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

  1. Apr 14, 2012 #1
    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.
     
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  3. Apr 14, 2012 #2

    Bacle2

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    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.
     
  4. Apr 15, 2012 #3

    lavinia

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    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.
     
  5. Apr 16, 2012 #4

    Bacle2

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    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.
     
  6. Apr 17, 2012 #5
    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 [itex]f:\mathbb{R}^n \rightarrow \mathbb{R}^m[/itex] is a continuous injection. Look at the restriction of f to some precompact open set [itex]U[/itex]. Since [itex]\overline{U}[/itex] is compact and [itex]\mathbb{R}^m[/itex] is Hausdorff, [itex]f|_{\overline{U}}[/itex] is a homeomorphism onto its image, and thus so is [itex]f|_{U}[/itex]. This contradicts invariance of domain.
     
  7. Apr 23, 2012 #6

    Bacle2

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    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.
     
  8. Apr 29, 2012 #7

    Bacle2

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    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.
     
  9. Apr 29, 2012 #8
    Re: Continously differentiable f: R^n --> R^m not 1-1?

    Inverse function theorem does it.
     
  10. Apr 29, 2012 #9

    mathwonk

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    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.
     
  11. Apr 29, 2012 #10
    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.
     
  12. May 1, 2012 #11
    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.
     
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