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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!

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By the way, I don't think your proposed definition for a retraction makes sense. It should be a map of the set of complex points of $X$ with the analytic topology (assuming this is what you mean when you talk about homotopy equivalence, etc.) What is $r(x)$ when $x$ is a closed point not lying in $X_0$? – user64687 Mar 3 '14 at 13:04
Why does my definition not work? Sorry, I do not understand your comment. I agree it is kind of silly and artificial, but it is a continuous map from $X$ to $X_0$, which is what I was after. – Brenin Mar 3 '14 at 17:59
If $x$ is a closed point, then $\{x\}^-=x$. So the intersection is empty. – user64687 Mar 3 '14 at 19:04
Now I see; the picture I had in mind was that of a point on the generic fiber of the family, and I attempted to define the retraction as the "flat limit" of that point on the special fiber. Doesn't this work even if I restrict to flat morphisms $X\to S=\Delta$ over the formal disk? – Brenin Mar 3 '14 at 19:10
Dear Brenin, I agree that in the situation of a family over the formal disk, your map is a well-defined retraction. I understood the original question as being about fibres over closed points of the base, with their natural topology as complex manifolds. In the situation over the formal disk, it seems as though we must give the fibres the Zariski topology, at which point notions such as homotopy equivalent seem (to me) to become less useful. But maybe someone else can be more helpful. – user64687 Mar 4 '14 at 9:07
up vote 2 down vote accepted

I'll stick with your question 2. By definition a fiber bundle $E\to X$ is locally homeomorphic to $U\times F$ where $U$ is open in $X$ and $F$ is the fiber. So in fact if $X$ is connected all the fibers of $E$ are homeomorphic. On the other hand if $X$ has distinct connected components the fibers over each one can differ arbitrarily, so that there's no homotopy theory to be done either way.

The appropriate homotopical analogue of a fiber bundle is what's called a fibration. This is a map $E\to^p X$ such that for any homotopy $Y\times I\to X$ such that $Y\times \{0\}$ can be lifted to $E$, the entire homotopy can be lifted to $E$. Fiber bundles are fibrations as long as the base space is reasonable-as you may have seen in the case of covering spaces. And it's elementary from the definition to show that fibrations over path-connected spaces (hint:spaces for which there's a homotopy of a point between any two points) have homotopy equivalent fibers. Fibrations are, sloganistically, the "good surjections" for homotopy theory.

All I'll say regarding your Question 1 is that "homotopy equivalent" is often the wrong thing to ask for in algebraic geometry since the Zariski topology is so coarse. You could specify, if you do want it, that you want to know whether the fibers are homotopy equivalent in the analytic topology, but I think that deformation equivalence plays something closer to the role of homotopy equivalence in algebraic geometry, and I believe the fibers of your map are deformation equivalent by the definition of the latter. Hopefully somebody else will come along to answer Question 1 more thoroughly.

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It's not at all clear to me that the two questions in Question 1 are asking the same thing, but in any case the answer to the first one is "definitely not".

For example, think about a flat family of elliptic curves degenerating to a nodal rational curve. The general fibre is topologically a torus, so has $H_1 \simeq \mathbf Z \oplus \mathbf Z$, but the special fibre is topologically a pinched torus, so has $H_1 \simeq \mathbf Z$. So they're not homotopy equivalent.

From a fancy point of view, this is the simplest instance of the phenomenon of vanishing cycles.

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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.)

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