From: kathleen-s@home.com (Kathleen) Subject: I don't think I've calculated these correctly (more homology groups) Date: 3 Feb 2001 00:40:44 -0500 Newsgroups: sci.math Summary: Sample calculations of homology groups If I let K be a complex whose underlying space is the Mobius band & I let K_o be its "edge", ab bc cd de ef fa d____e____f____a | /|\ | /| | / | \ | / | | / | \ | / | |/ | \|/ | ---------------- a b c d The homology groups H_i(K) & relative homology groups H_i(K,K_o) (over Z) are: H_j(K) = Z if j=0 Z if j=1 0 if j>1 H_j(K,K_o) = 0 if j=0 Z\2Z if j=1 0 if j>1 are these correct? :( I don't think I completely understand what I'm doing yet so I'm very skeptical of my answers ... ============================================================================== From: Robin Chapman Subject: Re: I don't think I've calculated these correctly (more homology groups) Date: Sat, 03 Feb 2001 10:36:00 GMT Newsgroups: sci.math In article , kathleen-s@home.com (Kathleen) wrote: [quote of previous article deleted --djr] The first is certainly correct. The Mobius strip is homotopy equivalent to the circle so its homology groups are the same. For the second, consider the long exact sequence of homology ... -> H_2(K_0) -> H_2(K) -> H_2(K, K_0) -> -> H_1(K_0) -> H_1(K) -> H_1(K, K_0) -> -> H_0(K_0) -> H_0(K) -> H_0(K, K_0) -> 0. Since K_0 is the circle this becomes ... -> 0 -> 0 -> H_2(K, K_0) -> -> Z -> Z -> H_1(K, K_0) -> -> Z -> Z -> H_0(K, K_0) -> 0. We need to identify the homomorphisms Z -> Z coming from the homomorphisms H_j(K_0) -> H_j(K), j = 0, 1. The H_0 homomorphism is an isomorphism (it always is, no matter what the space and nonempty subspace are), so that H_0(K,K_0) = 0. But in H_1 the generator of H_1(K_0) goes to twice a generator of H_1(K). Let L be the central circle of the Mobius strip. The map K -> L projecting to the central circle is a homotopy inverse of the inclusion L to K so induces an isomorphism of homology. The composition of the inclusion K_0 -> K with this map K -> L "winds" K_0 twice about L and so takes the generator of H_1(K_0) to twice that of H_1(L). So the map H_1(K_0) to H_1(K) hase zero kernel and cokernal of order 2. To summarize: your calculation of H_j(K,K_0) is also correct! -- Robin Chapman http://www.maths.ex.ac.uk/~rjc/rjc.html "His mind has been corrupted by colours, sounds and shapes." The League of Gentlemen Sent via Deja.com http://www.deja.com/