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

1. Nov 23, 2011

### mathsq

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

Thanks.

2. Nov 24, 2011

### 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.

3. Nov 24, 2011

### Citan Uzuki

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.

4. Nov 24, 2011

### Bacle2

Right. A small variant of Citan Uzuki's proof: use that a continuous bijection between compact and Hausdorff is a homeo., so that the image of any compact subset of Rn, say Snl restricts to a homeo between S^n and the Hausdorff subspace f(Sn) , but then you can use Citan Uzuki's argument, or Borsuk-Ulam that does not allow you to embedd S^n in nothing smaller than Rn+1.

5. Nov 24, 2011

### mathsq

Thank you! That makes sense.

6. Nov 24, 2011

### mathwonk

there are easier variants as well. any continuous map takes connected sets to connected sets. But the complement of a point p of R^2 is connected whereas its image in R is not by any injective map, since it is separated by f(p), if p is chosen semi carefully.