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\begin{document}

\section*{\bf Classification of Compact Surfaces}

\section*{ 0. Introduction }

 By the term `{\it surface}', we mean a 2-dimensional manifold. Here are some examples of surfaces :
\begin{figure}[h]
\begin{center}
\includegraphics{eg_2mfd}
\caption{{\bf Examples of 2-manifolds.}}
\label{fig:eg_2mfd}
\end{center}
\end{figure}

 {\bf Goal } : 
\begin{enumerate}
\item Classify compact surfaces assuming the existence of triangulation on the surfaces.
\item Show each compact surface can be obtained by identifying the boundary edges of a polygon.
\item Discuss a complete set of topological invariants to distinguish surfaces.
\end{enumerate}

\section*{1. Triangulation and Rado's theorem }
\begin{defn}
A {\bf triangulation} of a compact surface $M$ consists of a finite set of triangles $K = \{ T_1 , \ldots , T_n \} $ and homeomorphisms $\{ \phi_i : T_i \to T'_i \subset \Real^2 \} $, where $ T'_i = \phi_i ( T_i ) $ is a triangle in $\Real^2$ such that :
\begin{enumerate}
\item $ \cup_i T_i = M $
\item For $i \neq j$, $ T_i \cap T_j \neq \emptyset $ implies $T_i \cap T_j $ is either a single vertex or a single edge.
\end{enumerate}
\end{defn}

A vertex or an edge of $T_i$ is pre-image of a vertex or an edge of a triangle in $\Real^2$ by $\phi_i$

\begin{thm}{(Rado's theorem, 1925)}
There exists a triangulation on a compact surface.
\end{thm}

\begin{cl}
\label{cl1}
Each edge is an edge of exactly one or two triangles.
\end{cl}
\begin{cl}
\label{cl2}
Let $v$ be a vertex of a triangulation, which is not a boundary point of $M$. Then we can arrange triangles $T_1 , \ldots, T_n$ around $v$ such that $T_i$ and $T_i+1$ for $i=1,\ldots,n-1$, and $T_n$ and $T_1$ have an edge in common cyclically.
\end{cl}

\begin{proof}{ \ref{cl1}}
 Suppose there is an edge which is an edge of three or more triangles. Take two among those triangles. There is a coordinate chart $\psi$ defined on a neighborhood $U$ of $p$ on the edge.

 The space $S$ of two sheets of triangles is a topological disk. Consider a set $V \subset U \cap S $, which is relatively open in $S$, containing $p$, and homeomorphic to $\Real^2$. Let $\psi' : V \to \Real^2$ be the homeomorphism. Then $\psi'(V) \subset \Real^2$ is open. Applying invariance of domain theorem to $\psi \circ \psi'^{-1} : \Real^2 \to \Real^2$, we get $\psi( V ) \subset \Real^2 $ is open.
 
 Take a sequence $\{ x_n \} \subset U - S $ converging to $p$. Continuity of $\psi$ implies that $ \{ \psi(x_n) \} $ converges to $\psi(p)$. Since $\psi(V)$ is a open neighborhood of $\psi(p)$, $\psi(V)$ must contain $\psi(x_i)$, for some $i$, which is contradictory to the injectiveness of $\psi$.

\end{proof}


\begin{proof}{ \ref{cl2}}
 This follows from the claim \ref{cl1}, and the definition of triangulation.

\end{proof}
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