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Computing the homology of R^3 - S^1

  1. Jun 18, 2007 #1
    Compute the homology of R^3 - S^1.

    Actually a friend of mine asked me this question and I came up with the following way to solve this but I'm not sure if it's correct.

    My analysis:

    H_0 = Z (the integers) because it's path connected.
    H_1 = Z (the friend said so but I don't believe him)
    H_2 = ??


    R^3 - {point} = S^2 (= means homeomorphic to or homotopic to)
    R^3 - {line} = S^2 - {2 points} = R^2 - {1 point}

    So R^3 - S^1 = R^3 - {a line together with a point at infinity} = R^2 - {2 points} = figure eight

    Is this a valid reasoning? If so, then H_0 = Z, H_1 = Z direct sum Z, H_2 = 0.

    Thanks.
     
  2. jcsd
  3. Jun 18, 2007 #2
    I don't know much about this stuff, but I know that the homotopy type of R^3-S^1 actually depends on the choice of embedding of S^1. I remember my professor said they can use the homotopy of R^3-S^1 to study knots.

    So it depends which S^1. But I think it's clear that under the standard embedding of S^1 H_n=0 for n>0. I could be wrong about this though.
     
  4. Jun 19, 2007 #3

    matt grime

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    There are such things as excision formulae, you know. I think we can assume that S^1 means the natural copy of S^1 sitting in the x-y plane.

    It seems reasonably clear to me that H_1 is Z, since the fundamental group is Z (you just count the number of times you loop around the copy of S^1), and H_1 is the abelianization of the fundamental group. This just leaves H_2 to work out.
     
    Last edited: Jun 19, 2007
  5. Jun 19, 2007 #4
    But wouldn't looping around the copy of S^1 be trivial (since we can just pull the loop into the z-plane a little and then deform it to a point).

    EDIT - never mind, I'm an idiot.
     
    Last edited: Jun 19, 2007
  6. Jun 19, 2007 #5
    Thanks. You guys are fantastic! Yes, so if we take A = S^1 and X = R^3, then I got
    H_3 (X,A)=0,
    H_2 (X,A)= Z,
    H_1 (X,A)= Z,
    H_0 (X,A)= Z.

    Can someone explain to me why the map f: H_0(A) --> H_0(Z) must be a constant (the zero) map? If this is a constant map, then I was able to conclude (algebraically that) H_1 (X,A)= Z.

    H_1(X)=0 --> H_1(X,A) --> H_0(A) =Z --> H_0(X)=Z --> H_0(X,A) --> 0
     
  7. Jun 20, 2007 #6

    matt grime

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    It suffices to show that the map H_0(X)-->H_0(X,A) is the identity map (or an isomorphism, at any rate), to demonstrate that the map H_0(A)-->H_0(X) is the zero map. Can you do this (I've not thought about it, to be honest).
     
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