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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 $j<n$ 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

The object you describe can't actually be made into a simplicial object because the boundary maps are incompatible with being able to define degeneracies.

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 $S^n$".

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Let me put Tyler Lawson's answer into a general context. You can define the chain complex $K(A,n)$ by letting $$K(A,n)_i = \begin{cases} A & i = n \\ 0 & i \neq n \end{cases}$$ and with the zero differential. This is of course a well-defined chain complex: $0^2 = 0$ (you don't run into issues as for simplicial abelian groups). I suspect that intuitively this is what you wanted to do.

The Dold–Kan correspondence says that the category of chain complexes is equivalent to the category of simplicial abelian group: $$N : \mathsf{Ab}^{\Delta^{op}} \leftrightarrows \mathsf{Ch}_{\ge 0} : \Gamma.$$ Here $N$ is the functor of normalized chains and $\Gamma$ can be described explicitly. It turns out that if you apply $\Gamma$ to the chain complex defined above, you exactly get the simplicial abelian group described in Tyler Lawson's answer. A possible reference is the book Simplicial Homotopy Theory of Goerss and Jardine, Chapter III, Section 2. Explicitly, for a chain complex $C_*$, the simplicial abelian group $\Gamma(C)$ is given by: $$\Gamma(C)_n = \bigoplus_{[n] \twoheadrightarrow [k]} C_k$$ where the direct sum ranges over surjection from $[n] = \{ 0, \dots, n \}$ to $[k]$. The simplicial structure maps are defined through some kind of yoga with epi-mono factorizations. So for $C_* = K(A,n)_*$ you exactly get the simplicial abelian group of Tyler Lawson's answer.

It moreover follows from the construction (this is also explained in the book) of the equivalence $N \dashv \Gamma$ that the homology groups $H_*(C)$ of a chain complex $C_*$ are isomorphic to the homotopy groups $\pi_*(\Gamma(C_*))$; but of course the homology of the chain complex $K(A,n)$ is simply $A$ concentrated in degree $n$, thus $\Gamma(K(A,n))$ is an Eilenberg–MacLane simplicial abelian group.

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