# Continously differentiable f: R^n -> R^m not 1-1?

1. Apr 14, 2012

### ryou00730

Continously differentiable f: R^n --> R^m not 1-1?

My course is over with now, but I never could figure out this question. It's pretty much been haunting me ever since, and the internet has not given me a proof that convinces me. My problem is determining why:
A continuously di fferentiable function F: Rn → Rm is not 1-1 when n>m. I can understand why it would be true, I just can't seem to convincingly prove it. Anyone who is familiar with the subject, I would appreciate any feedback.

2. Apr 14, 2012

### Bacle2

Re: Continously differentiable f: R^n --> R^m not 1-1?

Look at the critical values of the map and use Sard's theorem to see what the image is like.

Re a continuous bijection alone, look at the issue of invariance of domain.

3. Apr 15, 2012

### lavinia

Re: Continously differentiable f: R^n --> R^m not 1-1?

Here are some ideas:

- Take the case of a map from the plane into the real line. At each point of the plane the Jacobian of the map,F, must have a non-trivial kernel. This is because the Jacobian is a linear map from R2 to R1,

F must be constant on any curve whose tangent lies if the kernel of the Jacoboan. If the Jacobian of F is not identically zero at a point - then it is not identically zero in a neighborhood of that point because it is continuous. Also,there is a well defined continuous field of directions(why?) in that neighborhood where the Jacobian of F is zero. Integrate this vector field of zero directions to get curves along which F is constant at every point in the neighborhood.

4. Apr 16, 2012

### Bacle2

Re: Continously differentiable f: R^n --> R^m not 1-1?

In the case of n=2, m=1, the image has measure zero in R^1, and, by continuity+connectedness of R^n+Sard's, the image f(R^n) is a connected subset of R^1 of measure-zero, which is...... Then generalize for other n>m.

For invariance of domain, use the fact that the inclusion map from R^m to R^n is a continuous injection, then compose.

5. Apr 17, 2012

### Citan Uzuki

Re: Continously differentiable f: R^n --> R^m not 1-1?

Bacle, I think you're confused. We're discussing the case where n>m, not n<m. Sard's theorem doesn't tell us anything useful about the image. Indeed, the image can even be all of R^m (consider the natural projection from R^2 to R^1).

We can use invariance of domain to prove the theorem. Indeed, suppose $f:\mathbb{R}^n \rightarrow \mathbb{R}^m$ is a continuous injection. Look at the restriction of f to some precompact open set $U$. Since $\overline{U}$ is compact and $\mathbb{R}^m$ is Hausdorff, $f|_{\overline{U}}$ is a homeomorphism onto its image, and thus so is $f|_{U}$. This contradicts invariance of domain.

6. Apr 23, 2012

### Bacle2

Re: Continously differentiable f: R^n --> R^m not 1-1?

Ouch, yes. Next time I'll try to read the actual question. I intended the opposite direction.

7. Apr 29, 2012

### Bacle2

Re: Continously differentiable f: R^n --> R^m not 1-1?

I realized I used the wrong theorem for this direction; it is Sard's in one direction and the inverse-value theorem+ Invariance of domain, in the opposite direction.

So, your function can clearly not be constant, so that df_x is not identically zero, so that there exists a local homeomorphism between an open set W in R^n, with an open set f(W) in R^m.

8. Apr 29, 2012

### deluks917

Re: Continously differentiable f: R^n --> R^m not 1-1?

Inverse function theorem does it.

9. Apr 29, 2012

### mathwonk

Re: Continously differentiable f: R^n --> R^m not 1-1?

i may be confused too but isn't bacle correct? i.e. if there exists a regular value, as guaranteed by sard, the locally the function is a projection by the implicit function theorem, hence not one to one.

10. Apr 29, 2012

### Citan Uzuki

Re: Continously differentiable f: R^n --> R^m not 1-1?

I don't think so. Sard's theorem guarantees the existence of a regular value, but not necessarily one in the image of the function.

11. May 1, 2012

### Citan Uzuki

Re: Continously differentiable f: R^n --> R^m not 1-1?

This topic may be getting old, but I just realized that there is a way of doing this that is close to your suggestion mathwonk. According to the constant rank theorem, the function will be locally a projection at any point where the rank of the derivative is constant in some neighborhood of the point. While there might not be a point where the derivative is surjective, there will be a point where the rank of the derivative attains a maximum, and at that point the rank will be constant in a neighborhood of the point, so your idea goes through.