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Can we characterize when a submersion $F:M \to N$ between two smooth manifolds has connected fibers? If this is too hard, what are some sufficient conditions?

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Jason's answer and Boyarksy's comment are nice examples. Does anyone know of an example where the fibers are not compact? –  David Carchedi Jun 24 '10 at 10:30
    
Also, it might be nice to do some class of examples which do not make $F$ into a fiber-bundle, so that the topologies of the fibers can vary. –  David Carchedi Jun 24 '10 at 10:59
    
Dear David, I don't know if you're still interested in this, but note that if we relax the condition that $F$ is a fibre-bundle, we pretty much lose all control. E.g. take $M$ to be an open subset of $\mathbb R^2$, take $N$ to be $\mathbb R$, thought of as (say) the $x$-axis of $\mathbb R^2$, and let $M \to N$ be the projection to the $x$-axis. Then the fibres of this map will be open subsets of $\mathbb R$, which can be pretty much anything. E.g. if $M$ is $\mathbb R^2$ minus one point, the fibres are $\mathbb R$ above all but one point, but above one point it is $\mathbb R$ minus a point –  Emerton Jan 3 '13 at 14:30
    
and so disconnected. Removing more points, or for example a curve, from $\mathbb R^2$, will give you examples with all kinds of behaviour. Regards, Matthew –  Emerton Jan 3 '13 at 14:31
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1 Answer 1

If $M$ and $N$ are both compact, then the submersion $F$ can be thought of as a fiber bundle map with fiber $F^{-1}(p)$ for any $p\in N$. Then one can apply the long exact sequence of homotopy groups of a fiber bundle to learn that if, for example, $M$ is connected and $N$ is 1-connected, that the fibers must be connected.

These sufficient conditions may be too specific, though.

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Can relax compactness of $M$ and $N$ to properness of $F$: still ensures the submersion $F$ is a fiber bundle map (Ehresmann), so $F_{\ast}(\mathbb{Z})$ is a locally constant sheaf of finite free $\mathbb{Z}$-modules whose fiber rank is the number of connected components in fibers. If $N$ is simply connected this locally constant sheaf is constant, yet its global sections are $\mathbb{Z}$ since $M$ is connected. Thus, all fibers are connected. The Mobius strip minus the zero-section is connected and fibers over $S^1$ with disconnected fibers, so the simply connectedness cannot be dropped. –  Boyarsky Jun 24 '10 at 4:56
    
Thanks, this is quite helpful. –  David Carchedi Jun 24 '10 at 10:24
    
@Jason: It is sufficient to assume that $\pi_1(N)$ has no non-trivial finite quotient. Indeed, since $F$ is a locally trivial submersion, the sets $\pi_0(F^{-1}(p))$ of connected components of fibers form a finite local system on $N$. Since $M$ is connected, this local is trivial. On the other hand, if $\pi_1(N) has a non-trivial finite quotient, this quotient will correspond to a finite covering $M$ of $N$. Then $M$ is compact and the projection to $N$ is a submersion with disconnected fibers. –  ACL Jan 3 '13 at 16:31
    
@Jason: It is sufficient to assume that $\pi_1(N)$ has no non-trivial finite quotient. Indeed, since $F$ is a locally trivial submersion, the sets $\pi_0(F^{−1}(p))$ of connected components of fibers form a finite local system on $N$. Since $M$ is connected, this local is trivial. On the other hand, if $\pi_1(N)$ has a non-−trivial finite quotient, this quotient will correspond to a finite covering $M$ of $N$. Then $M$ is compact and the projection to $N$ is a submersion with disconnected fibers. –  ACL Jan 3 '13 at 16:33
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