A \emph{(non-negative) chain complex}$C$ is a collection of abelian groups $C_n$ together with group homomorphisms $\del_n: C_{n+1}\to C_n$, which we call \emph{boundary homomorphisms}, such that $\del_n \circ\del_{n+1}=0$.
A \emph{(non-negative) chain complex}$C$ is a collection of abelian groups $C_n$ together with group homomorphisms $\del_n: C_n \to C_{n-1}$, which we call \emph{boundary homomorphisms}, such that $\del_n \circ\del_{n+1}=0$ for all $n \in\Np$.
\end{definition}
Thus graphically a chain complex $C$ can be depicted by the following diagram:
$$\cdots\to C_4\to C_3\to C_2\to C_1\to C_0. $$
\begin{center}
\begin{tikzpicture}
\matrix (m) [matrix of math nodes]{
\cdots& C_4 & C_3 & C_2 & C_1 & C_0 \\
};
\foreach\d/\i/\j in {5/1/2,4/2/3,3/3/4,2/4/5,1/5/6}\path[->] (m-1-\i) edge node[auto] {$\del_\d$} (m-1-\j);
\end{tikzpicture}
\end{center}
There are many variants to thie notion. For example, there are also unbounded chain complexes with an abelian group for each $n \in\Z$ instead of $\N$. In this thesis we will only need chain complexes in the sense of the definition above. Hence we will simply call them chain complexes, instead of non-negative chain complexes.
Of course we can make this more general by taking for example $R$-modules instead of abelian groups. We will later see which kind of algebraic objects make sense to use in this definition \todo{Ch: Will I discuss ab. cat. ?}. The boundary operators give rise to certain subgroups, because all groups are abelian, subgroups are normal subgroups.
There are many variants to this notion. For example, there are also unbounded chain complexes with an abelian group for each $n \in\Z$ instead of $\N$. In this thesis we will only need chain complexes in the sense of the definition above. Hence we will simply call them chain complexes, instead of non-negative chain complexes. Other variants can be given by taking a collection of $R$-modules instead of abelian groups. Of course not any kind of mathematical object will suffice, because we need to be able to express $\del_n \circ\del_{n+1}=0$, so we need some kind of \emph{zero object}. We will not need this kind of generality and stick to abelian groups.
In order to organize these chain complexes in a category, we should define what the maps are. The diagram above already gives an idea for this.
\begin{definition}
Let $C$ and $D$ be chain complexes, with boundary maps $\del^C_n$ and $\del^D_n$ respectively. A \emph{chain map}$f: C \to D$ consists of a family of maps $f_n: C_n \to D_n$, such that they commute with the boundary operators: $f_n \circ\del^C_{n+1}=\del^D_{n+1}\circ f_{n+1}$ for all $n \in\N$, i.e. the following diagram commutes:
\begin{center}
\begin{tikzpicture}
\matrix (m) [matrix of math nodes]{
\cdots& C_4 & C_3 & C_2 & C_1 & C_0 \\
\cdots& D_4 & D_3 & D_2 & D_1 & D_0 \\
};
\foreach\d/\i/\j in {5/1/2,4/2/3,3/3/4,2/4/5,1/5/6}\path[->] (m-1-\i) edge node[auto] {$\del^C_\d$} (m-1-\j);
\foreach\d/\i/\j in {5/1/2,4/2/3,3/3/4,2/4/5,1/5/6}\path[->] (m-2-\i) edge node[auto] {$\del^D_\d$} (m-2-\j);
\foreach\d/\i in {4/2,3/3,2/4,1/5,0/6}\path[->] (m-1-\i) edge node[auto] {$ f_\d$} (m-2-\i);
\end{tikzpicture}
\end{center}
\end{definition}
Note that if we have two such chain maps $f:C \to D$ and $g:D \to E$, then the levelwise composition will give us a chain map $g \circ f: C \to D$. Also taking the identity function in each degree, gives us a chain map $\id : C \to C$. In fact, this will form a category, we will leave the details (the identity law and associativity) to the reader.
\begin{definition}
$\Ch{\Ab}$ is the category of chain complexes with chain maps.
\end{definition}
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}
Given a chain complex $C$ we define the following subgroups:
\begin{itemize}
\item$Z_n(C)=\ker(\del: C_n \to C_{n-1})\nsubgrp C_n$, and
\item$Z_n(C)=\ker(\del_n: C_n \to C_{n-1})\nsubgrp C_n$, and
@ -30,7 +60,6 @@ Of course we can make this more general by taking for example $R$-modules instea
$$ H_n(C)= Z_n(C)/ B_n(C).$$
\end{definition}
\todo{CC: Chain maps}
\todo{CC: $H_n$ as a functor}
\subsection{The singular chain complex}
@ -78,7 +107,7 @@ We now have enough tools to define the singular chain complex of a space $X$.
This might seem a bit complicated, but we can pictures this in an intuitive way, as in figure~\ref{fig:singular_chaincomplex3}. And we see that the boundary operators really give the boundary of an $n$-simplex. To see that this indeed is a chain complex we have to proof that the composition of two such operators is the zero map.
\begin{figure}
\includegraphics{singular_chaincomplex3}
\caption{The boundary of a 2-simplex}
\caption{The boundary of a 2-simplex\todo{CC: update picture}}
Joshua Moerman
12 years ago