weil diviso but it helps you may assume that it is also a Cartier divisor, so that pullback makes sense.

** I would like to mention that this seems to be a standard technique in proving the Canonical Bundle Formula when the relative dimension is $1$. For example, this argument appears in the proof of Theorem 8.1 in the paper "Towards the second main theorem on complements" by Prokhorov and Shokurov (2009).

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1 Answer 1

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First, you have a multisection $D_1 = D \times_Y Y'$ for the family $X' \to Y'$, which is still generically finite of the same degree over $Y'$.

On the other hand, let $D' = D \times_{Y''} Y'$. Then the maps $D' \to D \to X$ and $D' \to Y'$ induce a map $D' \to X'$ which is generically finite over $Y'$ with connected fibers (by properties of the Stein decomposition). Hence it is generically one-to-one onto $Y'$. On the other hand, the image of $D'$ in $X'$ sits in $D_1$. Therefore, $D_1$ splits as the union of the image of $D'$ and an extra component, say $D_2$. Now $D_2$ is generically finite over $Y'$ of degree by 1 less than the degree of $D$. So, we can continue the procedure and be sure that it eventually stops.

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