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Feb 27, 2020 at 17:36 comment added Sasha If $f$ is a morphism of curves, there is no better bound than the degree of the morphism.
Feb 27, 2020 at 16:19 comment added Flyingpanda Yes, this is a good example.
Feb 27, 2020 at 16:18 comment added Bort 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 $\mathbf P^1$.
Feb 27, 2020 at 15:56 history edited Flyingpanda CC BY-SA 4.0
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Feb 27, 2020 at 15:53 comment added Flyingpanda But this is not sufficient... I expect this to be the number of $D_i$ or at least related to that...
Feb 27, 2020 at 15:49 comment added Sasha It is bounded by the degree of the morphism.
Feb 27, 2020 at 15:42 history asked Flyingpanda CC BY-SA 4.0