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This is exercise 11.4.C of Ravi Vail's note

Let $\pi :X\to Y$ be a surjective closed map of varieties with $Y$ irreducible and the fibre of $\pi$ are irreducible of the same dimension, then $X$ is irreducible.

and a solution can be found in this post Use irreducible fibers to show $X$ is irreducible.

I would want to ask is there any intuitive reasoning why this is true? Is this true for topological spaces in general?

Thanks in advance.

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chan kifung
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1 Answers1

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I like to intuitively think of this fact as capturing the coarse structure of a locally trivial fibration - it is believable that a vector bundle $E \to X$ (or rather a projective bundle to ensure that the map is closed) over an irreducible base should be irreducible. Indeed, in some sense it is just a "twisted" product $X \times \mathbb{P}^n \to X$, and for irreducible $X$ the product $X \times \mathbb{P}^n$ is irreducible.

Of course a general closed $Y \to X$ with irreducible fibers of the same dimension won't be locally trivial, but it is close enough for such kind of topological statements.

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