\section{The Dold-Kan correspondence}
\label{sec:Constructions}

Comparing chain complexes and simplicial abelian groups, one sees a certain similarity. Both concepts are defined as sequences of abelian groups with certain structure maps. At first sight simplicial abelian groups seem to have a richer structure. There are many face maps as opposed to only a single boundary homomorphism. Nevertheless, as we will show in this section, these two concepts give rise to equivalent categories.

\subsection{Unnormalized chain complex}
Given a simplicial abelian group $A$, we have a family of abelian groups $A_n$. For every $n>0$ we define a group homomorphism
$$\del_n = d_0 - d_1 + \ldots + (-1)^n d_n: A_n \to A_{n-1}.$$
\begin{lemma}
	Using $A_n$ as the family of abelian groups and the maps $\del_n$ as boundary maps gives a chain complex.
\end{lemma}
\begin{proof}
	We already have a collection of abelian groups together with maps, so the only thing to prove is $\del_{n-1} \circ \del_n = 0$. This can be done with a calculation.
	\begin{align*}
		\del_{n-1} \circ \del_n &= \sum_{i=0}^{n-1}  \sum_{j=0}^{n} (-1)^{i+j} d_i \circ d_j \\
			&\eqn{1} \sum_{i=0}^{n-1} \sum_{j=0}^{i} (-1)^{i+j} d_i \circ d_j + \sum_{i=0}^{n-1} \sum_{j=i+1}^{n} (-1)^{i+j} d_i \circ d_j \\
			&\eqn{2} \sum_{i=0}^{n-1} \sum_{j=0}^{i} (-1)^{i+j} d_i \circ d_j + \sum_{i=0}^{n-1} \sum_{j=i+1}^{n} (-1)^{i+j} d_{j-1} \circ d_i \\
			&\eqn{3} \sum_{i=0}^{n-1} \sum_{j=0}^{i} (-1)^{i+j} d_i \circ d_j - \sum_{i=0}^{n-1} \sum_{j=i}^{n-1} (-1)^{i+j} d_j \circ d_i \\
			&= \sum_{i=0}^{n-1} \sum_{j=0}^{i} (-1)^{i+j} d_i \circ d_j - \sum_{i=0}^{n-1} \sum_{j=0}^{i} (-1)^{i+j} d_i \circ d_j = 0
	\end{align*}
	In this calculation we did the following. We split the inner sum in two halves \refeqn{1} and we use the simplicial equations on the second sum \refeqn{2}. Then we do a shift of indices \refeqn{3}. By interchanging the roles of $i$ and $j$ in the second sum, we have two equal sums which cancel out. So indeed this is a chain complex.
\end{proof}

Thus, associated to a simplicial abelian group $A$ we obtain a chain complex $M(A)$ with $M(A)_n = A_n$ and the boundary operator as above. Following the book \cite{goerss} we will call the chain complex $M(X)$ the \emph{Moore complex} or \emph{unnormalized chain complex} of $X$. This construction defines a functor
$$ M: \sAb \to \Ch{\Ab} $$
by assigning $M(f)_n = f_n$ for a natural transformation $f: A \to B$. It follows from a nice calculation that $M(f)$ is indeed a chain map:
\begin{align*}
	f_{n-1} \circ \del &= f_{n-1} \circ (d_0 - d_1 + \ldots + (-1)^n d_n) \\
		&= f_{n-1} \circ d_0 - f_{n-1} \circ d_1 + \ldots + (-1)^n f_{n-1} \circ d_n \\
		&\eqn{1} d_0 \circ f_n - d_1 \circ f_n + \ldots + (-1)^n d_n \circ f_n \\
		&= (d_0 - d_1 + \ldots + (-1)^n d_n) \circ f_n = \del \circ f_n,
\end{align*}
where we used naturality of $f$ in step \refeqn{1}. This functor is in fact already used in the construction of the singular chain complex, where we defined the boundary maps (on generators) as $\del(\sigma) = \sigma \circ d_0 - \sigma \circ d_1 + \ldots + (-1)^{n+1} \sigma \circ d_{n+1}$. We will briefly come back to this in Section~\ref{sec:Homotopy}.

Let us investigate whether this functor $M$ can be part of an 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 degeneracy 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.

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
$$ C = \ldots \to 0 \to 0 \to \Z. $$
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 $M(K(C))_n$ is non-trivial for all $n \in \N$. For an equivalence we require a (natural) isomorphism: $M(K(C)) \tot{\iso} C$, this in particular means an isomorphism in each degree $n > 0$: $ 0 \neq M(K(C))_n \tot{\iso} C_n = 0 $, which is not possible. So the functor $M$, 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
\begin{align*}
	N(A)_n &= \bigcap_{i=1}^{n} \ker(d_i: A_n \to A_{n-1}), \quad\text{n > 0} \\
	N(A)_0 &= A_0.
\end{align*}
Now define group homomorphisms $\del: N(A)_n \to N(A)_{n-1}$ as
$$ \del = d_0|_{N(A)_n}. $$
\begin{lemma}
	The function $ \del $ is well-defined. Furthermore $ \del \circ \del = 0 $.
\end{lemma}
\begin{proof}
	Let $x \in N(A)_n$, then $d_i \del(x) = d_i d_0(x) = d_0 d_{i+1}(x) = d_0 (0) = 0$ for all $i < n$. So indeed $\del(x) \in N(A)_{n-1}$, because in particular it holds for $i > 0$. Using this calculation for $i = 0$ shows that $\del \circ \del = 0$. This shows that $N(A)$ is a chain complex.
\end{proof}

The chain complex $N(A)$ is called the \emph{normalized chain complex} of $A$.
\begin{lemma}
	The above construction gives a functor $N: \sAb \to \Ch{\Ab}$. Furthermore $N$ is additive.
\end{lemma}
\begin{proof}
	Given a map $f: A \to B$ of simplicial abelian groups, we consider the restrictions
	$$ f_n |_{N(A)_n}: N(A)_n \to B_n. $$
	Because $f_n$ commutes with the face maps we get
	$$ d_i(f_n(x)) = f_{n-1}(d_i(x)) = 0, $$
	for $i>0$ and $x \in N(A)_n$. So the restriction also restricts the codomain, i.e. $f_n |_{N(A)_n}: N(A)_n \to N(B)_n$ is well-defined. Furthermore it commutes with the boundary operator, since $f$ itself commutes with all face maps. This gives functoriality $N(f): N(A) \to N(B)$.

	Let $f, g: A \to B$ be two maps, then
	$$ N(f+g) = (f+g)|_{N(A)} = f|_{N(A)} + g|_{N(A)} = N(f) + N(g). $$
	By recalling that in both categories addition of maps was defined pointwise, we have additivity of $N$.
\end{proof}

\begin{example}
	We will look at the normalized chain complex of $\Z[\Delta[0]]$. Recall that it looks like
	$$ \Z[\Delta[0]] :=
	\begin{tikzpicture}[baseline=-0.5ex]
	\matrix (m) [matrix of math nodes] { 
		\Z & \Z & \Z & \cdots \\
	}; 

	\foreach \r in {-5, 5} \draw [raise line=\r, <-] (m-1-1) -> (m-1-2);
	\foreach \r in {0} \draw [raise line=\r, ->] (m-1-1) -> (m-1-2);

	\foreach \r in {-10, 0, 10} \draw [raise line=\r, <-] (m-1-2) -> (m-1-3);
	\foreach \r in {-5, 5} \draw [raise line=\r, ->] (m-1-2) -> (m-1-3);

	\foreach \r in {-15, -5, 5, 15} \draw [raise line=\r, <-] (m-1-3) -> (m-1-4);
	\foreach \r in {-10, 0, 10} \draw [raise line=\r, ->] (m-1-3) -> (m-1-4);
	\end{tikzpicture}.$$
	where all face and degeneracy maps are identity maps. Clearly the kernel of $\id$ is the trivial group. So $N(\Z[\Delta[0]])_i = 0$ for all $i > 0$. In degree zero we are left with $N(\Z[\Delta[0]])_0 = \Z$. So we can depict the normalized chain complex by
	$$ N(\Z[\Delta[0]]) = \cdots \to 0 \to 0 \to \Z. $$
	So in this example we see that the normalized chain complex is really better behaved than the unnormalized chain complex given by $M(\Z[\Delta[0]]$.
\end{example}

To see what $N$ exactly does there are some useful lemmas. These lemmas can also be found in \cite[Chapter~VIII~1-2]{lamotke}, but in this thesis more detail is provided. Some corollaries are provided to give some intuition, or so summarize the lemmas, these results can also be found in \cite[Chapter~8.2-4]{weibel}. For the following lemmas let $X \in \sAb$ be an arbitrary simplicial abelian group and $n \in \N$. For these lemmas we will need the subgroups $D_n(X) \subset X_n$ of degenerate simplices, defined as:
$$ D_n(X) = \sum_{i=0}^n s_i(X_{n-1}). $$

\begin{lemma}
	\label{le:decomp1}
	For all $x \in X_n$ we have:
	$$ x = b + c,$$
	where $b \in N(X)_n$ and $c \in D_n(X)$.
\end{lemma}
\begin{proof}
	Define the subgroup $P_n^k = \{ x \in X_n \I d_i x = 0 \text{ for all } i > k\}$. Note that by definition we have
	$$ N(X)_n = P_n^0 \subseteq P_n^1 \subseteq \ldots \subseteq P_n^{n-1} \subseteq P_n^n = X_n. $$
	We will prove with induction that for any $k \leq n$ we can write $x \in X_n$ as $x = b + c$, with $b \in P_n^k$ and $c \in D_n(X)$. For $k = n$ the statement is clear, because we can simply write $x = x$, knowing that $x \in P_n^n = X_n$.

	Assume the statement holds for $k > 0$, we will prove it for $k-1$. So for any $x \in X_n$ we have $x = b + c$, with $b \in P_n^k$ and $c \in D_n(X)$. Now consider $b' = b - s_{k-1} d_k b$. Now clearly for all $i > k$ we have $d_i b' = 0$. For $k$ itself we can calculate
	$$ d_k(b') = d_k(b - s_{k-1} d_k b) = d_k b - d_k s_{k-1} d_k b = d_k b - d_k b = 0, $$
	where we used the equality $d_k s_{k-1} = \id$. So $b' \in P_n^{k-1}$. Furthermore we can define $c' = s_{k-1} d_k b + c$, for which it is clear that $c' \in D_n(X)$. Finally conclude that
	$$ x = b + c = b - s_{k-1} d_k b + s_{k-1} d_k b + c = b' + c',$$
	with $b' \in P_n^{k-1}$ and $c' \in D_n(X)$.

	Doing this inductively gives us $x = b + c$, with $b \in P_n^0 = N(X)_n$ and $c \in D_n(X)$, which is what we had to prove.
\end{proof}
\begin{lemma}
	\label{le:decomp2}
	For all $x \in X_n$, if $s_i x \in N(X)_{n+1}$, then $x = 0$.
\end{lemma}
\begin{proof}
	Using that $s_i x \in N(X)_{n+1}$ means $0 = d_{k+1} s_i x$ for any $k \geq 0$ and by using using the simplicial identity: $d_{i+1} s_i = \id$, we can conclude $x = d_{i+1} s_i x = 0$.
\end{proof}

The first lemma tells us that every $n$-simplex in $X$ can be decomposed as a sum of something in $N(X)$ and a degenerate $n$-simplex. The latter lemma assures that there are no degenerate $n$-simplices in $N(X)$. So this gives us:

\begin{corollary}
	\label{cor:NandD}
	$X_n = N(X)_n \oplus D_n(X)$
\end{corollary}

We can extend the above lemmas to a more general statement.

\begin{lemma}
	\label{le:decomp3}
	For all $x \in X_n$ we can write $x$ as
	$$ x = \sum_\beta \beta^\ast (x_\beta), $$
	for certain $x_\beta \in N(X)_p$, where $\beta$ ranges over all surjective functions $\beta : [n] \epi [p]$.
\end{lemma}
\begin{proof}
	We will proof this using induction on $n$. For $n=0$ the statement is clear because $N(X)_0 = X_0$.

	Assume the statement is proven for $n$. Let $x \in X_{n+1}$, then from Lemma~\ref{le:decomp1} we see $x = b + c$. Note that $c \in D_n(X)$, in other words $c = \sum_{i=0}^{n-1} s_i c_i$, with $c_i \in X_n$. So with the induction hypothesis, we can write these as $c_i = \sum_\beta \beta^\ast c_{i, \beta}$, where the sum quantifies over $\beta: [n] \epi [p]$. Now $b$ is already in $N(X)_{n+1}$, so we can set $x_\id = b$, to obtain the conclusion.
\end{proof}
\begin{lemma}
	\label{le:decomp4}
	Let $\beta: [n] \epi [m]$ and $\gamma : [n] \epi [m']$ be two maps such that $\beta \neq \gamma$. Then we have $\beta^\ast(N(X))_m \cap \gamma^\ast(N(X))_{m'} = 0$.
\end{lemma}
\begin{proof}
	Note that $N(X)_i$ only contains non-degenerate $i$-simplices (and $0$). For $x \in \beta^\ast(N(X))_p \cap \gamma^\ast(N(X))_q$ we have $x = \beta^\ast y = \gamma^\ast y'$, where $y$ and $y'$ are non-degenerate. By Lemma~\ref{le:non-degenerate} we know that every $n$-simplex is \emph{uniquely} determined by a non-degenerate simplex and a surjective map. For $x \neq 0$ this gives a contradiction.
\end{proof}

Again the former lemma of these two lemmas proves the existence of a decomposition and the latter shows the uniqueness. So combining these gives:

\begin{corollary}
	\label{cor:decomp}
	For all $x \in X_n$ we can write $x = \sum_\beta \beta^\ast (x_\beta)$ in a unique way.
\end{corollary}

And by considering $X_n$ as a whole we get:

\begin{corollary}
	\label{cor:nondegN}
	$X_n = \bigoplus_{[n] \epi [p]} N(X)_p$.
\end{corollary}

Using Corollary~\ref{cor:decomp} we can prove a nice categorical fact about $N$, which we will use later on.
\begin{lemma}
	\label{le:fullyfaithful}
	The functor $N$ is fully faithful, i.e.
	$$ N: \Hom{\sAb}{A}{B} \iso \Hom{\Ch{\Ab}}{N(A)}{N(B)} \quad A, B \in \sAb. $$
\end{lemma}
\begin{proof}
	First we prove that $N$ is injective on maps. Let $f: A \to B$ and assume $N(f) = 0$, for $x \in A_n$ we know $x = \sum_\beta \beta^\ast x_\beta$, so
	\begin{align*}
		f(x) &= \textstyle f(\sum_\beta \beta^\ast (x_\beta)) \\
			&= \textstyle \sum_\beta f(\beta^\ast (x_\beta)) \\ 
			&= \textstyle \sum_\beta \beta^\ast (f (x_\beta)) \\ 
			&= \textstyle \sum_\beta \beta^\ast (N(f) (x_\beta)) = 0,	
	\end{align*}
	where we used naturality of $f$ in the second step, and the fact that $x_\beta \in N(A)$ in the last step. We now see that $f(x) = 0$ for all $x$, hence $f = 0$. So indeed $N$ is injective on maps.

	Secondly we have to prove $N$ is surjective on maps. Let $g: N(A) \to N(B)$, define $f: A \to B$ as
	$$ f(x) = \sum_\beta \beta^\ast g(x_\beta), $$
	again we have written $x$ as $x = \sum_\beta \beta^\ast x_\beta$. Clearly $N(f) = g$.
\end{proof}

If we reflect a bit on why the functor $M$ was not a candidate for an equivalence, we see that $N$ does a better job. We see that $N$ leaves out all degenerate simplices, so it is more carefully chosen than $M$, which included everything. In fact, Corollary~\ref{cor:NandD} exactly tells us $M(X)_n = N(X)_n \oplus D_n(X)$.


\subsection{From $\Ch{\Ab}$ to $\sAb$}
In this subsection we will construct a functor from chain complexes to simplicial abelian groups. We will do this in a fairly abstract way. There is, however, also an explicit description of this functor which will be given after proving the main equivalence.

Let $A$ be an additive category and $F: \sAb \to A$ an additive functor. We want to construct a functor $G: A \to \sAb$ which is right adjoint to $F$. For each $a \in A$ we have to specify $G(a): \DELTA^{op} \to \Ab$. Assume we already specified this, such that $G$ is the right adjoint, then by the additive Yoneda lemma we know
\begin{align*}
	G(a)_n &\iso \Hom{\sAb}{\Z[\Delta[n]]}{G(a)} \\
		&\iso \Hom{A}{F\Z[\Delta[n]]}{a}.
\end{align*}
This in fact can be used as the definition of $G$:
$$ G(a)_n = \Hom{A}{F\Z[\Delta[n]]}{a}. $$
To check that indeed $G(a) \in \sAb$ we only have to remind ourselves that we only composed two functors, namely
\begin{gather*}
	\DELTA \tot{\Delta[-]} \sSet \tot{\Z} \sAb \tot{F} A \quad\text{and} \\
	\Hom{A}{-}{a}: A^{op} \to \Ab
\end{gather*}
giving us $\Hom{A}{F\Z[\Delta[-]]}{a}: \DELTA^{op} \to \Ab$. Similarly $G$ itself is a functor, because it is defined using the $\mathbf{Hom}$-functor.

Many functors to $\sAb$ can be shown to have this description.\footnote{And also many functors to $\sSet$ are of this form if we leave out all additivity requirements.} In our case we can define a functor $K$ as
\begin{gather*}
	K: \Ch{\Ab} \to \sAb \\
	K(C) = \Hom{\Ch{\Ab}}{N\Z[\Delta[-]]}{C}.
\end{gather*}

This is a very abstract definition so we will first discuss what a chain map $N\Z[\Delta[n]] \to C$ looks like. Recall that the non-degenerate $m$-simplices of $\Delta[n]$ are exactly injective maps $\eta: [m] \mono [n]$ for all $m$. So $N\Z[\Delta[n]]$ consists of linear combinations of those non-degenerate simplices, as $N$ precisely gives us the non-degenerate elements. Note that $N\Z[\Delta[n]]_m$ are free groups, since $\Z[\Delta[n]]_m$ are free. In other words, when defining a chain map $N\Z[\Delta[n]] \to C$ it is sufficient to define it on the generators, i.e. on the injections $\eta: [m] \mono [n]$. This fact is used throughout the following proofs.

Furthermore the degeneracy maps $s_i: K(C)_{n-1} \to K(C)_n$ are given by precomposition of the induced map $\sigma_{i\ast}: N\Z[\Delta[n]] \to N\Z[\Delta[n-1]]$ which in their turn are given by postcomposition. More precisely this gives $s_i(f)_m(\eta) = f_m(\sigma_i \eta)$ for any $f \in K(C)_{n-1}$ and $\eta: [m] \mono [n]$. We will now have a closer look at the degenerate elements of $K(C)$.
\begin{lemma}
	\label{le:degen_k}
	Let $f: N\Z[\Delta[n]] \to C$ be a chain map then $f \in D_n(K(C))$ if and only if $f_r = 0$ forall $r \geq n$.
\end{lemma}
\begin{proof}
	If $f \in D_n(K(C))$ we have $f = \sum_{i=0}^n s_i(f^{(i)})$ for some maps $f^{(i)}: N\Z[\Delta[n-1]] \to C$. Since $N\Z[\Delta[n-1]]_r = 0$ as there are no injections $[r] \mono [n-1]$, we have $f^{(i)}_r = 0$ for all $r > n-1$.

	For the other direction let $f: N\Z[\Delta[n]] \to C$ be a chain map and $f_r = 0$ forall $r \geq n$. Define $f_m^{(i)}(\eta) = f_m(\delta_i \eta)$ for $\eta: [m] \mono [n]$. This gives a chain map $f^{(i)}: N\Z[\Delta[n-1]] \to C$ by a simple calculation:
	$$ \del(f_m^{(i)}(\eta)) = \del(f_m(\delta_i \eta)) \eqn{1} f_{m-1}(\del(\delta_i \eta)) \eqn{2} f_{m-1}(\delta_i \eta \delta_0) \eqn{2} f_{m-1}^{(i)}(\del(\eta)), $$
	where we used that $f$ is a chain map at \refeqn{1} and the definition of the boundary map of $N(-)$ \emph{and} the definition of face maps in $\Delta[-]$ at \refeqn{2}.

	Now let $\eta: [m] \mono [n]$ and $\eta \neq \id$ (we already know $f(\id_{[n]})=0$ by assumption) then by the epi-mono factorization we have $\eta = \delta_{i_a} \cdots \delta_{i_1}$ with $a>0$, so
	$$ f_m(\eta) = f_m(\delta_{i_a} \cdots \delta_{i_1}) \eqn{1} f_m^{(i_a)}(\delta_{i_{a-1}} \cdots \delta_{i_1}) \eqn{2} f_m^{(i_a)}(\sigma_{i_a} \delta_{i_a} \delta_{i_a} \cdots \delta_{i_1}) \eqn{3} s_{i_a}(f^{(i_a)})_m(\eta), $$
	where we used the definition of $f_m^{(i_a)}$ at \refeqn{1}, one of the simplicial identities at \refeqn{2} and the definition of degeneracy maps at \refeqn{3} as discussed earlier.

	By the fact that injections are generators this gives $f_m = \sum_{i=0}^n s_i(f^{(i)})_m$ for all $m$, i.e. $f = \sum_{i=0}^n s_i(f^{(i)})$. Hence $f \in D_n(K(C))$.
\end{proof}

We now have enough lemmas to prove the main equivalence quite easily. The most important lemma for the isomorphism $X \iso KNX$ will be the lemma stating that $N$ is fully faithful. For the other isomorphism we will use the above lemma to characterize the degenerated simplices in $K(C)$.

\begin{theorem}
	$N$ and $K$ form an equivalence.
\end{theorem}
\begin{proof}
	Let $X$ be a simplicial abelian group. Then we have the following natural isomorphisms of abelian groups:
	\begin{align*}
		X_n &\ison{1} \Hom{}{\Z[\Delta[n]]}{X} \\
			&\ison{2} \Hom{}{N\Z[\Delta[n]]}{NX} \\
			&\eqn{3} KN(X)_n
	\end{align*}
	Where we used the additive Yoneda lemma at \refeqn{1} (Lemma~\ref{le:yoneda_add}), then we use the fully faithfulness of $N$ at \refeqn{2} (Lemma~\ref{le:fullyfaithful}) and at \refeqn{3} we simply use the definition of $K$. Using naturality in $n$ we have established $X \iso KNX$ and by naturality in $X$ we have $\id \iso KN$, proving the first part of the equivalence.

	For the second part we will explicitly define an isomorphism as
	\begin{gather*}
		\phi_n: NK(C)_n \to C_n \\
		f \mapsto f_n(\id_{[n]}).
	\end{gather*}
	Note that this is well defined by the fact that $\id_{[n]}$ is a non-degenerate simplex. \todo{DK: Chain map + Naturality}. We will first show that $\phi_n$ is surjective. Let $x \in C_n$ define a chain map as
	\begin{align*}
		g_r(y) &= 0 \qquad \text{for } r \neq n, n-1\\
		g_n(\id_{[n]}) &= x \\
		g_{n-1}(\delta_i) &= \begin{cases}
			\del(x) \quad\text{if } i = 0 \\
			0 \quad\text{otherwise}
		\end{cases} \\
	\end{align*}
	Clearly $\phi_n(g) = x$ by definition and $g$ is a chain map as we defined it to commute with the boundary map. For proving injectivity consider $g \in \ker(\phi_n)$ then for trivial reasons we have $f_r = 0$ for all $r > n$ and $f_n(\id_{[n]}) = 0$ gives $f_n = 0$. Applying Lemma~\ref{le:degen_k} gives us $f \in D_n(K(C))$, but $f \in N(K(C))_n$. So by using Corollary~\ref{cor:NandD} we get $f = 0$. Thus $\phi_n$ is an isomorphism, which gives us $NK(C) \iso C$.

	We now have established two natural isomorphisms $\id_\sAb \iso KN$ and $NK \iso \id_\Ch{\Ab}$. Hence we have an equivalence $\Ch{\Ab} \simeq \sAb$.
\end{proof}

One might not be content with the abstract description of the functor $K$. In the remainder of this section a more explicit description will be given, and it will be indicated why the two descriptions coincide.

\begin{definition}
	For a chain complex $C$ define the abelian groups
	$$ K'(C)_n = \bigoplus_{\beta} C_p^\beta, $$
	where $\beta$ ranges over all surjections $\beta: [n] \epi [p]$ and $C_p^\beta = C_p$ ($\beta$ only acts as a decoration).
\end{definition}

Before we provide the face and degeneracy maps, one should see a nice symmetry with Corollary~\ref{cor:nondegN}. One can also prove the equivalence with this definition. The first isomorphism will be harder to prove, whereas the second isomorphism is easier, as we get the characterization given by Lemma~\ref{le:degen_k} almost by definition.

For a chain complex $C$ we will turn the groups $K'(C)_n$ into a simplicial abelian group by defining $K'$ on functions. Let $\alpha: [m] \to [n]$ be a function in $\DELTA$, we will define $K'(\alpha): K(C)_n \to K(C)_m$ by defining it on each summand $C_p^\beta$. Fix a summand $C_p^\beta$, by using the epi-mono factorization we know $\beta\alpha = \delta\sigma$ for some injection $\delta$ and some surjection $\sigma$. In the case $\delta = \id$, we make the following identification
$$ C_p^\beta \tot{=} C_p^\sigma \subset K'(C)_m. $$
In the case $\delta = \delta_0$ we use the boundary operator as follows:
$$ C_p^\beta \tot{\del} C_{p-1} \tot{=} C_{p-1}^\sigma \subseteq K'(C)_m. $$
In all the other cases we define the map $C_p^\beta \to K'(C)_m$ to be the zero map. We now have defined a map on each of the summands which gives a map $K'(\alpha): K'(C)_n \to K'(C)_m$.

We will not show that this functor $K'$ is isomorphic to our functor $K$ defined earlier, however we will indicate that it makes sense by writing out explicit calculations for $K(C)_0$ and $K(C)_1$. First we see that
$$ K(C)_0 = \Hom{\Ch{\Ab}}{N\Z^\ast\Delta[0]}{C} = \Bigg\{
\begin{tikzpicture}[baseline=-0.5ex]
	\matrix (m) [matrix of math nodes, row sep=1em, column sep=1em] { 
		\cdots  & 0 & 0 & \Z  \\
		\cdots  & C_2 & C_1 & C_0 \\
	}; 

	\foreach \x in {1, 2}
		\foreach \i/\j in {1/2, 2/3, 3/4} \draw[->] (m-\x-\i) -- (m-\x-\j);
	
	\foreach \i/\j in {2/2, 3/1, 4/0} \draw[->] (m-1-\i) -- node {$f_\j$} (m-2-\i);
\end{tikzpicture}
\Bigg\} \iso C_0 = K'(C)_0, $$
because for $f_1, f_2, \ldots$ there is now choice at all, and for $f_0: \Z \to C_0$ we only have to choose an image for $1 \in \Z$. In the next dimension we see
$$ K(C)_1 = \Hom{\Ch{\Ab}}{N\Z^\ast\Delta[1]}{C} = \Bigg\{
\begin{tikzpicture}[baseline=-0.5ex]
	\matrix (m) [matrix of math nodes, row sep=1em, column sep=1em] { 
		\cdots  & 0 & \Z & \Z^2  \\
		\cdots  & C_2 & C_1 & C_0 \\
	}; 

	\foreach \x in {1, 2}
		\foreach \i/\j in {1/2, 2/3, 3/4} \draw[->] (m-\x-\i) -- (m-\x-\j);
	
	\foreach \i/\j in {2/2, 3/1, 4/0} \draw[->] (m-1-\i) -- node {$f_\j$} (m-2-\i);
\end{tikzpicture}
\Bigg\} \iso C_1 \oplus C_0 = K'(C)_1, $$
because again we can choose $f_1$ anyway we want, which gives us $C_1$. But then we are forced to choose $f_0(x, x) = \del(f_1(x))$ for all $x \in \Z$, so we are left with choosing an element $c \in C_0$ for defining $f(1,-1) = c$. Adding this gives $C_1 \oplus C_0$.

