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.



Can a smooth manifold be realised as the image of a smooth function?
My comment turned answer:


Any smooth $m$-manifold $M$ admits a complete Riemannian metric (for example, as [this answer](https://mathoverflow.net/a/18851/172802) 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.


