\chapter{Introduction}

Abstracting and generalizing essential features of familiar objects
often lead to the
development of important mathematical ideas.  One goal of geometrical
analysis is
to describe the relationships and features that make up the essential
qualities of what we
perceive as our physical world. The strategy is to find ideas that we
view as central and then
to generalize those ideas and to explore those more abstract
extensions of what we perceive
directly.

Much of topology is aimed at exploring abstract versions of
geometrical objects in our world.
The concept of geometrical abstraction dates back at least to the
time of Euclid ({\emph{c}}. 225 B.C.E.)
The most famous and basic spaces are named for him, the Euclidean
spaces. All of the objects
that we will study in this course will be subsets of the Euclidean spaces.

\section{Basic Examples}

\begin{dfn}[\,$\R^n$]
\index{$\R^n$}%
\index{Real $n$-space}%
\index{Euclidean $n$-space}%
We define \emph{real} or \emph{Euclidean $n$-space}, denoted by
$\R^n$, as the set
\[
\R^n:=\{(x_1, x_2,\ldots,x_n)| x_i\in\R
{\textrm{\ for\ }} i=1,\ldots,n\}.
\]
\end{dfn}

We begin by looking at some basic subspaces of $\R^n$.

\begin{dfn}[standard $n$-disk]
\index{$\D^n$}%
\index{$n$-disk}%
\index{disk!$n$-dimensional}%
The \emph{\,$n$-dimensional disk}, denoted $\D^n$ is defined as
\begin{eqnarray*}
\D^n & := & \{(x_1,\ldots,x_n)\in\R^n| 0\leq x_i \leq 1{\textrm{\
for\ }} i=1,\ldots,n\ \} \\
    & \cong & \overbrace{[0,1]\times [0,1]\times\dots \times [0,1]}^{n
\textrm{\ times\
}}\subset\R^n.
\end{eqnarray*}
\end{dfn}

For example, $\D^1=[0,1]$. $\D^1$ is also
called the unit interval, sometimes denoted by $I$.

\begin{dfn}[standard $n$-ball, standard $n$-cell]
\index{$\B^n$}%
\index{$n$-ball}%
\index{$n$-cell}%
\index{ball!$n$-dimensional}%
The \emph{$n$-dimensional ball} or \emph{cell}, denoted $\B^n$, is
defined as:
\[
\B^n:=\{(x_1,\ldots,x_n)\in\R^n| x_1^2+\ldots+x_n^2\leq 1 \}.
\]
\end{dfn}


\begin{fact} The standard $n$-ball
and the standard $n$-disk are compact and
homeomorphic.
\end{fact}

\begin{dfn}[standard $n$-sphere]
\index{$\Sph^n$}%
\index{$n$-sphere}%
The \emph{$n$-dimensional sphere}, denoted $\Sph^n$, is defined as
\[
\Sph^n :=\{(x_0,\ldots,x_n)\in\R^{n+1}| x_0^2+\ldots+x_n^2 = 1 \}.
\]
\end{dfn}

\begin{note}
$\bd \B^{n+1}=\Sph^n$
\end{note}

As usual, the term $n$-sphere will apply to any space homeomorphic to
the standard $n$-sphere.

\begin{question} Describe $\Sph^0$, $\Sph^1$, and $\Sph^2$. Are they homeomorphic?
If not, are there any
properties that would help you distinguish between them?
\end{question}





\section{Simplices}

One class of spaces in $\R^n$ we will be studying will
be \emph{manifolds} or \emph{$k$-manifolds},
which are made up of pieces that locally look like
$\R^k$, put together in a ``nice'' way. In particular, we
will be studying manifolds that use triangles (or their higher-dimensional
equivalents) as the basic building blocks.

Since $k$-dimensional ``triangles'' in $\R^n$ (called \emph{simplices})
are the basic building blocks
we will be using, we begin by giving a vector description of them.

\begin{dfn}[$1$-simplex]  Let $v_0$, $v_1$ be two points in $\R^n$. If we
consider $v_0$ and $v_1$ as
vectors from the origin, then
$\sigma^1 = \{ \mu v_1 + (1-\mu )v_0 \mid 0\le \mu\le 1\}$ is the
straight line segment
between $v_0$ and $v_1$. $\sigma^1$ can be denoted by
$\{v_0 v_1\}$ or $\{v_1 v_0\}$ (the order the vertices
are listed in doesn't matter).
The set $\sigma^1$ is called a \emph{    $1$-simplex\/} or \emph{edge\/} %
\index{simplex!$1$}\index{edge}\index{$1$-simplex}%
    with vertices
    (or $0$-simplices) \index{$0$-simplex}\index{vertex}\index{simplex!$0$}%
    $v_0$ and $v_1$.
\end{dfn}


\begin{dfn}[$2$-simplex]
\label{thm:sigma2}
Let $v_0$, $v_1$, and $v_2$ be three non-collinear points in
$\R^n$. Then
\[
\sigma^2 = \left\{ \lambda_0 v_0+ \lambda_1 v_1 +
\lambda_2 v_2 \mid \lambda_0 + \lambda_1 +\lambda_2 = 1{\textrm{\ and\ }}0\le
\lambda_i \le 1 \forall i=0,1,2\right\}
\]  is a triangle with edges
$\{v_0 v_1\}$, $\{v_1 v_2\}$, $\{v_0 v_2\}$ and vertices $v_0$,
$v_1$, and $v_2$. The set $\sigma^2$ is a  \emph{$2$-simplex}
    \index{simplex!$2$}\index{$2$-simplex}%
with vertices $v_0$, $v_1$, and $v_2$ and edges $\{v_0 v_1\}$,
$\{v_1 v_2\}$, and $\{v_0 v_2\}$.
$\{v_0 v_2 v_2\}$ denotes the $2$-simplex $\sigma^2$ (where the order the vertices
are listed in doesn't matter).
\end{dfn}

Note that the plural of simplex is \emph{simplices}.


\begin{dfn}[$n$-simplex and face of a simplex]
\index{$n$-simplex} \index{simplex!$n$-dimensional}
\index{simplex!face of} \index{face of a simplex} Let $\{v_0,
v_2,\ldots, v_n\}$ be a set affine independent points in $\R^N$.
Then an $n$-\emph{simplex} $\sigma^n$ (of dimension $n$), denoted
$\{v_0v_1v_2\ldots v_n\}$,  is defined to be the following subset
of  $\R^N$:
\[
\sigma^n = \left\{ \lambda_0 v_0+ \lambda_1 v_1 + ... +
\lambda_n v_n \left| \sum_{i=0}^n \lambda_i = 1\right.;{\textrm{\ }}0\le
\lambda_i \le 1, \ i=0,1,2,\ldots, n\right\}.
\]
An $i$-simplex whose vertices are any subset of $i+1$ of the
vertices of $\sigma^n$ is an ($i$-dimensional) \emph{face} of
$\sigma^n$. The face obtained by deleting the $v_m$ vertex from
the list of vertices of $\sigma^n$ is often denoted by
$\{v_0v_1v_2\ldots\widehat{v_m}\ldots v_n\}$. (Note that it is an
$(n-1)$-simplex.)
\end{dfn}

\begin{exercise}
Show that the faces of a simplex are indeed simplices.
\end{exercise}

\begin{fact} The standard $n$-ball, standard $n$-disk and the
standard $n$-simplex are compact and
homeomorphic.
\end{fact}

We will use the terms $n$-disk, $n$-cell, $n$-ball interchangeably to
refer to any topological
space homeomorphic to the standard $n$-ball.
%We will usually reserve the use
%$n$-simplex in the context of bigger spaces made up of simplices.

\section{Simplicial Complexes}
%Before studying $2$-manifolds (or $n$-manifolds) in particular, we
%will define an object that is more general than that. You should try to
%construct a few examples that are not manifolds, as well as some that are.

Simplices can be assembled to create polyhedral subsets of $\R^n$ known
as complexes. These simplicial complexes are the principal
objects of study for this course.

\begin{dfn}[finite simplicial complex]
\index{finite simplicial complex}
\index{simplicial complex, finite}
\index{complex!finite simplicial}
Let $T$ be a finite collection of simplices in  $\R^n$ such
that  for every simplex $\sigma_i^{j}$ in $T$,
each face of  $\sigma_i^{j}$ is also a simplex in $T$ and
any two simplices in $T$ are either disjoint or
their intersection is a face of each. Then the subset $K$ of $\R^n$ defined by
$K=\bigcup\sigma_i^{j}$
running over all simplices $\sigma_i^{j}$ in $T$
is a \emph{finite simplicial complex} with \emph{triangulation} $T$,
\index{triangulation}%
 denoted $(K,T)$.
The set $K$ is often called the \emph{underlying space} of the simplicial complex.
If $n$ is the maximum dimension of all simplices in $T$, then
we say $(K,T)$ is of dimension $n$.
\end{dfn}
%\marginpar{\tiny{Do we define dim. of a simplicial complex?--C}}

\begin{example}
Consider $(K, T)$ to be the simplicial complex in the plane where
\begin{eqnarray*}
T & =  \left\{\right. &\{(0, 0)(0, 1)(1, 0)\}, \{(0, 0)(0, -1)\}, \{(0, -1)(1, 0)\}, \\
  &    &    \{(0, 0)(0, 1)\}, \{(0, 1)(1, 0)\}, \{(1, 0)(0, 0)\}, \\
  &    &   \left.\{(0, 0)\} , \{(0, 1)\}, \{(1, 0)\}, \{(0, -1)\}\right\}.
\end{eqnarray*}
So $K$ is a filled in triangle and a hollow triangle as pictured.
\end{example}

\begin{figure}[h]
\begin{center}
\includegraphics[height = 3cm]{ImagesChapter1/pdf/triangles}
\end{center}
\end{figure}


\begin{exercise}
Draw a space made of triangles that is \emph{not} a simplicial complex,
and explain why it is not a simplicial complex.
\end{exercise}


We have started by making spaces using simplices as building blocks. But what
if we have a space, and we want to break it up into simplices?
%This would
%mean that the space is homeomorphic to one that already has a triangulation:
If $J$ is a topological space homeomorphic to $K$
where $K$ is a the underlying space of a simplicial complex $(K,T)$ in $\R^m$,
then we say that $J$ is \emph{triangulable}.\index{triangulable space}

\begin{exercise}
Show that the following space is triangulable:

\begin{figure}[h]
\begin{center}
\includegraphics[height = 3cm]{ImagesChapter1/pdf/circledisk}
\end{center}
\end{figure}

by giving a triangulation of the space.
\end{exercise}


\begin{dfn}[subdivision]
\index{subdivision!of a finite simplicial complex}
Let $(K, T)$ be a finite simplicial complex.
Then $T'$ is a subdivision of $T$ if $(K, T')$ is a
finite simplicial complex,
 and each simplex in $T'$ is a subset of a simplex in T.
\end{dfn}


\begin{example}
The following picture illustrates a
finite simplicial complex and a subdivision of it.
%\marginpar{\tiny{Same example as above with each simplex
%subdivided.}}

\begin{figure}[h]
\begin{center}
\includegraphics[height = 3cm]{ImagesChapter1/pdf/subdivision}
\end{center}
\end{figure}

\end{example}

There is a standard subdivision of a triangulation that
later will  be useful:

\begin{dfn}[derived subdivision]
\index{subdivision!barycentric}%
\index{barycentric subdivision}%
\index{subdivision!derived}
{\rule{0cm}{0.1cm}}\newline\vspace*{-0.5cm} \be \item Let
$\sigma^2$ be a $2$-simplex  with vertices $v_0,v_1$, and $v_2$.
Then $p=\frac{1}{3} v_0 + \frac{1}{3} v_1 + \frac{1}{3} v_2$ is
the \emph{barycenter\/}
\index{barycenter}%
of $\sigma^2$.
\item Let $T$ be a triangulation of a
simplicial $2$-complex with $2$-simplices $\{ \sigma_i\}_{i=1}^k$. The
\emph{first derived  subdivision\/}
\index{subdivision!derived}%
\index{first derived subdivision}%
  of $T$, denoted $T'$, is the union of all vertices of $T$
  with the collection of $2$-simplices obtained
from $T$ by breaking each
$\sigma_i$ in $T$ into six pieces as shown, together with their edges
and vertices, and finally the edges and vertices obtained by breaking each edge that is not
a face of a $2$-simplex into
two edges. Notice that
the new vertices are the barycenter of each
$\sigma_i$ in $T$ and  the center of each edge in $T$. The
\emph{second derived
subdivision\/},
\index{subdivision!derived}%
\index{second derived subdivision}%
denoted $T''$, is $(T')'$, the first derived subdivision of $T'$, and so on.(See Figure \ref{barycentric})
\ee
\end{dfn}
%\marginpar{\tiny{Do general barycentric subdivision for $n$-simplices?}}

\begin{figure}[h]
\begin{center}
\includegraphics[height = 3cm]{ImagesChapter1/pdf/barycentric}
\caption{\label{barycentric}Barycentric subdivision of a $2$-simplex}
\end{center}
\end{figure}




\begin{example}
Figure \ref{secondbarycentric} illustrates a
finite simplicial complex and the second
derived subdivision of it.
%\marginpar{\tiny{(Same example as above with each simplex
%barycentrically subdivided, that is, the $2$-simplex divided
%into $6$ $2$-simplexes and the edges divided in half.)}}
\begin{figure}[h]
\begin{center}
\includegraphics[height = 3cm]{ImagesChapter1/pdf/secondbarycentric}
\end{center}
\caption {\label{secondbarycentric}Second barycentric subdivision of a $2$-simplex}
\end{figure}

\end{example}

\section[$2$-manifolds]{\mathversion{bold}$2$-manifolds\mathversion{normal}}

The concept of the real line and the Euclidean spaces produced
from the real line are fundamental to a large part of mathematics.
So it is natural to be particularly interested in topological
spaces that share features with the Euclidean spaces. Perhaps the
most studied spaces considered in topology are those that look
locally like the Euclidean spaces. The most familiar such space is
the $2$-sphere since it is modelled by the surface of Earth,
particularly in flat places like Kansas or the middle of the
ocean. If you are on a ship in the middle of the Pacific Ocean,
the surrounding terrain looks like the surrounding terrain if you
were living on a plane, which is Euclidean $2$-space or $\R^2$.
The concept of a space being locally homeomorphic to $\R^2$ is
sufficiently important that it has a name, in fact, two names. A
space locally homeomorphic to $\R^2$ is called a \emph{surface} or
\emph{$2$-manifold}. The $2$-sphere is a surface as is the
\emph{torus} (which looks like an inner-tube or the surface of a
doughnut).


\begin{dfn}[$2$-manifold or surface]
\index{$2$-manifold}%
\index{surface}%
A \emph{$2$-manifold} or \emph{surface} is a separable, metric
space $\Sigma^2$  such that for each $p\in \Sigma^2$, there is a
neighborhood $U$ of $p$ that is homeomorphic to $\R^2$.
\end{dfn}

\subsection{$2$-manifolds as simplicial complexes}

For now, we will restrict ourselves to $2$-manifolds that
are subspaces
of $R^n$ and that are triangulated.

\begin{dfn}[triangulated $2$-manifold]
 A \emph{triangulated compact $2$-manifold}
\index{manifold!$2$-dimensional!triangulated}%
\index{$2$-manifold!triangulated}%
\index{triangulated $2$-manifold}%
is a  space homeomorphic to a subset $M^2$ of $\R^n$ such that $M^2$
is the underlying space of a simplicial complex $(M^2,T)$.
\end{dfn}

%\begin{dfn}[triangulation]
%\index{triangulation!$2$-manifold}
%The set $T$ made up of $2$-simplexes $\{ \sigma_i\}_{i=1}^k$ and
%all their edges and vertices
%above is called a  \emph{triangulation\/}
%    \index{triangulation!$2$-manifold}%
%of the $2$-manifold.
%\end{dfn}

\begin{example}
The tetrahedral surface below, with triangulation
\begin{eqnarray*}
T & = & \left\{ \{v_0 v_1 v_2 \}, \{v_0 v_1 v_3\}, \{v_0 v_2 v_3\},
              \{v_1 v_2 v_3\}, \right.\\
  &    & \{v_0 v_1\},\{v_0 v_2\},\{v_0 v_3\},\{v_1 v_2\},\{v_1 v_3\},\{v_2 v_3\}, \\
  &     &\left. \{v_0\},\{v_1\},\{v_2\},\{v_3\}\right\}
\end{eqnarray*}
is a triangulated $2$-manifold (homeomorphic to $\Sph^2$).

%\newpage

\end{example}


\begin{figure}[h]
\begin{center}
\includegraphics[height = 3cm]{ImagesChapter1/pdf/tetrahedralsurface}
\end{center}
\caption{Tetrahedral surface}
\end{figure}

%\hfill Tetrahedral surface \hspace*{\fill}



The following theorem asserts that every compact $2$-manifold is triangulable, but its proof entails some
technicalities that would take
us too far afield.  So we will analyze triangulated $2$-manifolds and
simply note here without
proof that our results about triangulated $2$-manifolds actually hold
in the topological
category as well.

\begin{thm}  A compact, $2$-manifold is homeomorphic to a compact,
triangulated $2$-manifold, in other words, all compact $2$-manifolds
are triangulable.
\end{thm}


\begin{dfns}[$1$-skeleton and dual $1$-skeleton]{\rule{0cm}{0.1cm}}\newline
\vspace*{-0.5cm} \be \item The \emph{$1$-skeleton\/} of a
triangulation $T$ equals $\bigcup \{ \sigma_j\mid\sigma_j$ is a
$1$-simplex in $T\}$ and is  denoted $T^{(1)}$.
\index{triangulation!$1$-skeleton}%
\index{$1$-skeleton}%
%
\item The \emph{dual $1$-skeleton\/}
\index{triangulation!$1$-skeleton!dual}%
\index{$1$-skeleton!dual}%
\index{dual $1$-skeleton}%
of a triangulation $T$ equals
$\bigcup \{ \sigma_j \mid \sigma_j$ is an edge of a $2$-simplex in $T'$
and neither vertex of
$\sigma_j$ is a vertex of a $2$-simplex of
$T\}$. An edge in the dual $1$-skeleton has each of its ends at the
barycenters of
$2$-simplices of the original triangulation, that is, physically each
edge in the dual
$1$-skeleton is composed of two segments, each running from the
barycenter of a $2$-simplex to
the middle of the edge they share in the original triangulation.  So an edge in
the dual $1$-skeleton is the union of two $1$-simplices in $T'$.
\ee
\end{dfns}


\begin{examples} The following are triangulable $2$-manifolds:
\be
    \item[a.] $\Sph^2$
 
\begin{figure}[h]
\begin{center}
\includegraphics[height = 3cm]{ImagesChapter1/pdf/sphere}
\end{center}
\end{figure}

    \item[b.] $\T^2:=\Sph^1\times\Sph^1\subset\R^4$ or any other
space homeomorphic to the
boundary of a doughnut,
    \index{$\T^2$}%
    \index{torus}%
    the torus.

\begin{figure}[h]
\begin{center}
\includegraphics[height = 3cm]{ImagesChapter1/pdf/torus}
\end{center}
\end{figure}

    \item[c.] Double torus:(See Figure \ref{doubletorus})
    \index{double torus}%
    \index{torus!double}%
    
\begin{figure}[h]
\begin{center}
\includegraphics[height = 2.5cm]{ImagesChapter1/pdf/doubletorus}
\caption{\label{doubletorus}The double torus or surface of genus 2}
\end{center}
\end{figure}
\ee
The following example cannot be embedded in $\R^3$; however, it can be embedded in $R^4$.
\be
    \item[d.] The Klein bottle, denoted $\K^2$:(See Figure \ref{klein})
    \index{Klein bottle}%
    \index{$\K^2$}%

\begin{figure}[h]
\begin{center}
\includegraphics[height = 4cm]{ImagesChapter1/pdf/klein}
\caption{\label{klein}The Klein Bottle}
\end{center}
\end{figure}
\ee

There is another $2$-manifold  that  cannot be embedded
in $\R^3$ that we will study, which requires the
use of the quotient or identification topology (see Appendix~A):



\be
    \item[e.] The projective plane, denoted $\PP^2$,
    \index{projective plane!real}%
    \index{projective $2$-space!real}%
    \index{$\PP^2$}%
    $:=$ space of all lines through $\mathbf{0}$ in $\R^3$
    where the basis for the topology is the collection of open cones
    with the cone point at the origin.
\ee
\end{examples}

\begin{exercise} {\rule{0cm}{0.1cm}}\newline\vspace*{-0.5cm}
\be \item Show $\PP^2\cong\Sph^2/\langle x\sim -x\rangle$, that
is, the $2$-sphere with diametrically opposite points identified.
\item Show that $\PP^2$ is also homeomorphic to a disk with two
edges on its boundary (called a \emph{bigon}),\index{bigon}
identified as indicated in Figure \ref{rptwo}. 
\begin{figure}[h]
\begin{center}
\includegraphics[height = 2.3cm]{ImagesChapter1/pdf/rp2}
\caption{\label{rptwo} $\PP^2$}
\end{center}
\end{figure}


\item Show that
$\PP^2\cong$ M\"obius band with a disk  attached to its boundary (See Figure \ref{mobiusband}).

\begin{figure}[h]
\begin{center}
\includegraphics[height = 2.5cm]{ImagesChapter1/pdf/mobiusband}
\caption{\label{mobiusband} The M\"obius band}
\end{center}
\end{figure}

\ee
\end{exercise}

\begin{exercise} Show that $\T^2$ as defined above is homeomorphic to the
surface in $\R^3$
parametrized by:
\[
\left\{ (\theta, 1+\frac{1}{2} \cos\phi, \frac{1}{2}\sin\phi
\left| 0\leq\theta\leq 2\pi,  0\leq\phi\leq 2\pi \right. \right\}
\] in cylindrical coordinates.
\end{exercise}

There is a way of obtaining more $2$-manifolds by ``connecting'' two
or more together. For instance, the double torus looks like two tori
that have been joined together.
\begin{dfn}[connected sum]
\index{connected sum!surfaces}  Let $M_1^2$ and $M_2^2$ be two
compact, connected, triangulated 2-manifolds and let $D_1$ and
$D_2$ be $2$-simplices in the triangulations of $M_1$ and $M_2$
respectively. Paste $M_1^2 -\interior D_1$  and $M_2^2 -\interior
D_2$ along the boundaries of $\bd D_1$ and $\bd D_2$. The
resulting manifold is called the \emph{connected sum of $M_1^2$
and $M_2^2$}, and denoted by $M_1^2 \num M_2^2$. Similarly, define
the \emph{connected sum of $n$ $2$-manifolds} recursively.
\end{dfn}

This definition of connected sum can in fact be generalized to
the connected sum of any two $n$-manifolds. Can you see how to do it?

\begin{exercise}
Show that $\PP^2 \num \PP^2$ is homeomorphic to the Klein bottle.
\end{exercise}

\begin{exercise}
Show that $\T^2 \num \PP^2$, where $\T^2$ is the torus,
is homeomorphic $K^2\num \PP^2$, where $\K^2$ is the Klein bottle.
\end{exercise}

\subsection[$2$-manifolds as quotient spaces]{\mathversion{bold}$2$-manifolds as quotient spaces\mathversion{normal}}

There is another way of thinking of $2$-manifolds, as
the abstract spaces obtained from a particular kind
of quotients (see Appendix~A for a review of quotient
spaces).

The process of identifying all elements of an equivalence class to a
single one is often called
a \emph{gluing}%
\index{gluing} when the equivalence classes are mostly small, having
$1$ or $2$ or a finite number of
points in each.

In our case, we will be looking at the quotient spaces
obtained from polygonal disks, where all points of the
interior of the disk are in their own equivalence
class, the points on the interior of the edges are
in two-point equivalence classes, and the vertices
of the polygonal disks are in equivalence classes
with any number of other vertices. We think of
obtaining the $2$-manifold by gluing the edges
of the polygonal disk to each other pairwise, in some particular
pattern.

\begin{examples} In these examples the kind of arrow indicates which
edges are glued together, while the orientations of the arrows indicate
how to glue the two edges together. You should convince yourself
that any two gluing maps that
agree with the given orientations
will yield homeomorphic spaces.

\be
\item (torus)

\begin{figure}[h]
\begin{center}
\includegraphics[height = 3cm]{ImagesChapter1/pdf/torushomeo}
\end{center}
\end{figure}


\item (sphere) (See Figure \ref{spheresquare})


\begin{figure}[h]
\begin{center}
\includegraphics[height = 3cm]{ImagesChapter1/pdf/spheresquare}
\caption{\label{spheresquare} The sphere}
\end{center}
\end{figure}

\item (sphere) (See Figure \ref{spherewithgluing})

\begin{figure}[h]
\begin{center}
\includegraphics[height = 3cm]{ImagesChapter1/pdf/spherewithgluing}
\caption{\label{spherewithgluing} Another way to see the sphere}
\end{center}
\end{figure}

\item (double torus) (See Figure \ref{octogenustwo})

\begin{figure}[h]
\begin{center}
\includegraphics[height = 3cm]{ImagesChapter1/pdf/octogenustwo}
\caption{\label{octogenustwo} The double torus}
\end{center}
\end{figure}

\item (Klein bottle) (See Figure \ref{kleinsquare})

\begin{figure}[h]
\begin{center}
\includegraphics[height = 3cm]{ImagesChapter1/pdf/kleinsquare}
\caption{\label{kleinsquare} The Klein bottle}
\end{center}
\end{figure}

\item (projective plane) (See Figure \ref{projplane})

\begin{figure}[h]
\begin{center}
\includegraphics[height = 2.2cm]{ImagesChapter1/pdf/rp2}
\caption{\label{projplane} The projective plane}
\end{center}
\end{figure}

\item (projective plane) (See Figure \ref{rp2square})

\begin{figure}[h]
\begin{center}
\includegraphics[height = 3cm]{ImagesChapter1/pdf/rp2square}
\caption{\label{rp2square} Another version of the projective plane}
\end{center}
\end{figure}

\ee
You should check to see that alternative presentations of the same space
are homeomorphic. You should also check that these spaces are homoeomorphic to the
triangulable $2$-manifolds described in the previous subsection.
\end{examples}

The following theorem will be put off to chapter 4 (and stated in a slightly
different but equivalent way). Surprisingly, it is highly non-trivial to 
prove but not surprisingly it is incredibly useful. 

\begin{thm}[Jordan Curve Theorem]
Let $h: [0,1] \rightarrow \D^2$ be a topological embedding where
$h(0),h(1)\in \bd(\D^2)$. Then $h([0,1])$ separates $\D^2$ into exactly
two pieces.
\end{thm}

\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}


\section{Questions}

The most fundamental questions in topology are:

\begin{question}
How are spaces similar and different?  Particularly, which
are homeomorphic? Which aren't?
\end{question}

Showing two spaces are homeomorphic means we must construct a
homeomorphism between them. But
how do we show two spaces are \emph{not} homeomorphic?  When we are
confronted with the task of trying
to explore one space or to specify what is different about two
spaces, we must examine the
spaces looking for features or properties that are of topological significance.

\begin{question}
What features of the examples studied
are interesting either in their own
right or for the purpose of
distinguishing one from another?
\end{question}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\chapter{2-manifolds}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section[Classification of $2$-manifolds]{Classification of compact
$2$-manifolds}


A surface, or $2$-manifold, is locally homeomorphic to $\R^2$,
so we know how these spaces look locally.
But what are the possibilities for the global character of these spaces?
We have seen several examples (the $2$-sphere, the torus, the Klein bottle, $\PP^2$).
Now we seek to organize our understanding of the collection of all surfaces, that
is, to recognize, describe, and classify each surface as one from a simple list
of possible homeomorphism classes.
So we need to use the local Euclidean feature of $2$-manifolds
to help us describe the overall structure of these surfaces.

In working with these compact $2$-manifolds, we want to think of them as physical objects
made of simple building blocks, namely, triangles.
In fact, we will begin by considering just $2$-manifolds that reside in $\R^n$ and are made of triangles.
This investigation of these simple compact $2$-manifolds actually is
comprehensive since every compact $2$-manifold is homeomorphic to one made of finitely many triangles
which is embedded in $\R^n$.
The advantage of working with objects made from a
finite number of triangles is that we can use inductive procedures moving from triangle to triangle.


%\subsection[Preliminaries]{Compact, triangulated 2-manifolds: Preliminaries}


The main thing to have in mind at this point is that we should
view $2$-manifolds as concrete, physical objects that are constructed
from a finite number of
flat triangles (simplices) that fit together as specified: they overlap,
if at all, only along a shared edge or at a vertex of each.  This physical view of
$2$-manifolds will allow
us to understand them so clearly that we can describe an effective
method for determining the
global structure of the object by knowing the local structure.



The goal of the following two sections is to prove
(in two different ways) that
every compact, triangulated
$2$-manifold can be constructed by taking the connected sum of simple $2$-manifolds,
namely the sphere, torus,
and projective plane.

In the following section we
 proceed with a sequence of theorems
that show us that after
removing one disk, any compact, triangulated $2$-manifold is just a
disk with strips attached in
particularly simple ways.  We continually use the local structure of
the triangulated
$2$-manifold to see how the whole thing fits together.

The second proof of the classification theorem views
each compact, triangulated $2$-manifold
as the quotient space of
a polygonal disk with its
edges identified in pairs.

%\newpage

%\begin{dfn}[derived subdivisions of a triangulation]
%\index{subdivision!of a triangulation!barycentric}%
%\index{barycentric subdivision}%
%\index{subdivision!barycentric} {\rule{0cm}{0.1cm}}\newline\vspace*{-0.5cm}
%\be
%\item Let $\sigma^2$ be a $2$-simplex  with vertices
%$v_0,v_1$ and $v_2$. Then $p=\frac13 v_0 + \frac13 v_1 + \frac13 v_2$
%is the \emph{%barycenter\/}
%\index{barycenter}%
%of $\sigma^2$.
%\item Let $T$ be a triangulation of a
%triangulated, compact
%$2$-manifold $M^2$ with $2$-simplices $\{ \sigma_i\}_{i=1}^k$. The \emph{%first derived  subdivision\/}
%\index{subdivision!derived}%
%\index{first derived subdivision}%
%  of $T$, denoted $T'$, is the collection of $2$-simplexes obtained
%from $T$ by breaking each
%$\sigma_i$ in $T$ into six pieces as shown, together with their edges
%and vertices. Notice that
%the new vertices are the barycenter of each
%$\sigma_i$ in $T$ and  the center of each edge in $T$. The $2^{nd}$
%\emph{derived
%subdivision\/},
%\index{subdivision!derived}%
%\index{second derived subdivision}%
%denoted $T''$, is $(T')'$, and so on.
%\ee
%\end{dfn}

%\vspace*{1.5in}

%\hfill barycentric subdivision of a $2$-simplex  \hspace*{\fill}

\begin{dfn}[regular neighborhood]
Let $M^2$ be a $2$-manifold with  triangulation $T=\{
\sigma_i\}_{i=1}^k$. Let $A$ be a subcomplex of ($M^2$, $T$) . The
\emph{regular neighborhood\/}
\index{neighborhood!regular}%
\index{regular neighborhood}%
of $A$, denoted $N(A)$, equals $\bigcup \{
\sigma''_j \mid \sigma''_j \in T''$ and $\sigma''_j \cap A\ne
\emptyset\}$.
\end{dfn}


\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}

\subsection[Classification Proof I]{Classification of compact, connected
2-manifolds, I}

The basic idea of this proof is to show that
removing an open disk from a compact triangulated $2$-manifold
gives us a space
homeomorphic to a (closed) disk with some number of bands attached
to its boundary in a specified way. The number of bands, and how they
are attached then gives us the classification of the surface.

%\begin{thm}  Let each of $A_0$ and $A_1$ be a union of $2$-simplices
%such that each of $A_0$
%and $A_1$ is homeomorphic to a disk $\D^2$.  Suppose $A_0\cap A_1$ is
%homeomorphic to a PL arc
%on the boundary of each.  Then $A_0\cup A_1$ 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 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  Figure \ref{3handles1twist}.)
\item
Furthermore, the boundary of the disk with strips,
$D_1 \cup \bigl( \bigcup_{i=1}^k
H_i \bigr)$, is connected.
\ee

\begin{figure}[h]
\begin{center}
\includegraphics[height = 1.7in]{ImagesChapter2/pdf/3handles1twist}
\caption{\label{3handles1twist} A disk with four handles attached.}
\end{center}
\end{figure}

\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}

\begin{figure}[h]
\begin{center}
\includegraphics[width =2.2 in]{ImagesChapter2/pdf/twistsandentwined}
\caption{\label{twistsandentwined} Twisted strips and entwined strips}
\end{center}
\end{figure}


\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 $Y$.

\begin{figure}[h]
\begin{center}
\includegraphics[width =2.5in]{ImagesChapter2/pdf/xandy}
\caption{\label{xandy} These spaces are homeomorphic.}
\end{center}
\end{figure}

\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}


\begin{figure}[h]
\begin{center}
\includegraphics[height = 1 in]{ImagesChapter2/pdf/entwinedpair}
\caption{\label{entwinedpair} Entwining pair of strips}
\end{center}
\end{figure}

\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}

Notice that at this point we have shown that any compact, connected,
triangulated $2$-manifold
is a sphere, the connected sum of $n$ tori, or the connected sum of
$n$ projective planes; however, we
have not yet established that those possibilities are all
topologically distinct.  The
classification of $2$-manifolds requires us to prove our suspicions
that any two different
connected sums are indeed not homeomorphic. Before we develop tools
for confirming those
suspicions, we digress to develop another proof of this first part of
the classification
theorem.

\subsection[Classification Proof II]{Classification of compact,
connected 2-manifolds, II}

We now outline a different approach to proving that any compact,
connected, triangulated $2$-manifold is a
sphere, the connected sum of tori, or the connected sum of projective
planes. This approach uses
the quotient or identification topology described in the previous chapter.

Suppose that we are gluing the edges of a polygonal
disk to create a $2$-manifold. If we assign a unique letter
to each pair of edges that are glued together, and
we read the letters as we follow the edges along
the boundary of the disk (starting at a certain
edge) going clockwise, we get
a ``word'' made up of these letters. However, to
specify the gluing we need to know not only which edges
are glued together, but in what orientation. To keep track of
that, we will write the letter alone if the orientation
given on the edge agrees with the direction we're reading the edges in,
and the letter to the $-1$ power if it disagrees.
For example, $abca^{-1}dcb^{-1}d$ represents a gluing of the octagon as indicated,
so that the orientations
of two identified edges agree:

\begin{figure}[h]
\begin{center}
\includegraphics[height = 1 in]{ImagesChapter1/pdf/octogenustwo}
\caption{\label{octogenustwoagain} The genus two surface}
\end{center}
\end{figure}

\begin{dfn}[gluing of a $2n$-gon with edges identified in pairs]
\index{gluing $2n$-gon edges in pairs} An  expression
(\emph{word})\index{word} of $n$ letters, such as
$abca^{-1}dcb^{-1}d$, where each letter appears exactly twice,
represents the $2$-manifold obtained by gluing the edges of a
$2n$-gon in pairs as indicated by the sequence of letters. Notice
that a pair of edges with the same letter really has two different
possible gluings. To determine which gluing, we need to look at
the superscript or lack of subscript of each letter. A letter
without a subscript is viewed as oriented clockwise around the
$2n$-gon, while a superscript $-1$, as in $a^{-1}$, indicates that
that edge is oriented counterclockwise.  Then the identification
of the pair of edges respects those directions. So the equivalence
classes of the disk specified by such a $2n$ length string of $n$
letters consist of every singleton in the interior of the
$2n$-gon, pairs of points one from each interior of the edges with
the same label, and then equivalence classes of vertices as come
together when the edges are identified as specified. The
equivalence classes among vertices might have any number of
vertices in them, depending on the string of letters.
\end{dfn}

\begin{thm}{\rule{0cm}{0.1cm}}\newline\vspace*{-0.5cm}
\be
\item[1.] The bigon with edges identified by $aa^{-1}$ is
homeomorphic to $\Sph^2$.
\item[2.] The bigon with edges identified by $bb$ is homeomorphic to $\PP^2$.
\item[3.] The square with edges identified by $cdc^{-1}d^{-1}$ is
homeomorphic to $\T^2$.
\ee
\end{thm}

\begin{thm}[connected sum relation]
\index{surface!connected sum relation}\index{connected sum!surface, relation}
 The gluing of a square given by
  $ccdd$ is homeomorphic to $\PP^2\#\PP^2$ and the gluing of an octagon given by
$aba^{-1}b^{-1}cdc^{-1}d^{-1}$ is homeomorphic  to $\T^2\#\T^2$.
\end{thm}

\begin{question} Generalize the above to the connected sum of any two surfaces.
\end{question}

The next sequence of theorems will show us how to take a $2n$-gon
with edges identified in pairs and
modify the gluing prescription to find a canonical representation of
the same $2$-manifold.

\begin{thm} Let $Abb^{-1}C$ be a string of $2n$ letters where each
letter occurs twice, with or without a
superscript (so $A$ and $C$ should each be construed as being
comprised of many letters). Then the
$2$-manifold obtained by identifying a $2n$-gon following the gluing
$Abb^{-1}C$ is homeomorphic to the
$2$-manifold which is obtained by identifying a $(2n-2)$-gon following
the gluing given by
$AC$.
\end{thm}

\begin{thm} Suppose a $2$-manifold $M^2 \not\cong \Sph^2$ is represented by a $2n$-gon
with edges identified in pairs. Then
a homeomorphic $2$-manifold can be represented by a $2k$-gon with
edges identified in pairs where all the
vertices are in the same equivalence class, that is, all the vertices
are identified to each other.
\end{thm}


\begin{thm} Suppose a $2$-manifold $M^2 \not\cong \Sph^2$ is represented by a $2n$-gon
with edges identified in pairs. Then
a homeomorphic $2$-manifold can be represented by a $2k$-gon with
edges identified in pairs where all the
vertices are identified and every pair of edges with the same
orientation are consecutive.
\end{thm}

\begin{thm} Suppose a $2$-manifold $M^2 \not\cong \Sph^2$ is represented by a $2n$-gon
with edges identified in pairs. Then
a homeomorphic $2$-manifold can be represented by a $2k$-gon with
edges identified in pairs where all the
vertices are identified, every pair of edges with the same
orientation are consecutive, and all other edges
are grouped in disjoint sets of two intertwined pairs following the
pattern $aba^{-1}b^{-1}$.
\end{thm}

\begin{thm} The $2$-manifold represented by $aba^{-1}b^{-1}cc$ is
homeomorphic to the $2$-manifold
represented by $ddeeff$.
\end{thm}

\begin{question}
Re-state the above theorem in terms of connected sum.
\end{question}

\begin{thm} Any compact, connected, triangulated $2$-manifold is
homeomorphic to a $2n$-gon with edges
identified in pairs as specified in one of the three following ways:
$aa^{-1}$, or
$a_0a_0a_1a_1\ldots a_na_n$ (where $n\geq 0$) or
$a_0a_1a_{0}^{-1}a_{1}^{-1}\ldots a_{n-1}a_na_{n-1}^{-1}a_{n}^{-1}$
(where $n\geq 1$ is odd ).
\end{thm}

\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}

\newpage

\section{PL Homeomorphism}

Our goal is to organize connected, compact, triangulated
$2$-manifolds by homeomorphism type.  The concept
of topological homeomorphism does not reflect the triangulated
structure we have associated with these
objects, so here we present a natural way of equating two
triangulated $2$-manifolds that includes the
simplicial structure of them as well as the topological type.

The basic strategy is first to define an equivalence between
two triangulated $2$-manifolds
with triangulations $T_1$ and $T_2$
if the simplices of $T_1$ correspond to the simplices
of $T_2$ in a straightforward $1$-$1$ fashion.
Then we describe another idea of equivalence if the two $2$-manifolds
can be subdivided to find new triangulations that have this $1$-$1$ 
correspondence.


%\begin{dfn}[subdivision]
%\index{triangulation!subdivision}
%\index{subdivision of a triangulation}
%Let $M^2$ be a $2$-manifold with triangulation $T$. Then $(M^2,T')$ is a
%\emph{subdivision} of
%$(M^2,T)$ if every simplex of $T'$ is a subset of a simplex of $T$.
%\end{dfn}


\begin{dfn}[simplicial homeomorphism]
\index{simplicial homeomorphism}\index{triangulation!simplicially
homeomorphic} \index{homeomorphism!simplicial} We will say that
$M_1^2$ with triangulation $T_1$   is \emph{simplicially
homeomorphic} to $M_2^2$ with triangulation $T_2$ if and only if
there exists a homeomorphism from $M_1^2$ to $M_2^2$ that gives a
one-to-one correspondence between $T_1$ and $T_2$ in the following way:  the
homeomorphism maps each simplex in $T_1$ linearly to a single
simplex in $T_2$. So the vertices of $T_1$ go to the vertices of $T_2$ and the rest of the
homeomorphism is determined by extending the map on the vertices linearly over each simplex.
\end{dfn}


Of course, we have seen that a space can have many
different triangulations. Therefore, the concept of
a simplicial homeomorphism is too restrictive. An
underlying space with a triangulation and the same
space with its second derived subdivision
triangulation are not simplicially isomorphic.
So we can give a broader concept of equivalence:
\begin{dfn}[PL homeomorphism]
\index{PL homeomorphism}\index{homeomorphism!PL} $M_1^2$ with
triangulation $T_1$ is \emph{PL homeomorphic} to $M_2^2$ with
triangulation $T_2$ if and only if there exist subdivisions $T_1'$
and $T_2'$ of $T_1$ and $T_2$ respectively such that
$(M_1^2,T_1')$ is simplicially isomorphic to $(M_2^2,T_2')$.
\end{dfn}

The letters ``PL'' come from \emph{piecewise linear}, as the
correspondence described above gives a homeomorphism between
$M_1^2$ and $M_2^2$ that can be realized as a map that is linear
when restricted to each simplex of $T_1'$.

For $2$-manifolds you may assume without proof that a homeomorphism between
two manifolds induces a PL-homeomorphism. This however is not true for general 
$n$-manifolds. 

\section{Invariants}

One of our goals in studying topological spaces is to be able to
distinguish non-homeomorphic spaces from one another. A
fundamental strategy to tell the difference between two
topological spaces is to find some feature of one space that, on
the one hand, is preserved under homeomorphism and, on the other
hand, is not shared by the other space.  In distinguishing spaces
in a general topology course, we might look at topological
properties such as being normal, compact, or connected. However,
since we are now trying to distinguish among spaces all of which
are compact, metric spaces, we need to look for different types of
features that are invariant under homeomorphisms.  We use the word
\emph{invariant} to refer to any property of a space that is
shared by any homeomorphic space. That is, it is a property that
is preserved by homeomorphisms. So compactness, normality, and
connectedness are all invariants. The diameter of a $2$-manifold
embedded in $\R^3$, on the other hand, is not an invariant.


The crux of the whole course is to define and use invariants that are
useful for
distinguishing one space from another, especially invariants that can
help us distinguish rather
nice subsets of
$\R^n$ that might be constructed from a finite number of simplices.
We begin now by defining an
invariant that will help us distinguish one compact, connected,
triangulated $2$-manifold from some
others.

\subsection{Euler characteristic}

\begin{dfn}[Euler characteristic]
\index{$\chi$}
\index{Euler characteristic} Let $M^2$ be a 2-manifold with triangulation $T$.
Let
\begin{eqnarray*} v & = & \hbox{ number of vertices in $T$}\\
e & = & \hbox{ number of 1-simplices in $T$}\\
f & = & \hbox{ number of 2-simplices in $T$}
\end{eqnarray*}
and define the \emph{Euler  characteristic}, $\chi (M^2)$, of
$M^2$ by $\chi (M^2)= v-e+f$.
\end{dfn}

\begin{thm} Let $M^2$ be a connected, compact,
triangulated  2-manifold with triangulation $T$. Let $T'$
be a subdivision of $T$. Then
 $\chi  (M^2, T) = \chi (M^2,T')$.
%, and they are both
%orientable or both  non-orientable.
\end{thm}

In other words, for a triangulated, compact $2$-manifold,  the Euler
characteristic  is preserved
under subdivision.

\begin{thm} Let $M_1^2$ and $M_2^2$ be connected, compact,
triangulated  2-manifolds. If
$M_1^2$ is PL-homeomorphic to $M_2^2$, then $\chi  (M_1^2) = \chi
(M_2^2)$.
%, and they are both
%orientable or both  non-orientable.
\end{thm}

Since PL-homeomorphic manifolds must have the same Euler
characteristic, Euler characteristic helps to distinguish between
$2$-manifolds that are \emph{not} PL-homeomorphic.
%\begin{thm} The Euler characteristic for $\D^2$ is $1$.
%\end{thm}


\begin{thm} {\rule{0cm}{0.1cm}}\newline\vspace*{-0.5cm}
\be
\item $\chi (\Sph^2) =2$.
\item $\chi (\T^2)=0$.
\item $\chi (\PP^2)=1$.
\item $\chi (\K^2)=0$.
\ee
\end{thm}

\begin{thm}
Let $M_1^2$ and $M_2^2$ be two connected, compact,
triangulated  2-manifolds. Then
$\chi (M_1^2 \num M_2^2) = \chi (M_1^2) + \chi  (M_2^2) -2$.
\end{thm}

\begin{thm}
Let $\T_i^2$ be the torus for $i=1,\ldots,n$. Then
\[
\chi \biggl( \num_{i=1}^n \T_i^2\biggr) = 2- 2n\/.
\]
\end{thm}

\begin{dfn}[genus]
\index{genus} The \emph{genus} of $\Sph^2=0$. The \emph{genus} of
a $2$-manifold $\Sigma=\num_{i=1}^n \T^2$ is $n$.
\end{dfn}

\begin{thm} Let $\PP_i^2$ be the projective plane for $i=1,\ldots,n$.
Then
\[
\chi
\biggl( \num_{i=1}^n \PP^2\biggr) = 2-n\/.
\]
\end{thm}


%\begin{dfn}[genus]
%\index{genus!non-orientable surface} The \emph{genus} of $\PP^2=1$.
%The \emph{genus} of a
%$2$-manifold $\Sigma=\num_{i=1}^n \PP_i^2$ is $n$.
%\end{dfn}


\subsection{Orientability}

Euler characteristic is a useful invariant, in that it helps to
distinguish $2$-manifolds. However, it does not distinguish
between the torus and the Klein bottle, for example.  In fact, for
each \emph{even} number $\leq 0$ there are \emph{two}
non-homeomorphic compact, connected, triangulated $2$-manifolds of that Euler characteristic, one a
connected sum of tori, and one a connected sum of projective
planes. So although Euler characteristic is useful for
distinguishing non-homeomorphic surfaces, it does not differentiate
\emph{all} different surfaces.
%We say that
%$\chi$ is not a complete invariant, as it does not distinguish any
%$2$-manifolds.

There is a second invariant which, when combined with Euler
characteristic, will allow us to distinguish
between any two non-homeomorphic, compact, connected $2$-manifolds.
This invariant is orientability.

A surface is orientable if we can choose an ordered basis for the local Euclidean structure at each point of the surface in such a way
that the bases change smoothly as the point moves along
a path in the surface.


Note that orientability on its own is a
very coarse invariant: a $2$-manifold is
either orientable or non-orientable.
In other words, orientability divides the
set of all $2$-manifolds into two classes.
It turns out that the combination of orientability
and Euler characteristic is enough to differentiate any
two  compact, connected, triangulated $2$-manifolds.

We can explore the concept of orientability in
triangulated surfaces by considering
orderings of the vertices of each simplex.

First let us see what we mean by an orientation of a $0$-,
$1$-, and $2$-simplex.

\begin{dfns}[oriented simplices]
Let $\sigma^2$
be the $2$-simplex $\{v_0v_1v_2\}$,
$\sigma^1$ be the $1$-simplex $\{w_0w_1\}$, and $\sigma^0$ be the $0$-simplex $\{u_0\}$.

\be \item[1.] Two orderings of the vertices $v_0$, $v_1,\ldots
v_n$
of an $n$-simplex $\sigma^n$ are said to be \emph{equivalent}%
\index{vertices!equivalent!ordering}\index{ordering!equvalent}\index{equivalent!ordering}
if they differ by an even permutation. Thus $\{v_0, v_1,
v_2\}\sim\{v_1, v_2, v_0\}$. However, $\{v_0, v_1,
v_2\}\not\sim\{v_1, v_0, v_2\}$ since they differ by a single
$2$-cycle, which is an odd permutation. Note that this
equivalence relation produces precisely two equivalence classes of
orderings of vertices of an $n$-simplex for $n\geq 1$. An
equivalence class will be denoted by $[v_0 v_1\ldots v_n]$, where
$\{v_0, v_1, \ldots ,v_n\}$ is an element of the equivalence
class.

 \item[2.] An  \emph{orientation} of  the
$2$-simplex $\sigma^2$
    \index{orientation!$2$-simplex} \index{simplex!$2$!oriented}%
\index{$2$-simplex!oriented} is a one-to-one and onto function $o$
from the two equivalence classes of the orderings of the vertices of
$\sigma^2$ to $\{-1,1\}$. Note that there are two possible such
orientations for $\sigma^2$. Any vertex ordering that lies in the equivalence class whose image is
$+1$ will be called positively oriented or will be said to have a positive orientation, orderings in the other class will
be said to be negatively oriented or have a negative orientation. We can indicate the chosen positive orientation for $\sigma^2$ by
denoting $\sigma^2$ as $[v_0v_1v_2]$, where $[v_0v_1v_2]$ is in the positive equivalence
class. Note that $-o[v_0v_1v_2]=o[v_1v_0v_2]$. You can draw a circular arrow inside the $2$-simplex  in the direction indicated by any of the positively oriented orderings. That circular arrow (which will be either clockwise or counterclockwise on the page) indicates the choice of (positive) orientation for that $2$-simplex. 

\item[3.] An  \emph{orientation} of  the $1$-simplex $\sigma^1$ is%
\index{orientation!$1$-simplex}
\index{simplex!$1$!oriented}%
\index{$1$-simplex!oriented}%
 a one-to-one and onto function $o$
from the two orderings of the vertices of
$\sigma^1$ to$\{-1,1\}$. The ordering whose image is $+1$ has
the positive orientation, the other has a negative orientation. 
As with $\sigma^2$, note that there are only two possible
orientations for $\sigma^1$, and that $-o[w_0w_1]=o[w_1w_0]$.
 We think of $[w_0 w_1]$ as being the orientation that ``points'' from $w_0$
 to $w_1$.
\item[4.] Since $\sigma^0$ has a single equivalence class of
orderings of its vertex, we have a slightly different definition
of orientation for a $0$-simplex. An \emph{orientation} of a
$0$-simplex is a function $o$ from $\{[u_0]\}$ to $\{-1, 1\}$.
\ee
\end{dfns}

\begin{dfn}[induced orientation on an edge]
\index{orientation!induced!on an edge} \index{induced
orientation!on an edge} If we choose an orientation of a $2$-simplex, then there are associated orientations on each of the three edges called the induced orientations on the edges. If $[v_0 v_1 v_2]$ is the positive orientation of a $2$-simplex, then the \emph{orientations induced on the
edges} are
\be
\item[a.] $[v_1 v_2]$. 
\item[b.] $-[v_0 v_2]=[v_2 v_0]$.
\item[c.] $[v_0 v_1]$.
\ee
respectively.
\end{dfn}

\begin{note}
The definition of induced orientation is a natural one, since the
induced orientations on the edges give a directed cycle of edges
($v_0$ to $v_1$ to $v_2$ and back to $v_0$) which follow the selected positive
ordering of the vertices of the $2$-simplex.
\end{note}

\begin{exercise}
Show that the induced orientation on an edge of a $2$-simplex is
well defined; in other words, that it is independent of the choice
of positive equivalence class representative.
\end{exercise}

\begin{dfn}[induced orientation on a vertex]
\index{orientation!induced!on a vertex} \index{induced
orientation!on a vertex} The \emph{orientations induced on the
vertices} of $\sigma^1=[v_0 v_1 ]$ are
\be
\item[a.] $-[v_0]$.
\item[b.] $[v_1]$.
\ee
respectively.
\end{dfn}

We can now define what we mean by an orientable, triangulated
$2$-manifold. Intuitively, a triangulated $2$-manifold is orientable if it is possible to select orientations for each $2$-simplex in such a way that neighboring $2$-simplices have compatible orientations. The concept of `compatible' comes from the following observation. If you draw two triangles in the plane
that share an edge and orient them both in a
counterclockwise ordering, say, then the shared edge has induced
orientations from the two triangles that are opposite to each other.
In other words, when the orientations on both triangles are the same, then
the induced orientations on a shared edge are opposite. This observation gives rise to the definition of orientability.

\begin{dfn}[orientablity]
\index{orientablity!triangulated surface}  A triangulated
$2$-manifold $M^2$ is \emph{orientable\/} if and only if an
orientation can be assigned to each $2$-simplex $\tau$ in the
triangulation such that given any $1$-simplex
$e\subset\tau_1\cap\tau_2$, the orientation induced on $e$ by
$\tau_1$ is opposite to the orientation induced by $\tau_2$.
Otherwise, $M^2$ is \emph{non-orientable\/}.

A choice of orientations of the $2$-simplices of a triangulation
of $M^2$ satisfying the condition stated above is called an
\emph{orientation} of $M^2$.
\end{dfn}

\begin{note}
\end{note}

%\begin{thm}
%Show that for a triangulated $2$-manifold
%the simplicial definition of orientablity
%is equivalent to the general definition of orientability.
%\end{thm}

\begin{thm} Suppose $(M^2,T)$ is a $2$-manifold with triangulation
$T$ and $T'$ is a subdivision of $T$. Then if $(M^2,T)$ is
orientable, so is $(M^2,T')$.
\end{thm}

\begin{thm} Orientability is preserved under PL homeomorphism.
\end{thm}

\begin{thm}
$M^2$ is orientable if and only if it contains no M\"obius band.
\end{thm}


\begin{thm}
Let $M=M_1\num \ldots \num M_n$. Then
$M$ is orientable if and only if $M_i$
is orientable for each $i\in\{1,\ldots,n\}$.
\end{thm}

Compact, connected, triangulated $2$-manifolds are
determined by  orientability and
Euler characteristic. 

\begin{thm}[Classification of compact, connected $2$-manifolds]
\index{$2$-manifold!Classification Theorem!Euler characteristic}
If $M^2$ is a  connected, compact,
triangulated 2-manifold then:
\be
\item[(a)] if $\chi (M^2) = 2$, then $M^2 \cong \Sph^2$.
\item[(b)] if $M^2 $  is orientable and $\chi (M^2) = 2-2n$, for $n\geq 1$, then
\[ M^2 \cong \biggl(
\num_{i=1}^n T_i^2\biggr)\ .\]
\item[(c)] if $M^2$ is non-orientable  and $\chi (M^2) =2-n$, for $n\geq 1$, then
\[ M^2 \cong \biggl( \num_{i=1}^n  \PP_i^2\biggr)\ .
\]
\ee
\end{thm}

Notice that orientable connected, compact,  triangulated
2-manifolds must have even Euler
Characteristic.

\newpage

\begin{pr}
    Identify the following $2$-manifolds as a sphere, or a connected sum
of $n$ tori (specifying $n$), or a connected sum
of $n$ projective planes (specifying $n$).
    \be
    \item[a.]   $\T\#\PP$
    \item[b.]   $\K\#\PP$
    \item[c.]   $\PP\#\T\#\K\#\PP$
    \item[d.]   $\K\#\T\#\T\#\PP\#\K\#\T$
    \ee
\end{pr}

\newpage

\section{CW complexes}

Triangulating a surface in order to calculate the Euler
Characteristic can be quite tedious and
time-consuming. However, we don't need to divide a surface into
such small pieces in order to compute the Euler Characteristic. We can instead write the $2$-manifold as the union of much larger cells that fit together appropriately and use them to compute the Euler Characteristic. Our strategy for discovering an appropriate generalization of triangulation is to start with a triangulated surface and systematically enlarge the triangles and edges to produce other cell decompositions of the surface that will continue to reveal the Euler Characteristic.  In a way, this process of enlargement and amalgamation is the opposite of the subdivision process that we saw earlier preserved the Euler Characteristic.  We begin by amalgamating two adjacent 2-simplexes of a triangulation.

\begin{thm}
Let $(M^2,T)$ be a triangulated $2$-manifold. Suppose
$\sigma=\{uvw\}$ and $\sigma'=\{uvw'\}$ are two distinct
$2$-simplexes in $T$ that share the edge $e = \{uv\}$.  Then we can create a new structure for $M^2$ alternative to $T$, namely, $P$ where
$P$ = $T\cup\{\tau\}-\{\sigma,\sigma', $e$\}$, where $\tau=\sigma\cup\sigma'$
is the polygon formed by the union
of the two $2$-simplices along their shared edge. If $v'$,
$e'$, $f'$ are the numbers of vertices, edges, and polygons in $P$, then the Euler Characteristic
$\chi(M^2,T)=v'-e'+f'$ (see Figure \ref{cwcomp1}).
\end{thm}

\begin{figure}[h]
\begin{center}
\includegraphics[width = 3in]{ImagesChapter2/pdf/cwcomp1}
\caption{\label{cwcomp1} The basic idea of CW complexes }
\end{center}
\end{figure}

The previous theorem amalgamated two triangles together; however, we can continue in that vein by amalgamating polygonal disks together that we may have created. 

\begin{thm} Let $(M^2,T)$ be compact, triangulated $2$-manifold with Euler characteristic $\chi(M^2,T)$.
Suppose we create a polygonal structure $P$ on $M^2$ inductively as follows. Let $P_0$ = $T$. Suppose we have created $P_i$.  Suppose two $2$-dimensional objects $\sigma$ and $\sigma'$ in $P_i$ share a connected path of edges in the boundary of each
from vertex $u$ to $w$ ($v\neq w$). We create $P_{i+1}$ by removing $\sigma$ and $\sigma'$ from $P_i$, removing all the edges in the path from vertex $u$ to $w$, removing all vertices of the edges in that path except for $u$ and $w$, and putting in the single two dimensional object  $\sigma\cup \sigma'$. 
Then if $v$, $e$, $f$ are the
numbers of vertices, edges, and $2$-dimensional objects in $P_{i+1}$, then
$\chi(M^2,T)=v-e+f$ (see Figure \ref{pathremove}).
\end{thm}


\begin{figure}[h]
\begin{center}
\includegraphics[width = 3in]{ImagesChapter2/pdf/pathremove}
\caption{\label{pathremove} Removing a path from a CW complex }
\end{center}
\end{figure}

Notice that a $2$-dimensional object in $P_i$ may no longer be homeomorphic to a disk, but the `interior' of each is homeomorphic to an open disk.  We can continue our inductive definition of our new structure on $M^2$ by similarly reducing the number of $1$-dimensional objects.

\begin{thm} Let $(M^2,T)$ be compact, triangulated $2$-manifold with
a polygonal structure $P$ as defined inductively in the previous theorem. 
Suppose we substitute $P$ with a new structure obtained inductively as follows. Let $P=P_0$. If $P_i$ has an edge $e$ with a free vertex $v$, that is, $v$ is not the boundary of any other edge in $P_i$, then remove $v$ and $e$ from $P_i$ to create $P_{i+1}$.   If $P_i$ has a vertex $v$ that is one end of an edge $e$ in $P_i$ and one end of an edge $f$ in $P_i$ and $v$ is not on the end of any other edge, then remove $v$, $e$, and $f$ from $P_i$ and put in the new $1$-dimensional object $e\cup\ f$ to create $P_{i+1}$. 
Then if $v'$, $e'$, $f'$ are the
numbers of vertices, $1$-dimensional objects, and $2$-dimensional objects in an inductively defined $P$, then
$\chi(M^2,T)=v'-e'+f'$.
\end{thm}

\begin{figure}[h]
\begin{center}
\includegraphics[width = 3in]{ImagesChapter2/pdf/removeefv}
\caption{\label{removeefv} Removing edges, vertices, and faces from a CW complex}
\end{center}
\end{figure}

\begin{exercise} Start with a triangulation of $\Sph^2$ and carry out
the preceding process as far as
possible. What ``structure'' do you get? Confirm that you get the
right Euler Characteristic.
\end{exercise}

\begin{exercise} Start with a triangulation of $\T^2$ and carry out
the preceding process as far as
possible. What ``structure'' do you get? Confirm that you get the
right Euler characteristic.
\end{exercise}

We will now formalize what we have observed by defining a CW decomposition of a $2$-manifold.
\begin{dfn}[interior of a $0-$, $1-$, and $2$-simplex]
\index{interior!$2$-simplex} \index{interior!$1$-simplex}
\be
\item For each $2$-simplex $\sigma^2=\{v_0 v_1 v_2\}$
let $\interior \sigma^2=\{\lambda_0 v_0+\lambda_1 v_1+\lambda_0 v_2|
0<\lambda_i <1\}$.
\item For each $1$-simplex $\sigma^1=\{w_0 w_1\}$
let $\interior \sigma^1=\{\lambda_0 w_0+\lambda_1 w_1 |
0<\lambda_i <1\}$.
\item For each $0$-simplex $\sigma^0=\{u_0\}$
let $\interior \sigma^0=\sigma^0$.
\ee

\end{dfn}

\begin{thm}
Let $(M^2, T)$ be a compact, triangulated $2$-manifold
with triangulation $T$. Then $M^2$ equals the disjoint union of
the $\interior \sigma_i$ where $\sigma_i\in T$.
\end{thm}

\begin{dfn}[open $n$-cell from $T$]
\index{$n$-cell!open!from a triangulation}\index{open $n$-cell!from a triangulation}%
Let $(M^2,T)$ be a compact, triangulated $2$-manifold with
triangulation $T$. Suppose $C=\bigcup\{\interior
\sigma_i|\sigma_i\in T\}$ is homeomorphic to an open $k$-ball ($k\in\{0,1,2\}$). Then $C$ is an \emph{open $k$-cell from $T$}.
\end{dfn}

\begin{dfn}[cellular decomposition]
\index{cellular decomposition}%
Let $(M^2,T)$ be a compact, triangulated $2$-manifold with
triangulation $T$. If $M^2$ is the disjoint union of $C^k_i$
($k=0,1,2$ and $i=1,\ldots,n_k$), where each $C^k_i$ is an open
$k$-cell from $T$, then $S=\{C^k_i\}$ is a \emph{cellular
decomposition} of $M^2$.
\end{dfn}

These decompositions are called \emph{cellular decompositions} or \emph{CW decompositions}
 because the space can be viewed as constructed from the images of first vertices then
$1$-cells with their interiors mapped homeomorphically and their boundaries
mapped onto $0$-cells (points), and
then $2$-cells with their interiors mapped homeomorphically and their boundaries mapped
to the set of images
of the lower dimensional cells.

\begin{thm}
Let $S$ be a cellular decomposition of a compact, triangulated
$2$-manifold $(M^2,T)$.  If $v$, $e$, and $f$ are the number of $0$, $1$ and $2$
cells in $S$, then the Euler Characteristic $\chi(M^2,T)=v-e+f$.
\end{thm}

\begin{pr}
    Identify the following surfaces:
    \be
    \item[a.] The surface obtained by identifying the edges of
the octagon as indicated:

\begin{figure}[h]
\begin{center}
\includegraphics[height = 2 in]{ImagesChapter1/pdf/octogenustwo}
\caption{\label{octogenustwothrice} The genus two surface}
\end{center}
\end{figure}



    \item[b.] The surface obtained by identifying the edges of
the decagon as indicated (See Figure \ref{tensides}):

\begin{figure}[h]
\begin{center}
\includegraphics[height = 1.8 in]{ImagesChapter2/pdf/tensides}
\caption{\label{tensides} The decagon with edges indentified in pairs}
\end{center}
\end{figure}



    \ee
\end{pr}


\section{2-manifolds with boundary}


\begin{exercise}
What should be the definition of a connected, compact, triangulated $2$-manifold-with-boundary?
\end{exercise}

Your definition should be general enough to include the following examples
of $2$-manifolds-with-boundary:

\be
    \item[1.] $\D^2$
    \index{disk}%
    \index{$\D^2$}%

  \vspace*{1in}


    \item[2.] $\A^2:=\{(x_1,x_2)\in\R^2| \frac{1}{2} \leq x_1^2+x_2^2 \leq 1 \}$, the annulus (See Figure 
\ref{annulusagain})
    \index{annulus}%
    \index{$\A^2$}%

\begin{figure}[h]
\begin{center}
\includegraphics[height = 1.1 in]{ImagesChapter3/pdf/annulus}
\caption{\label{annulusagain} The annulus}
\end{center}
\end{figure}

    \item[3.] Pair of pants:
    \index{pair of pants}%

\begin{figure}[h]
\begin{center}
\includegraphics[height = 1 in]{ImagesChapter2/pdf/pairofpants}
\caption{\label{pairofpants} The pair of pants}
\end{center}
\end{figure}

    \item[4.] A disk with two intertwined handles attached, as shown in Figure
\ref{entwinedpairsagain}.


\begin{figure}[h]
\begin{center}
\includegraphics[height = 1.1 in]{ImagesChapter2/pdf/entwinedpair}
\caption{\label{entwinedpairsagain}}
\end{center}
\end{figure}

    \item[5.] M\"{o}bius band, see Figure \ref{mobiusagain}
    \index{M\"{o}bius band}%

\begin{figure}[h]
\begin{center}
\includegraphics[height = 1.1 in]{ImagesChapter1/pdf/mobiusband}
\caption{\label{mobiusagain}}
\end{center}
\end{figure}
\ee

\begin{figure}[h]
\begin{center}
\includegraphics[height = 1.2 in]{ImagesChapter2/pdf/ntwistedbands}
\caption{\label{ntwistedbands} a. n twisted bands}
\end{center}
\end{figure}

%    \be

\begin{figure}[h]
\begin{center}
\includegraphics[height = 1.2 in]{ImagesChapter2/pdf/1untwistedntwisted}
\caption{\label{1untwistedntwisted} b. $1$ untwisted band and $n-1$ twisted bands}
\end{center}
\end{figure}


\begin{exercise}
Formulate the necessary definitions and theorem statements that classify
compact, connected, triangulated $2$-manifolds-with-boundary. Prove
your theorems.
\end{exercise}

Once you have the above work done, you should be able to completely classify
and identify all
connected, compact, triangulated $2$-manifolds, with and without boundary.


To distinguish a compact manifold with no boundary from one with
topological boundary or to emphasize that a compact manifold has
no boundary the term ``closed manifold'' is often used. Beware: this term
does not mean topologically closed, as in ``the complement of an
open set', but rather it means ``a manifold that is compact \and without boundary''.
Both closed manifolds and compact manifolds-with-boundary are in fact closed subsets (in the topological sense)
of $\R^n$ and non-compact manifolds might be embedded as topologically closed subsets of $\R^n$. This unfortunate terminology is one of many examples of the use of a single
word to signify several different meanings. Context usually makes
the meaning clear.

\begin{pr}
    Identify the following surfaces made by two disks joined by
bands as indicated (See Figures \ref{ntwistedbands} and \ref{1untwistedntwisted}):
 %   \ee
\end{pr}

\vspace{.5 in}

\begin{exercise}
    Fill out the following table, using the connected sum decomposition. The
number of boundary components is  denoted by $|\partial|$.

\vspace*{.3in}

    \begin{tabular}{||c|l|l|l|l|l|l|l|l||}
        \hline \hline
        $\ \ |\partial|$ & \multicolumn{2}{c|}{$0$} &
                \multicolumn{2}{c|}{$1$} &
\multicolumn{2}{c|}{$2$} & \multicolumn{2}{c||}{$3$} \\
        \cline{2-9}

        \rule{0cm}{.7cm}$\chi\ \ \ $ & orient.      &
non-or.    &  orient.      &  non-or.  &
orient.      &  non-or.
                & orient.      &  non-or.   \\
        \hline \hline
            $2$ \rule{0cm}{.7cm}& & &  & & & & & \\
        \hline
            $1$ \rule{0cm}{.7cm}& & &  & & & & & \\
        \hline
        $0$ \rule{0cm}{.7cm}& & &  & & & & & \\
        \hline
        $-1$ \rule{0cm}{.7cm}& & &  & & & & & \\
        \hline
        $-2$ \rule{0cm}{.7cm}& & &  & & & & & \\
        \hline
        $-3$ \rule{0cm}{.7cm}& & &  & & & & & \\
        \hline
        $-4$ \rule{0cm}{.7cm}& & &  & & & & & \\
        \hline
        $-5$ \rule{0cm}{.7cm}& & &  & & & & & \\
        \hline \hline
        \end{tabular}

\end{exercise}

\section{*Non-compact surfaces}

The surfaces studied so far in this chapter are
all compact and connected $2$-manifolds, with or without boundary.
We can also consider non-compact $2$-manifolds, but we
will not do so in this class.
An interesting question to ask yourself is:
how do you extend all the concepts learned about compact spaces to non-compact ones?

For example, can you formulate and prove a classification theorem for non-compact, connected,
triangulated $2$-manifolds?
One of the difficulties that arises in the non-compact case is that we no longer have a
finite set of simplices in the triangulation.

The following exercises illustrate what some of the complications
of classifying non-compact $2$-manifolds may be, even when we
restrict to the orientable case:

\begin{exercise}
Below are some non-compact $2$-manifolds. Are any of these spaces
are homeomorphic? (Beware! It may be harder than you think!) Can
you prove whether they are or are not homeomorphic? 
(See Figure \ref{genusinfinity})
\end{exercise}

\begin{figure}
\begin{center}
\includegraphics[width = 4 in]{ImagesChapter2/pdf/genusinfinity}
\caption{\label{genusinfinity} Examples of non-compact surfaces with infinite genus}
\end{center}
\end{figure}

\begin{exercise}
Let $M$ be the non-compact $2$-manifold made by taking the two
parallel planes $\{(x,y,1)|x,y\in\R\}$ and $\{(x,y,0)|x,y\in\R\}$,
removing disks $\{(x,y,1)|(x-a)^2+(y-b)^<\frac{1}{4}, a, b\in\N\}$
and $\{(x,y,0)|(x-a)^2+(y-b)^<\frac{1}{4}, a, b\in\N\}$, and
finally gluing annuli $\{(x,y,z)|(x-a)^2+(y-b)^=\frac{1}{4}, a,
b\in\N, 0\leq z\leq 1\}$. Is this space homeomorphic to any of the
examples shown above?
\end{exercise}

\begin{figure}[h]
\begin{center}
\includegraphics[width= 6in]{ImagesChapter2/pdf/torustriangles}
\caption{\label{torustriangles} The torus triangulated}
\end{center}
\end{figure}



\begin{figure}[h]
\begin{center}
\includegraphics[width = 6 in]{ImagesChapter2/pdf/torussecondbary}
\caption{\label{torusbary} The torus triangulated by its second baricentric subdivision}
\end{center}
\end{figure}

\begin{figure}[h]
\begin{center}
\includegraphics[width = 6 in]{ImagesChapter2/pdf/torustriangles}
\caption{\label{kleintriangles} The Klein bottle triangulated}
\end{center}
\end{figure}

\begin{figure}[h]
\begin{center}
\includegraphics[width = 6 in]{ImagesChapter2/pdf/torussecondbary}
\caption{\label{kleinbary} The Klein bottle triangled by its second baricentric subdivision}
\end{center}
\end{figure}


%%%% {\rule{0cm}{0.1cm}}\newline\vspace*{-0.5cm}
%{\mathversion{bold}$2$\mathversion{normal}}
%%%%%%%%%%%%%
%\vspace*{1in}
%
%\hfill untwisted pairs \hfill twisted strips \hspace*{\fill}
%%%%%%%%%%%%%%