@ -109,9 +109,8 @@ Of course given two preadditive categories $\cat{C}$ and $\cat{D}$, not every fu
In other words the functor $F$ induces a group homomorphism: $F : \Hom{\cat{C}}{A}{B}\to\Hom{\cat{D}}{FA}{FB}$.
\end{definition}
\todo{CC: What to do with the example...}
\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 this section we will not be very precise, as it will only act as an motivation. However the intuition might be very useful later on, and so pictures are provided to give meaning to this construction.
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. We will form a chain complex from a topological space $X$. In this section we will not be very precise, as it will only act as a motivation. However the intuition might be very useful later on, and so pictures are provided to give meaning to this construction.
\begin{definition}
The topological space $\Delta^n$ is called the \emph{topological $n$-simplex} and is defined as:
@ -119,20 +118,22 @@ In order to see why we are interested in the construction of homology groups, we
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:
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. 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} for $n=2$. From the picture it is clear that the assignment $\sigma\mapsto\sigma\circ\delta^i$, gives one of the boundaries of $\sigma$. If we were able to add these different boundaries ($\sigma\circ\delta^i$, for every $i$), then we could assign to $\sigma$ its complete boundary at once. The free abelian group will enable us to do so. This gives the following definition.
Given a 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 a$(n+1)$-simplex $\sigma : \Delta^{n+1}\to X$ we can precompose with a face map to get a $n$-simplex $\sigma\circ\delta^i : \Delta^n \to X$, as shown in figure~\ref{fig:diagram_d} for $n=1$.
\begin{figure}
\begin{figure}[h!]
\includegraphics[scale=1.2]{singular_set}
\caption{The $2$-simplex $\sigma$ can be made into a $1$-simplex $\sigma\circ\delta^1$}
\label{fig:diagram_d}
\end{figure}
From the picture it is clear that the assignment $\sigma\mapsto\sigma\circ\delta^i$, gives one of the boundaries of $\sigma$. If we were able to add these different boundaries ($\sigma\circ\delta^i$, for every $i$), then we could assign to $\sigma$ its complete boundary at once. The free abelian group will enable us to do so. This gives the following definition.
\begin{definition}
For a topological space $X$ we define the \emph{$n$-th singular chain group}$C_n(X)$ as follows.
$$ C_n(X)=\Z[\Hom{\cat{Top}}{\Delta^n}{X}]$$
@ -140,7 +141,7 @@ For any space $X$, we will be interested in continuous maps $\sigma : \Delta^n \
This might seem a bit complicated, but we can picture this in an intuitive way, as in figure~\ref{fig:singular_chaincomplex}. 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.
This might seem a bit complicated, but we can picture this in an intuitive way, as in figure~\ref{fig:singular_chaincomplex}. 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.
\caption{The boundary of a 2-simplex, and a boundary of a 1-simple}
@ -148,24 +149,25 @@ This might seem a bit complicated, but we can picture this in an intuitive way,
\end{figure}
The above construction gives us a functor $C: \Top\to\Ch{\Ab}$ (we will not prove this). Composing with the functor $H_n: \Ch{\Ab}\to\Ab$, we get a functor:
$$ H^{sing}_n : \Top\to\Ab, $$
which assigns to a space $X$ its \emph{singular $n$-th homology group}$H^{sing}_n(X)$. A direct consequence of being a functor is that homeomorphic spaces have isomorphic singular homology groups. There is even a stronger statement which tells us that homotopic equivalent spaces have isomorphic homology groups. So from a homotopy perspective this construction is nice. In the remainder of this section we will give the homology groups of some basic spaces. It is hard to calculate these results from the definition above, so generally one gets these results by using theorem from algebraic topology. To calculate these examples one generally needs theorems from algebraic topology or homological algebra, which are beyond the scope of this thesis. So we simply give these results.
$$ H^\text{sing}_n : \Top\to\Ab, $$
which assigns to a space $X$ its \emph{singular $n$-th homology group}$H^\text{sing}_n(X)$\todo{CC: singular homology pictures (from presentation)}. A direct consequence of being a functor is that homeomorphic spaces have isomorphic singular homology groups. There is even a stronger statement which tells us that homotopic equivalent spaces have isomorphic homology groups. So from a homotopy perspective this construction is nice. In the remainder of this section we will give the homology groups of some basic spaces. It is hard to calculate these results from the definition above, so generally one proves these results by using theorems from algebraic topology or homological algebra, which are beyond the scope of this thesis. So we simply give these results.
\begin{example}
The following two examples show that the homology groups are reasonable.
\begin{itemize}
\item Let $\ast$ be the one-point space, its homology is given by:
$$ H^{sing}_n(\ast)\iso
$$ H^\text{sing}_n(\ast)\iso
\begin{cases}
\Z\text{ if } n = 0 \\
0 \text{ otherwise}
\end{cases}. $$
\item Let $S^k$ denote the $k$-sphere (for example $S^1$ is the circle). Its homology, for $n \in\Np$ is:
$$ H^{sing}_n(S^k)\iso
$$ H^\text{sing}_n(S^k)\iso
\begin{cases}
\Z\text{ if } n = 0 \text{ or } n = k \\
0 \text{ otherwise}
\end{cases}. $$
For $S^0$ (which consists of only two points) the first homology group is isomorphic to $\Z\oplus\Z$, and all other homology groups are trivial.
\item Note that $\R^n -\{0\}$ is homotopic equivalent to $S^n$, so their homology groups are the same. As a consequence $\R^n -\{0\}$ has the same homology groups as $\R^m -\{0\}$, only if $n=m$. Now if $\R^n$ is homeomorphic to $\R^m$, then also $\R^n -\{0\}\iso\R^m -\{0\}$, so this only happens if $n=m$.
\end{itemize}
We can use the latter example to proof a fact about $\R^n$ quite easily. Note that $\R^n -\{0\}$ is homotopic equivalent to $S^n$, so their homology groups are the same. As a consequence $\R^n -\{0\}$ has the same homology groups as $\R^m -\{0\}$, only if $n=m$. Now if $\R^n$ is homeomorphic to $\R^m$, then also $\R^n -\{0\}\iso\R^m -\{0\}$, so this only happens if $n=m$.
Joshua Moerman
11 years ago