TITLE: Are the fibers of a flat map homotopy equivalent?
QUESTION [7 upvotes]: At the end of the Wikipedia article on Deformation Retract, there is the following sentence:

Two spaces are homotopy equivalent if and only if they are both deformation retracts of a single larger space.

I was wondering if this has some meaning in Algebraic Geometry. For instance, consider the situation of a flat surjective map of complex schemes $f:X\to S$, where $S$ is a smooth curve, e.g. the formal disk. 
Question 1. Are the fibers of $f$ homotopy equivalent? Is every fiber $X_s$ a deformation retract of the total space $X$? 
I tried to show, first, that any fiber is a retract of the total space, but I am unsure about my solution, and in any case it is weaker than deformation retract. But it goes as follows: if $X_0$ is a particular fiber of $f$, we can define a retraction $r:X\to X_0$ as the identity on $X_0$, and by
$$r(x)=\{x\}^-\cap X_0,\,\textrm{for }x\in X\setminus X_0.\,\textrm{(this is the flat limit I guess)}$$
Question 2. In the topological category, say we have a topological fiber bundle $X\to S$. Are the fibers homotopy equivalent?
Thanks for any clue on this!

REPLY [4 votes]: Q1. No, it's not true that all fibers of a flat map are homotopy equivalent. For instance, the blow-up of $\mathbf P^1 \times \mathbf P^1$ at a point maps to $\mathbf P^1$. This map is flat and all fibers except for one are spheres; however, the remaining fiber is a wedge of two spheres. More generally, in a flat family of curves you can "collapse" a simple closed curve, and this example corresponds to collapsing the equator of $S^2$.
Here are two statements which are true: i) A smooth proper morphism of complex schemes is topologically a fiber bundle and all fibers are homeomorphic (Ehresmann's theorem). ii) In a flat family of subschemes of a projective space, all fibers have the same Hilbert polynomial.
Q2. In the only definition I know of a topological fiber bundle (it locally has the form $U \times F \to U$), it is immediate that all fibers are even homeomorphic. If you mean that $f$ is a fibration in the sense of homotopy theory, it is not true that all fibers are homeomorphic but it is still true that they are all homotopy equivalent. (The base is assumed connected throughout.)