@ -8,7 +8,7 @@ Before defining \emph{simplicial abelian groups}, we will first discuss the more
We define a category $\DELTA$, where the objects are the finite ordinals $[n]=\{0 < \dots < n\}$ for $n \in\N$ and maps are monotone functions: $\Hom{\DELTA}{[n]}{[m]}=\{ f : [n]\to[m]\I f(i)\leq f(j)\text{ for all } i < j \}$.
We define a category $\DELTA$, where the objects are the finite ordinals $[n]=\{0 < \dots < n\}$ for $n \in\N$ and maps are monotone functions: $\Hom{\DELTA}{[n]}{[m]}=\{ f : [n]\to[m]\I f(i)\leq f(j)\text{ for all } i < j \}$.
\end{definition}
\end{definition}
There are two special kinds of maps in $\DELTA$, the so called \emph{face} maps and \emph{degeneracy} maps. The \emph{$i$-th face maps}$\delta_i: [n-1]\to[n]$ is the unique injective monotone function which \emph{omits}$i$. More precisely, it is defined for all $n \in\Np$ as (note that we do not explicitly denote $n$ in this notation)
The category $\DELTA$ is sometimes referred to as the \emph{category of finite ordinals} or the \emph{cosimplicial index category}. There are two special kinds of maps in $\DELTA$, the so called \emph{face} maps and \emph{degeneracy} maps. The \emph{$i$-th face maps}$\delta_i: [n-1]\to[n]$ is the unique injective monotone function which \emph{omits}$i$. More precisely, it is defined for all $n \in\Np$ as (note that we do not explicitly denote $n$ in this notation)
$$\delta_i: [n-1]\to[n], k \mapsto\begin{cases} k &\text{if } k < i,\\ k+1&\text{if } k \geq i, \end{cases}\hspace{1.0cm}0\leq i \leq n. $$
$$\delta_i: [n-1]\to[n], k \mapsto\begin{cases} k &\text{if } k < i,\\ k+1&\text{if } k \geq i, \end{cases}\hspace{1.0cm}0\leq i \leq n. $$
The \emph{$i$-th degeneracy map}$\sigma_i: [n+1]\to[n]$ is the unique surjective monotone function which \emph{hits $i$ twice}. More precisely it is defined for all $n \in\N$ as
The \emph{$i$-th degeneracy map}$\sigma_i: [n+1]\to[n]$ is the unique surjective monotone function which \emph{hits $i$ twice}. More precisely it is defined for all $n \in\N$ as
@ -123,11 +123,11 @@ Note that because of the third equation, the degeneracy maps $s_i$ are injective
\begin{proof}
\begin{proof}
We will proof the existence by induction over $n$. For $n=0$ the statement is trivial, since all elements in $X_0$ are non-degenerate. Assume the statement is proven for $n$. Let $x \in X_{n+1}$. Clearly if $x$ itself is non-degenerate, we can write $x =\id^\ast x$. Otherwise it is of the form $x = s_i x'$ for some $x' \in X_n$ and $i$. The induction hypothesis tells us that we can write $x' =\beta^\ast y$ for some surjection $\beta: [n]\epi[m]$ and $y \in X_m$ non-degenerate. So $x = s_i \beta^\ast y =(\beta\sigma_i)^\ast y$.
We will proof the existence by induction over $n$. For $n=0$ the statement is trivial, since all elements in $X_0$ are non-degenerate. Assume the statement is proven for $n$. Let $x \in X_{n+1}$. Clearly if $x$ itself is non-degenerate, we can write $x =\id^\ast x$. Otherwise it is of the form $x = s_i x'$ for some $x' \in X_n$ and $i$. The induction hypothesis tells us that we can write $x' =\beta^\ast y$ for some surjection $\beta: [n]\epi[m]$ and $y \in X_m$ non-degenerate. So $x = s_i \beta^\ast y =(\beta\sigma_i)^\ast y$.
For uniqueness, assume $x =\beta^\ast y =\gamma^\ast z$ with $\beta: [n]\epi[m]$, $\gamma: [n]\epi[m']$ and $y \in X_m, z \in X_{m'}$ non-degenerate. Because $\beta$ is surjective there is an $\alpha:[m]\to[n]$ such that $\beta\alpha=\id$ and hence $y =\alpha^\ast\gamma^\ast z =(\gamma\alpha)^\ast z$. By the epi-mon factorization (Lemma~\ref{le:epimono}) we can write $\gamma\alpha=\delta_{i_a}\cdots\delta_{i_1}\sigma_{j_b}\cdots\sigma_{j_1}$, using that $y$ is non-degenerate we know that $\gamma\alpha$ is injective. So we have $\gamma\alpha: [m]\mono[m']$. Because of symmetry (of $y$ and $z$) we also have some map $[m']\mono[m]$, so $m = m'$. So $\gamma\alpha$ is also surjective, hence the identity function, thus $y = z$.
For uniqueness, assume $x =\beta^\ast y =\gamma^\ast z$ with $\beta: [n]\epi[m]$, $\gamma: [n]\epi[m']$ and $y \in X_m, z \in X_{m'}$ non-degenerate. Because $\beta$ is surjective there is an $\alpha:[m]\to[n]$ such that $\beta\alpha=\id$ and hence $y =\alpha^\ast\gamma^\ast z =(\gamma\alpha)^\ast z$. By the epi-mon factorization (Lemma~\ref{le:epimono}) we can write $\gamma\alpha=\delta_{i_a}\cdots\delta_{i_1}\sigma_{j_b}\cdots\sigma_{j_1}$, using that $y$ is non-degenerate we know that $\gamma\alpha$ is injective. So we have $\gamma\alpha: [m]\mono[m']$. Because of symmetry (of $y$ and $z$) we also have some map $[m']\mono[m]$, so $m = m'$. So $\gamma\alpha$ is also surjective, hence the identity function, thus $y = z$, meaning that the non-degenerate $m$-simplex is unique.
Now assume $x =\beta^\ast y =\gamma^\ast y$ with $\gamma, \beta: [n]\epi[m]$ such that $\beta\neq\gamma$, and $y \in X_m$ non-degenerate. Then we can find an $\alpha:[m]\to[n]$ such that $\beta\alpha=\id$ and $\gamma\alpha\neq\id$. With the epi-mono factorization write $\gamma\alpha=\delta_{i_a}\cdots\delta_{i_1}\sigma_{j_b}\cdots\sigma_{j_1}$, then by functoriality of $X$
Now assume $x =\beta^\ast y =\gamma^\ast y$ with $\gamma, \beta: [n]\epi[m]$ such that $\beta\neq\gamma$, and $y \in X_m$ non-degenerate. Then we can find an $\alpha:[m]\to[n]$ such that $\beta\alpha=\id$ and $\gamma\alpha\neq\id$. With the epi-mono factorization write $\gamma\alpha=\delta_{i_a}\cdots\delta_{i_1}\sigma_{j_b}\cdots\sigma_{j_1}$, then by functoriality of $X$
$$ y =\alpha^\ast\beta^\ast y =\alpha^\ast\gamma^\ast y = s_{j_1}\cdots s_{j_b} d_{i_1}\cdots d_{i_a} y. $$
$$ y =\alpha^\ast\beta^\ast y =\alpha^\ast\gamma^\ast y = s_{j_1}\cdots s_{j_b} d_{i_1}\cdots d_{i_a} y. $$
Note that $y$ was non-degenerate, so $s_{j_1}\cdots s_{j_b}=\id$, hence $d_{i_1}\cdots d_{i_a}=\id$. So $\gamma\alpha=\id$, which gives a contradiction. So $\beta$ is unique.
Note that $y$ was non-degenerate, so $s_{j_1}\cdots s_{j_b}=\id$, hence $d_{i_1}\cdots d_{i_a}=\id$. So $\gamma\alpha=\id$, which gives a contradiction. So $\beta=\gamma$, meaning that the surjection $\beta$ is also unique.
@ -186,22 +186,53 @@ Using Corollary~\ref{cor:decomp} we can prove a nice categorical fact about $N$,
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 $M(X)_n = N(X)_n \oplus D_n(X)$.
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 $\sAb$}
\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 $\Ab$.
\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}
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 $K$}
Now recall that we have a family of 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:
\todo{DK: work out the following in more detail (especially the naturalities)}
$$ F \Z^{\ast}\Delta: \DELTA\to A. $$
\begin{theorem}
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 this is a functor, because it is a composition of functors. Furthermore it is also functorial in the second argument, giving a functor
\begin{proof}
$$\Hom{A}{F \Z^{\ast}\Delta(-)}{-}: A \to\sAb$$
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 $\id\iso KN$.
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 $M, N : \sAb\to\Ch{\Ab}$, 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:
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 $K'(C)_1$.
\foreach\i/\j in {2/2, 3/1, 4/0}\draw[->] (m-1-\i) -- node {$f_\j$} (m-2-\i);
\foreach\i/\j in {2/2, 3/1, 4/0}\draw[->] (m-1-\i) -- node {$f_\j$} (m-2-\i);
\end{tikzpicture}
\end{tikzpicture}
\Bigg\}\iso C_0, $$
\Bigg\}\iso C_0 = K(C)_0, $$
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:
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:
\foreach\i/\j in {2/2, 3/1, 4/0}\draw[->] (m-1-\i) -- node {$f_\j$} (m-2-\i);
\foreach\i/\j in {2/2, 3/1, 4/0}\draw[->] (m-1-\i) -- node {$f_\j$} (m-2-\i);
\end{tikzpicture}
\end{tikzpicture}
\Bigg\}\iso C_1 \oplus C_0, $$
\Bigg\}\iso C_1 \oplus C_0 = K(C)_1, $$
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 $C_1\oplus C_0$. This pattern can be continued and gives the following result:
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 $C_1\oplus C_0$.
\begin{proposition}
For any chain complex $C$ we have $K(C)_n \iso\bigoplus_{[n]\epi[p]} C_p$.
\end{proposition}
\begin{proof}
\todo{DK: proposition about $K$}
\end{proof}
\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 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 $\Ch{\Ab}\simeq\sAb$.
Joshua Moerman
12 years ago