Dannidin, could you do one of two things for me?  Either post a proof
to the lemma from Vick of which this is an immediate corollary, or
formalize how to go from the proposition in Hatcher to which you refer
to a proof of the computation of H_n(S^3\X).  Also, to mathtalk and
dannidin, if one of you did not see my final comment for Computing
Homology Groups question below, please do so as I explain my rationale
for cancelling that question and the low asking price here.

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Clarification of Question by dubois-ga on 08 Dec 2002 14:58 PST

Actually, unless the Vick proof is straightforward and readily
comprehensible, I think showing how to go from Hatcher to a solution
would be preferable.  Thanks.

hi dubois,

The reduced homology groups are defined on page 110 of Hatcher. In
fact, for n>0 H_n-tilda is exactly equal (by definition) to H_n. For
n=0 the difference is that H_0 = C_0 / Im d_1  (d=my notation for
boundary sign)
whereas H_0-tilda = Ker e / Im d_1  (e = epsilon)
As Hatcher explains, this implies that H_0 is isomorphic to the direct
sum of H_0-tilda with Z.

Actually, in our specific case of S^3\X there is no need to think
about all this mumbo-jumbo: H_0 of a space is always the free abelian
group generated by the path components of the space, i.e. a direct sum
of copies of Z, one for each connected component. H_0(S^3\X) = Z since
obviously S^3\X is pathwise-connected. This was mathtalk's original
observation. For all n>0 the reduced homology groups are the same as
the ordinary homology groups.

Hope this helps. I might have some things to say about your other
question (the one about universal coverings of S_1xS_1) later on
today. The relative homology thing still scares me a bit, though...

dannidin

An excellent answer from Dannidin; extremely helpful and highly useful.

Hi, dubois:

I just wanted to add my comment that dannidan's answer agrees with my
geometric intution.  As I had mentioned, the "loops" or boundaries of
a 2-simplex can "entangle" the excised circle X, but larger
dimensional simplexes cannot become entangled in the same way.  In
this sense I expected H_0 = H_1 = Z and all the higher groups trivial.

I also went to look last night for my copy of Vick, but could only
must a copy of Munkres.

I'll be happy to take a look at Hatcher's chapter and see if I have
suggestions about the other part of your original problem, now
reposted separately.

regards, mathtalk-ga