A surjective flat (equals faithfully flat) map with smooth fibres is in fact a smooth morphism, and hence induces a submersion on the underlying manifolds obtained by passing
to complex points. Since the fibres are projective, it is furthermore proper (in the sense of algebraic geometry) [see the note added at the end; this is not a logical deduction from the given condition on the fibres, but nevertheless seems to be a reasonable reinterpretation of that condition], and hence proper (in the sense of topology). A theorem
of Ehresmann states that any proper submersion of smooth manifolds is a fibre bundle.
In particular, it is a fibration in the sense of homotopy theory, and the fibres are
diffeomorphic (thus also homeomorphic, homotopic, ... ).
Note: Your specific question is really about smooth morphisms (these are flat morphisms
with smooth fibres, although there are other definitions too, which are equivalent
under mild hypotheses on the schemes involved, and in particular, are equivalent for maps
of varieties over a field). One point about the notion of flat map is that it allows one
to consider cases in which the fibres over certain points degenerate, but still vary continuously (in some sense). It may well be a special feature of algebraic geometry
(and closely related theories such as complex analytic geometry) that one can have such a reasonable notion, a feature related to the fact that one can
work in a reasonable manner with singular spaces in algebraic geometry, because the singularities are so mild compared to what can occur in (say) differential topology.
[Added: I should add that I took a slight liberty with the question, in that I interpreted the condition that the fibres are projective stronger than is literally justified, in so far as I replaced it with the condition that the map is proper. As is implicit in Chris Schommer-Pries's comment below, we can find non-proper smooth surjections whose fibres are projective varieties: e.g. if, as in his example, we consider the covering of $\mathbb P^1$ by two copies of $\mathbb A^1$ in the usual way, then the fibres consists of either one or two points (one point for $0$ and the point at $\infty$, two points for all the others), and any finite set of points is certainly a projective variety.
Nevertheless, my interpretation of the question seems to have been helpful; hopefully, with the addition of this remark, it is not too misleading.]
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
- 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).
- 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
Best Answer
The answer here is a resounding no. I think the most important point is that you're applying the wrong topological intuition here. A variety shouldn't be thought of as like a manifold, but as like a complex manifold, and the corresponding theorem to the "submersion=fiber bundle" theorem in smooth manifolds is just false for complex manifolds. Just as an example, all elliptic curves are topologically the same, so the solutions to $x(x-1)(x-a)=y^2$ are a smooth fiber bundle over (most of) $\mathbb{C}$ with coordinate $a$, but all the fibers which aren't in the same orbit of $SL_2(\mathbb Z)$ on $\mathbb{C}$ are not isomorphic as complex manifolds.
The other way of saying this is that complex structures can exist in families; they have moduli. Moduli spaces exactly measure how theorems like "submersion=fiber bundle" fail since they measure continuous variation of structure.