bsc-thesis/thesis/4_Constructions.tex

\section{Constructions}
\label{sec:Constructions}

Comparing chain complexes and simplicial abelian groups, we see a similar structure. Both objects consists of a sequence of abelian groups, with maps in between. At first sight simplicial abelian groups have more structure, because there are maps in both directions. It is not clear how to make degeneracy maps given a chain complex, in fact it is already unclear how to define more maps (the face maps) out of one (the boundary map). Constructing a chain complex from a simplicial abelian group on the other hand seems doable.

\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 $\del_n : A_n \to A_{n-1}$:
$$\del_n = d_0 - d_1 + \ldots + (-1)^n d_n.$$
\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 proof is $\del_n \circ \del_{n+1} = 0$. 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}

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: Make this more readable...}

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.

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 $C(K(C))_n$ 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:
\begin{align*}
	N(A)_n &= \bigcap_{i=1}^{n} \ker(d_i : A_n \to A_{n-1}), \\
	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}

We will call this chain complex $N(A)$ 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 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 $N(f): N(A) \to N(B)$.

	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 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:
	$$ N(\Z[\Delta[0]]) = \cdots \to 0 \to 0 \to \Z. $$
	So in this example we see that the normalized chain complex is really better behaved than the unnormalized chain complex, given by $C$.
\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 $D(X)_n \subset X_n$ of degenerate simplices, defined as:
$$ D(X)_n = \sum_{i=0}^n s_i(X_{n-1}). $$

\begin{lemma}
	\label{le:decomp1}
	For all $x \in X_n$ we have:
	$$ x = b + c,$$
	where $b \in N(X)_n$ and $c \in D(X)_n$.
\end{lemma}
\begin{proof}
	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}
	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 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.

\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 $x_\beta \in N(X)_p$ and $\beta : [n] \epi [p]$.
\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 $c \in D(X)_n$, 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}
	Let $\beta : [n] \epi [m]$ and $\gamma : [n] \epi [m']$ be two maps such that $\beta \neq \gamma$. Then we have $\beta^\ast(N(X))_p \cap \gamma^\ast(N(X))_q = 0$.
\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 proofs the existence of a decomposition and the latter proofs the uniqueness. So combining this 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}
	$X_n = \bigoplus_{[n] \epi [p]} N(X)_p$.
\end{corollary}

Using corollary~\ref{cor:decomp} we can proof a nice categorical fact about $N$, which we will use later on.
\begin{lemma}
	The functor $N$ is fully faithful, i.e.:
	$$ N : \Hom{\sAb}{A}{B} \iso \Hom{\Ch{\Ab}}{N(A)}{N(B)}. $$
\end{lemma}
\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
	\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 $x_\beta \in N(X)_n$ 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 $\Ab$.

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 $\Ch{\Ab} \to \sAb$. Of course we will be interested in the one using $N$. So we define the functor:
$$ K(C) = \Hom{\Ch{\Ab}}{N\Z^\ast\Delta[-]}{C} \in \sAb. $$
This definitions is very abstract, but luckily we can also give a more explicit definition. By writing it out for low dimensions we see:

$$ 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, $$
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, $$

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 $K(C)_n \iso \bigoplus_{[n] \epi [p]} C_p$.
\end{proposition}
\begin{proof}
	\todo{C: proposition about $K$}
\end{proof}

\todo{C: 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 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 $\Ch{\Ab} \simeq \sAb$.
\end{proof}