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Basically a beginner type of topology question here, but I am trying to understand something and am a bit stuck on a definition.

According to J.P. May, a fibration is a map $p : E \to B$ such that for all spaces $Y$, that embed into $E$ by some map $f$, and have a compatible homotopy onto $B$ by $h: Y \times I \to B$, where $h(y, 0) = p(f(y))$ for all $y\in Y$; there is a unique extension of $h$ to a homotopy $\tilde{h} : Y \times I \to E$ such that, $p(\tilde{h}(y, t)) = h(y,t)$.

Now what I am trying to wrap my head around is what that actually means. What restrictions does this place on the space $E$ and the maps $p$? Are there examples of continuous maps $p$ which do not satisfy this property? If so, how is this supposed to generalize the usual notion of a fiber bundle, and what are the fibers constrained to be? Can the fibers be different dimensional spaces or have varying genus?

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2 Answers 2

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Fibrations are a generalization of fiber bundles. When fiber bundles occur in homotpoy theory, the primary usefulness is that they have the homotopy lifting property (aka covering homotopy property.) So the definition of "fibration" is purely a way to extend the notion of "fiber bundle" as broadly as possible and still be useful in homotopy theory.

The fibers are only constrained to be homotopy equivalents (assuming $B$ is path connected.) So the fiber over $x$ and the fiber over $y$ can have different dimensions, even one being a single point while the other is any contractible space.

There are plenty of maps which are not fibrations. For example, any map from a closed intarval of the reals numbers onto the circle is not a fibration.

A simple example of a fibration which is not a fiber bundle is to take an acute triangle for $E$ and one of its edges for $B$. Then define $\pi:E\rightarrow B$ as the orthonormal project of $E$ onto $B$. Then the fibers at the end points are just single points, while the fibers everywhere else are closed intervals.

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Thank you, your post cleared a lot up for me (I would give you an upvote, but I don't have enough karma). I think I understand how this works, but I am still uncertain exactly what this property is good for. I am guessing that it has something to do with wanting to be able to decompose spaces over short exact sequences/extensions, like is done in group theory, but this concept is a bit hazy still. I will think about it a bit more. –  Mikola Apr 4 '11 at 20:58

There is a weakening of the notion of a fibre-bundle. I'm not sure to what extent it has common name but I believe people tend to call it a homotopy-fibre bundle. The idea is just like in the definition of a fibre bundle, i.e. you have a continuous function $f : E \to B$ and for every $b \in B$ there is a neighbourhood $U \subset B$ of $b$ and a map:

$$\phi : f^{-1}(U) \to U \times F$$

which fits into a commutative diagram $\pi \circ \phi = f_{|f^{-1}(U)}$

where $\pi : U \times F \to U$ is projection onto the $B$-factor of the product.

The map $\phi$ is required to be a "fibrewise homotopy equivalence" meaning there is a map the other way $U \times F \to f^{-1}(U)$ (also preserving fibres) so that the two composites $U \times F \to U \times F$ and $f^{-1}(U) \to f^{-1}(U)$ are homotopic to the identity via fibre-preserving maps.

Anyhow, under fairly reasonable circumstances these are precisely what fibrations are (take your favourite proofs that fibre bundles are fibrations and adapt). I believe there's an exercise along these lines in Hatcher's textbook but it's been a while since I've looked at this material.

This gives you an idea of how to construct fibrations that aren't fibre bundles. For example, the map $h : X \to [0,1]$

where $X = [0,1] \times \{0\} \cup \{ (0,t) : t \in \mathbb R \}$. $h$ being projection onto the 1st coordinate, this is a fibration but not a fibre bundle.

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