Let $f:X \to Y$ be a finite surjective morphism from a $\mathbb{Q}$factorial variety to a smooth variety. Let $D_Y$ be a prime divisor on $X$ and let $\bigcup D_i$ be the inverse image of $D_Y$. Do we know anything about the number of connected components of $\bigcup D_i$? Does the number equal to the number of $D_i$'s? (i.e. $D_i$'s do not intersect with each other.) Thanks!
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2$\begingroup$ It is bounded by the degree of the morphism. $\endgroup$– SashaFeb 27, 2020 at 15:49

1$\begingroup$ But this is not sufficient... I expect this to be the number of $D_i$ or at least related to that... $\endgroup$– Hu ZhengyuFeb 27, 2020 at 15:53

2$\begingroup$ No, the number of connected components does not equal the number of irreducible components. For example take $X \rightarrow \mathbf P^2$ a double cover branched over a smooth quartic curve $C$. For a line $L \subset \mathbf P^2$ that is bitangent to $C$ the inverse image of $L$ is a union of 2 intersecting copies of $\mathbf P^1$. $\endgroup$– BortFeb 27, 2020 at 16:18

$\begingroup$ Yes, this is a good example. $\endgroup$– Hu ZhengyuFeb 27, 2020 at 16:19

1$\begingroup$ If $f$ is a morphism of curves, there is no better bound than the degree of the morphism. $\endgroup$– SashaFeb 27, 2020 at 17:36
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