From chain complex to simplicial abelian group

In many places, I have seen the slogan that "simplicial abelian group = chain complexes of abelian groups". These same sources usually tell me how to go in one direction. Namely, given a simplicial abelian group, one gets a chain complex whose $n^{\text{th}}$ term is the abelian group corresponding to the $n$-simplex and whose differential is the alternating sum of the face maps.

However, this process does not appear to me to be reversible. My question then is in what sense is the above slogan true? In other words, is there some procedure that takes a chain complex of abelian groups and produces a simplicial abelian group? If there is, in what sense is it an inverse to the above procedure?

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I'm surprised that none of the many places you mentioned used the term "Dold-Kan correspondence" to give you something to search for. People should make an effort to say the names of things, even when they're only side remarks, just in case someone wants to look them up. – Omar Antolín-Camarena Nov 25 '10 at 4:22

There are three standard functors from simplicial abelian groups to chain complexes. Let $C_{\ast}A$ be the one that you described. Let $D_{\ast}A$ be the sum of the images of all the degeneracy maps in $A$; then $D_{\ast}A$ is a contractible subcomplex of $C_{\ast}A$ and we define $N_{\ast}A=C_{\ast}A/D_{\ast}A$, which is chain homotopy equivalent (but not isomorphic) to $C_{\ast}A$. We also put $M_{\ast}A=\{a\in A: d_i(a)=0 \text{ for all } i>0\}$; this is a subcomplex of $C_{\ast}A$, and in fact $C_{\ast}A=M_{\ast}A\oplus D_{\ast}A$, so $M_{\ast}A\simeq N_{\ast}A$.
Now let $F(n)$ be the free simplicial abelian group on a generator in degree $n$, so the term in degree $k$ is the free abelian group on the set of nondecreasing maps $[n]\to [k]$. For any chain complex $U_{\ast}$ we let $L_nU_{\ast}$ denote the group of chain maps from $N_{\ast}F(n)$ to $U_{\ast}$. This defines a functor $L$ that is inverse to $N_*$.