CC: Made it better looking

thesis/2_ChainComplexes.tex

@@ -170,21 +170,21 @@ which assigns to a space $X$ its \emph{singular $n$-th homology group} $H^\text{
 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:
+	Let $\ast$ be the one-point space, its homology is given by:
 	$$ H^\text{sing}_n(\ast) \iso
 	\begin{cases}
 		\Z \text{ if } n = 0 \\
 		0 \text { otherwise}
 	\end{cases}. $$
-	\item Let $S^k$ denote the $k$-sphere (for example $S^1$ is the circle). Its homology, for $n \in \Np$ is:
+\end{example}
+\begin{example}
+	Let $S^k$ denote the $k$-sphere (for example $S^1$ is the circle). Its homology, for $n \in \Np$ is:
 	$$ H^\text{sing}_n(S^k) \iso
 	\begin{cases}
 		\Z \text{ if } n = 0 \text { or } n = k \\
 		0 \text { otherwise}
 	\end{cases}. $$
 	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.
-\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 $n=m$.
 \end{example}
+
+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 $n=m$.