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\begin{document}



\begin{thm} Any polygonal disk with edges identified in pairs is
homeomorphic to a compact, connected,
triangulated
$2$-manifold.
\end{thm}

\begin{thm} Any compact, connected, triangulated $2$-manifold is
homeomorphic to a polygonal disk with
edges identified in pairs.

\end{thm}

\begin{exercise}  The boundary of a tetrahedron is naturally
triangulated with a
triangulation $T$ consisting of four
$2$-simplexes together with their six edges and four vertices. On the
boundary of a tetrahedron
locate the first and second derived subdivisions of $T$, the
$1$-skeleton of $T$, the regular
neighborhood of the $1$-skeleton of $T$, the regular neighborhoods of
a vertex and an edge of
$T$, and the dual $1$-skeleton of $T$.
\end{exercise}

\begin{exercise}
On the accompanying pictures of the second derived subdivisions of triangulations of the torus and the Klein bottle,
find regular neighborhoods of subsets of the $1$-skeleton.
\end{exercise}

\begin{exercise}
Characterize graphs in the $1$-skeleton of $T$ for the triangulations
of the sphere, torus, and projective plane whose regular neighborhoods are
homeomorphic to a disk.
\end{exercise}

\begin{thm}  Let $M^2$ be a compact, triangulated
$2$-manifold with triangulation $T$. Let $S$ be a tree whose edges
are $1$-simplices in the
$1$-skeleton of $T$. Then $N(S)$, the  regular neighborhood of $S$,
is homeomorphic to
$\D^2$.
\end{thm}

\begin{thm}  Let $M^2$ be a compact, triangulated
$2$-manifold with triangulation $T$. Let $S$ be a tree equal to a
union of edges in the dual
$1$-skeleton of $T$. Then $\cup \{ \sigma''_j
\mid \sigma''_j \in T''$ and $\sigma''_j \cap S\ne \emptyset\}$ is
homeomorphic to $\D^2$.
\end{thm}

\begin{thm}  Let $M^2$ be a connected, compact, triangulated
$2$-manifold with triangulation $T$. Let $S$ be a tree in the
$1$-skeleton of $T$. Let $S'$ be the subgraph of the dual
$1$-skeleton of $T$ whose edges do not intersect $S$. Then $S'$ is connected.
\end{thm}

The following two theorems state that
$M^2$ can be divided into two pieces, one
a disk $D_0$, and the other a disk ($D_1$) with
bands (the $H_i$'s) attached to it.

\begin{thm}  Let $M^2$ be a connected, compact, triangulated
$2$-manifold. Then $M^2 = D_0 \cup D_1 \cup \bigl( \bigcup_{i=1}^k
H_i \bigr)$ where $D_0$, $D_1$,
and each $H_i$ is homeomorphic to $\D^2$,
$\hbox{\rm Int } D_0 \cap D_1 =\emptyset$, the $H_i$'s are disjoint,
$\bigcup_{i=1}^k \hbox{\rm Int }H_i \cap ( D_0 \cup D_1) =
\emptyset$, and for each $i$, $H_i \cap D_1$ equals
  $2$ disjoint arcs each
arc on the boundary of
each of $H_i$ and $D_1$.
\end{thm}

\begin{thm} Let $M^2$ be a connected, compact, triangulated
$2$-manifold. Then:
\be
\item There is a disk $D_0$ in $M^2$ such that $M^2 -
(\interior D_0)$ is
homeomorphic to the following subset of $\R^3$:  a disk
$D_1$ with a finite number of disjoint strips, $H_i$ for $i\in\{1,\ldots n\}$,
attached to boundary of
$D_1$ where each strip
has no twist or $1/2$ twist. (See  example below.)
\item
Furthermore, the boundary of the disk with strips,
$D_1 \cup \bigl( \bigcup_{i=1}^k
H_i \bigr)$, is connected.
\ee

\vspace*{1in}

\hfill untwisted pairs \hfill twisted strips \hspace*{\fill}

\end{thm}


\begin{exercise}
In the set-up in the previous theorem, any strip $H_i$ divides the boundary of $D_0$ into
two edges $e_i^1$ and $e_i^2$, where $H_i$ is \emph{not} attached.
Show that if a strip $H_j$ is attached to $D_0$ with
no twists, then there must be a strip $H_k$
that is attached to both $e_j^1$ \emph{and} $e_j^2$.
\end{exercise}

\begin{thm} Let $M^2$ be a connected, compact, triangulated
$2$-manifold. Then there is a disk $D_0$ in $M^2$ such that $M^2 -
\hbox{\rm Int } D_0$ is homeomorphic to a disk $D_1$ with strips
attached as follows: first
come a finite number of strips with $1/2$  twist each whose attaching
arcs are consecutive
along $\bd D_1$, next  come a finite number of pairs of untwisted
strips, each pair with
attaching arcs entwined as pictured with the four arcs from each pair
consecutive along $\bd
D_1$.
\end{thm}

\vspace*{1in}


\begin{thm} Let $X$ be a disk $D_0$ with one strip attached with a
$1/2$ twist with its
attaching arcs consecutive along $\bd D_0$ and one pair of untwisted
strips with attaching arcs
entwined as pictured with the four arcs consecutive along $\bd D_0$.
Let $Y$ be a disk $D_1$
with three strips with a $1/2$ twist each whose attaching arcs are consecutive
along $\bd D_1$. Then $X$ is homeomorphic to $Y$.

\vspace*{1in}

\hfill $X$ \hfill $Y$ \hspace*{\fill}

\end{thm}

\begin{thm}  Let $M^2$ be a connected, compact, triangulated
$2$-manifold. Then there is a disk $D_0$ in $M^2$ such that $M^2 -
\interior D_0$ is homeomorphic to one of the following:
\smallskip
\bi
\item[a)] a disk $D_1$,
\item[b)] a disk $D_1$ with $k$ $\frac{1}{2}$-twisted strips with
consecutive attaching arcs,
or
\item[c)] a disk $D_1$ with $k$ pairs of untwisted strips, each pair
in entwining position
with the four attaching arcs from each pair consecutive.
\ei
\end{thm}

\vspace*{1in}
\hfill entwining pair of strips \hspace*{\fill}


\begin{thm}[Classification of compact, connected $2$-manifolds]
\index{$2$-manifold!Classification Theorem}
Any connected, compact, triangulated $2$-manifold is
homeomorphic to the $2$-sphere
$\Sph^2$, a connected sum of tori, or a connected sum of projective planes.
\end{thm}


\end{document}