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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.

1LiterTears
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  • 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. – Ted Shifrin Aug 18 '13 at 03:45

2 Answers2

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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 $\mathcal X\neq\varnothing$.

Andreas Blass
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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.

user44441
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  • 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. – Ryan Reich Aug 18 '13 at 03:41
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    But Y is not necessarily path connected! – Shreya Sep 22 '18 at 19:31
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    @Shreya All connected manifolds are path connected. More generally, for every locally path connected topological space connectedness implies path connectedness. – WakeUp-X.Liu Jan 07 '22 at 04:47