@ -139,9 +139,9 @@ It can be shown that an equivalence $F: \cat{C} \tot{\simeq} \cat{D}$ is both a
The first definition of adjunction is useful when dealing with maps, since it gives an bijection between the $\mathbf{Hom}$-sets. However the second definition is useful when proving a certain construction is part of an adjunction, as shown in the following example.
\begin{example}
\emph{(The free abelian group)} There is an obvious functor $U: \Ab\to\Set$, which sends an abelian group to its underlying set, forgetting the additional structure. It is hence called a \emph{forgetful functor}. This functor is a right adjoint. The left adjoint $\Z[-]: \Set\to\Ab$is given by the \emph{free abelian group}; for a set $S$ define
\emph{(Free abelian groups)} There is an obvious functor $U: \Ab\to\Set$, which sends an abelian group to its underlying set, forgetting the additional structure. It is hence called a \emph{forgetful functor}. This functor has a left adjoint $\Z[-]: \Set\to\Ab$ given by the \emph{free abelian group functor}. For a set $S$ define
$$\Z[S]=\{\phi: S \to\Z\I\text{supp}(\phi)\text{ is finite}\}, $$
where $\text{supp}(\phi)=\{ s \in S \I\phi(s)\neq0\}$. The group structure on $\Z[S]$ is given pointwise. One can think of elements of this abelian group as formal sums, namely:
where $\text{supp}(\phi)=\{ s \in S \I\phi(s)\neq0\}$. The group structure on $\Z[S]$ is given by pointwise addition. One can think of elements of this abelian group as formal sums, namely:
It is clear that $U(\overline{f})\circ\eta= f$. We will leave the other details (naturality of $\eta$, $\overline{f}$ being a group homomorphism, and uniqueness w.r.t.$U(\overline{f})\circ\eta= f$) to the reader.
It is clear that $U(\overline{f})\circ\eta= f$. We will leave the other details (naturality of $\eta$, $\overline{f}$ being a group homomorphism, and uniqueness w.r.t.~$U(\overline{f})\circ\eta= f$) to the reader.
\end{example}
\subsection{The Yoneda lemma}
@ -167,10 +167,16 @@ So far we have only encountered definitions from category theory. However there
The functor $y$ is called the \emph{Yoneda embedding}.
\end{definition}
We will denote the set of natural transformation between two functors $F, G: \cat{C}\to\cat{D}$ as
Given a functor $F: \cat{C}\to\Set$ and any object $C \in\cat{C}$ there is a bijection
$$\mathbf{Nat}(y(C), F)\iso F(C), $$
which is natural in both $F$ and $C$, where $\mathbf{Nat}(G, G')$ denotes the set of natural transformation between $G$ and $G'$, in other words $\mathbf{Nat}=\mathbf{Hom}_{\Set^{\cat{C}^{op}}}$.
which is natural in both $F$ and $C$.
\end{lemma}
We will not provide a proof of this lemma, but we will give the function which can be proven to be a natural bijection. Given a natural transformation $\phi\in\mathbf{Nat}(y(C), F)$, we can consider the map $\phi_C : y(C)(C)\to F(C)$. Note that the codomain already is the right set, we only have to apply $\phi_C$ to the right object. The bijection is given by
$$\phi\mapsto\phi_C(\id_C). $$
We will use this lemma when we discuss simplicial abelian groups.
@ -41,6 +41,7 @@ Note that if we have two such chain maps $f:C \to D$ and $g:D \to E$, then the l
Note that we will often drop the indices of the boundary morphisms, since it is often clear in which degree we are working. The boundary operators give rise to certain subgroups, because all groups are abelian, subgroups are normal subgroups.
\begin{definition}
\label{def:cycles}
Given a chain complex $C$ we define the following subgroups:
\begin{itemize}
\item the subgroup of \emph{$n$-cycles}: $Z_n(C)=\ker(\del_n: C_n \to C_{n-1})\nsubgrp C_n$, and
@ -56,12 +57,12 @@ Note that we will often drop the indices of the boundary morphisms, since it is
It follows from $\del_n \circ\del_{n+1}=0$ that $\im(\del: C_{n+1}\to C_n)$ is a subset of $\ker(\del: C_n \to C_{n-1})$. Those are exactly the abelian groups $B_n(C)$ and $Z_n(C)$, so $ B_n(C)\nsubgrp Z_n(C)$.
\end{proof}
In general there is no inclusion in the other direction. This defect can be measured by a quotient and gives rise to the following definition. A motivation for this concept will be provided in section~\ref{sec:singular}.
In general there is no inclusion in the other direction. This defect can be measured by a quotient and gives rise to the following definition. A motivation for this concept will be provided in Section~\ref{sec:singular}.
\begin{definition}
Given a chain complex $C$ we define the \emph{$n$-th homology group}$H_n(C)$ for each $n \in\N$ as:
$$ H_n(C)= Z_n(C)/ B_n(C).$$
We will denote the class of an $n$-cycle $x \in Z_n(C)$ by $[x]$.
We will denote the class of an $n$-cycle $x \in Z_n(C)$ by $[x]$ and refer to it as the \emph{homology class of $x$}.
\end{definition}
\begin{lemma}
The $n$-th homology group gives a functor $H_n : \Ch{\Ab}\to\Ab$ for each $n \in\N$.
@ -156,9 +157,9 @@ Some geometric intuition for the boundary operator is provided by Figure~\ref{fi
\label{fig:singular_chaincomplex}
\end{figure}
The above construction gives us a functor $C: \Top\to\Ch{\Ab}$ (we will not prove this). Composing this with the functor $H_n: \Ch{\Ab}\to\Ab$ gives rise to the following definition.
The above construction defines a functor $C: \Top\to\Ch{\Ab}$ (we will not prove this) which sends a space $X$ to its \emph{singular chain complex}$C(X)$. So the terminology of Definition~\ref{def:cycles} applies to these chain complexes. Composing this with the functor $H_n: \Ch{\Ab}\to\Ab$ gives rise to the following definition.
\begin{definition}
The \emph{singular $n$-th homology group} of a space $X$ is defined as
The \emph{$n$-th singular homology group} of a space $X$ is defined as
Note that this is also the definition of the Yoneda embedding $\Delta[n]= y[n]$. In a moment we will see why the Yoneda lemma is useful to us. But let us first explicitly describe two of such standard $n$-simplices.
Note that $\Delta[-]: \DELTA\to\sSet$ is exactly the Yoneda embedding. In a moment we will see why the Yoneda lemma is useful to us, but let us first explicitly describe two examples of such standard simplices.
\begin{example}
We will compute how $\Delta[0]$ look like. Note that $[0]$ is an one-element set, so for any set $S$, there is only one function $\ast : S \to[0]$. Hence $\Delta[0]_n =\{\ast\}$ for all $n$. The face and degeneracy maps are now functions from $\{\ast\}$ to $\{\ast\}$. Again there is only one, namely $\id : \{\ast\}\to\{\ast\}$. This gives:
We will compute how $\Delta[0]$ look like. Note that $[0]$ is an one-element set, so for any set $S$, there is only one function $\ast : S \to[0]$. Hence $\Delta[0]_n =\{\ast\}$ for all $n$ and the face and degeneracy maps are necessarily the identity maps $\id : \{\ast\}\to\{\ast\}$. Thus, $\Delta[0]$ looks like:
$$\Delta[0] :=
\begin{tikzpicture}[baseline=-0.5ex]
\matrix (m) [matrix of math nodes] {
\{\ast\}&\{\ast\}&\{\ast\}&\cdots\\
};
\foreach\r in {-5, 5}\draw [raise line=\r, <-] (m-1-1) -> (m-1-2);
\foreach\r in {0}\draw [raise line=\r, ->] (m-1-1) -> (m-1-2);
Note that the only non-degenerate simplex is the unique $0$-simplex.
\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 function $f: [n]\to[m]$ is a composition of first applying degeneracy maps, and then face maps, i.e.: $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 complicated. We will describe the first three sets $\Delta[1]_0$, $\Delta[1]_1$ and $\Delta[1]_2$. We can use the fact that any monotone function $f: [n]\to[m]$ is a composition of first applying degeneracy maps, and then face maps, i.e.: $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 $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 $d_1$:
In this simplicial set there are three non-degenerate simplices. There is $\id\in\Delta[1]_1$, which clearly is non-degenerate, and the two $0$-simplices $\delta_0$ and $\delta_1$. One can think of this simplicial set as a line (the non-degenerate $1$-simplex) with its endpoints (the two $0$-simplices).
\end{example}
\subsection{Other simplicial objects}
Of course the abstract definition of simplicial abelian group can easily be generalized to other categories. For any category $\cat{C}$ we can consider the functor category $\cat{sC}=\cat{C}^{\DELTA^{op}}$. In this thesis we are interested in the category $\sAb=\Ab^{\DELTA^{op}}$ of simplicial abelian groups. So a simplicial abelian group $A$ is a collection of abelian groups $A_n$, together with face and degeneracy maps, which in this case means group homomorphisms $d_i$ and $s_i$ such that the simplicial equations hold.
\subsection{Simplicial objects in arbitrary categories}
Of course the definition of simplicial set can easily be generalized to other categories. For any category $\cat{C}$ we can consider the functor category $\cat{sC}=\cat{C}^{\DELTA^{op}}$. In this thesis we are interested in the category of simplicial abelian groups:
$$\sAb=\Ab^{\DELTA^{op}}. $$
So a simplicial abelian group $A$ is a collection of abelian groups $A_n$, together with face and degeneracy maps, which in this case means group homomorphisms $d_i$ and $s_i$ such that the simplicial equations hold.
Note that the set of natural transformations between two simplicial abelian groups $A$ and $B$ is also an abelian group. The proof that $\sAb$ is a preadditive category is very similar to the proof we saw in section~\ref{sec:Chain Complexes}. For two natural transformations $f,g: A \to B$ we simply define $f+g$ pointwise: $(f+g)_n = f_n + g_n$.
Note that the set of natural transformations between two simplicial abelian groups $A$ and $B$ is also an abelian group. The proof that $\sAb$ is a preadditive category is very similar to the proof we saw in Section~\ref{sec:Chain Complexes}. For two natural transformations $f,g: A \to B$ we simply define $f+g$ pointwise by$(f+g)_n = f_n + g_n$ and it is easily checked that this is a natural transformation.
As we are interested in simplicial abelian groups, it would be nice to make these standard $n$-simplices into simplicial abelian groups. We have seen how to make an abelian group out of any set using the free abelian group. We can use this functor $\Z[-] : \Set\to\Ab$ to induce a functor $\Z^\ast[-] : \sSet\to\sAb$ as shown in the following diagram.
As we are interested in simplicial abelian groups, it would be nice to make these standard $n$-simplices into simplicial abelian groups. We have seen how to make an abelian group out of any set using the free abelian group functor. We can use this functor $\Z[-] : \Set\to\Ab$ to induce a functor $\Z^\ast[-] : \sSet\to\sAb$ as shown in the following diagram.
\begin{figure}[h!]
\begin{tikzpicture}
\matrix (m) [matrix of math nodes]{
@ -195,20 +209,24 @@ As we are interested in simplicial abelian groups, it would be nice to make thes
\caption{The simplicial set $X$ can be made into a simplicial abelian group $X'$ by postcomposing with $\Z[-]$}
\caption{The simplicial set $X$ can be made into a simplicial abelian group $\Z^\ast[X]$ by postcomposing with $\Z[-]$.}
\label{fig:diagram_Z}
\end{figure}
This construction obviously defines a functor $\Z^\ast[-] : \sSet\to\sAb$. Similarly, postcomposition with the forgetful functor $U: \Ab\to\Set$ gives rise to a forgetful functor $U^\ast: \sAb\to\sSet$. Thus in formulas we have
This justifies that we may drop this extra decoration ($^\ast$) and write $\Z[-]$ (resp. $U$) instead of $\Z^\ast[-]$ (resp. $U^\ast$).
\begin{lemma}
The functor $\Z^\ast[-] : \sSet\to\sAb$ is a left adjoint, with $U^\ast: \sAb\to\sSet$ (the pointwise forgetful functor) as right adjoint.
The functor $\Z[-] : \sSet\to\sAb$ is a left adjoint, with $U: \sAb\to\sSet$ as right adjoint.
\end{lemma}
\begin{proof}
First we note that $U^\ast\Z^\ast[X]_n = U\Z[X_n]$ by definition, so pointwise we get (by the fact that $\Z$ and $U$ already form an adjunction):
First we note that $U\Z[X]_n = U\Z[X_n]$ by definition, so pointwise we get (by the fact that $\Z$ and $U$ already form an adjunction):
\begin{center}
\begin{tikzpicture}
\matrix (m) [matrix of math nodes]{
X_n & U^\ast\Z^\ast[X]_n &\Z[X_n]\\
X_n & U\Z[X]_n &\Z[X_n]\\
& U(A_n) & A_n \\
};
\path[->]
@ -219,41 +237,41 @@ As we are interested in simplicial abelian groups, it would be nice to make thes
(m-1-3) edge node[auto] {$\overline{f}$} (m-2-3);
\end{tikzpicture}
\end{center}
Then use naturality of $i$ (in $X_n$, thus in particular in $n$) to extend this to $i^\ast : X \to U^\ast\Z^\ast[X]$. Now if we're given a natural transformation $f: X \to U^\astA$ of simplicial sets we can again construct $\overline{f}: \Z^\ast[X]\to A$ pointwise. The reader is invited to check the details.
Then use naturality of $i$ (in $X_n$, thus in particular in $n$) to extend this to $i^\ast : X \to U\Z[X]$. Now if we are given a natural transformation $f: X \to UA$ of simplicial sets we can again construct $\overline{f}: \Z[X]\to A$ pointwise. The reader is invited to check the details.
\end{proof}
\begin{example}
We can apply this to the standard $n$-simplex $\Delta[1]$. This gives $\Delta[1]_0\iso\Z^2$, since $\Delta[1]_0$ has two elements, and $\Z^\ast[\Delta[1]]_1\iso\Z^3$, where the isomorphisms are taken such that:
We can apply this to the standard $n$-simplex $\Delta[1]$. This gives $\Delta[1]_0\iso\Z^2$, since $\Delta[1]_0$ has two elements, and $\Z^\ast[\Delta[1]]_1\iso\Z^3$, where the isomorphisms are taken such that
\begin{align*}
\delta_0 &\mapstot{\iso} (1, 0) \\
\delta_1 &\mapstot{\iso} (0, 1) \\
\sigma_0\delta_0 &\mapstot{\iso} (1, 0, 0) \\
\id&\mapstot{\iso} (0, 1, 0) \\
\sigma_0\delta_1&\mapstot{\iso} (0, 0, 1)
\delta_0 &\mapstot{\iso} (1, 0),\\
\delta_1 &\mapstot{\iso} (0, 1),\\
\delta_0\sigma_0 &\mapstot{\iso} (1, 0, 0),\\
\id&\mapstot{\iso} (0, 1, 0),\\
\delta_1\sigma_0&\mapstot{\iso} (0, 0, 1).
\end{align*}
The face maps from $\Z^\ast[\Delta[1]]_1$ to $\Z^\ast[\Delta[1]]_0$ under these isomorphisms are then given by:
The face maps from $\Z[\Delta[1]]_1$ to $\Z[\Delta[1]]_0$ under these isomorphisms are then given by
\begin{align*}
d_0(x, y, z) &= (x+y, z) \\
d_1(x, y, z) &= (x, y+z)
d_0(x, y, z) &= (x+y, z),\\
d_1(x, y, z) &= (x, y+z).
\end{align*}
\end{example}
\subsection{The Yoneda lemma}
Recall that the Yoneda lemma stated: $\mathbf{Nat}(y(C), F)\iso F(C)$, where $F:\cat{C}^{op}\to\Set$ is a functor and $C$ an object. In our case we consider functors $X: \DELTA^{op}\to\Set$ and objects $[n]$. So this gives us the natural bijection:
$$X_n \iso\Hom{\sSet}{\Delta[n]}{X}.$$
So we can regard $n$-simplices in $X$ as maps from $\Delta[n]$ to $X$. This also extends to the abelian case, where we get a natural isomorphism (of abelian groups):
\begin{lemma}\emph{(The abelian Yoneda lemma)}
Recall the statement of the Yoneda lemma from Section~\ref{sec:Category Theory}. In our case we consider functors $X: \DELTA^{op}\to\Set$ and objects $[n]$. So this gives us a natural bijection
$$\Hom{\sSet}{\Delta[n]}{X}\iso X_n$$
telling us that we can regard $n$-simplices in $X$ as maps from $\Delta[n]$ to $X$. This also extends to the case of simplicial abelian groups.
\begin{lemma}\emph{(The additive Yoneda lemma)}
Let $A$ be a simplicial abelian group. Then there is a group isomorphism
$$A_n \iso\Hom{\sAb}{\Z^\ast[\Delta[n]]}{A}, $$
$$\Hom{\sAb}{\Z[\Delta[n]]}{A}\iso A_n, $$
which is natural in $A$ and $[n]$.
\end{lemma}
\begin{proof}
By using the (non-abelian) Yoneda lemma and the fact that $\Z^\ast$ is a left adjoint, we already have a natural bijection:
The only thing that we need to check is that this bijection preserves the group structure. Recall that this bijection from $\Hom{\sAb}{\Z^\ast[\Delta[n]]}{A}$ to $A_n$ is given by (where $\id=\id_{[n]}$ is a generator in $\Z^\ast[\Delta[n]]$):
By using the (non-additive) Yoneda lemma and the fact that $\Z$ is a left adjoint, we already have a natural bijection:
The only thing that we need to check is that this bijection preserves the group structure. Recall that this bijection from $\Hom{\sAb}{\Z[\Delta[n]]}{A}$ to $A_n$ is given by (where $\id=\id_{[n]}$ is a generator in $\Z[\Delta[n]]$):
$$\phi(f)= f_n(\id)\in X_n \quad\text{ for } f: \Delta[n]\to X. $$
Now let $A$ be a simplicial abelian group and $f, g: \Z^\ast\Delta[n]\to A$ maps. Then we compute:
Now let $A$ be a simplicial abelian group and $f, g: \Z\Delta[n]\to A$ maps. Then we compute:
where we regard $\id\in\Delta[a]$ as en element $\id\in\Z^\ast\Delta[n]$, we can do so by the unit of the adjunction. So this bijection is also a group homomorphism, hence we have an isomorphism $A_n \iso\Hom{\sAb}{\Z^\ast[\Delta[n]]}{A}$ of abelian groups.
where we regard $\id\in\Delta[n]$ as an element $\id\in\Z\Delta[n]$, we can do so by the unit of the adjunction. So this bijection is also a group homomorphism, hence we have an isomorphism $\Hom{\sAb}{\Z[\Delta[n]]}{A}\iso A_n$ of abelian groups.
Comparing chain complexes and simplicial abelian groups, one sees a certain similarity. Both concepts are defined as sequences of abelian groups with certain structure maps. At first sight simplicial abelian groups seem to have a richer structure. There are many face maps as opposed to only a single boundary homomorphism. Nevertheless, as we will show in this section, these two concepts give rise to equivalent categories.
\subsection{Unnormalized chain complex}
Given a simplicial abelian group $A$, we have a family of abelian groups $A_n$. For every $n>0$ we define a group homomorphism $\del_n : A_n \to A_{n-1}$:
Given a simplicial abelian group $A$, we have a family of abelian groups $A_n$. For every $n>0$ we define a group homomorphism $\del_n : A_n \to A_{n-1}$
$$\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 prove is $\del_n\circ\del_{n+1}=0$. This can be done with a calculation.
We already have a collection of abelian groups together with maps, so the only thing to prove is $\del_{n-1}\circ\del_n=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.
\end{proof}
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.
\todo{DK: Make this more readable...}
This construction defines a functor $M : \sAb\to\Ch{\Ab}$. And in fact we already used it in the construction of the singular chain complex, where we defined the boundary maps (on generators) as $\del(\sigma)=\sigma\circ d_0-\sigma\circ d_1+\ldots+(-1)^{n+1}\sigma\circ d_{n+1}$. We will briefly come back to this in Section~\ref{sec:Homotopy}.
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.
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. $$
Now if we would construct a (non-trivial) simplicial abelian group $K(C)$ from this chain complex, we now know that $K(C)_n$ is non-trivial for all $n \in\N$. This means that $M(K(C))_n$ is non-trivial for all $n \in\N$. For an equivalence we require a (natural) isomorphism: $M(K(C))\tot{\iso} C$, this in particular means an isomorphism in each degree $n > 0$: $0\neq M(K(C))_n \tot{\iso} C_n =0$, which is not possible. So the functor $M$, as defined as above, will not give us the equivalence we wanted, although it is a very nice functor.
\subsection{Normalized chain complex}
To repair this defect we should be more careful. Given a simplicial abelian group, simply taking the same collection for our chain complex will not work (as shown above). Instead we are after some ``smaller'' abelian groups, and in some cases the abelian groups should completely vanish (as in the example above).
Given a simplicial abelian group $A$, we define abelian groups $N(A)_n$ as:
Given a simplicial abelian group $A$, we define abelian groups $N(A)_n$ as
Now define group homomorphisms $\del : N(A)_n \to N(A)_{n-1}$ as:
Now define group homomorphisms $\del : N(A)_n \to N(A)_{n-1}$ as
$$\del= d_0|_{N(A)_n}. $$
\begin{lemma}
The function $\del$ is well-defined. Furthermore $\del\circ\del=0$.
@ -45,14 +46,14 @@ $$ \del = d_0|_{N(A)_n}. $$
Let $x \in N(A)_n$, then $d_i \del(x)= d_i d_0(x)= d_0 d_{i+1}(x)= d_0(0)=0$ for all $i < n$. So indeed $\del(x)\in N(A)_{n-1}$, because in particular it holds for $i > 0$. Using this calculation for $i =0$ shows that $\del\circ\del=0$. This shows that $N(A)$ is a chain complex.
\end{proof}
We will call this chain complex $N(A)$ the \emph{normalized chain complex} of $A$.
The chain complex $N(A)$ is called the \emph{normalized chain complex} of $A$.
\begin{lemma}
The above construction gives a functor $N: \sAb\to\Ch{\Ab}$. Furthermore $N$ is additive.
\end{lemma}
\begin{proof}
Given a map $f: A \to B$ of simplicial abelian groups, we consider the restrictions:
Given a map $f: A \to B$ of simplicial abelian groups, we consider the restrictions
$$ f_n |_{N(A)_n} : N(A)_n \to B_n. $$
Because $f_n$ commutes with the face maps we get:
Because $f_n$ commutes with the face maps we get
$$ d_i(f_n(x))= f_{n-1}(d_i(x))=0, $$
for $i>0$ and $x \in N(A)_n$. So the restriction also restricts the codomain, i.e. $f_n |_{N(A)_n} : N(A)_n \to N(B)_n$ is well-defined. Furthermore it commutes with the boundary operator, since $f$ itself commutes with all face maps. This gives functoriality $N(f): N(A)\to N(B)$.
@ -62,48 +63,62 @@ We will call this chain complex $N(A)$ the \emph{normalized chain complex} of $A
\end{proof}
\begin{example}
We will look at the normalized chain complex of $\Z[\Delta[0]]$. Recall that it looked like:
$$\Z[\Delta[0]]=\Z\to\Z\to\Z\to\cdots, $$
where all face and degeneracy maps are identity maps. Clearly the kernel of $\id$ is the trivial group. So $N(\Z[\Delta[0]])_i =0$ for all $i > 0$. In degree zero we are left with $N(\Z[\Delta[0]])_0=\Z$. So we can depict the normalized chain complex by:
We will look at the normalized chain complex of $\Z[\Delta[0]]$. Recall that it looks like
$$\Z[\Delta[0]] :=
\begin{tikzpicture}[baseline=-0.5ex]
\matrix (m) [matrix of math nodes] {
\Z&\Z&\Z&\cdots\\
};
\foreach\r in {-5, 5}\draw [raise line=\r, <-] (m-1-1) -> (m-1-2);
\foreach\r in {0}\draw [raise line=\r, ->] (m-1-1) -> (m-1-2);
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 $C$.
So in this example we see that the normalized chain complex is really better behaved than the unnormalized chain complex given by $M(\Z[\Delta[0]]$.
\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:
$$ D(X)_n =\sum_{i=0}^n s_i(X_{n-1}). $$
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 $D_n(X)\subset X_n$ of degenerate simplices, defined as:
$$ D_n(X)=\sum_{i=0}^n s_i(X_{n-1}). $$
\begin{lemma}
\label{le:decomp1}
For all $x \in X_n$ we have:
$$ x = b + c,$$
where $b \in N(X)_n$ and $c \in D(X)_n$.
where $b \in N(X)_n$ and $c \in D_n(X)$.
\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 $x \in P^n = X_n$.
Define the subgroup $P_n^k =\{ x \in X_n \I d_i x =0\text{ for all } i > k\}$. Note that by definition we have
We will prove with induction that for any $k \leq n$ we can write $x \in X_n$ as $x = b + c$, with $b \in P_n^k$ and $c \in D_n(X)$. For $k = n$ the statement is clear, because we can simply write $x = x$, knowing that $x \in P_n^n = X_n$.
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
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 $b \in P_n^k$ and $c \in D_n(X)$. 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 $b' \in P_n^{k-1}$. Furthermore we can define $c' = s_{k-1} d_k b + c$, for which it is clear that $c' \in D_n(X)$. 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 $c' \in D(X)_n$.
with $b' \in P_n^{k-1}$ and $c' \in D_n(X)$.
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.
Doing this inductively gives us $x = b + c$, with $b \in P_n^0= N(X)_n$ and $c \in D_n(X)$, 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 $x = d_{i+1} s_i x =0$.
Using that $s_i x \in N(X)_{n+1}$ means $0= d_{k+1} s_i x$ for any $k \geq0$ and by using using the simplicial identity: $d_{i+1} s_i =\id$, 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 assures that there are no degenerate $n$-simplices in $N(X)$. So this gives us:
\begin{corollary}
\label{cor:NandD}
$X_n = N(X)_n \oplus D(X)_n$
$X_n = N(X)_n \oplus D_n(X)$
\end{corollary}
We can extend the above lemmas to a more general statement.
@ -117,17 +132,17 @@ We can extend the above lemmas to a more general statement.
\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.
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 $c \in D_n(X)$, 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 $\beta^\ast(N(X))_p\cap\gamma^\ast(N(X))_q=0$.
Let $\beta : [n]\epi[m]$ and $\gamma : [n]\epi[m']$ be two maps such that $\beta\neq\gamma$. Then we have $\beta^\ast(N(X))_m\cap\gamma^\ast(N(X))_{m'}=0$.
\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 $x \neq0$ this gives a contradiction.
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.
\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:
Again the former lemma of these two lemmas proves the existence of a decomposition and the latter shows the uniqueness. So combining these gives:
\begin{corollary}
\label{cor:decomp}
@ -140,27 +155,27 @@ And by considering $X_n$ as a whole we get:
$X_n =\bigoplus_{[n]\epi[p]} N(X)_p$.
\end{corollary}
Using corollary~\ref{cor:decomp} we can proof a nice categorical fact about $N$, which we will use later on.
Using Corollary~\ref{cor:decomp} we can prove a nice categorical fact about $N$, which we will use later on.
\begin{lemma}
The functor $N$ is fully faithful, i.e.:
$$ N : \Hom{\sAb}{A}{B}\iso\Hom{\Ch{\Ab}}{N(A)}{N(B)}. $$
$$ N : \Hom{\sAb}{A}{B}\iso\Hom{\Ch{\Ab}}{N(A)}{N(B)}\quad A, B \in\sAb. $$
\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
First we prove that$N$ 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 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.
where we used naturality of $f$ in the second step, and the fact that $x_\beta\in N(A)$ 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:
Secondly we have to prove$N$ 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{DK: is this clear?}
again we have written $x$ as $x =\sum_\beta\beta^\ast x_\beta$. Clearly $N(f)= g$.
\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 careful than $C$, which included everything. In fact, corollary~\ref{cor:NandD} exactly tells us $C(X)_n = N(X)_n \oplus D(X)_n$.
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 $M(X)_n = N(X)_n \oplus D_n(X)$.
\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 $\Ab$.
@ -175,7 +190,7 @@ where we are supposed to fill in the second argument first, leaving us with a si
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:
We've already seen homology in chain complexes. We can of course now translate this notion to simplicial abelian groups, by assigning a simplicial abelian group $X$ to $H_n(N(X))$. But there is a more general notion of homotopy for simplicial sets, which is also similar to the notion of homotopy in topology. We will define the notion of homotopy groups for simplicial sets.
We have already seen homology in chain complexes. We can of course now translate this notion to simplicial abelian groups, by assigning a simplicial abelian group $X$ to $H_n(N(X))$. But there is a more general notion of homotopy for simplicial sets, which is also similar to the notion of homotopy in topology. We will define the notion of homotopy groups for simplicial sets.
When dealing with homotopy in a topological space $X$ we always need a base-point $\ast\in X$. This is also the case for homotopy in simplicial sets. We will notate the chosen base-point of a simplicial set $X$ with $\ast\in X_0$. Note that it is a $0$-simplex, but in fact the base-point is present in all sets $X_n$, because we can consider its degenerate simplices $s_0(\ldots(s_0(\ast))\ldots)\in X_n$, we will also denote these elements as $\ast$. Of course in our situation we are concerned about simplicial abelian groups, where there is an obvious choice for the base-point, namely$0$.
When dealing with homotopy groups in a topological space $X$ we always need a base-point $\ast\in X$. This is also the case for simplicial sets. We will notate the chosen base-point of a simplicial set $X$ with $\ast\in X_0$. Note that it is a $0$-simplex, but in fact the base-point is present in all sets $X_n$, because we can consider its degenerate simplices $s_0(\ldots(s_0(\ast))\ldots)\in X_n$, we will also denote these elements as $\ast$. Of course in our situation we are concerned with simplicial abelian groups, where there is an obvious choice for the base-point given by the neutral element$0$.
\subsection{Homotopy groups}
\begin{definition}
Given a simplicial set $X$ with base-point $\ast$, we define $Z_n(X)$ to be the set of $n$-simplices with the base-point as boundary, i.e.:
Given a simplicial set $X$ with base-point $\ast$, we define $Z_n(X)$ to be the set of $n$-simplices with the base-point as boundary, i.e.
$$ Z_n(X)=\{ x \in X_n | d_i(x)=\ast\text{ for all } i \leq n \}. $$
For two $n$-simplices $x, x' \in Z_n(X)$, we define $x \sim x'$ if there exists $y \in X_{n+1}$ such that:
For two $n$-simplices $x, x' \in Z_n(X)$, we define $x \sim x'$ if there exists $y \in X_{n+1}$ such that
\begin{align}
d_0(y) &= x \\
d_1(y) &= x' \\
d_0(y) &= x,\\
d_1(y) &= x',\\
d_i(y) &= \ast\text{ for all } i > 1.
\end{align}
We will call $y$ the \emph{homotopy} and notate $y: x \sim x'$.
@ -54,7 +54,7 @@ Before proving this, one should have a look at figure~\ref{fig:simplicial_eqrel}
\emph{Transitivity}. Let $x_0, x_1, x_2\in Z_n(X)$ with $y: x_0\sim x_1$ and $z: x_1\sim x_2$. Define $w = y + z - s_0 x_1$. By linearity we have $d_0 w = x_0+ x_1-x_1= x_0$, similarly $d_1 w = x_2$. Again for $i>1$ we have $d_i w =0$.
\emph{Addition}. Let $y: x_0\sim x_1$ and $z: x_2\sim x_3$. Then by linearity $y +g: x_0+ x_2\sim x_1+ x_3$ and $-y: -x_0\sim-x_1$.
\emph{Addition}. Let $y: x_0\sim x_1$ and $z: x_2\sim x_3$. Then by linearity $y +z: x_0+ x_2\sim x_1+ x_3$ and $-y: -x_0\sim-x_1$.
\end{proof}
\begin{definition}
@ -91,17 +91,17 @@ Note that this is an abelian group, because $Z_n(X)$ is a subgroup of $X_n$, and
\end{proof}
\subsection{Topology}
In section~\ref{sec:Constructions} we saw that we can construct a functor $G: \cat{C}\to\sSet$ if we are provided a functor the other way around. If we can define a functor $F: \DELTA\to\Top$, then for any space $X$ we have a simplicial set $\Hom{\Top}{F-}{X}: \DELTA^{op}\to\Set$. In section~\ref{sec:Chain Complexes}, we already defined the \emph{topological $n$-simplex}$\Delta^n$ and face maps $\delta^i : \Delta^n \mono\Delta^{n+1}$. We can similarly define degeneracy maps $s^i: \Delta^n \to\Delta^{n-1}$ as:
In Section~\ref{sec:Constructions} we saw that we can construct a functor $G: \cat{C}\to\sSet$ if we are provided a functor the other way around. If we can define a functor $F: \DELTA\to\Top$, then for any space $X$ we have a simplicial set $\Hom{\Top}{F-}{X}: \DELTA^{op}\to\Set$. In Section~\ref{sec:Chain Complexes}, we already defined the \emph{topological $n$-simplex}$\Delta^n$ and face maps $\delta^i : \Delta^n \mono\Delta^{n+1}$. We can similarly define degeneracy maps $s^i: \Delta^n \to\Delta^{n-1}$ as:
The reader is invited to check the cosimplicial identities himself and conclude that we now have a functor $F: \DELTA\to\Top$, and hence we have a functor $S: \Top\to\sSet$ given by:
$$\text{Sing}(X)_n =\Hom{\Top}{\Delta^n}{X}. $$
Recall construction of the singular chain complex in section~\ref{sec:Chain Complexes}:
Recall construction of the singular chain complex in Section~\ref{sec:Chain Complexes}:
$$ C_n(X)=\Z[\Hom{\cat{Top}}{\Delta^n}{X}]. $$
Where the boundary map was given as an alternating sum. Looking more closely we see that this construction decomposes as:
where the last functor is the \emph{unnormalized chain complex}. All the categories involved have a notion of homotopy. In topological spaces this is the known notion where $f, g:X \to Y$ are homotopic if there exists a homotopy $H:I \times X \to Y$ with the appropriate properties. In simplicial sets (or simplicial abelian groups) we only saw the notion of homotopy groups, but there exists a more general notion of homotopy, as discussed in the overview of Friedman \cite{friedman}. And finally in chain complexes we saw homology groups, but this category also has a more general notion of chain homotopy, which can be found in any book on homological algebra such as in the book of Rotman \cite{rotman}.
It is known that for any simplicial abelian group both the normalized and unnormalized chain complex have the same homology groups. More precisely for any simplicial abelian group $X$ we have:
$$ H_n(N(X))\iso H_n(C(X))\quad\text{for all } n \in\N. $$
$$ H_n(N(X))\iso H_n(M(X))\quad\text{for all } n \in\N. $$
This is for example proven in \cite[Theorem 4.1]{eilenberg}. So this assures that the homology groups of the singular chain complex of a space are really the homotopy groups of the simplicial abelian group which is in the background.
Joshua Moerman
12 years ago