Why should I prefer bundles to (surjective) submersions? - MathOverflow

Why should I prefer bundles to (surjective) submersions?

2010-03-11T05:32:00Z

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 ${\rm d}f|_y : {\rm T}_y Y \to {\rm T}_{f(y)}X$ 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).

Answer by Ryan Budney for Why should I prefer bundles to (surjective) submersions?

2010-03-11T06:00:35Z

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.

Answer by some guy on the street for Why should I prefer bundles to (surjective) submersions?

2010-03-11T06:05:24Z

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?

Answer by Deane Yang for Why should I prefer bundles to (surjective) submersions?

2010-03-11T15:05:58Z

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.

Answer by Ben Webster for Why should I prefer bundles to (surjective) submersions?

2010-03-11T18:03:58Z

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.

Answer by Sergei Ivanov for Why should I prefer bundles to (surjective) submersions?

2010-03-11T19:23:56Z

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.

Answer by Sean Tilson for Why should I prefer bundles to (surjective) submersions?

2010-08-03T20:13:15Z

It probably won't matter which concept you use due to the theorem of Ehresmann. See: http://en.wikipedia.org/wiki/Ehresmann%27s_theorem

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

Answer by John Klein for Why should I prefer bundles to (surjective) submersions?

2011-01-23T18:57:46Z

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.