MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is a surjective submersion between two manifolds always a fibration?

share|cite|improve this question
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

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)).

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.