# Can connectedness of fibers of a smooth map be checked on a dense set?

Suppose $f: M\to N$ is a smooth map between two smooth manifolds, with $M$ compact and connected, and suppose there is a dense subset of $f(M)$ where each fiber is connected, then each fiber of $f$ is connected.

If it helps, you can just consider the case where the set of regular values is dense in $f(M)$ and the fiber of each regular value is connected, and you want to prove every fiber of $f$ is connected.

• It might be good to rephrase this as an actual question and to change to title so that it reflects what you are actually asking. – Greg Stevenson Nov 4 '09 at 5:24
• I take your advice, thank you; but why do you think the title is inappropriate? – Wayne Nov 4 '09 at 5:37
• @Wayne: if somebody reads this title on the front page, they have no idea what the question is. Perhaps you could change it to "If the generic fibers of a smooth map are connected, are all fibers connected?" – Anton Geraschenko Nov 4 '09 at 5:39
• @Wayne: When you look at the homepage mathoverflow.net (rather than from the questions page), you can't see the statement of the question, just the title. So you should use the title to communicate the gist of the question. It makes it much more pleasant to browse the site if you can easily spot the questions you find interesting. – Anton Geraschenko Nov 4 '09 at 5:51
• I insist . – Anton Geraschenko Nov 4 '09 at 6:34

Perhaps I've misunderstood the question, but it looks like it's false.

Let M={(x,y)∈ℝ²|(x,y)≠(0,0)}, N=ℝ, and define f(x,y)=x. This is a smooth map of smooth manifolds, with the fibers over ℝ-{0} connected, but the fiber over 0 disconnected.

Edit: Wayne has added the hypothesis that M is compact. I think the statement is true under this hypothesis. Here's a sketch proof. Suppose f-1(x) is disconnected, then I'd like to prove that there is an open neighborhood of x where the fibers are disconnected. Since manifolds are normal, there are two non-empty disjoint open sets U and V in M covering f-1(x). Now prove a generalization of the hotdog lemma, which will say that there is an open neighborhood W of x such that U∪V covers f-1(W). Since U and V are disjoint, this will show that the fibers over points of W are disconnected. To prove the generalized hotdog lemma. use the fact that smooth maps locally "look like products", choose a cover of f-1(x) by "box shaped" open sets contained in U∪V. You can choose a finite number of these by compactness of f-1(x) (it's a closed subset of a compact space), and take W to be the intersection of all of their images in N.

More Edit: The above proof doesn't work (see comments below and Richard Kent's post). Apparently, I'm confused about the meaning of Ehresmann's_theorem, because it looks to me like the map f:S2⊂ℝ3→ℝ given by f(x,y,z)=z is smooth, but it doesn't look like a trivial fibration around the poles. The algebro-geometric analogue says that a smooth morphism X→Y always factors as X→AnY→Y, where the first map is etale. But an algebraic geometer would say that the map S2→ℝ is not smooth.

• M isn't compact in your example. – Darsh Ranjan Nov 4 '09 at 6:03
• Yes, I know. The compactness condition wasn't there when I answered the question. – Anton Geraschenko Nov 4 '09 at 6:08
• Hi,Anton. Nice proof. I don't think you need Ehresmann's theorem that smooth maps look like a product locally. It is enough to use that f is closed:then the closed image of M-UuV will miss x and the open complement of this image is your W with f^(-1)(W) included in U u V. So the result is true if you only assume f proper (=closed with quasi-compact fibres). By the way, what is this appetizing hot-dog lemma? – Georges Elencwajg Nov 4 '09 at 8:51
• Your proof breaks on Richard's example at the step "Since U and V are disjoint, this will show that the fibers over points of W are disconnected." However, I think it might work when the set of regular values is connected (e.g., the singular values live in codimension >= 2). – Reid Barton Nov 5 '09 at 2:06
• As the Wikipedia entry states, Ehresmann's theorem assumes the map to be a submersion (as you basically note, algebro-geometric language would suggest that a "smooth" map ought to be a submersion; however in differential topology "smooth" of course just means C^infty) – Mike Usher Nov 5 '09 at 7:52

Can't you just modify Ryan's example with boundary to get a counterexample with closed manifolds?

Let $M$ be the unit sphere in $\mathbb{R}^3$. Project $M$ to the x-axis.

Then compose this projection with the universal covering $\mathbb{R} \to N = S^1$ whose fundamental domain is the interval $[-1,1]$.

Then every fiber is a "longitude" except for one, which is a pair of poles.

• You're absolutely right. I wanted to say "this map isn't smooth" because I do algebraic geometry, but for differential geometers, the map is smooth. – Anton Geraschenko Nov 5 '09 at 2:42
• How is the map (x,y,z) --> x not smooth? It's a polynomial (linear) function. Do you want to rephrase your question for algebraic maps between complex algebraic varieties or something like that? – Ryan Budney Nov 5 '09 at 4:25
• @Ryan: It's not my question; I just edited the title (see mathoverflow.net/revisions/4054/list). In algebraic geometry, smooth maps are required to be submersions (en.wikipedia.org/wiki/Submersion_(mathematics)), and this map (as a map from the sphere to the line) is not a submersion at the poles. If you express the sphere as an algebraic variety (as CP^1, say), this map is not smooth; it's not even algebraic. – Anton Geraschenko Nov 5 '09 at 14:51

I assume you want the manifolds to be boundaryless -- otherwise you'd have to modify your question yet again. Consider the example of the function $e^{ix}$, from the reals to the unit circle, restricted to the interval $[0,2\pi]$, all fibers except the one over $1$ is connected.

• I edited out the distracting, vague and ultimately doomed-to-failure 2nd argument I put in the above comment. – Ryan Budney Nov 5 '09 at 2:18

I think in general it can not be checked on a dense set. Consider the normalization map from a smooth curve to a nodal curve, for example, a natural smooth surjection you can come up with from S^1 to the figure eight. The fiber over the nodes of the figure eight has two points but every other fiber is a single point.

• The normalization map is not smooth. – Anton Geraschenko Nov 4 '09 at 21:27
• Wayne, you should assume that f is surjective in the question, otherwise it's obviously false: Just take any smooth immersion of the circle in the plane that's in general position and isn't an embedding. Like a parameterization of a figure 8 lying in the plane. (So M = S^1 and N = R^2.) – Autumn Kent Nov 5 '09 at 1:20
• Wait, in differential geometry land the normalization map is smooth! In algebraic geometry land (where I live), it's not smooth because it's not flat. The problem here is that the nodal curve is not smooth. – Anton Geraschenko Nov 5 '09 at 2:41
• @Richard Kent: I do not assume the surjectivity of f in the question. As you said, if we take a parametrization of a figure 8 lying in the plane by the circle, i.e., M = S^1 and N = R^2, then we can see it is false in general. – Wayne Nov 5 '09 at 3:09
• @Anton Geraschenko: Thank you for pointing out - here the nodal curve is not smooth, but we can take N=R^2, as Richard said in his comment above. Sorry I should not have used the word "normalization". You are right in algebraic geometry the normalization map of a nodal curve is not flat. Here in the 1-dimensional real case, it is evident that the map from S^1 to the figure 8 is smooth in the differential context. – Wayne Nov 5 '09 at 3:22