This construction gives a functor $C : \sAb\to\Ch{\Ab}$. And in fact we already used it in the construction of the singular chaincomplex, where we defined the boundary maps as $\del(\sigma)=\sigma\circ\delta^0-\sigma\circ\delta^1+\ldots+(-1)^{n+1}\sigma\circ\delta^{n+1}$ (on generators). The terms $\sigma\circ\delta^i$ are the maps given by the $\mathbf{Hom}$-functor from $\Top$ to $\Set$, in fact this $\mathbf{Hom}$-functor can be used to get a functor $Sing : \Top\to\sSet$, applying the free abelain group pointwise give a functor $\Z^\ast : \sSet\to\sAb$, and finally using the functor $C$ gives the singular chain complex.
This construction gives a functor $C : \sAb\to\Ch{\Ab}$\todo{C: prove this? Is it a adjunction?}. And in fact we already used it in the construction of the singular chaincomplex, where we defined the boundary maps as $\del(\sigma)=\sigma\circ\delta^0-\sigma\circ\delta^1+\ldots+(-1)^{n+1}\sigma\circ\delta^{n+1}$ (on generators). The terms $\sigma\circ\delta^i$ are the maps given by the $\mathbf{Hom}$-functor from $\Top$ to $\Set$, in fact this $\mathbf{Hom}$-functor can be used to get a functor $Sing : \Top\to\sSet$, applying the free abelain group pointwise give a functor $\Z^\ast : \sSet\to\sAb$, and finally using the functor $C$ gives the singular chain complex.
\todo{C: is this a nice thing to add?}
\todo{C: Note that we cannot do $\Ch{\Ab}\to\sAb$ this simple, as we need monomorphisms}
Let us investigate whether this functor can be used for our sought equivalence. For a functor from $\Ch{\Ab}$ to $\sAb$ we cannot simply take the same collection of abelian groups. This is due to the fact that the degenracy maps should be injective. This means that for a simplicial abelian group $A$, if we know $A_n$ is non-trivial, then all $A_m$ for $m > n$ are also non-trivial.
\todo{C: Note that hence $C$ will not work as an equivalence}
But for chain complexes it \emph{is} possible to have trivial abelian groups $C_m$, while there is a $n < m$ with $C_n$ non-trivial. Take for example the chain complex $ C =\ldots\to0\to0\to\Z$. Now if we would construct a (non-trivial) simplicial abelian group $K(C)$ from this chain complex, we now know that $K(C)_n$ is non-trivial for all $n \in\N$. This means that $C(K(C))_n$ is non-trivial for all $n \in\N$. For an equivalence we require a (natural) isomorphism: $C(K(C))\tot{\iso} C$, this in particular means an isomorphism in each degree $n > 0$: $0\neq C(K(C))_n \tot{\iso} C_n =0$, which is not possible. So the functor $C$, as defined as above, will not give us the equivalence we wanted, although it is a very nice functor.
\subsection{Normalized chain complex}
To repair this defect we should be more careful. Given a simplicial abelian group, simply taking the same collection for our chain complex will not work (as shown above). Instead we are after some ``smaller'' abelian groups, and in some cases the abelian groups should completely vanish (as in the example above).
Given a simplicial abelian group $A$, we define abelian groups $N(A)_n$ as: