# complex algebraic morphisms as topological maps: every morphism is a topological fibration on a Zariski dense open subset?

In writing up a paper, we need references (and help) for the following facts which are probably well-known. They concern morphisms of complex algebraic varieties as continious maps in complex topology. (Here by 'complex topology' I mean the topology induced by the metric on $\Bbb C$).

(i) Is every algebraic morphism of complex algebraic varieties necessarily a fibration in the (non-noetherian) complex topology on a Zariski-open Zariski-dense subset ? That is, does there exists a Zariski open subset $Y^0$ of $Y$ such that $f_{\Bbb C}: f_{\Bbb C}^{-1}(Y^0(\Bbb C)) \longrightarrow Y^0(\Bbb C)$ is a fibration in the (non-noetherian) complex topology ?

(ii) If $f:W\rightarrow Y$ is a dominant rational map of irreducible complex varieties, with Y normal, then the index of the image of $\pi_1(W) \rightarrow \pi_1(Y)$ divides the number of irreducible components of a generic fibre.

Does something like this holds in prime characteristic ? (In char 0 this appears in [Janos Kollar, "Shaferevich maps and automorphic forms", Lemma 2.10.2]; is there a more standard reference as I find it hard to follow the proof there, not being an algebraic geometer.)

(iii) Stein factorisation. Every proper morphism $f:X\rightarrow Y$ of algebraic varieties decomposes as $X\rightarrow^{f_1} X' \rightarrow^{f_2} Y$ such that $f_1$ has connected fibres and is a fibration in complex topology over a Zariski open dense subset, and $f_2$ is finite and etale on an Zariski open dense subset ?

The last question only makes sense if (i) is not always true; without the bit about complex topology Stein factorisation appears in EGA. These questions came up when trying to define a noetherian "Zariski-type" topology on the universal covering space of a complex algebraic variety that is weaker than the complex analytic topology, sort of a model of etale topology...

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For (i) does "generic smoothness" not suffice? This is in Hartshorne, the chapter on smooth morphisms. –  Daniel Loughran Feb 2 '11 at 22:43
If the initial space is smooth, then yes, certainly. But it didn't seem that mmm was assuming this. –  Donu Arapura Feb 3 '11 at 0:21
The result in Hartshorne (Lemma 10.5) only assumes that you have a dominant morphism of integral schemes of finite type over an algebraically closed field of charactertistic zero. Indeed, in this case there is a open dense subset where your schemes are non-singular varieties. If the morphism is also proper then we can use Ehresmann's theorem to deduce that it is a fibration, but I guess you need more in the non-proper case. –  Daniel Loughran Feb 3 '11 at 11:17