# Does there always exist a continuous/smooth map from $\mathbb{R}^n$ onto a manifold $M^n$?

I want to know if we can get away with only having one set of coordinates for a manifold if we allow multiple coordinates to map to the same point.

## 2 Answers

An obvious requirement is that $$M$$ must be connected. This is also sufficient.

To get a smooth surjection $$f:\mathbb{R}^n\to M$$ for any connected $$n$$-manifold $$M$$, cover $$M$$ by countably many open sets $$(U_k)_{k\in\mathbb{Z}}$$ which are diffeomorphic to balls in $$\mathbb{R}^n$$ (and such that these diffeomorphisms can be extended to smooth maps on a neighborhood of $$\overline{U_n}$$). Now let $$f$$ map $$(2k,2k+1)\times\mathbb{R}^{n-1}$$ diffeomorphically to $$U_k$$ for each $$k$$. On sets of the form $$[2k+1,2k+2]\times\mathbb{R}^{n-1}$$, we just interpolate smoothly (near the boundary, we use the fact that our diffeomorphisms extend to a neighborhood of $$\overline{U_k}$$, and then in between we use connectedness of $$M$$ to let $$f$$ follow some path between a point of $$U_k$$ and a point of $$U_{k+1}$$).

For continuous maps, we can do even better: we can get a continuous surjection $$\mathbb{R}\to M$$. The construction is the similar, except that instead of diffeomorphisms to coordinate charts we use space-filling curves.

Another construction goes through Riemannian metrics. Assume that $M$ is connected. If you put a complete riemannian metric on your manifold, the exponential map $\exp_p:T_pM\rightarrow M$ will be surjective. The idea is that any two points are connected by a geodesic. You can view the world standing in one point!