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let $G$ be an abelian torsion group. let $z\in H_i(K(G,1))$. throughout the coefficient group is $\mathbb Q$.

1) Why there exists a FINITE subcomplex $X$, such that $j:X\hookrightarrow K(G,1),\, j_*(x)=z$ for some $x\in H_i(X)$ ?

2) Why the image of $\pi_1(X)$ in $\pi_1(K(G,1))=G$ is finite?

MY guess:

1) $G=\oplus_p G_p$ where $G_p$ ranges over all $p$-subgroups of $G$ so by Kunneth $H_i(K(G,1))=\oplus_{i_1+\cdots+i_l=i}H_{i_1}(K(G_{p_1},1) \otimes \cdots\otimes H_{i_l}(K(G_{p_l},1 ) $ and I don't know what does it mean for $z$ to be in this decomposition!!

2) if $X$ is a finite CW that implies that its $\pi_1$ is finitely generated and not finite !!! and what finite CW means here anyway: having a finite number of cells at each dimension or having no cells beyond a dimension $n$ so that $X$ equals its $n$-skeleton?

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    I only made that comment because it was *not* clear to me…2011-05-16

1 Answers 1

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Pick the usual simplicial model for $K(G,1)$. If $x\in H_i(K(G,1))$, then $x$ is the class of a cycle which involves finitely many simplices, which in turn involve finitely many group elements. Let $H$ be the subgroup generated by all these. Then $x$ is in the image of $H_i(K(H,1))\to H_i(K(G,1)).$ Moreover, if $X$ is the $n$-skeleton of $K(H,1)$ with $n>i+1$, then $x$ is in the image of the composition $H_i(X)\to H_i(K(H,1))\to H_i(K(G,1)).$

Since $H$ is finitely generated and torsion, it is finite, so $X$ is finite.

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    I stuck at the same question reading the paper which seems is the same as OP's, and found this answer really clear. Thanks.2017-08-17