# What's stopping me from choosing the nth Eilenberg Mac Lane space to be the following simplicial abelian group?

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

$F_n(X)_j=0$ for all $j<n$ 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. –  Mariano Suárez-Alvarez Dec 27 '10 at 14:21
@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)? –  KCB Dec 27 '10 at 14:31
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. –  KCB Dec 27 '10 at 14:40

Let's examine $n=0$. Then you're going to need to define a degeneracy map $s_0: X \to X$ that satisfies the simplicial identities $d_0 s_0 = d_1 s_0 = id$. However, substituting in your value for $d_1$, this says $id = 0$.
More generally, if you have a simplicial object which is zero in degrees less than $n$ and $X$ in degree $n$, then the degeneracies give rise in degree $m$ to - at least - one summand isomorphic to $X$ per surjection of ordered sets $\{0\ldots m\} \twoheadrightarrow \{0\ldots n\}$.
However, you can define a simplicial object which, in degree $m$, is $$\bigoplus_{\{0\ldots m\} \twoheadrightarrow \{0\ldots n\}} X$$ with appropriate boundary maps, and this does give you an Eilenberg-Mac Lane space for $X$. This is some kind of "direct sum of copies of $X$ indexed by the simplices of $S^n$".