# Is there a C^infty map that is one to one from R^n to R

Is there a C^infty map that is one to one from R^n to R?

Thanks.

Bacle2
Some comments; maybe someone else can extend:

I think no, but I have no proof. By invariance of domain, the map cannot be onto, since ℝn is open in itself, and, if it where onto ℝ, you could compose it with the inclusion (an inclusion, actually) into ℝn (any n>1 will do ) , concluding that there is a copy of
ℝ that is open in ℝn, i.e., assume f(ℝn)=ℝ, and then let i be the standard inclusion into ℝn, i.e, i(x)=(x,0,0,...,0). Then i°f is a map whose image is a copy of ℝ, which is open in ℝn.

Or, you can use Sard's theorem (we had a similar problem here a while back): Since every point in ℝn is a critical point of the differentiable map f, the image of ℝn under this map must have measure zero in ℝ.

There isn't even a continuous injection from $\mathbb{R}^n \rightarrow \mathbb{R}$ when n>1. For suppose there were. Let f be the restriction of this map to, say, the closed unit ball. If A is a closed subset of the closed unit ball, then A is compact, so f(A) is compact and therefore closed. Hence f is a closed map and therefore a homeomorphism onto its image. This implies the unit ball in $\mathbb{R}^n$ is homeomorphic to a subset of $\mathbb{R}$. But this is impossible, since any subset of $\mathbb{R}$ that has at least two points is either disconnected or can be made disconnected by removing a single point, whereas the closed unit ball in $\mathbb{R}^n$ is connected and remains so if any single point is removed.

Bacle2