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Given an abelian group $X$, let $F_n(X)$ denote the simplicial abelian group defined as follows:

$F_n(X)_j=0$ for all $j and $F_n(X)_j=X$ for all $j\geq n$

with the appropriate zero and identity maps between them so the normalization (normalized moore complex) gives the appropriate homology (say, $d_i=0$ for every $i\geq 1$ (obviously for the parts higher than $n$) and $d_0=id_X$). Then by the Hurewicz theorem, this simplicial abelian group seems like it should have homotopy concentrated in degree $n$ with $n$th component isomorphic to $X$ by computing the homology of the normalization.

Since this thing is a Kan complex (because it's a simplicial abelian group), isn't this a representative for the Eilenberg-Mac Lane space $\kappa(X,n)$?

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    That the homology is concentrated in one degree does not mean that the homotopy is concentrated in one degree.2010-12-27
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    @Mariano Suárez-Alvarez: Doesn't the Hurewicz theorem say that $(\forall j\geq 0)\pi_j(S)\cong H_j(N(S))$ for any simplicial abelian group $S$ (where $N$ denotes the normalization)?2010-12-27
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    It may not be the Hurewicz theorem. Anyway it's proven in Goerss-Jardine _Simplicial Homotopy Theory_ (Remarks following Lemma 2.6). It follows by the Eckmann-Hilton argument.2010-12-27

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