Google Answers Logo
View Question
Q: For Dannidin Only ( Answered 5 out of 5 stars,   1 Comment )
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.
Subject: Re: For Dannidin Only
Answered By: dannidin-ga on 09 Dec 2002 01:00 PST
Rated:5 out of 5 stars
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...

dubois-ga rated this answer:5 out of 5 stars and gave an additional tip of: $36.00
An excellent answer from Dannidin; extremely helpful and highly useful.

Subject: Re: For Dannidin Only
From: mathtalk-ga 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 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

Important Disclaimer: Answers and comments provided on Google Answers are general information, and are not intended to substitute for informed professional medical, psychiatric, psychological, tax, legal, investment, accounting, or other professional advice. Google does not endorse, and expressly disclaims liability for any product, manufacturer, distributor, service or service provider mentioned or any opinion expressed in answers or comments. Please read carefully the Google Answers Terms of Service.

If you feel that you have found inappropriate content, please let us know by emailing us at with the question ID listed above. Thank you.
Search Google Answers for
Google Answers  

Google Home - Answers FAQ - Terms of Service - Privacy Policy