From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Re: Cell complex Date: 2 Jul 1999 03:54:35 GMT Newsgroups: sci.math Keywords: Every set of abelian groups can be the homology of a complex In article <7l83g6\$ad2\$1@metro.ucc.usyd.edu.au>, Kundan Misra <"kundan"@maths.usyd.edu.au, kundanmisra@hotmail.com> wrote: >Let H_i be a finitely generated abelian group, for 1 <= i <= n. > >Show that there is a connected finite cell complex X (with all cells of >dimension at most n+1) such that H_q(X) isomorphic to H_q, for 1 <= q <=n. Odd -- no response has yet shown up here. First note that it suffices to assume H_i = {0} for all 0 S^n of degree k. Here you'll need to recognize the boundary of E as being homeomorphic to S^n as well; with this identification made you can explicitly give such a map f with formulas if that appeals to you -- take the points in S^n as being those in R^(n+1) = C x R^(n-1) whose distance from the origin is 1, and let f be the restriction of F : C x R^(n-1) -> C x R^(n-1) to the unit sphere, where F(z,v) = (wz,v) and w is a primitive k-th root of unity. [As I said: "if that appeals to you" ...]. If the term "suspension" means anything to you, you may simply perform this construction with n=1, and then suspend n-1 times. The same approach works with homotopy groups instead of homology groups, except we need to use direct products instead of wedge products. This reduces to the need to construct spaces K(C,n) with only one nonzero homotopy group pi_n(K) = C. These spaces exist, but are almost never finite-dimensional. They're called Eilenberg-MacLane spaces. K(Z,1) is the circle, for example, but K(Z,2) is not the two-sphere; as noted recently in this group pi_3(S^2) isn't {0}. Be thankful you're working in the simpler homology setting! dave