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.
 A: 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.
A: 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.
A: 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. 
A: 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.
