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Sphere as an embedded submanifold

  1. Apr 1, 2010 #1
    hello,
    i already have 2 ways using 2 theorems to prove that a sphere is an embedded submanifold.
    i just want to check if i can prove it using this specific theorem:
    let M and N be manifolds
    if \phi : M --> N is an embedding, then \phi (M) is an embedded submanifold.
    because i am having trouble finding this \phi.
    thank you.
     
  2. jcsd
  3. Apr 1, 2010 #2
    Submanifold of what? It makes no sense to speak of a submanifold without referring to what manifold it's a submanifold of. For instance in your theorem you assert that [itex]\phi(M)[/itex] is a submanifold of N.

    If you don't care what manifold it's a submanifold of you can just use the rather trivial example of letting [itex]\phi : \mathbb{S}^n \to \mathbb{S}^n[/itex] be the identity, and then getting that [itex]\mathbb{S}^n[/itex] is a submanifold of itself (or do you perhaps only want to deal with proper submanifolds?).
     
  4. Apr 1, 2010 #3
    yeah sorry, i forgot to mention in the theorem, a submanifold of N, as you said.
    and i do want it a proper submanifold,
    i guessed N might be R^3, for instance.
     
  5. Apr 1, 2010 #4
    Well in that case you know the n-sphere is a subset of the (n+1)-dimensional Euclidean space [itex]\mathbb{S}^n \subseteq \mathbb{R}^{n+1}[/itex]. [itex]\mathbb{R}^{n+1}[/itex] is of course a manifold so we just define [itex]\phi : \mathbb{S}^n \to \mathbb{R}^{n+1}[/itex] to be the inclusion [itex]\phi(x)=x[/itex]. This is an embedding.
     
  6. Apr 6, 2010 #5
    " Well in that case you know the n-sphere is a subset of the (n+1)-dimensional Euclidean space .S^n is of course a manifold so we just define Phi: S^n -->R^(n+1) to be the inclusion Phi(x)=x . This is an embedding." (Sorry,I don't know well how to use
    quoting function)

    I think this is somewhat tautological, i.e., that if A is a subset of X, A given
    the subspace topology, then the inclusion map of A in X is an embedding
    into X, but I may be wrong:

    Given a space X, and a subspace A of X, the inclusion map i:A-->X is always an
    embedding:

    i) Let U be open in X. Then i^-1(U)=U/\A , open in subspace of A in X.

    ii) Let V open in (A, subspace) . Then V=W/\A ; W open in X
    then i(V) is open in the subspace topology of X.
     
  7. Apr 6, 2010 #6
    This is correct. I'm sorry if I somehow conveyed with my post that it was a non-trivial fact. jem05 was asking for an embedding from the n-sphere into some larger topological space, so I just provided the simplest one I could think of.
     
  8. Apr 6, 2010 #7
    ramshop wrote, in part:

    " This is correct. I'm sorry if I somehow conveyed with my post that it was a non-trivial fact. jem05 was asking for an embedding from the n-sphere into some larger topological space, so I just provided the simplest one I could think of. "

    No problem. It is the only one I can think of myself, other than maybe the cube, or
    minor variations of it.
     
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