\todo{C: As an example calculate $N(\Z[\Delta[0]])$}
\todo{C: As an example calculate $N(\Z[\Delta[0]])$}
\todo{C: The exciting part: $\Ch{\Ab}\to\sAb$}
\subsection{From $\Ch{\Ab}$ to $\sAb$}
For the other way around we actually get a functor for free, via abstract nonsense. Let $F : \sAb\to A$ be any functor, where $A$ is an abelian category. We are after a functor $G : A \to\sAb$, this means that if we are given $C \in A$, we are looking for a functor $G(C) : \DELTA^{op}\to\Ab$. Fixing $C$ in the second argument of the $\mathbf{Hom}$-functor gives: $\Hom{A}{-}{C} : A^{op}\to\Ab$, because $A$ is an abelian category. We see that the codomain of this functor already looks good, now if we have some functor from $\DELTA^{op}$ to $A^{op}$, we can precompose, to obtain a functor from $\DELTA^{op}$ to $\Ab$.
Now recall that we have a family of protoype simplicial sets $\Delta[n]$, which are given by the functor $\Delta : \DELTA\to\sSet$. We can apply the free abelian group pointwise, which gives a functor $\Z^{\ast} : \sSet\to\sAb$. And finally we have our functor $F : \sAb\to A$. Composing these gives:
$$ F \Z^{\ast}\Delta : \DELTA\to A. $$
We can formally regard this functor as a functor from $\DELTA^{op}$ to $A^{op}$. Now combining this with the $\mathbf{Hom}$-functor gives:
Now this is a functor, because it is a composition of functors. Furthermore it is also functorial in the second argument, giving a functor
$$\Hom{A}{F \Z^{\ast}\Delta(-)}{-} : A \to\sAb$$
where we are supposed to fill in the second argument first, leaving us with a simplicial abelian group.
Now we know that $\Ch{\Ab}$ is an abelian group and we have actually two functors $C, N : \sAb\to\Ch{\Ab}$, so we now have functors from $\Ch{\Ab}\to\sAb$.