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Consider, $M$, a smooth $m$ dimensional submanifold of $\mathbf R^n$. Does there exist a smooth map $X: \mathbf{R}^m\to\mathbf R^n$ such that $M=X(\mathbf R^m)$?

$X$ may have points at which the Jacobian is singular, which means that $M$ doesn't have to be diffeomorphic to $\mathbf{R}^m$. Furthermore, the stereographic map shows that there exists a smooth map between manifolds of different topology, so that $M$ and $\mathbf{R}^m$ apparently don't even have to be homoeomorphic. This then raises the question whether any $M$ can be represented thus.

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    $\begingroup$ If the only condition you put on $X$ is being smooth (and $M$ being connected I guess), then you can give $M$ a complete Riemannian metric, and then the exponential map based at any point of $M$ gives you a surjective map $\mathbb{R}^m\to M$ (see for example theorem 2.7 of Do Carmo's "Riemannian Geometry") $\endgroup$
    – Saúl RM
    Commented Sep 27, 2022 at 18:11
  • $\begingroup$ @SaúlRM Why isn't this an answer? $\endgroup$
    – dennis
    Commented Sep 27, 2022 at 18:27
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    $\begingroup$ Crossposted to math.SE: math.stackexchange.com/questions/4540286/… $\endgroup$
    – Qiaochu Yuan
    Commented Sep 27, 2022 at 18:36
  • $\begingroup$ @dennis because I thought maybe you were thinking of adding some extra condition about $X$, but if not I will post it as an answer $\endgroup$
    – Saúl RM
    Commented Sep 27, 2022 at 20:23
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    $\begingroup$ E.g. for m=2 consider a bouquet $B$ of countably many balloons with their strands, like those they sell at the amusement parks. One can easily map $\mathbb R^2$ smoothly onto $B$, and then $B$ smoothly onto any 2-dimensional connected manifold $M$ with countable atlas - just map the base of the bouquet to a base point of $M$, each balloon to a domain of a chart, and each strand to a suitable connecting path. $\endgroup$
    – Pietro Majer
    Commented Sep 27, 2022 at 21:17

1 Answer 1

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My comment turned answer:

Any smooth $m$-manifold $M$ admits a complete Riemannian metric (for example, as this answer says, any manifold embeds into some Euclidean space as a closed subset by Whitney embedding theorem).

So if we endow any connected smooth manifold $M$ with a complete metric and choose any point $p\in M$, then the exponential map $exp_p:T_pM\to M$ gives a smooth map (Prop 5.7c of [1]) which is surjective, because any two points in $M$ are joined by a geodesic (Cor 6.15 of [1]).

[1]: John M. Lee. Riemannian Manifolds. An introduction to curvature. Springer, 1997.

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