Definition of fiber bundle in algebraic geometry

If we have a map p: X --> Y of topological spaces, we can make a definition expressing that the topological type of the fibers of p varies continuously (edit: better to say "locally constantly", thanks Dave) with the base: we can say that p is a fiber bundle.

My question is, can we capture this notion algebro-geometrically, in the case where X and Y are varieties over a field of characteristic zero and p is a map of varieties? I'm looking for a definition hopefully having the following properties (side question: do these seem reasonable?):

1) If X and Y are over the complex numbers, then p is an algebro-geometric fiber bundle if and only if it is a topological fiber bundle on complex points;

2) If f: X-->Y is arbitrary then there is an algebraic stratification of Y such that over each stratum f is a fiber bundle.

Examples should include smooth maps having smooth proper compactifications for which the boundary divisors are in strict normal crossings position, but I would rather the definition not be along these lines because, for instance, I don't want to need resolution of singularities to check that the structure map to the ground field is a fiber bundle.

Edit: In response to several comments, yes, another example would be the normalization map for a cuspidal singularity. In fact I would like the definition to be "topological", in the sense that it factors through h-sheafification.

Edit 2: Whoops, it looks like I used some bad terminology, which probably led to misinterpretations. Sorry folks! To fix things I've replaced all instances of "fibration" with "fiber bundle".

Any thoughts are appreciated!

-
Andrea, not exactly: if you take a smooth fiber bundle and remove some random subvarieties from the total space it will still be smooth, but probably no longer a fiber bundle; also you can have non-smooth fiber bundles, like the projection off an arbitrary variety (maybe singular). –  Dustin Clausen Dec 17 '10 at 22:44
Dustin -- interesting question. My guess would be that the analog of a locally trivial bundle in topology is an \'etale locally trivial bundle, but I'm not sure. –  algori Dec 17 '10 at 23:37
A small quibble about terminology that might help clarify things: I think "fiber bundle" (in topology and geometry) usually means locally trivial --- so the topology doesn't vary at all with the base. On the other hand, a "fibration" is a map satisfying the homotopy lifting property, etc, and might be more in lines with what Donu seems to be suggesting. (There is some work by Meigniez on criteria for a submersive fibration to be a fiber bundle, generalizing the fact Sándor refers to.) –  Dave Anderson Dec 18 '10 at 2:43
Do you want the definition to cover things like the resolution of a cusp? Topologically a homeomorphism (over $\mathbb C$), so topologically a fiber bundle, but maybe not what you meant? –  Tom Goodwillie Dec 18 '10 at 4:27
P.S. I do think that it's possible to formalize the notion that the etale homotopy types are locally constant on the base: probably something like, there is an associated map of etale infinity-topoi p: X--->Y; then we can formally make a left adjoint p_! to p^*, with the proviso that it really lands in pro-objects; then apply it to the terminal object up top and ask if the profinite completion of the thing below is a formal inverse limit of finite locally constant sheaves of homotopy types... but I'd really prefer something more manifestly geometric :) –  Dustin Clausen Dec 18 '10 at 16:40

OK, let me venture to give a definition. Say that a morphism $f:X\to Y$, of varieties over a field, is an algebraic fibration if there exists a factorization $X\to \overline{X}\to Y$, such that that the first map is an open immersion, and the second map is proper and there exists a partition into Zariski locally closed strata $\overline{X}=\coprod\overline{X}_i$, such that restrictions $\overline{X}_i\to Y$ are smooth and proper. $X$ should be a union of strata. Perhaps, one should also insist that this is Whitney stratification.

Shenghao's comment made me realize that my original attempt at an answer was problematic. Rather than trying to fix it, let me make a fresh start. Let us say that $f:X\to Y$ is an algebraic fibration if there exist a simplicial scheme $\bar X_\bullet$ with a divisor $D_\bullet\subset \bar X_\bullet$ such that

1. There is map $\bar X_\bullet -D_\bullet\to X$ satisfying cohomological descent, in the sense of Hodge III, for the classical topology (over $\mathbb{C}$) or etale topology (in general).
2. The composite $\bar X_n\to Y$ is smooth and proper, and $D_n$ has relative normal crossings for each $n$.

These conditions will ensure that $R^if_*\mathbb{Z}$ (resp. $R^if_*\mathbb{Z}/\ell \mathbb{Z}$) are locally constant etc.

I think that this would also apply Shenghao's question

What would be a characteristic-$p$ analogue for $C^{\infty}$-fiber bundles?

Although I won't claim that this is in any sense optimal.

Oh, and I forgot to say that when $Y=Spec k$ is a point, every $X$ can be seen to be fibration (as it should) by De Jong

-
Algori, I probably didn't convey my definition very clearly. Over $\mathbb{C}$, by Ehresmann the restrictions of $f$ $f_i\overline{X_i}\to Y$ are $C^\infty$ fibre bundles, so the sheaves $R^*f_{i*}\Z$ are locally constant. At least under appropriate Whitney conditions, this should imply by devissage, $R^if_*\Z$ is locally constant. So unless I'm mistaken, this would be a strong condition. –  Donu Arapura Dec 18 '10 at 3:33
Donu -- yes, I see it now. I've just deleted a comment which resulted from not having read the posting properly. –  algori Dec 18 '10 at 4:09
Donu - thank you, I think this pretty much captures what I was looking for. I was hoping for a different kind of answer -- but maybe that was hoping for too much! –  Dustin Clausen Dec 31 '10 at 7:59
Dear Donu, I have a question about your definition. Does $X_i$ stand for something and $\overline{X}_i$ is its closure? If $C$ is a singular variety, I guess the projection map $Y\times C\to Y$ would be called a fiber bundle, but it doesn't seem to fit into your definition: if one restricts to the strata then the map may not be proper, and if one restricts to the closures then it may not be smooth... –  shenghao Aug 9 '11 at 12:04
Donu - What do you think of the following reformulation, to avoid talking about specific realizations?: X --> Y is a fiber bundle if, locally on Y in the h-topology, it is, locally on X in the h-topology, of your form X-bar minus D. –  Dustin Clausen May 29 '12 at 23:29

EDIT: This was an answer to the original question about fibrations and not fiber bundles. For the latter this is much less relevant.

I am not sure what you expect from 1) exactly. That sounds like a definition itself.

On the other hand a flat morphism has many of the requirements you ask for. In particular, 2) follows from flattening stratification.

For schemes of finite type over a field a flat morphism with geometrically regular, equidimensional fibers is smooth. In that case, if the fibers are also compact (and we are over $\mathbb C$), then they are diffeomorphic, so even a little better than what you wanted.

-
Sandor, thank you for the suggestion, but won't, say, the map projection map from A^2 minus the origin down to A^1 be flat with geometrically regular equidimensional fibers, but not a fiber bundle? –  Dustin Clausen Dec 17 '10 at 22:46
For 1) the analogous thing I have in mind is that a map of varieties over C is proper in the algebro-geometric sense if and only if it is proper in the topological sense, or ditto "connected" for a single variety (these are true, right?) –  Dustin Clausen Dec 17 '10 at 22:50
Yes, sorry, I left out compact. –  Sándor Kovács Dec 17 '10 at 22:56
Oh, I see. Yeah, I agree that a smooth proper map should be a fiber bundle (though a general flat proper map shouldn't, if you think of a curve degenerating). But I was wondering if one could say something more general. –  Dustin Clausen Dec 17 '10 at 22:58
My objection to #1 is similar to Sándor's: the normalization map of a cusp curve os topologically a homeomorphism, so do you want to count it as a "fibration"? –  Laurent Moret-Bailly Dec 18 '10 at 12:18