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I hope this question isn't too open-ended for MO --- it's not my favorite type of question, but I do think there could be a good answer. I will happily CW the question if commenters want, but I also want answerers to pick up points for good answers, so...

Let $X,Y$ be smooth manifolds. A smooth map $f: Y \to X$ is a bundle if there exists a smooth manifold $F$ and a covering $U_i$ of $X$ such that for each $U_i$, there is a diffeomorphism $\phi_i : F\times U_i \overset\sim\to f^{-1}(U_i)$ that intertwines the projections to $U_i$. This isn't my favorite type of definition, because it demands existence of structure without any uniqueness, but I don't want to define $F,U_i,\phi_i$ as part of the data of the bundle, as then I'd have the wrong notion of morphism of bundles.

A definition I'm much happier with is of a submersion $f: Y \to X$, which is a smooth map such that for each $y\in Y$, the differential map ${\rm d}f|_y : {\rm T}_y Y \to {\rm T}_{f(y)}X$ is surjective. I'm under the impression that submersions have all sorts of nice properties. For example, preimages of points are embedded submanifolds (maybe preimages of embedded submanifolds are embedded submanifolds?).

So, I know various ways that submersions are nice. Any bundle is in particular a submersion, and the converse is true for proper submersions (a map is proper if the preimage of any compact set is compact), but of course in general there are many submersions that are not bundles (take any open subset of $\mathbb R^n$, for example, and project to a coordinate $\mathbb R^m$ with $m\leq n$). But in the work I've done, I haven't ever really needed more from a bundle than that it be a submersion. Then again, I tend to do very local things, thinking about formal neighborhoods of points and the like.

So, I'm wondering for some applications where I really need to use a bundle --- where some important fact is not true for general submersions (or, surjective submersions with connected fibers, say).

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Doesn't the definition of a smooth manifold demand existence of structure without any uniqueness? (This isn't a rhetorical question - I'm honestly not sure.) – Qiaochu Yuan Mar 11 '10 at 5:42
@Qiaochu. No, you have to specify (say) an equivalence class of atlases to define a smooth manifold. So R with the chart x --> x^3 is a different smooth manifold to R with the obvious chart (though diffeomorphic to it). More interestingly, the action of the homeomorphism group of S^7 on its smooth atlases has 28 orbits. – Tim Perutz Mar 11 '10 at 5:57
Theo, you answered your own question by saying that you like to work locally. Submersions don't have global structure. Take a smooth fibre bundle and delete any closed subset; it's still a submersion. Now try to say something interesting about its topology. Or integrate a vector field on it. – Tim Perutz Mar 11 '10 at 6:02
@Qiaochu: One way to say "smooth manifold" is to talk about maximal atlases, and these are unique. I guess I could use the same device to talk about bundles. So maybe that's not a complaint against them, but it's not a reason to like them any better either. – Theo Johnson-Freyd Mar 11 '10 at 6:30
I see where you're coming from about not liking existential quantifiers in certain definitions, but if they're of a local nature (ie there exists a cover such that on each piece blah blah blah), which they are in the case of bundles, then they're really well behaved! This is the whole point of sheaf theory! – JBorger Mar 11 '10 at 9:46
up vote 24 down vote accepted

One would be that a fibre bundle $F \to E \to B$ has a homotopy long exact sequence

$$ \cdots \to \pi_{n+1} B \to \pi_n F \to \pi_n E \to \pi_n B \to \pi_{n-1} F \to \cdots $$

This isn't true for a submersion, for one, the fibre in a submersion does not have a consistent homotopy-type as you vary the point in the base space.

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This is probably making a hash of the earlier answers, but bundles are special fibrations; specifically, they are fibrations with (not canonically) isomorphic fibers. And we all like fibrations, right?

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Good answer! You can lift any curve in the base into the total space of a bundle, but you can't lift it into the total space of a submersion. – Konrad Waldorf Mar 11 '10 at 8:28
Konrad Waldorf, I do not understand. Take the boundary of the mobius band, i.e., the nontrivial Z_2 bundle over S^1. There is no section for the projection. – AndrewLMarshall May 2 '10 at 5:55
@Andrew, perhaps Konrad is liberal with the term "curve". I certainly didn't mean you can lift maps with arbitrary domain --- fibration only means you can lift whole homotopies that already lift at one end. – some guy on the street May 4 '10 at 17:48

There's no reason I can see for preferring bundles over submersions, unless you need bundles. If you don't need the extra global structure implied by a bundle, then by all means stick to submersions.

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I'm tempted to accept this answer, as it's closest to what I really believe. But I think Ryan most accurately answered my question as asked. In any case, everyone should vote up Deane. – Theo Johnson-Freyd Mar 31 '10 at 2:44

There's also the cohomology version of Ryan's answer: the Leray-Serre spectral sequence, which tells you some very nice things about the cohomology of a bundle, and essentially nothing useful about the cohomology of a submersion. You can consider this a particular instance of Tim's comment.

In general, algebraic geometers and homotopy theorists work with bundles (or more generally, fibrations), every day of their lives, and will extremely rarely encounter submersions. Even if you don't want to work in such fields, their existence is a good reason to distinguish bundles from submersions.

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There is the Leray spectral sequence of a map. It's just much better behaved for a fibration. – Ryan Budney Mar 11 '10 at 21:04
It's a bit unfair to say "nothing useful," but at the same time, I'm not very good at taking cohomology of random constructible sheaves on a space, as opposed to the local systems that show up in Serre for a bundle. – Ben Webster Mar 11 '10 at 21:10
Ben, I disagree with your statement that "algebraic geometers [...] will extremely rarely encounter submersions". We just call them "smooth morphisms". Also, the Leray spectral sequence behaves quite nicely already for flat morphisms, you don't even need smoothness. – Sándor Kovács Jan 24 '11 at 1:21
I'm not sure what you mean by "the Leray spectral sequence works quite nicely for flat morphisms." If you take an arbitrary flat morphism (say the inclusion of a curve minus a point into a curve), a naive interpretation of Leray-Serre gives nonsense; of course this can be fixed, as Ryan points out, but at a significant cost in terms of complication. Of course, things work beautifully for proper smooth maps, but I would call those fibrations; they are in the analytic topology over $\mathbb C$, and behave like them over other fields. – Ben Webster Jan 24 '11 at 4:46
Ben, sorry, I did indeed have proper flat in mind with respect to that comment about Leray. And I am happy to call those fibrations. However, the original question was about bundles. But the main point of my comment was that I think that we still do see submersions on a regular basis, just don't call them that. – Sándor Kovács Jan 24 '11 at 5:16

Consider co-dimension 0. In this case, bundles are covering maps, with all the goodies that they bring. And submersions are just local homeomorphisms - not very exciting compared to coverings.

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It probably won't matter which concept you use due to the theorem of Ehresmann. See:

It states something like most surjective submersions are in fact fibre bundles (most meaning that this is the case if the surjective submersion is proper, and I am not sure how dense proper maps are). Is there an approximation theorem for proper maps?

So i think the answer is that you don't have to. Also, (smooth?) fibrant replacement can be done to any map so that you get a LES in homotopy (although this map may no longer be a submersion.).

hope this helps, sean

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Well, for maps like $\mathbb C \setminus \{0\} \to \mathbb R$ given by projection onto the x-axis, it matters. – Ryan Budney Aug 4 '10 at 0:51
This is an interesting comment, it leads me to wonder if it would be good to think of submersions as bundles with singularities. But I guess that is obvious since everything we are working with is a manifold and so locally it would always look like a product. – Sean Tilson Aug 4 '10 at 2:00

You write:

So, I'm wondering for some applications where I really need to use a bundle --- where some important fact is not true for general submersions (or, surjective submersions with connected fibers, say).

Actually, I am going to play devil's advocate here: sometimes it's better to have a submersion! This point comes up in a very relevant way in the classical smoothing theory of topological manifolds. Siebenmann (cf. Kirby and Siebenmann's book) defines a moduli space of smoothings of a topological manifold $M$ to be the space of $$(N,f)$$ such that $N$ is smooth and $f: N \to M$ is a homeomorphism.

Siebenmann chooses to topologize this in what seems a funny way: a $k$-simplex of such things is a pair $(N,f)$, where now $N \to \Delta^k$ is a smooth submersion (not necessarily proper if $M$ isn't compact!) and $f: N \to M \times \Delta^k$ is a homeomorphism which is compatible with projection to $\Delta^k$. This gives a $\Delta$-space (a simplicial set w/o degeneracies). Call its geometric realization $\text{Sm}(M)$.

Why doesn't he just topologize families as fiber bundles?

Here's why:

Let ${\cal O}_M$ be the poset of open subsets of $M$ which are abstractly homeomorphic to open balls. The fundamental theorem of smoothing theory asserts that the contravariant functor $\text{Sm} : {\cal O}_M \to \text{Top}$ given by $$ U \mapsto \text{Sm}(U) $$ is a "homotopy sheaf" if $\dim M \ge 5$, i.e., the (restriction) map $$ \text{Sm}(M) \to \underset{U \in {\cal O}_M} {\text{holim}}\quad \text{Sm}(U) $$ is a homotopy equivalence. This would not be the case if we had defined the families as bundles (rather than as submersions). Note: we cannot appeal to Ehresmann here as the submersions which are used in the define $k$-simplices in $\text{Sm}(U)$ are not assumed to be proper.

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