Let $p:Y \to X$ be a proper, flat (see later why) surjective map between smooth connected complex varieties $Y,X$, esp. $X$ unibranch. Assume that there exist a Zariski open (esp. dense) $U \subset X$ over which every fibre $p^{-1}(x)$ is connected (resp. irreducible).
Then there is a topological version of Zariski's connectedness thm. (see eg 5.6 U5, in Algebraic Geometry II by Mumford & Oda, or 3.24 in Mumford's AG I: Complex projective varieties) that assures that then every fibre of $p$ (so not only over $U$) is connected.
The argument uses interplay of Zariski topology with complex analytic topology of induced map $p(\Bbb C): X(\Bbb C) \to Y(\Bbb C)$. Let's stick on fibres over closed points only, so focus on fibres of $p(\Bbb C)$.
The map $p(\Bbb C)$ is topologically/ differential geometrically a proper, smooth surjective submersion (... there are pathological cases with $p$ surjective, but not submersion; lets exclude these; guess to assume additionally flatness for $p$ would resolve it) as we can treat complex manifolds $Y(\Bbb C), X(\Bbb C)$ as smooth manifolds.
What I'm wondering about, cannot we by density of $\Bbb C$-valued points reduce the statements about connectedness & irreducibility of the fibres reduce to analogous statements about fibres wrt $p(\Bbb C)$? If yes, cannot we reasoning that way also deduce not only that all fibres are connected, but even irreducible?
I know that the latter cannot be the case, as there are concrete counterexamples with families where general irreducible fibers degenerate to reducible fibres. But this raises then the question where the in following presented "argumentation" - which if I'm not overlooking something implies even that the fibres irreducible - would concretely break down?
My point was why cannot we also argue here via Ehresmann Lemma? For proof see, eg Prop 6.2.2, Complex Geometry, Huybrechts.
If not, why? if yes, wouldn't this imply even stronger statement that all fibres are irreducible assuming fibres over $U$ are? (..see below why I think so)
Clearly, this requires some clarification as Ehresmann works in differential geometric setting, and says that such map $f(\Bbb C)$ as above is locally a trivial fibration in differential geometric setting.
Attention: As these local trivializations are given not complex analytically, but differential geometrically, ie there exist open $V \subset X(\Bbb C)$ (wrt smooth topol) such that $(p(\Bbb C))^{-1}(V)$ is diffeomorphic to $(p(\Bbb C))^{-1}(x)) \times V$. Note, that this completely forgets about complex structure of the maps.
But we still get as important consequence that this implies especially that all fibres $(p(\Bbb C))^{-1}(x) $ are diffeomorphic, and so homeomorphic (a priori wr analytic topology), esp the topological features like connectedness and irreducibility are consequently preserved in all fibres.
Btw.: The historical side of the issue I would like to discuss here. See for details the marked "Remark" part there motivating essentially why I relate ZCT here with Ehresmann lemma.
This raises two questions:
(Q 1): Does this reasoning via Ehresmann give also a correct proof of above quoted Zariski connectedness thm? (of course only in complex setting)
The crucial point which confuses me here is that note, that ZCT is about connectedness, but the Ehresmann argument presented above seemingly implies irreducibility in complex analytic topology of the fibres of $p(\Bbb C)$ which are in turn dense in fibres of $p$.
Of course, note that the proof is about $p(\Bbb C)$ in complex analytic topology,
but as Zariski-connectedness and complex analytic connectedness are equivalent we get Zariski-connectedness; compare with remarks after proof of above qouted (3.24) in Mumfords CPV.
(Q 2): The issue with irreducibility of fibres in light of the Ehresmann lemma appears more subtle as the above presented proof implies not only connectedness, but even irreducibility too, but on the other hand it seems to be "to strong" as usually families admit reducible members.
Here, we should maybe be more careful with two involved topologies:
Recall, a top. space is irreducible iff a dense subset of it is irreducible.
So again the irreducibiliy of fibres of $p$ is then equivalent to irreducibility of fibres of $p(\Bbb C)$ beeing dense in them. But here we should be careful, as wrt analytic topology such a fibre is almost never irreducible.
But we can simply impose Zariski topology of each fibre $(p(\Bbb C))^{-1}(x) $ as topology induced by subsets $(p(\Bbb C))^{-1}(x) \cap U(\Bbb C)$ where $U \subset X$ Zariski open, since as sets we have inclusion $X(\Bbb C) \subset X$. Note that Zariski topology is subtopology of finer analytic topology (...algebr functions are analytic).
So the question becomes if the homeomorphisms between fibers $(p(\Bbb C))^{-1}(x)$ with resp analytic topology given by Ehresmann induce continous (neccess bijective) maps wr Zariski topology. If yes, this would imply that if there exist a Zariski open $U \subset X$ over which every fibre $p^{-1}(x)$ is irreducible, then every fibre of $p$ is irreducible.
But this appears wrong; there are families where general irreducible fibers degenerate to reducible fibres and it is "natural" in algebraic geometry that families have some reducible members. So it would be unplausible to expect this. But then, what is wrong with my reasoning via application of Ehresmann lemma above which seemingly implies that all would be even irreducible?
So my motivation is to develop better intuition how much "non algebraicness & pure topology" sits in connectedness & irreducibility of varying fibers. The at least for me the cumbersome point is that in most contemporary literature one deduces ZCT - a statement purely about topology - as consequence of intrinsically algebraic Zariski's Main Theorem resp. prorties of maps of Stein decomposition. And I'm wondering if this statement - which is purely about topology - cannot be deduced with "almost" pure topological methods.
