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Relative Singular Homology

  1. Feb 20, 2008 #1
    Let X be a topological space and let y be a point in X. Denote H'_n(X) be the reduced homology of X and Denote H_n(X,y) the relative homology of X and the one point space {y}.

    Prove: H'_n(X) is isomorphic to H_n(X,y).

    I have proven that it is true for n>0. But, for n=0, I get H_n(X) isomorphic to H_n(X,y) and since the reduced homology of X is H_n(X) mod integers, I have run into a contradiction. Wikipedia gives my result as a property of relative homology, but my teacher has something different. (Note H_n(X)=H'_n(X) for n>0).

  2. jcsd
  3. Feb 20, 2008 #2


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    Hrm. When I worked it out, I got the exact sequence

    [tex]0 \to \mathbb{Z} \to H_0(X) \to H_0(X, y) \to 0.[/tex]

    Can you show your work?

    Incidentally, note talk page for that wikipedia article. I'm going to edit the wikipedia page to remove confusion.
    Last edited: Feb 20, 2008
  4. Feb 20, 2008 #3
    Yeah, I got that exact sequence. But, somehow I wealsed to get the wrong, wikipedia answer, so I probably need to re-check my work! Thanks.
  5. Feb 20, 2008 #4
    ok got it: just use the first isomorphism theorem and that H'_0(X)=H_0(X) modulo integers.
    Last edited: Feb 21, 2008
  6. Feb 21, 2008 #5


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    Just to make sure -- there are a lot of ways the integers can be mapped into an abelian group. You have to make sure you actually have the same subgroup for the kernel in both!

    e.g. the kernel of Z --> Z/2 and Z --> Z/4 are both abelian groups isomorphic to the integers.
  7. Feb 21, 2008 #6
    Actually should it just be

    [tex]H_1(X,y) \to \mathbb{Z} \to H_0(X) \to H_0(X, y) \to 0.[/tex]

    If not how did you get the zero?
  8. Feb 21, 2008 #7
    [tex]0 \to H_1(X) \to H_1(X,y) \to \mathbb{Z} \to H_0(X) \to H_0(X, y) \to 0.[/tex]

    So, I am still stuck. No matter how you word it, it boils down to the map from H_1(X) to H_1(X,y) being surjective and the map from H_0(X) to H_0(X,y) being injective. No matter what you need those facts and so I think you have to go the chain complexes and look at how the relative homology and the differential of the chain C_n(X)/C_n(y) are defined. We do know that

    C_m(y) iso C_n(y) iso Integers for all m,n>=0. This surely must tell you something about the differential but I have no idea what.
  9. Feb 21, 2008 #8


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    I was half-thinking cohomology and had some of the arrows reversed. :frown:

    Those can't both be true; otherwise the sequence wouldn't be exact at Z.

    Anyways, there is a direct description of [itex]H_0(\cdot)[/itex]; doesn't it allow you to conclude that [itex]H_0(y) \to H_0(X)[/itex] is monic?
    Last edited: Feb 21, 2008
  10. Feb 22, 2008 #9
    Ok, I got it:

    [tex]0 \to H_1(X) \to H_1(X,y) \to \mathbb{Z} \to H_0(X)=C_0(X) \to H_0(X, y)=C_0(X)/C_0(y)=C_0(X)/\mathbb{Z} \to 0.[/tex]

    So, right away we have H'_0(X)=H_0(X) module integers = H(X,y). Furthermore, we know that the kernal of the map from is C_0(y) = integers which implies the map from the integers to H_0(X) is injective which implies the image of the map from H_1(X,y) to the integers is 0 which implies we have the exact sequence

    [tex]0 \to H_1(X) \to H_1(X,y) \to 0[/tex]

    i.e., H'_1(X) iso to H_1(X,y). the proof for n>1 is trivial.
  11. Feb 23, 2008 #10


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    Again, I'd like to point out that "H_0(X) modulo integers" is not well-defined -- for most groups, there are lots of ways to map Z into it. For example, for appropriate choices of a map Z -> Z, each the following sequences are exact:

    [tex]0 \to \mathbb{Z} \to \mathbb{Z} \to 0[/tex]
    [tex]0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z} / 2 \to 0[/tex]
    [tex]0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z} / 3 \to 0[/tex]

    and for most choices of the map Z -> Q, the following sequence is not exact:

    [tex]0 \to \mathbb{Z} \to \mathbb{Q} \xrightarrow{\pi} \mathbb{Q} / \mathbb{Z} \to 0[/tex]

    (in fact, it's usually not even a complex!) ([itex]\mathbb{Z}[/itex] is an honest to goodness subgroup of [itex]\mathbb{Q}[/itex], so [itex]\mathbb{Q} / \mathbb{Z}[/itex] makes sense. The map [itex]\mathbb{Q} \xrightarrow{\pi} \mathbb{Q} / \mathbb{Z}[/itex] in the above sequence is the projection map. Other maps could fit there, though!)

    So, the fact that you have two exact sequences

    [tex]0 \to \mathbb{Z} \to H_0(X) \to H_0(X, y) \to 0[/tex]

    [tex]0 \to \mathbb{Z} \to H_0(X) \to H'_0(X) \to 0[/tex]

    does not imply that you have an isomorphism [itex]H_0(X, y) \cong H'_0(X)[/itex].
    Last edited: Feb 23, 2008
  12. Feb 23, 2008 #11
    I agree with you the fact that H_0(X,y) iso H'_0(X) is trivially done by observation--i.e.

    H_0(X,y)=C(X)/Z and H_0(X)=C(X) and H'_0(X)=H_0(X)/Z.
  13. Feb 23, 2008 #12


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    I'm saying exactly the opposite.

    I'm saying:
    (1) You cannot conclude H_0(X,y) is isomorphic from H'_0(X) given just the information you presented.

    (2) The expressions C(X)/Z and H_0(X)/Z aren't even well defined. It's like writing 0/0 in ordinary arithmetic!

    Quotient groups G/H are defined for when H is a subgroup of G. Z is not a subgroup of H_0(X). Z is isomorphic to (many) subgroups of H_0(X); if you pick a specific map [itex]f:\mathbb{Z} \to H_0(X)[/itex], then you can talk about the group [itex]H_0(X) / \im f[/itex]. But if you pick another map [itex]g:\mathbb{Z} \to H_0(X)[/itex], then it is possible that [itex]
    H_0(X) / \mathrm{im\ } f \not\cong H_0(X) / \mathrm{im\ } g[/itex].
    Last edited: Feb 23, 2008
  14. Feb 23, 2008 #13
    Yes, yes, all that is true. So, lets concentrate on (2) because that is were the confusion is.

    Theorem: H_0(X) = Z + Z + .... +Z (n direct sums of copies of Z where n is the number of connected components of X), and H'_0(X) = H_0(X)/Z (i.e. n-1 direct sums of copies of Z where n is the number of connected components of X).

    So, in this way, H_0(X)/Z is well-defined.

    Now, consider a point y in X, C_n({y}) is the free abelian group generated by all continuous functions of the form f:delta^n --> {y} where delta^n is the standard n-simplex. In other words C_n({y}) is the free abelian group with one generator. In other words, C_n({y}) iso Z.

    So, we have a canonical isomorphism

    H_0(X,y)=C_0(X)/C_0(y) = H_0(X)/C_0(y) iso H'_0(X)

    as H_0(X) iso n direct sums of Z and C_0(y) iso Z; n direct sums of Z modulo Z iso n-1 direct sums of Z iso H'_0(X).
  15. Feb 23, 2008 #14


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    Not really; you've not produced any sort of quotient group. If you chose a specific copy of Z, then you will have a well-defined quotient group H_0(X)/Z -- but that goes back to what I've been saying the whole time: you have to specify the map Z --> H_0(X) before H_0(X)/Z makes sense!

    You have not shown that.

    Let's assume you are right in saying that [itex]H_0(X, y) \cong H_0(X) / \mathop{\mathrm{image}} \left(C_0(y) \to H_0(X))[/itex], where [itex]C_0(y) \to H_0(X)[/itex] is the map induced by the chain map induced by the inclusion [itex]y \to X[/itex].

    You have not shown just what that image is. Suppose that X is path connected, so that [itex]H_0(X) \cong \mathbb{Z}[/itex]. What if the map [itex]C_0(y) \to H_0(X)[/itex] is isomorphic to the "multiplication by 2" map [itex]\mathbb{Z} \xrightarrow{\cdot 2} \mathbb{Z}[/itex]? In that case, we'd have [itex]H_0(X, y) \cong \mathbb{Z} / 2[/itex]. What if it's the zero map? Then we'd have [itex]H_0(X, y) \cong \mathbb{Z}[/itex]. Neither of those are isomorphic to [itex]H'_0(X)[/itex].
    Last edited: Feb 23, 2008
  16. Feb 25, 2008 #15
    You're right!

    I don't think you can prove it this way unless I am missing something. I think you have to show there is a chain homotopy equivalence from the augmented chain complex of C(X) to the chain complexC(X)/C(y). This will induce a isomorphism in homology. I haven't tried this technique yet.
  17. Feb 25, 2008 #16


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    Your basic idea is sound; in order to apply it, you simply need to figure out exactly what the map C_0(y) --> H_0(X) is. (and that shouldn't be too difficult)
  18. Feb 26, 2008 #17


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    Hope not to distract too much from 00_99's question. Just a general question
    on relative homology. Unfortunately, I haven't done much more than simplicial
    homology for a while.

    I am reading Griffiths and Harris' book:

    Given complexes (K*,d) and (J*,d) (with std. meaning: a complex is

    a seq. of Abelian groups with a differential d such that d^2=0 ), such

    that (K*,d) is a subcomplex of (J*,d) , i.e, for all non-neg. integer p,

    all groups J^p<K^p , for all Abelian groups K^p in (K*,d).

    Then G&H define the quotient complex :

    K*/J* , with "the obvious differential" . What is this "obvious


    Now, the only actual case I can think of is that of simplicial

    homology, and relative homology, with A<B , where A is a simplicial

    subcomplex of B . Then a relative cycle c_p is given


    C_p(A/B)=C_p(A)/C_p(B)= (by Abelianness)

    C_p(A)-C_p(B) (A/B is just meant to stand for A relative to B,

    not a topological quotient.)

    [c_p]=c_p(+)C_p-1(B) , i.e. just the coset rep. of c_p in the quotient

    where C_p-1 is the group of (p-1)-chain maps in B.

    And the relative differential d_r s:


    with d the differential of the original complex.

    Is this the generalized formular for relative homology groups, differentials, etc. ?

  19. Feb 26, 2008 #18


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    A map of complexes, by definition, commutes with the differential. So if you have a map [itex]f : A^\cdot \to B^\cdot[/itex], then you have commutative squares

    A^n & \xrightarrow{d} & A^{n-1} \\
    {}^{{}_f}\!\!\!\downarrow & & {}^{{}_f}\!\!\!\downarrow \\
    B^n & \xrightarrow{d} & B^{n-1} \\

    i.e. for any element [itex]a \in A^n[/itex], f(da) = df(a)

    In particular, if A is a subcomplex of B, then this applies to the inclusion [itex]A \to B[/itex] and also to the projection [itex]B \to B/A[/itex]. (i.e. if denotes the equivalence class of b, then [itex]d = [db][/itex])
    Last edited: Feb 26, 2008
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