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Relationship betwen homotopy groups?

  1. Jul 28, 2010 #1

    quasar987

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    Is there a relationship between the homotopy groups of a pair (X,A) and of the quotient X/A ? It feels like they should be equal under mild hypothesis.

    More precisely, I am interested in the case where X is a smooth manifold and A a submanifold.

    Thx
     
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  3. Jul 29, 2010 #2
    I don't know much homotopy theory, but your intuition is correct in homology: The homology of X/A is isomorphic to the homology of the pair (X,A) provided there is an open neighborhood of A in X that deformation retracts to A.

    I apologize for not answering your question if you already knew that.
     
    Last edited: Jul 29, 2010
  4. Jul 31, 2010 #3
    By homotopy of a pair, do you mean homotopy classes of maps of cubes, or disks, which have their boundaries contained in A?

    I could imagine that your intuition is correct, given certain conditions.

    I did notice that if [tex] p :E \rightarrow B [/tex] is a fibration with fibre F, then
    [tex] p_* : \pi_n (E,F,f_0 ) \rightarrow \pi_n (B,b_0) [/tex]

    is an isomorphism. So maybe you just need that the quotient map is a fibration, which it is in your case isn't it?
     
  5. Jul 31, 2010 #4

    quasar987

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    I don't know. Is it? It looks that way though.
     
    Last edited: Jul 31, 2010
  6. Aug 1, 2010 #5
    Actually, I'm not 100%, I think that it will probably vary very delicately on the submanifold.

    I know that if you have a closed subgroup (H, say) of a Lie group (G, say) then H is a submanifold of G and G-->G/H is a fibre bundle (and hence a fibration).

    So this obviously wouldn't have been formulated if it worked for all submanifolds.

    I guess that this won't be too helpful then, unless your submanfold sits "nicely" inside your other manifold :frown:
     
  7. Aug 1, 2010 #6
    Maybe this is helpful for you, from the wiki page for fibre bundles:

     
  8. Aug 1, 2010 #7
    Sorry, ignore everything I am saying, I am being an idiot.

    The fibre map for Lie groups makes more identifications than the one you gave, it is the coset space.

    Sorry about that ^^
     
  9. Aug 1, 2010 #8
    Just as a quick counter example, X could be the sphere, A the sphere as well, so that the homotopy groups are unchanged for (X,A) but X/A is the single point set, maybe this is more helpful :D
     
  10. Aug 1, 2010 #9

    quasar987

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    Heh, thx Jamma.
     
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