\section{Chain Complexes}
\label{sec:Chain Complexes}
\begin{definition}
	A chain complex $C$ is a collection of abelian groups $C_n$ together with boundary operators $\del_n: C_{n+1} \to C_n$, such that $\del_n \circ \del_{n+1} = 0$. The collections of all such objects will be denoted by $\Ch{\cat{Ab}}$.
\end{definition}

In other words a chain complex is the following diagram.
$$ \cdots \to C_4 \to C_3 \to C_2 \to C_1 \to C_0 $$

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.

\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 $B_n(C) = \im(\del: C_{n+1} \to C_n) \nsubgrp C_n$.
	\end{itemize}
\end{definition}
\begin{lemma}
	Given a chain complex $C$ we have for all $n \in \N$:
	$$ B_n(C) \nsubgrp Z_n(C).$$
\end{lemma}
\begin{proof}
	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}
\begin{definition}
	Given a chain complex $C$ we define the \emph{$n$-th homology group} $H_n(C)$:
	$$ H_n(C) = Z_n(C) / B_n(C).$$
\end{definition}

\subsection{The singular chain complex}
In order to see why we are interested in the construction of homology groups, we will look at an example from algebraic topology. We will see that homology gives a nice invariant for spaces. So we will form a chain complex from a topological space $X$. In order to do so, we first need some more notions.
\begin{definition}
	The topological space $\Delta^n$ is called the \emph{topological $n$-simplex} and is defined as:
	$$ \Delta^n = \{x \in \R^{n+1} \I x_i \geq 0 \text{ and } x_0 + \ldots + x_n = 1 \}.$$
	The topology on $\Delta^n$ is the subspace topology.
\end{definition}

In particular $\Delta^0$ is simply a point, $\Delta^1$ a line and $\Delta^2$ a triangle. There are nice inclusions $\Delta^n \mono \Delta^{n+1}$ which we need later on. For any $n \in \N$ we define:
\begin{definition}
	For $i \in \{0, \ldots, n+1\}$ the $i$-th face map $\delta^i : \Delta^n \mono \Delta^{n+1}$ is defined as:
	$$ \delta^i (x_0, \ldots, x_n) = (x_0, \ldots, x_{i}, 0, x_{i+1}, \ldots, x_n) \text{ for all } x \in \Delta^n.$$
\end{definition}

For any space $X$, we will be interested in continuous maps $\sigma : \Delta^n \to X$, such a map is called a $n$-simplex. Note that if we have any continuous map $\sigma : \Delta^{n+1} \to X$ we can precompose with a face map to get $\sigma \circ \delta^i : \Delta^n \to X$, as shown in figure~\ref{fig:diagram_d}. This will be used for defining the boundary operator. We can make pictures of this, and when concerning continuous maps $\sigma : \Delta^{n+1} \to X$ we will draw the images in the space $X$, instead of functions.

\begin{figure}
	\begin{tikzpicture}
		\matrix (m) [matrix of math nodes]{
			\Delta^{n+1} & X \\
			\Delta^n & \\
		};
		\path[->]
		(m-1-1) edge node[auto] {$ \sigma $} (m-1-2)
		(m-2-1) edge node[auto] {$ \delta^i $} (m-1-1)
		(m-2-1) edge node[auto] {$ $} (m-1-2);
	\end{tikzpicture}
	\caption{The $n$-simplex $\sigma$ can be made into a $n+1$-simplex $\sigma \circ \delta^i$}
	\label{fig:diagram_d}
\end{figure}

\todo{Ch: Make some pictures here}

We now have enough tools to define the singular chain complex of a space $X$.

\begin{definition}
	For a topological space $X$ we define an abelian group $C_n(X)$ as follows.
	$$ C_n(X) = \Z[\Hom{\cat{Top}}{\Delta^n}{X}] $$
	The boundary operator $\del : C_{n+1}(X) \to C_n(X)$ is defined on generators as:
	$$ \del(\sigma) = \sigma \circ \delta^0 - \sigma \circ \delta^1 + \ldots + (-1)^{n+1} \sigma \circ \delta^{n+1}.$$
\end{definition}

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}
	\label{fig:singular_chaincomplex3}
\end{figure}

\todo{Ch: Proposition: $C(X) \in \Ch{\cat{Ab}}$}
\todo{Ch: Example homology of some space}
\todo{Ch: Show that $\Ch{\Ab}$ is an ab. cat. At least show functoriality $\Hom{\Ch{\Ab}}{-}{-}$}