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What are some "interesting" examples of submersions that are not surjective?

Usually, the notion of submersion comes with a prefix of surjective. Most of the maps we come across when we do differential geometry are surjective submersions.

It arises a question, are these two properties necessarily be combined always?

One might wonder if there are surjective maps that are not submersions. There are many such maps, but one that immediately comes to mind is the map $f:\mathbb{R}\rightarrow \mathbb{R}$ defined as $x\mapsto x^3$.

Next question is, are there submersions that are not surjective? Quick checking of "some" examples discussed in a first course tells that, when ever we talk about submersion, the property of surjectivity comes with it. It does not mean there are none.

Consider a manifold $M$ and an open subset $U$ of $M$. Consider the inclusion map $i:U\rightarrow M$. It is easy to see that this map is a submersion. But, if $U$ is a proper open subset then, this inclusion map is not a surjective map. Thus, there are submersions that are not surjective.

What are some "interesting" examples of submersions that are not surjective which are not of the form $U\rightarrow M$ for an open subset $U$ of $M$?

Please do not assume the reader knows any theorems other than the definition of surjective maps, submersions.

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    $\begingroup$ For any submersion $N\rightarrow U$, the composition $N\rightarrow U\rightarrow M$ is a non-surjective submersion. This gives many examples, but I don't know if I would call them interesting. $\endgroup$
    – Jason DeVito - on hiatus
    Jun 3, 2020 at 16:27
  • $\begingroup$ @JasonDeVito :) As you have said, it does not seem to be interesting... +1 for your example though... $\endgroup$
    – Praphulla Koushik
    Jun 3, 2020 at 16:44
  • $\begingroup$ Submersions are open maps, so the only way to get an example that you seek is to have a connected component of the range manifold which is not in the image. $\endgroup$
    – Ted Shifrin
    Jun 3, 2020 at 17:32
  • $\begingroup$ @TedShifrin that is an answer... :) Please see if you can add it as an answer with an example.. $\endgroup$
    – Praphulla Koushik
    Jun 3, 2020 at 17:53
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    $\begingroup$ Consider $f: \Bbb{R} \to \Bbb{R}$, $f(x) = \arctan(x)$. Then, $f$ is a submersion (in fact a diffeomorphism onto its image which is $(-\pi/2, \pi/2)$), but not surjective. Is this interesting enough? $\endgroup$
    – peek-a-boo
    Jun 3, 2020 at 18:20

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It is an exercise (following from the fact that linear projections are open maps) that all submersions are open maps. If $f\colon X\to Y$ is a submersion and $X$ is compact, then $f(X)$ will be both open and closed in $Y$, hence a connected component of $Y$. (So, for an example, take $X$ to be any compact manifold, and let $Y$ be two disjoint copies of $X$; let $f$ map to the first copy by the identity map. I could make slightly more interesting examples by taking cylinders for $X$, mapping to the base of the cylinder ... or $S^3\to S^2$ by the Hopf map, then map to $S^2\times S^2$ by mapping only to the first factor, etc.)

Now, if $X$ is not compact, you already said that you could give examples by including open subsets ...

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