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This is in some sense a follow-up to my question on submersions.

Let $f\colon\thinspace M\to N$ be a generic smooth map between closed manifolds of dimensions $m$ and $n$. Assume that the codimension $k = n-m$ is negative.

I am interested in the topology of the point fibers $f^{-1}(y)\subseteq M$ for $y\in N$.

Sard's Theorem implies that for almost all $y\in N$ the fiber $f^{-1}(y)$ is a closed submanifold of $M$ of dimension $-k$. If $y$ is not a regular value, then $f^{-1}(y)$ is a priori just a compact subset of $M$ which needn't be a manifold. However one might hope that (under some conditions on the map $f$ which are satisfied in a dense subset of $C^\infty(M,N)$) it is close to being a manifold. For example, it might always be a pseudomanifold, or manifold with singularities.

Can anything general be said about the spaces $f^{-1}(y)\subseteq M$ occurring as fibers of generic maps?

I am particularly interested in the dimensions of the singular fibers (where dimension has to be interpreted appropriately). In all the examples I can picture (ie critical level sets of Morse functions) the dimensions of the singular fibers seem to be $\le -k$.

In fact I would really like to ask the following precise question about homotopy types:

Let $f\colon\thinspace M\to N$ be a generic smooth map between closed manifolds of dimensions $m$ and $n$, with $m>n$. Is every fiber $f^{-1}(y)$ dominated by a CW-complex of dimension $\le m-n$?

The genericity assumption is there because I would like to be able to approximate an arbitrary map $g\colon M\to N$ by a map with singular fibers having the above nice property.

Probably what I am asking is covered in the literature on singularity theory or surgery theory, however I couldn't find any references on this particular point. Any answers or pointers would be appreciated.

Update: In this paper of Gromov (on page 3) it is stated that if $f\colon\thinspace M\to N$ is a smooth generic map with $m+1\ge n$ then for a critical value $y\in N$ the number of singularities of $f$ on the level set $f^{-1}(y)$ is at most $n$. (A weaker claim is made on page 115 of Golubitsky and Guillemin (Lemma 1.9), namely that the same conclusion holds if $f$ is infinitesimally stable = stable).

This implies that $f^{-1}(y)$ has the structure of a smooth $(m-n)$-dimensional submanifold of $M$ away from a finite number of isolated singular points. So my questions now become:

  • Does this imply that $f^{-1}(y)$ has the homotopy type of a CW-complex of dimension $m-n$?

  • Can anyone supply a proof or a reference for the statement of Gromov quoted above?

Update II: Thanks to Tom Goodwillie, who has answered both of the questions in my update. However I'm still left wondering about the two original questions. My gut feeling is that the Hawaiian earing (or some related pathology) should not appear as the fiber of a generic smooth map, and perhaps the fibers should generically be manifolds with singularities, or something similar.

But I'd be happy to be proved wrong!

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I'm pretty sure the answer is yes and follows from the Thom-Boardman-Mather stratification. If nobody answers this I should be able to write up a proper answer in a week or so. It's the end of final exams here at the moment. –  Ryan Budney Apr 18 '12 at 14:59
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Manifold except at one point does not imply CW homotopy type. Think of the Hawaiian earring. –  Tom Goodwillie Apr 21 '12 at 12:26
    
In the statement quoted from Gromov, the number of singularities in a fiber should be at most $n$, not $m$. –  Tom Goodwillie Apr 21 '12 at 12:51
    
@Tom: you're right. I'll edit it. –  Mark Grant Apr 23 '12 at 5:53
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3 Answers

up vote 5 down vote accepted

Fibers of a generic smooth map are polyhedra (so in particular CW-complexes) by the triangulation conjecture of Thom, proved by Andrei Verona [Stratified Mappings - Structure and Triangulability, Springer LNM vol. 1102]. I'm not sure that the full strength of the triangulation conjecture is really needed here - hopefully Ryan's projected answer will clarify this. A more general triangulation conjecture was proved by Masahiro Shiota [Thom’s conjecture on triangulations of maps, Topology, 39 (2000), 383–399], who also has further results on this subject on the arXiv.

Verona's theorem says that any proper, topologically stable smooth map between smooth manifolds $M$ and $N$ is "triangulable", i.e. equivalent to a PL map by a topological change of coordinates in $M$ and in $N$. On the other hand the set of proper topologically stable smooth maps $M\to N$ is dense in $C^\infty(M,N)$ by the Thom-Mather theorem.

So don't worry, there are no Hawaiian earrings or other nightmares hiding in the fibers. To ensure that nothing obstructs easy sleep, we want the fibers of a generic smooth map $M^m\to N^n$ to be polyhedra of dimension $\le\max(m-n,0)$. Indeed, forgetting about the "polyhedra" part for a moment, we know the dimension estimate from multijet transversality (as sketched by Tom). We can now achieve "polyhedra" and "of dimension $\le\max\(m-n,0)$" to hold simultaneously because maps generic in two ways are still generic. (That is, the intersection of two open and dense subsets of $C^\infty(M,N)$ is open and dense. I guess this needs $M$ to be compact, and the general case follows by writing $M$ as a union of an increasing chain of compact submanifolds and applying Baire's theorem.)

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Thanks Sergey, you are the man! I will sleep well tonight. –  Mark Grant May 1 '12 at 10:28
    
By the way, do you know a reference for the (presumably trivial) fact that the fibers of a PL map are all polyhedra? –  Mark Grant May 1 '12 at 10:29
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Any standard PL topology text will do (Rourke-Sanderson, Zeeman, Hudson, Stallings or Glaser, some of them downloadable from Andrew Ranicki's webpage). Given a point $p\in N$, you can triangulate $M$ and $N$ so that $f$ is simplicial and $p$ is a vertex of the triangulation of $N$. Then $f^{-1}(p)$ will be triangulated by a subcomplex of the triangulation of $M$. I might also get a bit of untroubled sleep now - it's quite comforting that a bit of PL thinking can't seemingly be avoided in dealing with smooth maps :) –  Sergey Melikhov May 1 '12 at 11:34
    
Right. The trivial thing I kept forgetting was that corresponding fibers of topologically equivalent maps are topologically equivalent. Thanks again! –  Mark Grant May 2 '12 at 5:54
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I can explain the statement of Gromov (in which, as I said in a comment, the final $m$ should be $n$).

We have $f:M\to N$ with $m+1\ge n$.

The basic idea is that the set $S(f)$ of all points in $M$ at which $f$ is not a submersion should generically have codimension $m-n+1$, because the space of all $n\times m$ matrices of rank less than $n$ has that codimension. Thus $S$ should have dimension $n-1$. And therefore when $f$ maps $S(f)$ to $N$ it should do so in such a way that $n+1$ points never go to one point.

To make this precise you could use Mather's multi-jet transversality theorem. If you want, I'll try to explain that later.

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@Tom: This makes sense, thanks. Should the codimension of $S(f)$ in $M$ be $m-n+1$? –  Mark Grant Apr 23 '12 at 6:18
    
Oops, right. Fixed now –  Tom Goodwillie Apr 23 '12 at 10:15
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No, this isn't right. Take $N=\mathbb{R}^n$ and $M=\{(v,L):v\in L\in \mathbb{R}P^{n-1}\}$ and $f(v,L)=v$. Here $M$ is a one-dimensional vector bundle over $\mathbb{R}P^{n-1}$ and so has dimension $n$, the same as $N$. All preimages $f^{-1}\{v\}$ are single points except that $f^{-1}\{0\}=\mathbb{R}P^{n-1}$, which (for cohomological reasons) cannot be dominated by a CW complex of dimension less than $n-1$.

The manifold $M$ can also be described as the blow-up of $N$ at the point $0$. If you want an example with closed manifolds, you can just blow up your favourite closed $n$-manifold at a point. If you want an example where the codimension is strictly negative, just consider the composite $S^1\times M\xrightarrow{\text{proj}} M\xrightarrow{f} N$.

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@Neil: I may have worded my question badly. My intention was to ask if the property of having fibers dominated by CW-complexes of dimension $\le k$ was a generic property amongst all smooth maps between manifolds of codimension $-k$. The counter-examples you describe seem rather special. If we perturb $f$ slightly, does it remain a counter-example? –  Mark Grant Apr 18 '12 at 15:24
    
My guess is that you cannot perturb this behaviour away, but I do not see a proof of that at the moment. –  Neil Strickland Apr 18 '12 at 15:55
    
(1) This isn't a map of compact manifolds. (2) I think you can perturb away even Neil's example -- the perturbation won't be a proper homotopy, but it will be a 1-parameter family in the compact-open topology, just not in the uniform topology. –  Ryan Budney Apr 18 '12 at 21:10
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