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Homology of Hawaiian Earrings

  1. Mar 19, 2009 #1
    I would like to verify that H1(Hawaiian Earrings) is uncountable. Let's call the Hawaiian Earrings X and build it as a quotient space of [0,1] by the set {0,1, 1/2, 1/3, 1/4, 1/5, ...}. The map [tex]f:[0,1] \rightarrow X [/tex] with f(x)=[1-x] is a continuous map from [tex]\Delta^1[/tex] into X. Explicitly, it hits every ring exactly once in the order of the radii of the rings. Similarly, every ordering [tex](n_i)[/tex] of the positive integers corresponds to a continuous map if the ith ring that the map traverses is the one of length [tex]1/n_i - 1/(n_1+1)[/tex], with 1 being mapped to 0.

    It is easy to verify that this set of maps is uncountable. Now of course I have to verify that (ideally) none of the maps are homologous, and this is where I am stuck. It is straight forward to show that these maps are not pairwise homotopic, but I have no idea how to show that they are in different homology classes. Any hints or ideas?
    Last edited: Mar 19, 2009
  2. jcsd
  3. Mar 20, 2009 #2
    Edit: I realized that there is some fundamental flaw in the setup above (many if not most of the maps would be homologous), but I have a different approach in post #4.
    Last edited: Mar 20, 2009
  4. Mar 20, 2009 #3

    matt grime

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    Isn't the first homology group the abelianisation of the fundamental group? If so, you've demonstrated that the fundamental group is the free product of Z with itself uncountably many times, hence H_1 is the direct product of Z with itself uncountably many times.
  5. Mar 20, 2009 #4
    Thanks, but I am afraid it's not quite as easy as that because the fundamental group is a lot more complicated than a free group on countably many generators. According to literature, H_1 is [tex]\mathbb{Z}^{|\mathbb{N}|} \times \mathbb{Q}^{\mathbb{|R|}} \times (\text{some term I forgot})[/tex]

    I am just trying to find an uncountable subset of H_1.

    Different approach: Use the same setup as in post #1, but replace the orderings of the natural numbers by the power set of the natural numbers. That's definitely uncountable, and unless I am missing some fundamental point, every map should be in a different homology class because it includes or excludes circles from a previous map.
  6. Mar 20, 2009 #5

    matt grime

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    Well, that (horrible) group has still got an uncountable abelianization: Q^R is uncountable and abelian already.
  7. Mar 20, 2009 #6
    The group above *is* the abelianization of the fundamental group. The actual fundamental group looks a lot worse.

    I already know that the group is uncountable. I was just trying to convince myself that it is *very* uncountable by finding an actual uncountable subset, explicitly. One would think that it shouldn't be too hard to find an uncountable subset of Q^R, but I started with several really dumb approaches. I think I do have a fair number of examples now that should work.
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