@ -10,14 +10,23 @@ $$\del_{n-1} = d_0 - d_1 + \ldots + (-1)^n d_n \text{ for every } n > 0.$$
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 proof is $\del_n \circ\del_{n+1}=0$.
We already have a collection of abelian groups together with maps, so the only thing to proof is $\del_n \circ\del_{n+1}=0$. This can be done with a calculation:
We split the inner sum in two halves and we use the simplicial equations on the second sum. Then we do a shift of indices and change the roles of $i$ and $j$ in the second sum, so that the sums have an equal range and cancel out. So indeed this is a chain complex.
\end{proof}
This construction gives a functor $C : \sAb\to\Ch{\Ab}$\todo{C: prove this? Is it a adjunction?}. And in fact we already used it in the construction of the singular chaincomplex, 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 abelain group pointwise give a functor $\Z^\ast : \sSet\to\sAb$, and finally using the functor $C$ gives the singular chain complex.
This construction gives a functor $C : \sAb\to\Ch{\Ab}$\todo{C: prove this? Is it an adjunction?}. And in fact we already used it in the construction of the singular chaincomplex, 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 abelain group pointwise give a functor $\Z^\ast : \sSet\to\sAb$, and finally using the functor $C$ gives the singular chain complex.
\todo{C: is this a nice thing to add?}
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 degenracy 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.
@ -28,21 +37,32 @@ But for chain complexes it \emph{is} possible to have trivial abelian groups $C_
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:
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 $\del\circ\del=0$, hence $N(A)$ is a chain complex.
The function $\del$ is well-defined. Furthermore $\del\circ\del=0$.
\end{lemma}
\begin{proof}
\todo{C: This is easy}
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}
\todo{C: As an example calculate $N(\Z[\Delta[0]])$}
\todo{C: define $D(X)_n$}
\todo{C: work out following lemmas}
We will call this chain complex $N(A)$ the \emph{normalized chain complex} of $A$.
\todo{C: functoriality}
\begin{example}
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 $C$.
\end{example}
To see what $N$ does exactly there are some lemmas. For the following lemmas let $X \in\sAb$ be an arbitrary simplicial abelian group and $n \in\N$.
To see what $N$ does exactly there are some lemmas. 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:
$$ D(X)_n =\sum_{i=0}^n s_i(X_{n-1}). $$
\begin{lemma}
\label{le:decomp1}
@ -51,40 +71,53 @@ To see what $N$ does exactly there are some lemmas. For the following lemmas let
where $b \in N(X)_n$ and $c \in D(X)_n$.
\end{lemma}
\begin{proof}
define $P^k =\{ x \in X_n \I d_i x =0, i > k\}$, then do induction (from $k$ to 0).
gives $x = b+c$ with $b \in P^0$, $c \in D(X)_n$.
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.
\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}
Simply calculate using the simplicial equations: $0= d_{i+1} s_i x = x$.
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 $x= d_{i+1} s_i x =0$.
\end{proof}
The first lemma tells us that every $n$-simplex in $X$ can be written as something in $N(X)$plus a degenerate $n$-simplex. The latter lemma asures that there are no degenerate $n$-simplices in $N(X)$. So this gives us:
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:
\begin{corollary}
\label{cor:NandD}
$X_n = N(X)_n \oplus D(X)_n$
\end{corollary}
We can extend the above lemmas to a more general statement. \todo{C: figure out what $\ast$ exactly is.}
We can extend the above lemmas to a more general statement.
\todo{C: define somewhere what $\beta^\ast$ exactly is.}
\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 $x_\beta\in N(X)_n$ and $\beta : [n]\epi[p]$.
for certain $x_\beta\in N(X)_p$ and $\beta : [n]\epi[p]$.
\end{lemma}
\begin{proof}
induction using the first lemma
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.
\end{proof}
\begin{lemma}
\label{le:decomp4}
For $\beta\neq\gamma$ we have $\beta^\ast(N(X))_p \cap\gamma^\ast(N(X))_q =0$.
\end{lemma}
\begin{proof}
?
Follows from $x =\beta^\ast y$ uniquely for a non-degenerate $y$. \todo{C: proof this in chapter about sAb}
\end{proof}
Again the former lemma of these two lemmas proofs the existence of a decomposition and the latter proofs the uniqueness. So combining this gives:
@ -108,18 +141,20 @@ Using corollary~\ref{cor:decomp} we can proof a nice categorical fact about $N$,
\begin{proof}
First we proof $N$ 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
where we used naturality of $f$ in the last step. We now see that $f(x)=0$ for all $x$, hence $f =0$. So indeed $N$ is injective on maps.
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.
Secondly we have to proof $N$ 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 $N(f)= g$. \todo{C: is this clear?}
\end{proof}
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 $C(X)_n = N(X)_n \oplus D(X)_n$.
\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 $\Ab$.
Joshua Moerman
12 years ago