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Is a surjective submersion between two manifolds always a fibration?

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Yes, if the map is also proper (this goes back to Ehresmann), otherwise there are easy counterexamples. – Donu Arapura Apr 8 '13 at 15:19
If the submersion is proper then this is Ehresmann's theorem. See's_theorem – Damian Rössler Apr 8 '13 at 15:19
Take, for example the surjective submersion $[0,2) \sqcup (1,3] \to [0,3]$ – David Roberts Apr 9 '13 at 0:20
"Algebraic geometry" does not seem like the right tag for this question. If it is, then the answer is no, I guess; one could have a family of elliptic curves where all the fibers are nonisomorphic. – Allen Knutson Apr 9 '13 at 1:10

2 Answers 2

No. To see this, take a bona fide fibration such as $\operatorname{pr}_2:\mathbf{R}^2\to\mathbf{R}$ and remove a point from the domain.

Meigniez on p. 3778 lists a number of sufficient conditions that a submersion $f$ be a fibration. The best known, already noted by Damian and Donu, is that $f$ be proper as in Ehresmann's Theorem (proved e.g. in Bröcker and Jänich, (8.12)).

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Suppose that $f:X\rightarrow Y$ is a surjective submersion of manifolds. Let $n=\dim(X)$ and $m=\dim(Y)$. Suppose that $y\in Y$. In suitable local coordinates $(x_1,\ldots,x_n)$ on $X$ and $(y_1,\ldots,y_m)$ on $Y$ at $y$, $f$ has the form $(a_1,\ldots,a_n)\mapsto (a_1,\ldots,a_m)$. Hence, you should obtain a nice $(n-m)$-dimensional fibre. This should be made rigorous, but I think it gives a decent place to start.

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