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.
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.
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.
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. 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:
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
$$ C =\ldots\to0\to0\to\Z. $$
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 $M(K(C))_n$ 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.
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).
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.
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 $N(f): N(A)\to N(B)$.
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
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:
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
Using that $s_i x \in N(X)_{n+1}$ means $0= d_{k+1} s_i x$ for any $k \geq0$ and by using using the simplicial identity: $d_{i+1} s_i =\id$, we can conclude $x = d_{i+1} s_i x =0$.
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:
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.
Let $\beta: [n]\epi[m]$ and $\gamma : [n]\epi[m']$ be two maps such that $\beta\neq\gamma$. Then we have $\beta^\ast(N(X))_m \cap\gamma^\ast(N(X))_{m'}=0$.
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 $x \neq0$ this gives a contradiction.
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.
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 $M(X)_n = N(X)_n \oplus D_n(X)$.
where $\beta$ ranges over all surjections $\beta: [n]\epi[p]$ and $C_p^\beta= C_p$ ($\beta$ only acts as a decoration).
\end{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:
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 $K(\alpha): K(C)_n \to K(C)_m$.
\todo{DK: functoriality of $K(C)$, functoriality of $K$}
\todo{DK: work out the following in more detail (especially the naturalities)}
\begin{theorem}
$N$ and $K$ form an equivalence.
\end{theorem}
\begin{proof}
Let $X$ be a simplicial abelian group. Consider $X_n$, by the additive Yoneda lemma this is naturally isomorphic to $\Z^\ast[\Delta[n]]\to X$, which is by the fully faithfulness and additivity of $N$ naturally isomorphic to $N\Z^\ast[\Delta[n]]\to NX$. The latter is exactly the definition of $KNX$. So, by naturality in $n$, we have established $X \iso KNX$. Hence, by naturality in $X$, we have $\id\iso KN$.
By the previous proposition we have $K(C)_n \iso\bigoplus_{[n]\epi[p]} C_p$. For the summands $C_p$ with $p < n$, we clearly see that $C_p \subset D_n(K(C))$, so $N$ gets rid of these. Then the only summand left is $C_n$ (with the surjection $\id : [n]\epi[n]$). So we see $NKC_n \iso C_n$, furthermore the boundary map is preserved. Hence $NKC \iso C$. And this was natural in $C$, so we get $NK \iso\id$.
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 explicit description of the functor $K$. There is a more abstract way of constructing a functor $\Ch{\Ab}\to\sAb$ from a functor $\sAb\to\Ch{\Ab}$. We will briefly discuss this and show how the functor $K$ would be derived from $N$ by this abstract construction, with two examples it will be illustrated why the two descriptions are the same.
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 could have defined our functor $K$ as
$$ K'(C)=\Hom{\Ch{\Ab}}{N\Z[\Delta[-]]}{C}. $$
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 $K'(C)_1$.
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:
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 $C_1\oplus C_0$.
