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Thesis: Functor Ch -> sAb

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Joshua Moerman 12 years ago
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      thesis/4_Constructions.tex

14
thesis/4_Constructions.tex

@ -40,4 +40,16 @@ $$ \del = \delta^0|_{N(A)_n}. $$
\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:
$$ \Hom{A}{F \Z^{\ast} \Delta (-)}{C} : \DELTA^{op} \to \Ab. $$
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 $\Ch{\Ab} \to \sAb$.