@ -67,21 +67,21 @@ Of course the maps $\delta_i$ and $\sigma_i$ in $\DELTA$ satisfy certain equatio
Because a simplicial abelien group $A$ is a contravariant functor, these equations (which only consist of compositions and identities) also hold in its image. For example the first equation would look like: $ A(\delta_i)A(\delta_j)= A(\delta_{j-1})A(\delta_i)$ for $ i < j$ (again note that $A$ is contravariant, and hence composition is reversed). This can be used for an explicit definition of simplicial abelien groups. In this definition a simplicial abelian group $A$ consists of a collection abelian groups $(A_n)_{n}$ together with face and degeneracy maps (which are grouphomomorphisms) such that the simplicial equations hold. More precisely:
\begin{definition}
\emph{(Explicitly)} An simplicial abelian group $A$ consists of a collection abelian groups $A_n$ together with face maps $\delta^i : A_n \to A_{n-1}$ and degenracy maps $\sigma^i : A_n \to A_{n+1}$ for $0\leq i \leq n$ and $n \in\N$, such that:
\emph{(Explicitly)} An simplicial abelian group $A$ consists of a collection abelian groups $A_n$ together with face maps $d_i : A_n \to A_{n-1}$ and degenracy maps $s_i : A_n \to A_{n+1}$ for $0\leq i \leq n$ and $n \in\N$, such that:
\begin{align}
\delta^i\delta^j &= \delta^{j-1}\delta^i \hspace{0.5cm}\text{ if } i < j,\\
\delta^i\sigma^j &= \sigma^{j-1}\delta^i \hspace{0.5cm}\text{ if } i < j,\\
\delta^j\sigma^j &= \delta^{j+1}\sigma^j = \id,\\
\delta^i\sigma^j &= \sigma^j\delta^{i-1}\hspace{0.5cm}\text{ if } i > j+1,\\
\sigma^i\sigma^j &= \sigma^{j+1}\sigma^i \hspace{0.5cm}\text{ if } i \leq j.
d_i d_j &= d_{j-1} d_i \hspace{0.5cm}\text{ if } i < j,\\
d_i s_j &= s_{j-1} d_i \hspace{0.5cm}\text{ if } i < j,\\
d_j s_j &= d_{j+1} s_j = \id,\\
d_i s_j &= s_j d_{i-1}\hspace{0.5cm}\text{ if } i > j+1,\\
s_i s_j &= s_{j+1} s_i \hspace{0.5cm}\text{ if } i \leq j.
\end{align}
\end{definition}
It is already indicated that a functor from $\DELTA^{op}$ to $\Ab$ is determined when the images for the face and degeneracy maps in $\DELTA$ are provided. So gives this a way of restoring the first definition from this one. Conversely, we can apply functorialty to obtain the second definition from the first. So these definitions are the same \todo{sAb: is it ok not to prove this?}. So from now on we will denote $A([n])$ by $A_n$, $A(\sigma_i)$ by $\sigma^i$ and $A(\delta_i)$ by $\delta^i$, whenever we have a simplicial abelien group $A$.
It is already indicated that a functor from $\DELTA^{op}$ to $\Ab$ is determined when the images for the face and degeneracy maps in $\DELTA$ are provided. So gives this a way of restoring the first definition from this one. Conversely, we can apply functorialty to obtain the second definition from the first. So these definitions are the same \todo{sAb: is it ok not to prove this?}. So from now on we will denote $A([n])$ by $A_n$, $A(\sigma_i)$ by $s_i$ and $A(\delta_i)$ by $d_i$, whenever we have a simplicial abelien group $A$.
When using a simplicial abelian group to construct another object, it is often handy to use this second definition, as it gives you a very concrete objects to work with. On the other hand, constructing this might be hard (as you would need to provide a lot of details), in this case we will often use the more abstract definition.
\todo{sAb: Note that $\sigma^i$ is a monomorphism because of (3)}
\todo{sAb: Note that $s_i$ is a monomorphism because of (3)}
\subsection{Other simplicial objects}
Of course the abstract definition of simplicial abelian group can easilty be generalized to other categories. For example $\Set^{\DELTA^{op}}=\sSet$ is the category of simplicial sets. There are very important simplicial sets:
@ -99,27 +99,27 @@ Note that indeed $\Hom{\DELTA}{X}{[n]} \in \Set$, because the collection of morp
\end{example}
\begin{example}
$\Delta[1]$ is a bit more interesting, but still not too hard. We will compute the first three abelian groups $\Delta[1]_0$, $\Delta[1]_1$ and $\Delta[1]_2$. We can use the fact that any monotone increasing map $f: [n]\to[m]$ is a composition of first applying degeneracy maps, and then face maps, ie.: $f: [n]\tot{\sigma^{i_0}\cdots\sigma^{i_M}}[k]\tot{\delta^{j_0}\cdots\delta^{j_N}}[m]$, where $k \leq m, n$.
$\Delta[1]$ is a bit more interesting, but still not too hard. We will compute the first three abelian groups $\Delta[1]_0$, $\Delta[1]_1$ and $\Delta[1]_2$. We can use the fact that any monotone increasing map $f: [n]\to[m]$ is a composition of first applying degeneracy maps, and then face maps, ie.: $f: [n]\tot{\sigma_{i_0}\cdots\sigma_{i_M}}[k]\tot{\delta_{j_0}\cdots\delta_{j_N}}[m]$, where $k \leq m, n$.
For $\Delta[1]_0$ we have to consider maps from $[0]$ to $[1]$, we cannot first apply degeneracy maps (there is no object $[-1]$). So this leaves us with the face maps: $\Delta[1]_0=\{\delta_0, \delta_1\}$. For $\Delta[1]_1$ we of course have the identity function and two functions $\delta_0\sigma_0, \delta_1\sigma_0$. Now $\Delta[1]_2$ are the maps from $[2]$ to $[1]$.
We will compute the two face maps $\delta^0$ and $\delta^1$ from $\Delta[1]_1$ to $\Delta[1]_0$. Recall that the $\mathbf{Hom}$-functor in the first argument (the contravariant argument) works with precomposition. So this gives:
We will compute the two face maps $d_0$ and $d_1$ from $\Delta[1]_1$ to $\Delta[1]_0$. Recall that the $\mathbf{Hom}$-functor in the first argument (the contravariant argument) works with precomposition. So this gives:
Where we in the first calculation used the identity law. In the second and third line we used the third simplicial equation, asserting that $\sigma_0\delta_0=\id$. Similarly we can calculate the face map $\delta^1$:
Where we in the first calculation used the identity law. In the second and third line we used the third simplicial equation, asserting that $\sigma_0\delta_0=\id$. Similarly we can calculate the face map $d_1$:
@ -5,7 +5,7 @@ Comparing chain complexes and simplicial abelian groups, we see a similar struct
\subsection{Unnormalized chain complex}
Given a simplicial abelian group $A$, we have a family of abelian groups $A_n$. We define a grouphomomorphism $\del_{n-1} : A_n \to A_{n-1}$ as:
$$\del_{n-1}=\delta^0-\delta^1+\ldots+(-1)^n \delta^n \text{ for every } n > 0.$$
$$\del_{n-1}=d_0- d_1+\ldots+(-1)^n d_n \text{ for every } n > 0.$$
\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}
@ -17,7 +17,7 @@ $$\del_{n-1} = \delta^0 - \delta^1 + \ldots + (-1)^n \delta^n \text{ for every }
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 $\del(\sigma)=\sigma\circ\delta^0-\sigma\circ\delta^1+\ldots+(-1)^{n+1}\sigma\circ\delta^{n+1}$ (on generators). The terms $\sigma\circ\delta^i$ 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 a adjunction?}. And in fact we already used it in the construction of the singular chaincomplex, where we defined the boundary maps as $\del(\sigma)=\sigma\circd_0-\sigma\circd_1+\ldots+(-1)^{n+1}\sigma\circd_{n+1}$ (on generators). The terms $\sigma\circd_i$ 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.
@ -28,9 +28,9 @@ 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:
Joshua Moerman
11 years ago