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Q: For Dannidin Only ( Answered ,   1 Comment )
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 Subject: For Dannidin Only Category: Science > Math Asked by: dubois-ga List Price: \$2.00 Posted: 08 Dec 2002 14:24 PST Expires: 07 Jan 2003 14:24 PST Question ID: 121502
 ```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.``` 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```
 dubois-ga rated this answer: and gave an additional tip of: \$36.00 `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```