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Help with universal theorem of coefficients for homology

  1. Sep 9, 2007 #1
    *note before reading: pretend that the superscripts are really subscripts, for some reason , Latex is making the subscripts into superscripts*

    I'm not sure if I'm interpreting the universal coefficient theorem for homology correctly. Let's say I have a Homology group H(X;Z). This is a homology group using ring of integers. Now, let's say I want to compute Homology over the Field [tex]Z[/tex][tex]_{2}[/tex] (integers modulo 2), in order to simplify matters. By the way, I am trying to calculate the Betti numbers for a simplicial complex. Now, let's say I have a matrix M whose nullspace is isomorphic to the homology group. Now, I want to make all of the entries in the matrix modulo 2, if possible, in order to greatly simplify calculations.

    So far we have:
    kerM = H(X; Z)

    Now, the universal theorem gives us this: 0[tex]\rightarrow[/tex][tex]H[/tex][tex]_{k}[/tex]([tex]X[/tex]) [tex]\otimes[/tex]G[tex]\rightarrow[/tex][tex]H[/tex][tex]_{k}[/tex](G)[tex]\rightarrow[/tex]Tor(H[tex]_{k+1}[/tex]([tex]Z[/tex]),G)[tex]\rightarrow[/tex]0

    Basically, in this case, we can replace G with [tex]Z[/tex][tex]_{2}[/tex]. Also, the torsion should go to zero, since we have a free group. Now, since kerM is isomorphic to [tex]H[/tex][tex]_{k}[/tex]([tex]X[/tex]), can I say that ker(M)[tex]\otimes[/tex] [tex]Z[/tex][tex]_{2}[/tex] is isomorphic to [tex]H[/tex]([tex]X[/tex])[tex]\otimes[/tex] Z2= H(X;Z[tex]_{}2)[/tex][tex]Z[/tex][tex]_{2}[/tex] ? Can i also say that Ker(M) [tex]\otimes[/tex] Z[tex]_{}2[/tex] is the same as Ker(M[tex]\otimes[/tex]Z[tex]_{}2[/tex]) ?

    Back to my goal again, I want to make the entries in the matrix M, integers modulo 2. However, I realize that the nullspace of M modulo 2 and and the nullspace of M are different. By applying the universal coefficient theorem, can I say that these two nullspaces are isomorphic? If this universal coefficient theorem is correct, can I just convert the entries of the new matrix ([tex]M[/tex][tex]\otimes[/tex][tex]Z[/tex][tex]_{2}[/tex]) into 1's and 0's? It seems I would need to tensor the matrix [tex]M[/tex] with [tex]Z[/tex][tex]_{2}[/tex], but what would [tex]Z[/tex][tex]_{2}[/tex] in matrix form be? The identity matrix with modulo 2 entries? (basically just ones on the diagonal?). If I am totally wrong, can anyone suggest a method of making the nullspace of M and the nullspace of M modulo 2 isomorphic?
     
    Last edited: Sep 10, 2007
  2. jcsd
  3. Sep 10, 2007 #2

    Hurkyl

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    You need to put consecutive symbols all within the same [ tex ] ... [ /tex ] tag to get the formatting right. To make it line up nicely with texts in paragraphs, you should use [ itex ] ... [ /itex ] instead of [ tex ] ... [ /tex ].
     
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