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\begin{document}
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 \section*{Compact surface from polygon}

 \begin{thm}
 $M$ is a connected closed(=compact and without boundary) surface.
 Then $M$ is homeomorphic to a finite polygon with edge identified
 in pairs.
 \end{thm}
 \begin{proof1}
 \\$1^{st} Step$ : We can number triangles as $T_1, T_2,\cdots,T_n$ so
 that each $T_i$ has an edge $e_i$ in common with at least one of
 the triangles $T_1,\cdots,T_{i-1}$.\\
 $\because$ Choose inductively. Suppose $A=\{T_1,\cdots,T_k\}$ are
 chosen and no triangles of $B=\{T_{k+1},\cdots,T_n\}$ has an edge
 in common with a triangles in $A$. Then $T_1\bigcup\cdots\bigcup
 T_k$ and $T_{k+1}\bigcup\cdots\bigcup T_n$ are disjoint closed sets
 in $M$. This is contradiction to connectedness of $M$.\\
 $2^{nd} Step$ : We can arranged $T_1,\cdots,T_n$ and $e_2,\cdots,e_n$($e_i$ is
 $T_i$'s edge for all $i=1,\cdots,n$) in the order as in the $1^{st}$ Step. Then we may
 assume that $\phi_i(T_i)=T_i'$ are all disjoint in $\mathbb{R}^2$
 for some homeomorphism $\phi_i$.\\
 \textbf{Claim} $P=\coprod_{i=1}^{n} T_i'/\sim$ is topologically a disk, where
 $\sim$ is defined by $\phi_i(p)\sim\phi_{i-1}(p)$,
 $\forall p\in e_i<T_i$, $i=2,\cdot,n$. This claim is clearly true by induction.\\
 Now $\hspace*{3em} P=\coprod_{i=1}^{n} T_i'/\sim$   $\overset{\psi=\bigcup\phi_i^{-1}}{\longrightarrow}$  $M$.\\
 $\hspace*{8em} \searrow \hspace*{3em} \nearrow \overline{\psi}$\\
 $\hspace*{10em} P/\sim$\\
 Here another '$\sim$' defined on $P$ is a side pairing. Since $P$ is compact, $P/\sim$ is also
 compact. Therefore $\overline{\psi}$ is a continuous bijective map from
 compact space to Hausdorff space. Hence $\overline{\psi}$ is a
 homeomorphism.
 \end{proof1}


 \begin{thm}
 A closed surface $M$ is homeomorphic to a surface obtained from $S^2$ from which
 finite number of discs are removed and for which either a
 "handle"(=torus with a hole) or a "cross-cap"(=M\"{o}bius band) is
 attached along the boundary of the disc, i.e., $M$ is a connected
 sum of tori and projective planes.
 \end{thm}
 \begin{proof1}
 \\\textbf{Step 1} By theorem 1, there is a polygon with edges identified in pairs homeomorphic to $M$.
 We can make sequence of edges like $\cdots abca\cdots$. The rule of naming edges is that edges that have same name are
 identified and that if clockwise edge's name is '$a$', counterclockwise edge's name is '$a^{-1}$'. Edges like '$aa^{-1}$'
 can be deleted.\\

 \textbf{Step 2} All the vertices of the polygon D may be assumed to be identified
 to a single point in $M$.\\
 $\because$ Suppose that $A\neq B$ in $M$.\\
 %\parpic{\includegraphics[width=0.45\textwidth]{figure1.eps}}\\
 \\\\ �׸� \\\\\\\\
 $[A]_D =\{$vertices in D that are identified to A$\}$\\
 $\Rightarrow\sharp[A]_{D'} =\sharp[A]_D -1$ and $\sharp[B]_{D'} =\sharp[B]_D +1$\\
 Therefore, by induction, we can make $\sharp[A]=0$ in a certain modified polygon
 $D$.\\
 ($A$�� �Ѱ� ������ ���� Step1���� ���ư���.)\\

 \textbf{Step 3} For the type of edges of the form $XaYa$, we can
 cancel $a$ and introduce $dd$ by cutting and pasting, i.e., make of
 the form $XY^{-1}dd$. \\
 %\parpic{\includegraphics[width=0.45\textwidth]{figure2.eps}}\\
  \\\\ �׸� \\\\\\

 \textbf{Step 4} Suppose we have $\cdots a\cdots a^{-1}\cdots$. Then there exists $b$ such that $\cdots a\cdots b\cdots a^{-1}\cdots b^{-1}\cdots$.
 There is no edges of the form $\cdots a\cdots b\cdots a^{-1}\cdots b\cdots$ by step 3. Suppose not.\\
 %\parpic{\includegraphics[width=0.45\textwidth]{figure5.eps}}\\
  \\\\ �׸� \\\\\\
  Therefore we must identify edges in $A$ with edges in $A$. Then we can not identify all the vertices to a single point. This is a contradiction to step 2.\\

 \textbf{Step 5} We can change $\cdots a\cdots b\cdots a^{-1}\cdots b^{-1}\cdots$ to $\cdots cdc^{-1}d^{-1}\cdots$ by the following sequence of cut and paste.\\
 %\parpic{\includegraphics[width=0.9\textwidth]{figure3.eps}}\\
  \\\\ �׸� \\\\\\
  Note that through step 5, $\cdots aa\cdots$ or $\cdots
 aba^{-1}b^{-1}\cdots$ is not destroyed.\\

 \textbf{Conclusion} We have only successive edges of the form
 $\cdots aa\cdots$ or $\cdots bcb^{-1}c^{-1}\cdots$ in the clockwise
 sequencing.
 \end{proof1}

 HW. Identify $abacb^{-1}c$.

 \begin{note}
 $M$, $N$ : connected closed surface
 \begin{enumerate}
 \item $M\sharp N$ is well-defined.\\
 $\because$ Let $x,y\in M$ with small open disc neighborhood $U$ and $V$ respectively.
 Then it can be shown that there exists a homeomorphism
 $\phi$ on $M$ such that $\phi(x)=y$ with $\phi(U)=V$. Then $M\sharp N$ obtained by
 deleting $U$ and $M\sharp N$ obtained by deleting $V$ are homeomorphic. (why?)
 
 ��� �� homeomorphism $\phi$ �� �����Ѵٴ� ���� �����ϱ� ���ؼ��� ���� ��ǵ��� �ʿ��ϴ�.
 \begin{itemize}
 \item[(1)] (Sch\"onflies Thm): Let $C$ be a simple closed curve in $\mathbb{R}^2$. Then $\exists$ a homeomorphism $h:\mathbb{R}^2\longrightarrow\mathhbb{R}^2$ such that $h$ carries $C$ onto the unit circle. (Ref: Moise, ``Geometric topology in dim 2 and 3'')
 \item[(2)] (Annulus Thm): Let $A$ be an annulus region in $\mathbb{R}^2$ bounded by two simple closed curve. Then $A$ is homeomorphic to the standard annulus:
     $$
     \{x=(x_1,x_2)\in\mathbb{R}^2\,|\,1\leq||x||\leq 2\}.
     $$
 \end{itemize}
 OHW. 
  \begin{itemize}
 \item[(1)] Prove Annulus Thm from Sch\"onflies Thm.
 \item[(2)] Improve Annulus Thm so that the homeomorphism can be chosen as an ambient homeomorphism on \mathbb{R}^2.
 \item[(3)] Improve Sch\"onflies Thm so that the homeomorphism $h$ can be chosen identity outside a big disk.
 
 \end{itemize}

 OHW. connected sum�� �����ϰ� �����ȴ�. �� ����� ���� �����ϰ� �����Ϸ��� ������ ����� �� �̿ܿ� � �͵��� ������ �ϳ�? \\
 (1) connected sum�� gluing homeomorphism�� isotopy class ���� depend�Ѵ�.\\
 (2) gluing homeomorphism�� orientation�� �����ϴ�.

 \item $M\sharp N=N\sharp M$
 \item $(M\sharp N)\sharp L=M\sharp(N\sharp L)$
 \end{enumerate}
 \end{note}

 \begin{thm}
 $M$ is a closed surface and
 $M'$ is a closed surface obtained from $M$ by replacing 3 cross-caps
 with a handle and a cross-cap.
 Then, $M\cong M'$, i.e., $P^2\sharp P^2\sharp P^2\cong T^2\sharp
 P^2$.
 \end{thm}

 \begin{proof1} Recall that $P^2\sharp P^2=K$.\\
 Therefore, if we can show $K\sharp P^2=T^2\sharp P^2$, we can prove this theorem. \\
 $K\sharp P^2=T^2\sharp P^2$ is true by following figure.\\
 %\parpic{\includegraphics[width=0.9\textwidth]{figure4.eps}}
 \\\\\\\\�׸�\\\\\\\\\\\\\\
 \end{proof1}

 \begin{cor}
 If $M$ has a cross-cap, we can replace all the handle by cross-caps,
 i.e., $M=P^2\sharp M^1\Rightarrow M=P^2\sharp\ldots\sharp P^2$. In
 this case, one handle is replaced by two cross-caps.
 \end{cor}

 \begin{cor}
 $M$ : closed surface \\
 $\Rightarrow$ $M=$
 \begin{cases}
 $T^2\sharp\ldots\sharp T^2=\sharp^g T^2 &\text{$(=\Sigma_g)$}\\
 $P^2\sharp\ldots\sharp P^2=\sharp^k P^2 &\text{$(=N_k)$}
 \end{cases}
 \end{cor}

\end{document}
