Sphere as an embedded submanifold

  • Thread starter jem05
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  • #1
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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.
 

Answers and Replies

  • #2
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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?).
 
  • #3
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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.
 
  • #4
430
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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.
 
  • #5
662
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" 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.
 
  • #6
430
3
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:
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.
 
  • #7
662
1
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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