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F is diffeomorphism implys df is injective?

  1. Oct 2, 2011 #1
    Let U be a non-empty open set in Rn, if f:U->Rm is a diffeomorphism onto its image, show that df(p) is injective for all p in U. How can I attack this problem?
     
  2. jcsd
  3. Oct 2, 2011 #2

    quasar987

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    Differentiate the relation f-1 o f = id
     
  4. Oct 2, 2011 #3
    Note that [itex]\mathrm df(p)[/itex] is a linnear transformation between vectorial spaces. The tangent space to [itex]R^n[/itex] at [itex]p[/itex] is [itex]R^n[/itex]. The tangent space to [itex]f(U)[/itex] at [itex]f(p)[/itex] is again [itex]R^n[/itex], so, by Dimension Theorem, [itex]\dim(\ker(\mathrm df(p))=0[/itex], and then, [itex]\mathrm df(p)[/itex] is inyective.
     
  5. Oct 2, 2011 #4

    quasar987

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    felper, note that the statement of the problem does not assume that m=n. Moreover, in your argument, you seem to be using the rank-nullity formula + the fact that df(p) is surjective. So how would you prove that df(p) is surjective?
     
  6. Oct 2, 2011 #5
    [itex]\mathrm df(p)[/itex] is onto its image. Also, if you think [itex]f(U)[/itex] as a manifold, its totally clear it's tangent space is [itex]R^n[/itex].
     
  7. Oct 2, 2011 #6
    so I get df-1odf=I, and df(p) is invertible at every p, and the linear transformation x->df(p)x is injective, is that the logic?
     
  8. Oct 2, 2011 #7
    A diffeomorphism gives you an isomorphism between the respective tangent spaces.

    So you get an injective linear map between TpU and Tf(p)Rn, but, by dimension reasons, the map cannot be injective.

    Still, I think, by, e.g., invariance of domain, you cannot have a diffeomorphic embedding

    of U in R^n with f(U) in R^m , unless n=m; otherwise, you get a homeo. between an open

    subset of Rn and an open subset (as the image of a diffeomorphism, which sends open sets to open sets) of Rm, contradicting invariance of domain . Notice that the continuous image of an interval is

    a curve (and space-filling curves are seriously-non-injective) ; the continuous image

    of a square ( an I^2) is a surface, etc. , so continuous maps preserve dimension;

    I don't have an actual proof for this, but I think this is a result.
     
  9. Oct 2, 2011 #8
    Phyalan, I was wondering if you were precisely asking what I assumed, i.e., that diffeomorphisms between manifolds give rise to isomorphisms between the respective
    tangent spaces. If so, Quasar's answer gives you a proof, since (FoG)*=F*oG*, as Quasar said.
     
  10. Oct 2, 2011 #9

    quasar987

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    Well, you do get df-1odf=I, and generally speaking, if you have two maps such that f o g = id, then this is the same as saying g is injective.

    Of course, you can also differentiate the other realtion, that is, f o f-1 = id, and this tells you that df is surjective.

    So it turns out that df is actually an isomorphism between R^n and R^m, and since you know that those exists only btw vector spaces of the same dimension, this allows you to conclude to the important fact that diffeomorphisms exist only between open subsets of euclidean space of the same dimension.

    In fact, the same is true if you replace "diffeomorphism" simply by "homeomorphism". I.e., if f:U-->R^m is a homeomorphism onto its image, then f(U) is open, and m=n. This is the Brouwer invariance of domain theorem that Bacle is talking about in his post.
     
  11. Oct 9, 2011 #10
    Here I have got a proof of the question, I am not very sure why bother to do it this way, is it because the image of f, f(U), is not necessarily an open set so the inverse may not exist and the method that I posted above doesn't work?
    Pf:
    f is di ffeomorphic onto its image f(U) means that if we write f = i o f1 where i: f(U)→ Rm is the inclusion map, then  f1: U → f(U) is a diff eomorphism. So, there is an open neighborhood (denoted by V ) of f(U) in Rm and a smooth map
    g : V → Rn
    which extends f1-1.Now, if we write f = ivo f2 where iv : V → Rm is the inclusion map, then
    f2 : U → V
    is a smooth map. By the chain rule, we have Tpf = Tf(p)iv o Tpf2 = Tpf2.
    Since we have
    g o f2 = 1U
    where 1U: U → U is the identity map, by the chain rule, we have
    Tf(p)g o Tpf2 = 1 : TpU→ TpU:
    Then Tpf2, hence Tpf, is injective.
     
  12. Oct 9, 2011 #11

    quasar987

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    Maybe rewrite this without the random characters, but it sounds like you are overcomplicating things. Plus, I gave you the whole solution in post #9 (first two sentences)...
     
  13. Oct 9, 2011 #12

    lavinia

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    this is just the Chain Rule
     
  14. Oct 9, 2011 #13
    To rewrite Quasar's idea,a bit differently, the functoriality properties of the induced map :

    i)(FoG)*=F*oG*
    ii)(Id)*=Id*

    Then, since F is a diffeo, there is a differentiable inverse F-1, so that:

    (FoF-1)*=F*oF-1*=Id*. This F-1* is both a right- and left- inverse to F*.
     
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