

Subject:
For Dannidin Only
Category: Science > Math Asked by: duboisga List Price: $2.00 
Posted:
08 Dec 2002 14:24 PST
Expires: 07 Jan 2003 14:24 PST Question ID: 121502 

Subject:
Re: For Dannidin Only
Answered By: dannidinga on 09 Dec 2002 01:00 PST Rated: 
hi dubois, The reduced homology groups are defined on page 110 of Hatcher. In fact, for n>0 H_ntilda 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_0tilda = Ker e / Im d_1 (e = epsilon) As Hatcher explains, this implies that H_0 is isomorphic to the direct sum of H_0tilda with Z. Actually, in our specific case of S^3\X there is no need to think about all this mumbojumbo: 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 pathwiseconnected. 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 
duboisga
rated this answer:
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An excellent answer from Dannidin; extremely helpful and highly useful. 

Subject:
Re: For Dannidin Only
From: mathtalkga on 08 Dec 2002 17:04 PST 
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 2simplex 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, mathtalkga 
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