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.
Given a simplicial abelian group $A$, we have a family of abelian groups $A_n$. For every $n>0$ we define a group homomorphism $\del_n : A_n \to A_{n-1}$:
We already have a collection of abelian groups together with maps, so the only thing to prove is $\del_n \circ\del_{n+1}=0$. 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.
This construction gives a functor $C : \sAb\to\Ch{\Ab}$\todo{DK: prove this? Is it an adjunction?}. And in fact we already used it in the construction of the singular chain complex, where we defined the boundary maps as $\del(\sigma)=\sigma\circ d_0-\sigma\circ d_1+\ldots+(-1)^{n+1}\sigma\circ d_{n+1}$ (on generators). The terms $\sigma\circ d_i$ 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 abelian group pointwise give a functor $\Z^\ast : \sSet\to\sAb$, and finally using the functor $C$ gives the singular chain complex.
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 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 $C(K(C))_n$ 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:
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.
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 restircts 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 $N(f): N(A)\to N(B)$.
We will look at the normalized chain complex of $\Z[\Delta[0]]$. Recall that it looked like:
$$\Z[\Delta[0]]=\Z\to\Z\to\Z\to\cdots, $$
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:
$$ N(\Z[\Delta[0]])=\cdots\to0\to0\to\Z. $$
So in this example we see that the normalized chain complex is really better behaved than the unnormalized chain complex, given by $C$.
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 $D(X)_n \subset X_n$ of degenerate simplices, defined as:
Define the subgroup $P^k =\{ x \in X_n \I d_i x =0\text{ for all } i > k\}$. Note that $P^0= N(X)_n$ and $P^n = X_n$. We will prove with induction that for any $k \leq n$ we can write $x \in X_n$ as $x = b + c$, with $b \in P^k$ and $c \in D(X)_n$.
For $k = n$ the statement is clear, because we can simply write $x = x$, knowing that $x \in P^n = X_n$.
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 $b \in P^k$ and $c \in D(X)_n$. 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 $b' \in P^{k-1}$. Furthermore we can define $c' = s_{k-1} d_k b + c$, for which it is clear that $c' \in D(X)_n$. Finally conclude that
$$ x = b + c = b - s_{k-1} d_k b + s_{k-1} d_k b + c = b' + c',$$
with $b' \in P^{k-1}$ and $c' \in D(X)_n$.
Doing this inductively gives us $x = b + c$, with $b \in P^0= N(X)_n$ and $c \in D(X)_n$, which is what we had to prove.
Using that $s_i x \in N(X)_{n+1}$ means $0= d_{k+1} s_i x$ for any $k > 0$ and by using using the simplicial equations: $d_{i+1} s_i =\id$, 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 asures that there are no degenerate $n$-simplices in $N(X)$. So this gives us:
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 $c \in D(X)_n$, 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 $\beta^\ast(N(X))_p \cap\gamma^\ast(N(X))_q =0$.
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 $x_\beta\in N(X)_n$ 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 previous functor $C$ 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 carefull than $C$, which included everything. In fact, corollary~\ref{cor:NandD} exactly tells us $C(X)_n = N(X)_n \oplus D(X)_n$.
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 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 $\Ch{\Ab}\to\sAb$. Of course we will be interested in the one using $N$. So we define the functor:
\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, $$
because for $f_1, f_2, \ldots$ there is now choice at all, and for $f_0 : \Z\to C_0$ we only have to choose an image for $1\in\Z$. In the next dimension we see:
\matrix (m) [matrix of math nodes, row sep=1em, column sep=1em] {
\cdots& 0 &\Z&\Z^2 \\
\cdots& C_2 & C_1 & C_0 \\
};
\foreach\x in {1, 2}
\foreach\i/\j in {1/2, 2/3, 3/4}\draw[->] (m-\x-\i) -- (m-\x-\j);
\foreach\i/\j in {2/2, 3/1, 4/0}\draw[->] (m-1-\i) -- node {$f_\j$} (m-2-\i);
\end{tikzpicture}
\Bigg\}\iso C_1 \oplus C_0, $$
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 $C_1\oplus C_0$. This pattern can be continued and gives the following result:
\begin{proposition}
For any chain complex $C$ we have $K(C)_n \iso\bigoplus_{[n]\epi[p]} C_p$.
Let $X$ be a simplicial abelian group. Consider $X_n$, by the abelian 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 $\Ch{\Ab}\simeq\sAb$.
