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This question is inspired by the para2 of this post Is the inverse image of an irreducible variety under the natural projection irreducible (in the setting of homogeneous spaces)? There the author said ": let $\eta$ be the generic point of $X′$. Since all fibers of $p$ over closed points are irreducibel we can conclude that $p^{−1}(η)$ is also irreducible" (in the scheme setting)

Question 1: How can we get the irreducibility of the "generic fiber" from that of "closed fibers" in the scheme setting as that of the above post?

Question 2: Can we get the same result in the analytic setting?

More presicely:
Let $p: X \to Y$ be a holomorphic map to an irreducible complex analytic varieties $Y$. let $\eta$ be the generic point of $Y$. If we know that all fibers of $p$ over closed points are irreducible. Can we get that $p^{-1}(\eta)$ is also irreducible?

Thanks in advance. Also very appreciated it if some possible references were given.

Lelong Wang
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    What have you tried? Many of your recent questions in the algebraic-geometry tag have been a little light on context, and it would be nice if you could address this. – KReiser Oct 21 '21 at 18:50
  • @KReiser Thank you very much for your kindful advice. I am sorry for my style of asking questions. In the future, I will try to explain more background or personal thinking in my questions. On this present question, I have reedited it (but a pity that I I don't have any clues on this question). – Lelong Wang Oct 22 '21 at 07:44
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    With the typical definition of complex-analytic varieties, there are no generic points, so you'll need to explain what you mean a bit more. As for question 1, the key is to understand what the fiber over the generic point is. Try proving that irreducible components of the generic fiber are exactly the irreducible components of $X$ passing through the generic fiber, then think about what this means with regards to fibers over the closed points. – KReiser Oct 22 '21 at 08:06
  • @KReiser Thank you. It seems that I misunderstood the definition of "generic fiber" (in the algebraic setting).
    I want to ask that: If we adopt the definition that a point in an irreducible complex analytic variety is called the generic point if its analytic Zariski closure is the whole variety, then you mean by " there are no generic points" that there maybe
    no generic point? Why? In algebraic setting, the generic point corresponds the $0$-ideal in some sense. Do you give your claim "no generic points" by this philosophy in AG?     – Lelong Wang Oct 22 '21 at 08:42
  • This is the sort of complex-analytic variety that I mean. There are no generic points (most of the time) because the space is locally modeled on a locally closed subset of $\Bbb C^n$ in the standard topology. If you mean something else by the term "complex-analytic variety", you should explain that in your post. – KReiser Oct 22 '21 at 08:59
  • @KReiser Yeah, what I mean by "complex-analytic variety" is just as you say, i.e. "Complex analytic variety" in the Wiki. But for this variety, there is the "analytic Zariski topology"--taking the analytic subset as the closed set-- which is different with the complex topology-the topology induced by "the standard topology" in your expression （which is Hausdorff). So do you think that there is no generic point in an irreducible complex analytic variety with respect to the analytic Zariski topology? Thank you. – Lelong Wang Oct 22 '21 at 09:41
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    Such a space does not have generic points outside of trivial cases. It is easy to verify that any single point in such a space is closed: you can find a finite collection of analytic functions which vanish at just that point. It's just like equipping $k^n$ with the Zariski topology - no generic points (except for trivial cases). – KReiser Oct 22 '21 at 18:27

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