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Math Help - Homology Groups

  1. #1
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    Computing Homology Groups

    Hey, I've been reading Hatcher and doing the exercises, couldn't answer this one:

    (p.132 ex. 17)



    Tried various directions but I'm pretty much lost. Any help will be appreciated!

    Edit: I did get that since X/A is the wedge of two tori, that H_i(X,A)=\tilde{H}_i(X/A) =   \left\{<br />
     \begin{array}{lr}<br />
       \mathbb{Z}^2 &  i=0\\<br />
       \mathbb{Z}^4 & i=1\\<br />
       \mathbb{Z}^2 & i=2<br />
     \end{array}<br />
   \right.
    Last edited by Defunkt; December 11th 2010 at 08:38 AM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Defunkt View Post
    Hey, I've been reading Hatcher and doing the exercises, couldn't answer this one:

    (p.132 ex. 17)



    Tried various directions but I'm pretty much lost. Any help will be appreciated!

    Edit: I did get that since X/A is the wedge of two tori, that H_i(X,A)=\tilde{H}_i(X/A) =   \left\{<br />
     \begin{array}{lr}<br />
       \mathbb{Z}^2 &  i=0\\<br />
       \mathbb{Z}^4 & i=1\\<br />
       \mathbb{Z}^2 & i=2<br />
     \end{array}<br />
   \right.
    I am very, very novice at homology theory so take what I am about to say as probably wrong (and possibly gobbeldy-gook). That said, since no one has answered your question I'll take a stab at my interpretation (see below) of it.


    So, what is the A in H_n(X,A)? Is it the cellular decomposition (I have Hatcher, but not with me. And, I haven't read it yet)? If so, have you discussed the fact yet that the homology is independent of the decomposition? If you have then I think I can help with a) (maybe). Why can't you decompose \mathbb{T}^2 into a zero-cell, two one-cells, and one two-cell in the natural way so that our chain complex becomes 0\to\mathbb{Z}\to\mathbb{Z}\oplus\mathbb{Z}\to\mat  hbb{Z}\to 0. I think (giving the outward normal orientation of course) it's easy to show that \partial_0:\mathbb{Z}\to 0,\partial_1:\mathbb{Z}\oplus\mathbb{Z},\partial_2  :\mathbb{Z}\to\mathbb{Z}\oplus\mathbb{Z} and \partial_3:0\to\mathbb{Z} are all trivial. From there it's pretty easy to calculate that

    H_n(\mathbb{T}^2)=\begin{cases}\mathbb{Z} & \mbox{if}\quad n=0,2\\ \mathbb{Z}\oplus\mathbb{Z} & \mbox{if}\quad n=1\\ 0 & \mbox{otherwise}\end{cases}
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  3. #3
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    Haven't gone through cellular homology yet, so only used singular homology, exact sequences and excision for these:

    for a), note that (X,A) is a good pair, and so H_n(X,A) \cong \tilde{H}_n(X /A). The quotient's 1-skeleton has only one path-component, and therefore H_0(X,A) \cong \tilde{H}_0(X/A) = 0.

    For n=1, use the long exact sequence of the pair X,A:

    ...\rightarrow \tilde{H}_1(A) \rightarrow \tilde{H}_1(S^2) \rightarrow H_1(S^2,A) \rightarrow \tilde{H}_0(A) \rightarrow \tilde{H}_0(S^2)\rightarrow ...

    \tilde{H}_1(S^2) = 0 and \tilde{H}_0(S^2) = 0 and \tilde{H}_0(A) = \mathbb{Z}^{|A|-1}. The first two imply that the map H_1(X,A) \rightarrow \tilde{H}_0(A) is an isomorphism and from the latter we have the result for n=1.

    For n=2, S_2(A) = 0 and therefore H_2(S^2,A) = H_2(S^2) = \mathbb{Z}.

    For any n>2, the result is obviously 0.

    Similar arguments give us the same H_0(X,A),H_2(X,A) for X=S^1 \times S^1. For the first homology group, the long exact sequence we used to calculate H_1(S^2,A) is
    ...\rightarrow 0 \rightarrow \mathbb{Z}^2 \rightarrow H_1(X,A) \rightarrow \mathbb{Z}^{|A|-1} \rightarrow 0 \rightarrow ...
    and so we get H_1(S^1 \times S^1, A) = \mathbb{Z}^{|A|+1}

    To finish (b), we note that X/B is homotopy equivalent to a torus with 2 points identified, and since we already know that homotopic spaces have the same homology groups, and since (X,B) is a good pair, we get:

    H_n(X,B) \cong \tilde{H}_n(X/B) \cong \tilde{H}_n(\mathbb{T}^2/{\{x_0,y_0\}) \cong {H}_n(\mathbb{T}^2, \{x_0,y_0\})

    But we already calculated this one in (a).
    Last edited by Defunkt; December 13th 2010 at 05:50 PM.
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