Let $\mathcal{X}$ and $\mathcal{Y}$ be compact manifolds and let $\mathcal{Y}$ be connected. Prove that a submersion $F : \mathcal{X} \to \mathcal{Y}$ must be surjective.

I don't have much thought on this question, except for if $F$ is not surjective, then its degree is zero.

May I request for some hints? Thank you.

I am trying to fill in Andreas Blass' excellent answer:

Given $F$ a submersion, according to local submersion theorem, there exist local coordinate around $x$ and $y$ such that $F(x_1, \dots, x_k) = (x_1, \dots, x_l), k > l$. Hence the image of $F$ is open.

Local Submersion Theorem. Suppose that $f: X \to Y$ is a submersion at $x$, and $y = f(x)$. Then there exist local coordinate around $x$ and $y$ such that $f(x_1, \dots, x_k) = (x_1, \dots, x_l), k > l$. That is, $f$ is locally equivalent to the canonical submersion at $x$.

Meanwhile, given $\mathcal{X}$ is compact, because continuous function maps compact space to compact space, the image of $F$ is closed.

Therefore, the image is either all of $\mathcal{Y}$ or empty. Assuming $\mathcal{X}$ is not empty, we proved that $F$ is surjective.

  • $\begingroup$ Degree doesn't make sense, most likely. This is an exercise in the first 15 pages of G&P and you really need to be able to do such problems on your own, without our doing them for you. $\endgroup$ – Ted Shifrin Aug 18 '13 at 3:45

Use "submersion" to conclude that the image of $F$ is open. Use "compact" to conclude that this image is closed. Use "connected" to conclude that this image is either all of $\mathcal Y$ or empty. Now either you're lucky and the source of this problem requires manifolds to be nonempty, or you need to add to the problem a hypothesis that $\mathcal X\neq\varnothing$.

  • $\begingroup$ Thank you so much Andreas!!! If $\mathcal{X} = \varnothing$, I guess it is "vacuously true." =D $\endgroup$ – 1LiterTears Aug 18 '13 at 3:29
  • 2
    $\begingroup$ There's a unique map $\emptyset \to \mathcal{Y}$ which is vacuously a submersion, but it's not surjective. $\endgroup$ – Anthony Carapetis Aug 18 '13 at 4:51
  • $\begingroup$ Oh right, I added assuming $\mathcal{X}$ is nonempty, hope it work? Thank you @AnthonyCarapetis. $\endgroup$ – 1LiterTears Aug 18 '13 at 5:14

If $F$ is not surjective, then one can find a point $y \in F(X)$ and a path in $Y$ (since $Y$ is connected) such that the path starts at $y$ and the rest of it lies in $Y\backslash F(X)$. In particular, the velocity vector of this path is not in the image of the differential $dF_{x} : T_x X \rightarrow T_y Y$, for any $x \in F^{-1}(y)$. This contradicts the definition of submersion. This is the idea; you have to clean it up and fill in the details. Hope that helps.

  • $\begingroup$ This will work, but one very important detail is to show that you can choose your path not to be tangent to the boundary of $F(X)$. This makes it a little less easy. $\endgroup$ – Ryan Reich Aug 18 '13 at 3:41
  • 1
    $\begingroup$ But Y is not necessarily path connected! $\endgroup$ – Mojojojo Sep 22 '18 at 19:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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