\chapter{Fundamental group and covering spaces}
\setcounter{example}{0}

We do not need to calculate the Euler characteristic
of a torus and a sphere to intuit that they are not homeomorphic.
The difference between them (or between them and the surface
of a two-holed doughnut) is associated with something we describe as a ``holes''.
We need to make this concept precise. What do we mean by a ``hole''?

There are two ways to make this concept definite, both
developed by Henri Poincar\`e at the turn of
the 20$^{\textrm{th}}$ century. The first of these methods
to capture the intuitive idea of holes in a space is called
the {\em fundamental group} of a space.
Unlike the Euler characteristic, which
is a numerical invariant, and
orientation, which is a parity ($+$ or $-$) invariant,
the fundamental group is an algebraic group associated to
the space. One would expect this more complex invariant to carry
more information about the space, and indeed it often does.

Intuitively, the basic goal of the fundamental group
is to recognize holes in a space. If we think about a racetrack,
we will have a good idea of how the fundamental group works. From
 a starting point, a car measures its progress by counting laps.
Going once around is different from going twice around.  The Indy
500 involves going many times around the track.  Going backward
around the track is frowned upon in competitive races, but if a
car did that, we would know how to count such a feat using negative
numbers. Counting number of times around the track is the most basic
feature of the fundamental group. But another feature of the racetrack
also suggests a basic idea of the fundamental group, namely, when a car
has completed a lap, the exact path of the car is not important as long
as it stays on the track.  To make the analogy exact, we will insist that
the car does return to the exact point where it started.

Let's become a little more mathematical in our description of the fundamental
group of the racetrack.  A racetrack is known mathematically as an annulus,
which we could describe as $\A^2= \{(x,y) \in \R^2 | 1/2 \leq x^2 + y^2 \leq 1\}$ (See Figure \ref{annuls}).

\begin{figure}
\begin{center}
\includegraphics[width= 3 cm]{ImagesChapter3/pdf/annulus} 
\caption{\label{annuls} The Annulus }
\end{center}
\end{figure}





We choose a point $x_0$ on the annulus to be the base point.
Then consider any continuous function $f$ from a simple closed curve
($\cong \Sph^1$) into the annulus.
We will choose a point on $\Sph^1$ to start, call it $z$, and require that $f(z) = x_0$.
Intuitively, that map $f$ of $\Sph^1$ into the annulus `goes around' the annulus some number of times.
Physically, we can think of $f$ as laying a rubber band around the annulus with the point
$z$ placed on top of $x_0$. If we think about the intuitive concept of how many times
the map goes around the annulus, we soon see that, in the rubber band model,
if we could distort the rubber band to a new position without lifting it out
of the annulus, then it would still go around the same number of times.
Sliding around and distorting a map are really giving a continuous family
of maps of $\Sph^1$ into the annulus.  Such a continuous family of maps is called
a homotopy.  The fundamental group just makes mathematically precise the idea
of putting maps of a circle into a space into equivalence classes via the idea
of homotopy. To get to the idea of fundamental group, we'll first develop the idea of a homotopy between maps from any space to any other space and then specialize that idea to define the fundamental group.

\section{Fundamental group}

We all have a familiarity with the idea of a homotopy, because we have all watched movies. Let's think of the physical objects that are being projected onto the screen as the domain and think of the screen as the plane $\R^2$.  So, our domain might be two people, a dog, and a table.  We will assume that all of these objects are projected at every moment of our movie and that we are thinking of an idealized projection system that projects that image at every instant of time. So, this movie has uncountably many frames per second.  We are also thinking of the camera as being fixed throughout. When the scene opens at time $0$, the people, dog, and table are shown posed in some no doubt interesting tableau.  Every point in each person, dog, and table is mapped to a point on the screen.  Notice that this map is not $1$--$1$. Points on the inside of each object certainly are mapped to places where other points map as well.  At time $0$, the action commences.  The people move about, gesticulating animatedly.  The dog barks and wags its tail.  The table just sits there.  At every instant of the film, each point in the people, dog, and table are mapped via projection to a point in $\R^2$, also known as the silver screen.  At time $1$, the movie is over.  It ends with the points of the people, dog, and table being mapped to points on the screen.  The first scene of the move, which was a function from the domain (people, dog, table) into the range ($\R^2$), was transformed via a continuous family of maps (the scenes at each moment) into the last scene of the movie, which was another function from the domain (people, dog, table) into the range ($\R^2$).  The beginning scene and final scene of our movie illustrate the idea of homotopic maps, which we now define formally.

\begin{dfn}[homotopic maps]
\index{homotopy!maps}
Let $f$, $g:X\rightarrow Y$ be two continuous functions. $f$ and $g$ are said to be
{\em homotopic} if there is a continuous map $F:X\times [0,1]\to Y$ such that $F(x,0)=f(x)$ and $F(x,1)=g(x)$
for all $x\in X$. We denote that maps $f$ and $g$ are homotopic by writing $f\simeq g$. The map $F$ is called a {\em homotopy}
between $f$ and $g$.

If $g$ is a constant map (mapping all points in $X$ to
a single point in $Y$) and $f\simeq g$, then we say $f$ is {\em null homotopic}.\index{null homotopic}
\end{dfn}

\begin{thm}
Given topological spaces $X$ and $Y$, $\simeq$ is an equivalence relation
on the set of
all continuous functions from $X$ to $Y$.
\end{thm}

We can think of a homotopy of two maps as a
continuous $1$-parameter family of maps $F_t$ from $X$ to $Y$
``deforming'' $f$ into $g$ ({\em i.e.}, $F_0=f$ and $F_1=g$).

Notice that in the gripping movie that we described earlier, the table remained fixed throughout. So at each moment of the film, the function when restricted to the table was the same for the whole duration of the movie.  That consistency on a subset of the domain gives rise to the idea of a relative homotopy.

\begin{dfn}[relative homotopy]
\index{relative homotopy}%
\index{homtopy!relative}
Given topological spaces $X$ and $Y$ with $S\subset X$, then two continuous
functions $f$, $g:X\to Y$ are {\em homotopic relative to\/} $S$ if and only if
there is a continuous function $H: X\times  [0,1]\to Y$ such that
\[
\begin{array}{lclcll}
H(x,0) & = & f(x) &  &      & \forall x\in X \\
H(x,1) & = & g(x) &  &      & \forall x\in X \\
H(x,t) & = & f(x) &= &g(x)\ & \forall \ x\in S\ \&\ t\in [0,1]
\end{array}
\]
\end{dfn}

In other words, $H$ is a $1$-parameter family of maps $H_t: X\to Y$ ($t\in [0,1]$)
which continuously
deforms $f$ into $g$
while keeping the images of points in the set $S$ fixed.

\begin{thm}
Given topological spaces $X$ and $Y$ with $S\subset X$, being
homotopic relative to $S$ is an equivalence relation
on the set of
all continuous functions from $X$ to $Y$.
\end{thm}

One of the motivations for studying homotopic maps is to capture the idea of holes in a space. Our method of recognizing a hole is to think about going around the hole. So this focus on marching around holes gives special interest to the paths that we follow in going from place to place in the space and the loops that may go around holes.  These paths can be thought of as maps from the interval into a space, so such maps are given a special name.

\begin{dfn}[paths and loops]
\index{path}%
\index{loop}%
A continuous function
$\alpha : [0,1]\to X$ is a {\em path\/}. If $\alpha (0) = \alpha (1)
=x_0$, then $\alpha$ is a {\em loop\/} (or {\em closed path\/}) {\em based
at\/} $x_0$.
\end{dfn}

Notice that a path is a function rather than a subset of the topological space in which the image of the path sits. Two paths from the same starting point to the same ending point are equivalent if, keeping the end points fixed at all times, we can find a continuous family of paths that `morph' one path into the other.  The morphing is formalized as a homotopy relative to the endpoints of $[0,1]$.

\begin{dfn}[path equivalence]
Two paths $\alpha,\beta$ are
{\em equivalent\/}\index{path!equivalence}, denoted  $\alpha\sim\beta$, if and only if $\alpha$ and
$\beta$ are homotopic relative to $\{ 0,1\}$. Denote the equivalence
class of paths equivalent to $\alpha$ by $[\alpha]$ (See Figure \ref{alphabeta}).
\end{dfn}

\begin{figure}
\begin{center}
\includegraphics[height = 4 cm]{ImagesChapter3/pdf/alphabeta}
\caption{\label{alphabeta} Path equivalence}
\end{center}
\end{figure}

The concept of path equivalence applies to loops as well---since loops are paths.  We will actually be most concerned with equivalences of loops, so we give them special attention. Technically, a loop is a path, that is, a function from $[0,1]$, whose endpoints are mapped to the same place, but intuitively, a loop is a map from $\Sph^1$ into the space.  That intuition is formalized using the following wrapping map.

\begin{dfn}[standard wrapping map]
\index{wrapping map, standard}
\index{standard wrapping map}
The map $\omega:\R^1\rightarrow\Sph^1\subset \R^2$ defined by
$t\mapsto (\cos 2\pi t,\sin 2\pi t)$ is called the
{\em standard wrapping map} of $\R^1$ to $\Sph^1$.
\end{dfn}

\begin{thm}
Let $\alpha$ be a loop into the topological space $X$.  Then $\alpha =\beta \circ \omega$ where $\omega$ is the standard wrapping map and $\beta$ is a continuous function from $\Sph^1$ into $X$.
\end{thm}

The above theorem allows us to think of a loop as a map from a circle when it is useful for us to do so. This description also allows us to state a useful characterization of triviality of a loop.

\begin{dfn}[homotopically trivial loop]
Let $X$ be a topological space. A loop $\alpha$ is {\em homotopically trivial} or is a {\em trivial loop} if $\alpha$ is equivalent to the {\em constant path}\index{constant path}\index{path!constant} $e_{\alpha(0)}$ where $e_{\alpha(0)}$ takes $[0,1]$ to $\alpha(0)$.
\end{dfn}

\begin{thm}
Let $X$ be a topological space and let $p$ be a point in $X$. Then a loop  $\alpha =\beta \circ \omega$ (where $\omega$ is the standard wrapping map and $\beta$ is a continuous function from $\Sph^1$ into $X$) is homotopically trivial if and only if $\beta$ can be extended to a
continuous function from $\B^2$ into $X$.
\end{thm}

We return now to the exploration of paths. The physical idea of walking from point $a$ to point $b$ and then proceeding from there to point $c$ yields the natural idea of how to combine paths.

\begin{dfn}[path product]
\index{path!product}
Let $\alpha$, $\beta$ be paths with $\alpha (1) = \beta (0)$.
Then their {\em product\/}, denoted $\alpha\cdot\beta$, is the
path that first moves along $\alpha$, followed by moving along $\beta$.
$\alpha\cdot\beta$ is defined by:
\[
\alpha \cdot \beta (t) =
\left\{
\begin{array}{ll}
\alpha (2t),& \quad 0\le t \le\frac12\\
\beta(2t-1),& \quad \frac12 <t \le 1
\end{array}
\right.
\]
\end{dfn}

Notice the need to speed up in order to accomplish both the paths $\alpha$ and $\beta$ during the prescribed $1$ unit of time allotted for a path.

\begin{figure}
\begin{center}
\includegraphics[width = 10 cm]{ImagesChapter3/pdf/pathproduct} 
\caption{\label{pathproduct} Path Product}
\end{center}
\end{figure}


\begin{thm}
If $\alpha\sim \alpha'$ and $\beta\sim\beta'$,
then $\beta\cdot\alpha\sim\beta'\cdot\alpha'$.
\end{thm}

Thus products  of paths can be extended to products of equivalence
classes by defining $[\alpha]\cdot[\beta]:=[\alpha\cdot\beta]$.  Products of paths and products of equivalence classes of paths enjoy the associative property.

\begin{thm}
Given $\alpha$, $\beta$, and $\gamma$, then $(\alpha\cdot\beta)\cdot
\gamma \sim\alpha\cdot (\beta\cdot \gamma)$ and $([\alpha]\cdot[\beta])\cdot
[\gamma] \sim[\alpha]\cdot ([\beta]\cdot [\gamma])$.
\end{thm}

If we think of a path $\alpha$ as taking us from $\alpha (0)$ to $\alpha (1)$, then traversing that same trail in reverse is the inverse path.

\begin{dfn}[path inverse]
\index{path!inverse}
Let $\alpha$ be a path, then its {\em path inverse\/}
$\alpha^{-1}$ is the path defined by
$\alpha^{-1} (t) = \alpha (1-t)$.
\end{dfn}

If we take a path and then take its inverse, that combined path is equivalent to not moving at all.

\begin{thm}
Let $\alpha$ be a path with $\alpha(0)=x_0$, then
$\alpha\cdot
\alpha^{-1} \sim e_{x_0}$, where $e_{x_0}$ is the
constant path $e_{x_0}: [0,1]\rightarrow x_0$. Stated differently, if $\alpha$ is a path, then $\alpha\cdot
\alpha^{-1}$ is homotopically trivial.
\end{thm}

We now have all the ingredients to associate a group with a topological space.  This group has been designed to try to capture the idea of holes in the space.

\begin{dfn}[fundamental group]
\index{$\pi_1(X,x_0$}
\index{Fundamental group}
\index{base point}
Let $x_0\in X$, a topological space. Then the set of equivalence classes
of loops based at $x_0$ with binary operation $[\alpha][\beta] =
[\alpha\cdot\beta]$ is a called the {\em fundamental group of\/} $X$
{\em based at\/} $x_0$ and is denoted $\pi_1 (X,x_0)$. The point $x_0$ is called
the {\em base point} of the fundamental group.
\end{dfn}

\begin{thm}
The fundamental group $\pi_1 (X,x_0)$ is a group. The identity element is the class of homotopically trivial loops based at $x_0$.
\end{thm}

The fundamental group is defined for a space $X$ with a specified base point selected.  However, for many spaces the choice of base point is not significant, because the fundamental group computed using one base point is isomorphic to the fundamental group using any other point.  In particular, path connected spaces enjoy this independence of base points.

\begin{thm}
If $X$ is path connected, then $\pi_1(X,p)\cong \pi_1 (X,q)$ for
any points $p,q\in X$.
\end{thm}

Since in path connected spaces the fundamental group
is independent of the base point (up to
isomorphism), for such spaces $X$ we sometimes just write
$\pi_1 (X)$ for the fundamental group without specifying the base point.

(Note. A corollary is a theorem whose truth is an immediate consequence of the statement of a preceding theorem.  A scholium is a theorem whose truth is an immediate consequence of the proof of a preceding theorem, but does not follow immediately from the statement of the preceding theorem.)

\emph{Scholium}.  Suppose $X$ is a topological space and $p$, $ q\in X$ lie in the same path
component. Then $\pi_1(X,p)$ is isomorphic to $\pi_1(X,q)$.

We have now defined the fundamental group
of a space, so let's find the fundamental groups of some spaces.  We begin with several example of spaces that have trivial fundamental groups.

\begin{exercise}
We will use $1$ to denote the trivial group:
\be
\item[1.] $\pi_1 ([0,1])\cong 1$.
\item[2.] $\pi_1 (\Sph^0, 1)\cong 1$ where $\Sph^0$ is the zero-dimensional sphere
    $\{-1,1\}\subset\R^1$.
\item[3.] $\pi_1({\textrm {convex\ set}})\cong 1$.
\item[4.] $\pi_1({\textrm{cone}})\cong 1$.
\item[5.] $\pi_1$(cone over Hawaiian earring) $\cong 1$.
\item[6.] $\pi_1$($\R^n$) $\cong 1$ for $n$ for $n\geq 1$.
\item[7.] $\pi_1(\Sph^2) \cong 1$.
\ee
\end{exercise}

A space whose fundamental group is trivial is called `simply connected.'

\begin{dfn}[simply connected]
\index{simply connceted}
\index{$1$-connected}
\index{connected!simply}
A path-connected topological space with trivial fundamental group is said to be {\em simply connected\/}
or {\em $1$-connected}.
\end{dfn}

Of course, the fundamental group would not serve a useful purpose if all spaces were simply connected. The first example we will consider of a space with non-trivial fundamental group is the circle.  The following theorem will require some significant work to prove.

\begin{thm}
The fundamental group of the circle $\Sph^1$ is infinite cyclic,
that is, $\pi_1(\Sph^1)\cong\Z$.
\end{thm}

\subsection{Cartesian products}

To add to the spaces whose fundamental groups we can compute, let us
now look at the Cartesian products of spaces and observe that the fundamental group of a product of topological spaces is just the product of the fundamental groups of the factors .

\begin{thm}
Let $(X, x_0)$, $(Y, y_0)$ be path connected spaces. Then $\pi_1 (X~\times~Y,~(x_0, y_0)) \cong \pi_1(X, x_0)\times \pi_1 (Y, y_0)$.
\end{thm}

\begin{exercise}
Find:
\be
\item[1.] $\pi_1 (\T^2\cong\Sph^1\times\Sph^1)$;
\item[2.] $\pi_1 (\D^2\times\Sph^1) $;
\item[3.] $\pi_1 (\Sph^2\times\Sph^1) $;
\item[4.] $\pi_1 (\Sph^2\times\Sph^2) $;
\item[5.] $\pi_1 (\Sph^2\times\Sph^2\times\Sph^2) $;
\item[6.] $\pi_1 (\Sph^{p_1}\times\ldots\times\Sph^{p_k}) $ where $p_j\geq 2$
for $1\leq j\leq k$.

\ee
\end{exercise}

The last $3$ sections of the previous exercise
present us with many examples of simply-connected spaces. We will see later that Cartesian products of different collections of spheres yield topologically different spaces; however, at this point in the course, it is not obvious how we are going to detect the differences in these spaces.  These examples raise the question
of what additional ideas beyond the fundamental group we will need to show that locally homeomorphic spaces aren't actually homeomorphic. For now, we leave this tantalizing question, but we will return to it in a later chapter. For the moment we will be content with having established the fundamental groups of quite a few spaces.

\subsection{Induced homomorphisms}

One of the standard techniques of mathematics is to explore the question of how structure on one mathematical object is transported to another mathematical object via a map. We have now defined the fundamental group for a space. Topological spaces are mapped to one another via continuous functions.  So we can ask how the fundamental group of one space is carried to a target space via a continuous function.

\begin{dfn}[induced homomorphism]
\index{homomorphism!induced!$\pi_1$}
\index{induced homomorphism}
\index{$f_*$!$\pi_1(X)$}
Let $f:X\to Y$ be a continuous function. Then $f_* :\pi_1 (X,x_0)\to
\pi_1 (Y,f(x_0))$ defined by $f_* ([\alpha]) = [f\circ\alpha]$ is called
the {\em induced homomorphism on fundamental groups\/}.
\end{dfn}

\begin{dfn}[well-defined function]
\index{well-defined function}
\index{function!well-defined}
Let $f:X\to Y$ and suppose that $X$ has a partition $X^*$
of equivalence classes with equivalence relation $\equiv$.
Then $f_*:X^*\to Y$ given by $f_*([x])=f(x)$
is a {\em well-defined function} if and only if $f(x_0)=f(x_1)$ for all
$x_0\equiv x_1$.
\end{dfn}

In other words a function defined on a set of equivalence classes is well-defined if
its image is independent of the choice of representative of the equivalence class.

\begin{exercise}
Check that for a continuous function $f:X\to Y$, the induced homomorphism on the fundamental group
$f_*$ is well-defined.
\end{exercise}

\begin{thm}
If $g: (X, x_0)\to (Y, y_0)$, $f:(Y, y_0)\to (Z, z_0)$ are continuous functions, then $(f\circ g)_*
= f_* \circ g_*$.
\end{thm}

\begin{thm}
If $f$, $g: (X, x_0)\to (Y, y_0)$ are continuous functions and $f$ is homotopic to $g$ relative to $x_0$,
then $f_*: = g_*$ .
\end{thm}

\begin{thm}
If $h:X\to Y$ is a homeomorphism then $h_* :\pi_1 (X,x_0)\to
\pi_1 (Y,f(x_0))$ is a group isomorphism. So homeomorphic spaces have naturally isomorphic
fundamental groups.
\end{thm}

The above theorem tells us that the fundamental group of a path connected space is a {\em topological invariant},
and hence we can establish that two path connected spaces are not homeomorphic if we can show that they have different (non-isomorphic)
fundamental groups.  Thus, the fundamental group helps to distinguish among spaces, but it
does not always detect the differences between spaces, as we have already seen among our examples of spaces with trivial fundamental groups.

The correspondence between a topological space and an algebraic group is extremely useful, because we can use algebra to answer topological questions ({\em e.g.}, are two spaces not homeomorphic?) and, as we shall see later, we can
use topology to answer algebraic questions.


\section{Retractions and fixed points}

A retraction is a map that maps a space $X$ into a subset $A$, keeping
the points in $A$ fixed.

\begin{dfn}[retraction, retract]
\index{retract}
\index{retraction}
Let $A\subset X$. A continuous function $r:X\to A$ is a {\em
retraction\/} if and only if
for every $a\in A$, $r(a) = a$.
If $r:X\to A$ is a retraction, then $A$ is a {\em retract\/} of $X$.
\end{dfn}

\begin{thm}
Let $A$ be a retract of $X$ and let $i:A\hookrightarrow X$ be the inclusion map.
Then $i_*:\pi_1(A)\to \pi_1(X)$ is injective.
\end{thm}

\begin{question}
Give an example to show why the conclusion of the previous theorem does not follow merely from the assumption that $A$
is a subset of $X$.
\end{question}

\begin{thm}
Let $A$ be a retract of $X$, $a_0\in A$ and $r:A\hookrightarrow X$ the retraction.
Then $r_*:\pi_1(X,a_0)\to \pi_1(A,a_0)$ is surjective.
\end{thm}

\begin{thm}[No Retraction Theorem for $\D^2$]
\index{No Retraction Theorem!$\D^2$}
There is no retraction from $\D^2$ to its boundary.
\end{thm}

\begin{thm}
The identity map $i:\Sph^1\to \Sph^1$ is not null homotopic.
\end{thm}

\begin{thm}
The inclusion map $j:\Sph^1\to \R^2-{\mathbf{0}}$ is not null homotopic.
\end{thm}

A fixed point free map on a disk would imply the existence of retraction of a disk to its boundary, which we know to be impossible.  Thus the No Retraction Theorem for $\D^2$ can be used to deduce the Brouwer Fixed Point Theorem for $\D^2$.

\begin{thm}[Brouwer Fixed-Point Theorem for $\D^2$]
\index{Brouwer Fixed-Point Theorem!$\D^2$}
Let $f:\D^2\to\D^2$ be a continuous map, then there is some $x\in \D^2$
for which $f(x)=x$.
\end{thm}


\begin{dfn}[strong deformation retract]
\index{strong deformatioin retract}
\index{deformation retract!strong}
\index{retract!strong deformation}
Let $A\subset X$. A continuous function $r:X\to A$ is a {\em strong deformation
retraction\/} if and only if there is a homotopy $R:X\times [0,1]\to X$ such that
\[
\begin{array}{lclll}
R(x,0) & = & x &       & \forall x\in X \\
R(x,1) & = & r(x) &       & \forall x\in X \\
R(a,t) & = & a &  \forall \ a\in A\ \&\ t\in [0,1].
\end{array}
\]
If $r:X\to A$ is a strong deformation retraction, then $A$ is a {\em strong deformation retract\/} of $X$.
\end{dfn}

We can re-word the definition in terms of homotopies:
a strong deformation retraction $r:X\to A$ is a retraction
that is homotopic to the identity on $X$ relative to $A$.

\begin{exercise}
Show that $\R^2-\{2$ points$\}$ strong deformation retracts onto the wedge of two circles.
In addition, show that $\R^2-\{2$ points$\}$ strong deformation 
retracts onto a theta curve. Are the wedge of two circles and the theta curve homeomorphic?
\end{exercise}

\vspace*{1in}

\hfill wedge of two circles \hfill theta curve \hspace*{\fill}

\begin{thm}
If $r:X\to A$ is a strong deformation retraction and $a\in A$, then
$\pi_1 (X,a)\cong \pi_1(A,a)$.
\end{thm}

\begin{exercise}
Compute the fundamental groups of the following spaces:
{\rule{0cm}{0.1cm}}\newline\vspace*{-0.5cm}
\be
\item $\pi_1(\textrm{solid\ torus}\cong \D^2\times\Sph^1)\cong$
\item $\pi_1(\R^2 - {\mathbf{0}}) \cong$
\item $\pi_1(\textrm{house\ with\ 2\ rooms})\cong $ \index{house with $2$ rooms}

\vspace*{1in}

\hfill House with 2 rooms \hspace*{\fill}

\item $\pi_1(\textrm{dunce's\ hat})\cong$ \index{dunce's hat}

\vspace*{1in}

\hfill Dunce's hat \hspace*{\fill}

\ee
\end{exercise}



\begin{dfn}[contractible space]
\index{contractible}
The topological space $X$ is {\em contractible\/} if and only if the identity map
on $X$ is null homotopic.
\end{dfn}

\begin{thm}
A contractible space is simply connected.
\end{thm}

\begin{thm}
A retract of a contractible space is contractible.
\end{thm}

\begin{thm}
The House with Two Rooms is contractible.
\end{thm}

\begin{thm}
The Dunce's Hat is contractible.
\end{thm}

\newpage
%
%\begin{exercise}
%Show that given a non-vanishing vector-field $v(x)$ on $\D^2$, there are points of
%$\bd (\D^2)=\Sph^1$ where the vector points outward, and points where it points inward.
%\end{exercise}

%\begin{hint}
%Hint:  Define vector field $v:\D^2\to \R^2-{\mathbf{0}}$
%by $v(x)=f(x)-x$. Let $w$ be the vector field restricted to $\Sph^1$.
%What is $w$ homotopic to? You should find a contradiction.
%\end{hint}


\section{Van Kampen's Theorem, I}

At this stage, $\Sph^1$ is the only source of examples of spaces with non-trivial fundamental groups.
It seems a lot of work for not much payoff! So the question now 
becomes: How can we compute the fundamental group of more complex spaces, in particular,
spaces with $\pi_1(X)\neq 1$. So far, we only have the ability to compute the fundamental
group of spaces two ways: taking the Cartesian product of spaces whose fundamental groups we already know,
and taking strong deformation retracts involving one space whose fundamental group we already know.
The first allowed us to compute the fundamental group of products of $\Sph^n$, and, in particular, 
the fundamental group of the torus and the annulus, and the second allowed us to
calculate the fundamental group of the annulus a second way, and also the fundamental group of 
other spaces such as the dunce's hat and the house with two rooms. 
But at this point we can't compute the fundamental groups of any surfaces other than the torus, let alone more general spaces.

To expand the number of spaces whose fundamental groups we can compute, our strategy will be to think about breaking up spaces into pieces whose fundamental groups we know and analyzing
how the fundamental group of the whole relates to the fundamental groups of the pieces.

\begin{thm}
Let $X=U\cup V$, where $U$ and $V$ are open and $U\cap V$ is path connected, 
and let $p\in U\cap V$.  Then any element of $\pi_1(X,p)$ has a representative
$\alpha_1\beta_1\cdots\alpha_n\beta_n$, where each $\alpha_i$ is a loop in $U$ based at
$p$ and each $\beta_i$ is a loop in $V$ based at $p$.
\end{thm}

In its full generality, Van Kampen's Theorem will allow us to compute the fundamental group of any space that, as in the theorem above, is broken up into two open and path connected subspaces with non-empty path connected intersection. One of the challenges of computing $\pi_1(X)$ is that the representative $\alpha_1\beta_1\cdots\alpha_n\beta_n$ is not necessarily unique, even allowing for the
choice of representatives of $\alpha_i$ in $\pi_1(U)$ and $\beta_i$ in $\pi_1(V)$.

Therefore, before we see the theorem in full generality, let us look at some special cases of the theorem
which are not only useful, but will allow us to understand the underlying ideas used in the general version.  In the case where $U\cap V$ is simply connected, the representative  $\alpha_1\beta_1\cdots\alpha_n\beta_n$ is in a certain way unique, making the calculation of $\pi_1(X)$ quite simple.


\subsection{Van Kampen's Theorem: simply connected intersection case}

There
are many ways of creating new groups from given groups. One
of them, which is useful
in the context of fundamental groups,
is the free product. Here is how to think about the free
product of two groups $G$ and $H$. Each element of the
free product $G*H$ is just a finite string of letters from $G$
and $H$ juxtaposed. So elements (called words)
of $G*H$ are of the form
$g_1h_1g_2h_2\ldots g_nh_n$, $h_1g_2h_2\ldots g_nh_n$,
$g_1h_1g_2h_2\ldots h_{n-1}g_n$
or $h_1g_1h_2g_2\ldots g_{n-1}h_n$.
To multiply two such strings, we just juxtapose them. If
there are two $g$'s next to each other, we multiply them in $G$
and replace them by the answer. We do the same procedure
with two $h$'s
next to each other. If any $g$ or $h$ is the identity in its
respective group, then we erase it. This means that the
identities are really considered the same identity, which is the
identity 
of $G*H$. And that's it.

Here is a more formal definition using group presentations:
\begin{dfn}[free product of groups]
\index{free product of groups}\index{group!free product}
\index{product!free}
Let $G$ and $H$ be two groups.
The {\em free product} of $G$ and $H$, denoted $G*H$ is
the group generated by all the elements of $G$ and $H$,
subject only to the relations for $G$ and $H$.
In other words, if
\[
G=\langle g_1,\ldots,g_n | r_1, \ldots , r_u \rangle
\]
and
\[
H=\langle h_1,\ldots,h_m | t_1, \ldots , t_w \rangle,
\]
then
\[
G*H=\langle g_1,\ldots,g_n, h_1,\ldots,h_m | r_1, \ldots , r_u, t_1, \ldots , t_w \rangle
\]
\end{dfn}




The following version of Van Kampen's Theorem tells us that when $X$ can
be decomposed into two open subspaces that are path connected and whose
intersection is path connected and simply connected, then the fundamental
group of the entire space is the free product of the fundamental group 
of the two subspaces.

\begin{thm}[Van Kampen's Theorem, simply connected intersection case]
\index{Van Kampen's Theorem!simply connected intersection}
Let $X=U\cup V$, where $U,V$ are open, path connected subsets of $X$,
$U\cap V$ is path connected and  simply connected, and $x\in U\cap V$.
Then $\pi_1 (X,x)\cong \pi_1(U,x) * \pi_1(V,x)$.
\end{thm}


\begin{dfn}[wedge of two spaces]
\index{wedge}
Let $A$ and $B$ be two disjoint spaces, with points $p\in A$ and $q\in B$. Then
the {\em wedge of $A$ and $B$}, denoted $A\vee B$, is defined as the quotient space
$A\cup B/ p\sim q$. In other words, we glue $p$ to $q$.
\end{dfn}

\begin{cor}
Let $\infty$ denote the wedge of two circles. Then $\pi_1(\infty) \cong \Z *\Z$.\
\end{cor}

\begin{question}
Let $X$ be the wedge of $n$ circles. What is $\pi_1(X)$?
\end{question}


\begin{thm}
If $A$ and $B$ are each connected, then $A\vee B$ is connected. If $A$ and $B$ are each path
connected, then $A\vee B$ is path connected.
\end{thm}

The following example shows the necessity of
the hypotheses that $U$ and $V$ are open
 in the above theorem.

\begin{thm}
Let $X$ be the wedge of two cones over two Hawaiian earrings, where they are identified at the points
of tangency of the circles of each Hawaiian earring, as
in the figure below. Then $\pi_1 (X) \not\cong 1$.

\vspace*{1in}

\hfill Wedge of cones over Hawaiian earrings \hspace*{\fill}

\end{thm}


\begin{question}
State conditions
that suffice to ensure that $\pi_1 (A\vee B)\cong \pi_1 (A) * \pi_1(B)$.
\end{question}


[We state the following theorem about the fundamental group of the Hawaiian earring just for interest, but it does not fit into the flow of the current discussion. 

\begin{thm*}
Show that $\pi_1($Hawaiian earring$)$ is not finitely generated, in fact, 
$\pi_1($Hawaiian earring$)$ is not countably generated.]
\end{thm*}

\subsection{Van Kampen's Theorem: simply connected pieces case}

Sometimes, one can break up a space into two pieces 
that are each simply connected.

\begin{thm}[Van Kampen's Theorem, simply connected pieces case]
\index{Van Kampen's Theorem!simply connected parts}
Let $X= U\cup V$ where $U$ and $V$ are open, path connected, and simply
connected subsets of $X$ and $U\cap V$ is path connected. Then $X$ is
simply connected.
\end{thm}

\begin{exercise}{\rule{0cm}{0.1cm}}\newline\vspace*{-0.5cm}
\be
\item[1.] $\pi_1 (\Sph^2) = 1$
\item[2.] $\pi_1 (\Sph^n) = 1$
\ee
\end{exercise}

A good strategy when presented with a theorem is to ask yourself if the hypotheses
can be further weakened. A good habit when learning a new theorem is to try to 
generate counterexamples that illustrate the limits of theorems and
the necessity of each hypothesis.

\begin{question}
Can you find an example where $U$ and $V$ are simply connected, 
but $X=U\cup V$ is not simply connected?
\end{question}

\section{Fundamental groups of surfaces}

Throughout this course, surfaces have been our motivating examples. It is thus
natural that we would like to find
the fundamental group of all compact, connected surfaces. In particular, we would like to know if
the fundamental group is an invariant that distinguishes them. Of course, we will
eventually also want to be able to calculate $\pi_1$ for other spaces.

We have seen that the strong deformation retract of a space has the same fundamental group
as the space.  Our strategy is to show that a punctured surface  strong deformation retracts
to a space whose fundamental group we know. We will then find $\pi_1$ of closed surfaces if we
can understand what effect ``capping off'' the puncture with a disk has on the fundamental group.

A second strategy will be to find the fundamental group of the connected sum of two surfaces given
that we know each of their fundamental groups. 


\begin{exercise}
Describe a strong deformation retract, together with its fundamental group,
of a once-punctured compact, connected, triangulated
$2$-manifold.
\end{exercise}

The answer to the above exercise gives us the fundamental group of a once-punctured surface.

\begin{exercise}
Let $M^2$ be a compact, connected, triangulated $2$-manifold, and assume we write $M^2$ as
$U\cup D^2$, where $U$ and $D^2$ are open subspaces of $M^2$, $D^2$ is an open disk,
 $U\cap D^2\simeq A^2$ is 
an open annulus, and $p \in A^2$. Describe the non-trivial elements of $\pi_1(U,p)$ that are trivial in $\pi_1(M^2,p)$. 
\end{exercise}

\begin{exercise}
State and prove a theorem that 
allows you to calculate $\pi_1(M^2)$ for any compact, connected, triangulated $2$-manifold $M^2$. 
\end{exercise}

\begin{exercise}
{\rule{0cm}{0.1cm}}\newline\vspace*{-0.5cm}
\be
\item[1.]Describe a group presentation of $\num_{i=1}^k\T^2_i$.
\item[2.]Describe a group presentations of $\num_{i=1}^k\PP^2_i$.
\ee
\end{exercise}

There is a problem with group presentations, namely, it can be extremely difficult
to see whether two
presentations represent the same group or not. 
So, how do we know whether the groups we find above are or are not
isomorphic? Well, one strategy for detecting differences among 
the fundamental groups of surfaces is to {\em abelianize}\index{abelianization!$\pi_1(X)$}
the groups and see what we get. If the abelianizations are not isomorphic, then neither were the original groups.

\begin{exercise}
Explicitly determine, using the Classification of Finitely Generated Abelian Groups, what
the abelianizations of the fundamental groups found in the previous exercises are. What, if anything,
distinguishes orientable from non-orientable surfaces? Are any of these abelianized
groups isomorphic? Is this invariant (the abelianized fundamental group) a complete invariant for closed
surfaces---{\em i.e.}, is it sufficient to distinguish between any two surfaces?
\end{exercise}

 
Thus far, we have succeeded in finding a representation of the fundamental group of a surface by viewing the surface a a punctured surface union a disk.  But our previous analysis of surfaces involved connected sums, so it is natural for us to think about how the fundamental group of a connected sum is related to the fundamental groups of the two punctured surfaces whose union makes it up.  

\begin{exercise}
Suppose that $M^2=\T_1 \# \T_2$ where $\T_1$ and $\T_2$ are tori and $M^2=U \cup V $ where $U$ is an open set of $\T_1$ homeomorphic to $T_1-$(a disk),  $V$ is an open set of $\T_2$ homeomorphic to $T_2-$(a disk), and $U \cap V$ is homeomorphic to an open annulus. Let $p \in U \cap V$.  We  know from a previous exercise that $\pi_1(U,p)$ is generated by two loops $\alpha$ and $\beta$. Likewise, $\pi_1(V,p)$ is generated by two loops $\gamma$ and $\delta$. Consider the loop $\mu$ that generates $\pi_1(U\cap V,p)$. Represent $\mu$ in terms of the generators of  $\pi_1(U,p)$. Now represent $\mu$
in terms of the generators of  $\pi_1(V,p)$. So the single loop $\mu$ is equivalent to two different loops in $M^2$. $\pi_1(M^2,p)$ is generated by $\{\alpha,\beta,\gamma,\delta\}$. What relations exist among these generators?  Give a presentation of $\pi_1(M^2,p)$ whose generators are $\{\alpha,\beta,\gamma,\delta\}$.
\end{exercise}


The exercise above gives the basic insight into how we can deduce the fundamental group of a union of two pieces from the fundamental groups of the two pieces.


\newpage

\section{Van Kampen's Theorem, II}

 Van Kampen's Theorem says that if we can split a
space $X$ into two parts $U$ and $V$ satisfying certain conditions, and we know what $\pi_1(U)$
and $\pi_1(V)$ are, then we can find $\pi_1(X)$ (up to isomorphism, as always).
We have already seen two special cases of this result.
Let us now see Van Kampen's Theorem in full generality.

Group presentations are useful, concrete ways of representing $\pi_1$ (although not
completely trouble-free, as it is hard to tell when two presentations represent the
same group). Van Kampen's Theorem can be stated in the language of group presentations.

\begin{thm}[Van Kampen's Theorem; group presentations version]
\index{Van Kampen's Theorem!general!group presentations}
Let $X=U\cup V$, where $U,V$ are open and path connected and $U\cap V$
is path connected and non-empty. Let $x\in U\cap V$.

Let
$\pi_1 (U,x)=\langle g_1,\ldots,g_n | r_1, \ldots , r_m \rangle$,
 $\pi_1 (V,x)=\langle h_1,\ldots,h_t | s_1, \ldots , s_u \rangle$ and
$\pi_1 (U\cap V,x)=\langle k_1,\ldots,k_v | t_1, \ldots , t_w \rangle$
then
\begin{eqnarray*}
\pi_1 (X,x) & =  \left\langle g_1,\ldots,g_n , h_1,\ldots,h_t \right.|& r_1, \ldots , r_m, s_1, \ldots , s_u,\\
             &                         & \left. i_*(k_1)=j*(k_1),\ldots , i_*(k_v)=j*(k_v)\right\rangle
\end{eqnarray*}
where $i$, $j$ are the inclusion maps
of  $U\cap V$ into $U$ and $V$ respectively.
\end{thm}

Without the language of group presentations, Van Kampen's Theorem is stated as follows:

\begin{thm}[Van Kampen's Theorem]
\index{Van Kampen's Theorem!general}
Let $X=U\cup V$ where  $U,V$ are open and path connected and $U\cap V$
is path connected and non-empty. Let $x\in U\cap V$.  Then
\[
\pi_1 (X,x) \cong \frac{\pi_1(U,x) * \pi_1(V,x)}{N}
\]
where $N$ is the
smallest normal subgroup containing
$\{ i_* (\alpha)j_* (\alpha^{-1})
\}_{\alpha\in\pi_1 (U\cap V,x)}$ and $i$, $j$ are the inclusion maps
of $U\cap V$ in $U$ and $V$ respectively. Note that $N$ is the set of products of conjugates of $i_*(\alpha)j_*(\alpha^{-1})$.

\end{thm}

\begin{exercise}
Use Van Kampen's theorem to explicitly calculate the group presentation of the double torus
$\T^2\num \T^2$.
\end{exercise}



\section{3-manifolds}

In the preceding chapter and
in this chapter we have studied
$2$-manifolds in some detail. In point of fact,
we only need Euler characteristic and
orientability to completely classify compact $2$-manifolds. But what
happens when we look at higher-dimensional manifolds?
There, $\chi$ is no longer a useful invariant
(orientability is always of interest). For any closed, compact, oriented
$3$-manifold $M^3$, $\chi (M^3)=0$
(it's zero for any odd-dimensional closed, compact, oriented
manifold).
$3$-manifolds
are not as well understood as $2$-manifolds, but the fundamental group can be a useful invariant in the
study of $3$-manifolds.

Let us look at two specific examples: lens spaces, which are compact $3$-manifolds with no boundary, and
knot exteriors, which are compact $3$-manifolds with boundary.

\subsection{Lens spaces}

Every compact, connected $3$-manifold can be constructed by taking familiar $3$-manifolds with boundaries
(such as solid tori or solid double tori, or, in general, solid $n$-tori)
and gluing a pair of them together along their boundaries.
The $3$-manifolds that can be obtained by gluing a pair of
solid tori together along their boundaries are called lens spaces.
These relatively simple $3$-manifolds are completely classified.

Let $p$ and $q$ be relatively prime
natural numbers. A $(p,q)$-lens space, denoted $L(p,q)$, indexed by $p/q\in \Q$, can be defined
in several different ways.
    \index{lens space}\index{$L(p,q)$}%
$L(p,q)$ was first defined as an identification space that started with a $3$-ball drawn
in the shape of a lens (hence the name).  The top and bottom hemispheres of this lens
are each divided into $p$ triangle-like wedges.
Each triangle from the top hemisphere is identified
with a triangle in the bottom hemisphere that is a
certain specified number (relatively prime to $p$) of triangles around the equator.
The resulting quotient space is a lens space (See Figure \ref{lensspace}).

\begin{figure}
\begin{center}
\includegraphics{ImagesChapter3/pdf/lensspace}
\caption{\label{lensspace} Lens space as a quotient of a lens}
\end{center}
\end{figure}


\begin{dfn}[isotopy]
A homotopy $H_t: X\to Y$ ($t\in I$) is an {\em isotopy} if and only if for every $t$ in
$I$, $H_t$ is an embedding.
\end{dfn}

\begin{exercise}
Show that isotopies form an equivalence relation on the set of all embeddings of
$X$ into $Y$.
\end{exercise}

\begin{dfn}[meridian]
Let $V\cong \D^2\times\Sph^1$ be a solid torus. A
simple closed curve $J$ on $\bd V$ is a {\em meridian}
if  and only if it bounds a disk in $V$ (See Figure \ref{solidtorus}).
\end{dfn}


\begin{figure}
\begin{center}
\includegraphics{ImagesChapter3/pdf/solidtorus}
\caption{\label{solidtorus} Solid torus with meridian}
\end{center}
\end{figure}


\begin{dfn}[longitude]
Let $V\cong \D^2 \times \Sph^1$. A
simple closed curve $K$ on $\bd V$
is a {\em longitude} or {\em longitudinal curve}
if and only if $K$ represents the
generator of $\pi_1 (V)$, that is,
$K$ goes once around $V$.  A longitude can
be isotoped to intersect a meridianal curve
once.
\end{dfn}

The meridian of a solid torus is unique up to isotopy, but 
the longitude is not, since a longitude can spiral around 
the torus a number of full turns as it goes around.
A choice of longitude is
called a $framing$.

\begin{lem}
Two simple closed curves $\alpha$ and $\beta$ in
$\T^2\cong\Sph^1\times\Sph^1$ are homotopic
if and only if they
are isotopic.
\end{lem}

\begin{lem}
Given a meridian $\mu$ and longitude
$\lambda$, $\{[\mu],[\lambda]\}$
forms a basis for $\pi_1(\bd(\D^2\times\Sph^1))$.
\end{lem}

\begin{lem}
Let $\{[\mu],[\lambda]\}$ be
a basis for $\pi_1(\bd(\D^2\times\Sph^1))\cong \Z\times\Z$.
Then $q[\mu]+p[\lambda]$ has a simple closed curve
representative if and only if $p$ and $q$ are relatively prime.
\end{lem}

We can therefore use $q[\mu]+p[\lambda]$ (where $p$ and $q$
are relatively prime) to mean a simple closed curve
representative of that class. Two different representatives
that are simple will be isotopic.

\begin{dfn}[lens space]
    \index{lens space}%
    \index{$L(p,q)$}%
Let $V_1$ and $V_2$ be two solid tori with meridians $\mu_i$ and chosen longitudes
$\lambda_i$ respectively.
Let $h:\bd (V_1)\to\bd (V_2)$ be a homeomorphism
such that $h(\mu_1)$ goes to a curve in the isotopy class
$q\mu_2 +p\lambda_2$ (where p,q are coprime). Then the quotient
space $V_1\cup_h V_2$ is the $(p,q)$-{\em lens space}
 $L(p,q)$.
\end{dfn}

Lens spaces can also be defined other ways, for
example, as quotient spaces of $\Sph^3$ under certain group actions.
Beware that some authors use $L(p,q)$ to mean $L(q,p)$.

\begin{exercise}
Use Van Kampen's Theorem to explicitly calculate a group presentation of $\pi_1(L(p,q))$.
\end{exercise}

\subsection{Knots in $\Sph^3$}

\begin{dfn}[knot]
\index{knot}\index{knot complement}\index{knot exterior}
Let $i:\D^2\times\Sph^1\to\Sph^3$ be a PL embedding, that
is, an injective continuous function that is linear on
each simplex of a triangulation of $\D^2\times\Sph^1$. Then
$K=i({\mathbf{0}} \times\Sph^1)\cong\Sph^1$ is a {\em knot} in $\Sph^3$, and
$N(K)= (\interior(D^2 \times \Sph^1))$ is an open regular neighborhood of $K$.
The {\em knot complement} or {\em knot exterior} of $K$ is
$M_K= \Sph^3 - N(K)$.
\end{dfn}

If we look at a knot from above, we see a curve with
crossings where it goes over or under itself.
%\marginpar{\tiny{Should we say anything about getting rid
%of all but the double points?}}
For example, this
picture is a picture of a trefoil knot (See Figure \ref{trefoil}):

\begin{figure}
\begin{center}
\includegraphics[height =2 in]{ImagesChapter3/pdf/trefoil}
\caption{\label{trefoil} The trefoil knot}
\end{center}
\end{figure}


If we are given a picture of a projection of a knot $K$ into $\R^2$
where gaps
indicate where undercrossings occur and where all 
crossings are transverse crossings of two arcs, then we can use the
pictures, along Van Kampen's Theorem to produce a presentation of
$\pi_1(M_K)$. Roughly speaking, each arc on the picture gives
a generator and each crossing represents a relation.

For each arc in a knot projection, draw a labeled
perpendicular arrow as shown (See Figure \ref{trefoilwitharrows}):

\begin{figure}
\begin{center}
\includegraphics[height =2 in]{ImagesChapter3/pdf/trefoilwitharrows}
\caption{\label{trefoilwitharrows} The arrows for the arcs of a trefoil knot}
\end{center}
\end{figure}


The arrow $a_i$,
for example, represents the loop in $M_K$
obtained by starting well above the knot (at the base point
chosen for $\pi_1(M_K)$),
going straight down to the tail of $a_i$, then going along $a_i$
under the knot, and finally returning to the starting point going straight
from the head of $a_i$.

\begin{lem}
Every loop in $M_K$ is homotopic in $M_K$ to a product of $a_i$'s.
In other words, the loops
$\{a_i\}$ generate $\pi_1(M_K)$.
\end{lem}

\begin{figure}
\begin{center}
\includegraphics{ImagesChapter3/pdf/crossings}
\caption{\label{crossings} The arrows around a crossing}
\end{center}
\end{figure}



\begin{lem}
At every crossing, such as that illustrated in Figure \ref{crossings}, the
following relation holds: $acb^{-1}=c$
or $acb^{-1}c^{-1}=1$.
\end{lem}

\begin{thm}
Let $K$ be a knot in $\Sph^3$ and let $\{a_i\}$ be
the set of loops consisting of one loop for each arc in a knot projection
of $K$ as described above. Then $\pi_1(M_K)=\{a_1,a_2,\ldots,a_n| a_ia_ja_k^{-1}a_j^{-1}$
where there is one relation of the form $a_ia_ja_k^{-1}a_j^{-1}$ for
each crossing in the knot projection$\}$.
\end{thm}

\begin{exercise}
Find the fundamental group of the complement of the unknot (See Figure \ref{unknot}).


\begin{figure}
\begin{center}
\includegraphics{ImagesChapter3/pdf/unknot}
\caption{\label{unknot} The unknot}
\end{center}
\end{figure}
\end{exercise}

\begin{exercise}
Find the fundamental group of the  complement of the trefoil knot.
\end{exercise}

\begin{exercise}
Find the fundamental group of the  complement of the figure-$8$ 
knot (See Figure \ref{figeight}).
\end{exercise}

\begin{figure}
\begin{center}
\includegraphics{ImagesChapter3/pdf/fig8}
\caption{\label{figeight} The figure-$8$ knot}
\end{center}
\end{figure}



\section{Homotopy equivalence of spaces}

\begin{dfn}[homotopy equivalence]
\index{homotopy!equivalence}
\index{equivalence!homotopy}
Two spaces $X$ and $Y$ are said to be {\em homotopy equivalent} or to 
have the same {\em homotopy type} if there
exist maps $f: X\to Y$ and $g:Y\to X$ such that 
$g\circ f\simeq i_X$ and $f\circ g\simeq i_Y$, where $i_X$ denotes the
identity on $X$ and $i_Y$ denotes the
identity on $Y$. We write $X\sim Y$ to mean that $X$ and $Y$ 
are homotopy equivalent.
\end{dfn}

\begin{thm}
If $X$ is a strong deformation retract of $Y$, then $X$ and $Y$ are homotopy equivalent.
\end{thm}

\begin{thm}
If $X\sim Y$ then $\pi_1(X)\cong\pi_1(Y)$.
\end{thm}

In other words $\pi_1$ doesn't distinguish spaces that have the same homotopy type.

\section{Higher homotopy groups}

The fundamental group of a space $X$ is the set of homotopy equivalence classes of maps of $\Sph^1$ into
$X$ with certain constraints. This idea
can be generalized to maps of $\Sph^n$ into $X$, giving the {\em higher
homotopy groups}.

Recall that a loop can be thought of as either a map
$\alpha$ from $[0,1]=\D^1$ to $X$ where $\alpha(\bd\D^1)=x_0$ for some point
$x_0\in X$,  or as a map $\Sph^1\to X$, and in fact we used the two ways interchangeably.
We will use the second way of looking at the higher-dimensional analogues of paths:

\begin{dfn}[product of homotopy classes]
Let $X$ be a topological space and ${\mathbf{x_0}}\in X$. Let $f$,
$g:(\D^n,\partial\D^n)\to(X,{\mathbf{x_0}})$, that is
$f$ and $g$ are continuous maps that take $\partial\D^n \mapsto {\mathbf{x_0}}$.
Let $[f]$ and $[g]$ denote the respective homotopy classes of these maps rel $\partial\D^n$.
Then we define $[f]\cdot[g]$ to be the homotopy class of:
\[
f \cdot g (x_1,x_2,\ldots,x_n) =
\left\{
\begin{array}{ll}
\alpha (2x_1,x_2,\ldots,x_n),& \quad 0\le t \le\frac12\\
\beta(2x_1-1,x_2,\ldots,x_n),& \quad \frac12 <t \le 1
\end{array}
\right.
\]
where $(x_1,x_2,\ldots,x_n)\in \D^n$
\vspace*{1in}

\begin{figure}
\begin{center}
\includegraphics[height = 1.7 in]{ImagesChapter3/pdf/higherhomotopy}
\caption{\label{higherhomotopy} Two spheres with the same basepoint}
\end{center}
\end{figure}
\end{dfn}

\begin{exercise}
The set of homotopy classes of maps
$f:(\D^n,\partial\D^n)\to(X,{\mathbf{x_0}})$, where $\partial\D^n\mapsto {\mathbf{x_0}}$,
with the product defined above, forms a group.
\end{exercise}

\begin{dfn}[higher homotopy groups]
\index{homotopy groups}
\index{$\pi_n(X)$}
The above mentioned group is called the {\em $n^{\textrm{th}}$ homotopy group of\/} $X$
{\em based at\/} $x_0$ and is denoted $\pi_n (X,x_0)$. The point $x_0$ is called
the {\em base point} of the homotopy group.
\end{dfn}

\begin{thm}
Homotopy equivalent spaces have the same homotopy groups.
\end{thm}


Homotopy groups are generally hard to compute (even for $\Sph^n$).
In the next chapter we will develop the study of
homology groups, which turn out to be easier to compute and,
hence, are generally more useful than the higher homotopy groups
in distinguishing higher dimensional topological spaces from one another.
 
\newpage

\section{Covering spaces}


\begin{thm}
\hspace*{3em}
\be
\item Any loop $\alpha :I\to \Sph^1$ can be written $\alpha =\omega\circ\wt{\alpha}$,
where $\wt{\alpha}: I\to \R^1$, $\wt{\alpha}(0)=0$ and $\omega$ is the standard wrapping map.
\item $\wt{\alpha}(1)$ is an integer. 
\item $\alpha _1$ and $\alpha _2$ are equivalent loops in $\Sph^1$ if and only if $\wt{\alpha _1}(1)=\wt{\alpha _2}(1)$.
\item $\pi_1(\Sph^1)=\Z$.
\ee
\end{thm}

The preceding theorem outlines a method to compute $\pi_1(\Sph^1)$.  The
pre-image of any small open arc in $\Sph^1$ under the standard wrapping map $\omega$ is a collection of open intervals in
$\R^1$ and $\omega$ is a homeomorphism when restricted to any single one of these open intervals.
This map $\omega$ and its use in the above theorem can be generalized to create the concept that is the subject of this section. From the idea that the real line is covering the simple closed curved via the wrapping map,  the term ``covering space" is used to refer to these generalized wrapping maps.  Just as the wrapping map was useful in our method of computing the fundamental group of the circle, covering spaces in general are useful for understanding the structure of the fundamental groups of spaces.

\begin{dfn}[covering space]
\index{covering space}
Let $X$, $\wt X$ be connected, locally path connected spaces and let ${p:
\wt X\to X}$ be a continuous function. Then the pair $(\wt X,p)$ is
a {\em covering space of\/} $X$ if and only if  for each $x\in X$ there exists an open
set $U$ containing $x$ such that $p$ restricted to each component
of $p^{-1} (U)$ is a homeomorphism onto $U$.  When $(\wt X,p)$ is a covering space of $X$, we refer to the space $\wt X$ as a cover of $X$ and $p$ as a covering map.
\end{dfn}

\begin{example}
Let $X = \Sph^1$, $\wt X =\R^1$, and $p:\R^1\to \Sph^1$ be defined by
$p(t) = (\cos t,\sin t)$. Then $(\R^1,p)$ is a covering space of $\Sph^1$.
\end{example}


\begin{example}
Let $X = \Sph^1$, $\wt X =\Sph^1$, and $p:\Sph^1\to \Sph^1$ be defined by
$p(z) = z^n $, where $z\in \C$ is a complex number with $|z|=1$. Then $(\Sph^1,p)$ is a covering space of $\Sph^1$.
\end{example}

\begin{example}
If $X\cong$ the wedge of two circles, $\wt X$'s are as Figure \ref{coversofw2c}, and $p$'s are the maps indicated, then in each case, $(\wt X,p)$ is a covering space of X.

\begin{figure}
\begin{center}
\includegraphics[width = 4 in]{ImagesChapter3/pdf/coversofw2c}
\caption{\label{coversofw2c} Several coverings of the wedge of two circles }
\end{center}
\end{figure}



\end{example}


\begin{example}
Let $\Sigma^2$ be a PL non-separating, two-sided, properly embedded surface
in a connected 3-manifold $M^3$.  Gluing two copies of $M^3-N(\Sigma^2)$
together gives a covering space of $M^3$. See Figure \ref{nonseparating}.

\begin{figure}
\begin{center}
\includegraphics[width = 3 in]{ImagesChapter3/pdf/nonseparating}
\caption{\label{nonseparating}}
\end{center}
\end{figure}


\end{example}

\begin{thm}
Let $(\wt X,p)$ be a covering space of $X$. If $x$, $y\in X$, then
$|p^{-1} (x)| = |p^{-1} (y)|$.
\end{thm}

\begin{dfn}[$n$-fold covering]
\index{covering, $n$-fold} \index{covering, degree}
If $(\wt X, p)$ is a covering space of a space $X$ and $n= |p^{-1}(x)|$
for some $x\in X$, then $(\wt X,p)$ is called an {\em $n$-fold\/} covering of
$X$. We also say $\wt{X}$ is a cover of {\em degree $n$}.
\end{dfn}

\begin{example}
The example above, namely, $p:\Sph^1\to \Sph^1$ defined by
$p(z) = z^n $, where $z\in \C$, is an $n$-fold covering of $\Sph^1$
by itself.
\end{example}

\begin{exercise}
{\rule{0cm}{0.1cm}}\newline\vspace*{-0.5cm}
\be
\item Describe two non-homeomorphic 2-fold covers of the Klein bottle.
\item Describe all non-homeomorphic 2-fold covers of the wedge of two circles.
\item Describe all non-homeomorphic 3-fold covers of the wedge of two circles.
\ee
\end{exercise}

There is a quick way of
eliminating many possibilities from the potential $n$-fold
coverings of a surface.

\begin{thm}
Let $F$ be a compact connected surface and $p_n:\wt{F}\to F$ be
an $n$-fold covering of $F$. Then $F$ is a
compact surface and $\chi(\wt{F})=n\chi(F)$.
\end{thm}

\begin{thm}
Let $F$ be a compact connected orientable manifold and $p: \wt{F}\to F$ be an $n$-fold covering of $F$. Then $\wt{F}$ is orientable.
\end{thm}

\begin{exercise}
{\rule{0cm}{0.1cm}}\newline\vspace*{-0.5cm}
\be
\item Describe all non-homeomorphic 3-fold covers of the Klein bottle.
\item Describe all non-homeomorphic 2-fold covers of $\T^2\num\T^2$.
\item Describe all non-homeomorphic 3-fold covers of $\T^2\num\T^2\num\T^2$.
\item Describe all non-homeomorphic 3-fold covers of $\PP^2$.
\ee
\end{exercise}

\begin{dfn}[lift of a function]
\index{lift of a function}
\index{cover!lift to}
Given a covering space $(\wt X,p)$ of $X$ and a continuous function
${f:Y\to X}$,
then a continuous function $\wt f: Y\to \wt X$ is called a {\em lift of\/}
$f$ if $p\circ \wt f =f$.

$$\xymatrix{
& \widetilde{X} \ar[d]_{p} \\
Y \ar[ur]^{\tilde{f}} \ar[r]_{f} & X\\ 
}$$


\end{dfn}

\begin{thm}
Let $(\R^1,\omega)$ be the standard wrapping map covering of $\Sph^1$.  Then any path $f:[0,1]\to\Sph^1$ has a lift.
\end{thm}

\begin{thm}
If $(\wt X,p)$ is a cover of $X$, $Y$ is connected, and $f$, $g:Y\to \wt X$
are continuous functions such that $p\circ f= p\circ g$, then $\{ y\mid
f(y) = g(y)\}$ is empty or all of $Y$.
\end{thm}

\begin{thm}
Let $(\wt X,p)$ be a cover of $X$ and let $f$ be a path in $X$. Then for
each $x_0\in \wt X$ such that $p(x_0) = f(0)$, there exists a unique
lift $\wt f$ of $f$ satisfying $\wt f(0) = x_0$.
\end{thm}

\begin{question}
Let $p$ be a $k$-fold covering of $\Sph^1$ by itself and $\alpha$
a loop in $\Sph^1$ which when lifted to $\R^1$ by the standard lift
has $\wt{\alpha}(0)=0$ and $\wt{\alpha}(1)=n$. For which integers $n$ does $\alpha$ lift to a loop in
the $k$-fold covering?
\end{question}

\begin{thm}[Homotopy Lifting Lemma]
\index{Homotopy Lifting Lemma}
Let $(\wt X,p)$ be a cover of $X$ and $\alpha$, $\beta$ be two paths in $X$.
If $\wt \alpha$, $\wt \beta$ are lifts of $\alpha$, $\beta$ satisfying
$\wt\alpha (0) =\wt\beta (0)$, then $\wt \alpha\sim\wt\beta$ if and only if
$\alpha\sim\beta$.
\end{thm}


\begin{thm}
If $(\wt X,p)$ is a cover of $X$, then $p_*$ is a monomorphism ({\em i.e.},
1--1 or injective) from $\pi_1(\wt X)$ into $\pi_1(X)$.
\end{thm}

The previous theorem implies that the fundamental group of a cover of $X$ is isomorphic to a subgroup of the fundamental group of the space $X$.

\begin{thm}
Let $(\wt X,p)$ be a cover of $X$, $\alpha$ a loop in $X$, and $\wt x_0
\in\wt X$ such that $p(\wt x_0) =\alpha (0)$. Then $\alpha$ lifts to a
loop based at $\wt x_0$ if and only if $[\alpha]\in p_* (\pi_1(\wt X,\wt {x}_0))$.
\end{thm}

\begin{exercise}
Restate the proof of the fact that $\pi_1(\Sph^1)\cong \Z$ in terms of covering spaces.
\end{exercise}


\begin{thm}
Let $(\wt X,p)$ be a covering space of $X$ and let $x_0\in X$. Fix $\wt x_0\in p^{-1}(x_0)$. Then a subgroup $H$ of $\pi_1(X,x_0)$
is in $\{ p_* (\pi_1(\wt X,\wt x))\}_{p(\wt x)=x_0}$ if and only if $H$ is a
conjugate of $p_* (\pi_1(\wt X,\wt x_0))$.
\end{thm}

\begin{thm}
Let $(\wt X,p)$ be a covering space of $X$.
Choose $x\in X$, then $|p^{-1}(x)| = [ \pi_1(X): p_* (\pi_1(\wt{X}))]$.
\end{thm}

So, the index of the subgroup of $\pi_1(X)$ corresponding
to a finite covering $(\wt{X},p)$ equals the degree of the covering.

\begin{exercise}
Describe a 3-fold cover $(\wt X,p)$ of $\Sph^1$ and the
subgroup
$p_* (\pi_1(\wt X))$ of $\pi_1(\Sph^1)$.
\end{exercise}

\begin{thm}
Let $(\wt X,p)$ be a covering space of $X$ and $\wt x_0\in\wt X$, $x_0\in X$
with $p(\wt x_0)=x_0$. Also let $f: Y\to X$ be continuous where $Y$ is
connected and locally path connected and $y_0 \in Y$ such that
$f(y_0) =x_0$. Then there is a lift $\wt f:Y\to \wt X$ such that
$p\circ \wt f=f$ and $f(y_0) = \wt x_0$ if and only if $f_* (\pi_1(Y,y_0))\subseteq
p_* (\pi_1(\wt X,\wt x_0))$. Furthermore, $\wt f$ is unique.

\end{thm}

\begin{exercise}
Let $X= \Sph^1$, $\wt X = \R$, $(\wt X,\omega)$ be the covering space of $X$
given by the standard wrapping map,
and $Y$ as in Figure \ref{counterexample}. When does a map $f:Y\to X$ not have a lift? Why is this example here?
\end{exercise}

\begin{figure}
\begin{center}
\includegraphics[width = 2 in]{ImagesChapter3/pdf/counterexample}
\caption{\label{counterexample} Consider this picture}
\end{center}
\end{figure}



\begin{dfn}[cover isomorphism]
\index{cover isomorphism}
Let $(\wt X_1,p_1)$ and $(\wt X_2,p_2)$ be covering spaces of $X$.
Then a map ${f:\wt X_1\to \wt X_2}$ such that $f$ is a homeomorphism and $p_2\circ f=p_1$ is called
a {\em cover isomorphism\/}.
\end{dfn}

\begin{thm}
Let $(\wt X_1,p_1)$ and $(\wt X_2,p_2)$ be covering spaces of $X$.
Let $\wt x_1\in \wt X_1$ and $\wt x_2\in \wt X_2$ such that $ p_1(\wt x_1)
= p_2(\wt x_2)$. Then there is cover isomorphism $f:\wt X_1\to \wt X_2$
with $f(\wt x_1)= \wt x_2$ if and only if $p_* (\pi_1 (\wt X_1,\wt x_1)) = p_*
(\pi_1(\wt X_2,\wt x_2))$.
\end{thm}

\begin{dfn}[covering transformation]
\index{covering transformation}
Let $(\wt X,p)$ be a covering space. Then a cover isomorphism from $\wt X$
to itself is called a {\em covering transformation\/}. The set of
covering transformations, denoted ${\mathcal{C}}(\wt X,p)$, is a group where the group operation is composition.
\end{dfn}

\begin{exercise}
What is ${\mathcal{C}}(\wt X,p)$ for the covering space of the figure eight shown in Figure \ref{counterexample}.

\begin{figure}
\begin{center}
\includegraphics[width = 2 in]{ImagesChapter3/pdf/4foldcw2c}
\caption{\label{counterexample} A covering of the figure 8 or wedge of two circles}
\end{center}

\end{figure}

\end{exercise}

\begin{thm}
If $(\wt X,p)$ is a covering space of $X$ and $f\in {\mathcal{C}}(\wt X,p)$,
then $f= Id_{\wt X     }$ if and only if $f$ has a fixed point.
\end{thm}

\begin{dfn}[regular cover]
\index{regular covering}
\index{covering!regular}
Let $(\wt X,p)$ be a covering space of $X$. If $p_* (\pi_1(\wt X) )
\triangleleft \pi_1(X)$, then $(\wt X,p)$ is a {\em regular\/}
covering space.
\end{dfn}

\begin{question}

Consider the second three-fold covering space of the figure eight in Example 3. Find an element of $p_*(\pi _1(\wt X))$ which, when conjugated, is not in $p_*(\pi _1(\wt X))$.
\end{question}

\begin{thm}
If $(\wt X,p)$ is a regular covering space of $X$ and $x_1,x_2\in \wt X$ such
that $p(x_1) = p(x_2)$, then there exists a unique $h\in {\mathcal{C}}(\wt X,p)$
such that $h(x_1)= x_2$.
\end{thm}

\begin{question}
The preceding theorem tells us that for a regular covering space,
there is a (unique) covering transformation carrying
any point in the set $p^{-1}(x)$ to any other point in the same set.
Is this true of an irregular covering space?
\end{question}

\begin{thm}
A covering space is regular if and only if for every loop either all its lifts are loops or all its lifts are paths that are not loops.
\end{thm}


\begin{exercise}
{\rule{0cm}{0.1cm}}\newline\vspace*{-0.5cm}
\be
\item Describe all regular 3-fold covering spaces of a figure eight.
\item Describe all irregular 3-fold covering spaces of a figure eight.
\item Describe all regular 4-fold covering spaces of a figure eight.
\item Describe all irregular 4-fold covering spaces of a figure eight.
\item Describe all regular 3-fold covering spaces of a wedge of 3 circles.
\item Describe all regular 4-fold covering spaces of a wedge of 3 circles.
\ee
\end{exercise}


%\begin{thm}
%If $(\wt X,p)$ is a regular covering space of $x$ and $f$ is a
%homomorphism from $\wt X$ to $\wt X$, then  $f\in {\mathcal{C}}(\wt X,p)$.
%\end{thm}

There is an important correspondence between the covering transformations of regular
covers of $X$ and the normal subgroups of $\pi_1(X)$.

\begin{thm}
Let $(\wt X,p)$ be a regular covering space of $X$. Then ${\mathcal{C}}(\wt X,p)
\cong \pi_1 (X)/p_* (\pi_1(\wt X))$. In particular, ${\mathcal{C}}(\wt X,p)
\cong \pi_1 (X)$ if $\wt{X}$ is simply connected.
\end{thm}


\begin{exercise}
Observe that the standard wrap map is a regular covering map of $\Sph^1$ by $\R^1$.  Describe the covering transformations for this covering space. Describe the covering map that maps $\R^2$ to the torus $\T^2$ and describe the covering transformations for this covering space.
\end{exercise}


\begin{dfn}[semi-locally simply connected]
\index{semi-locally simply connected}
\index{simply connected!semi-locally}
A space $X$ is called {\em semi-locally simply connected\/} if and only if every
$x\in X$ is contained in an open set $U$ such that every loop in $U$
based at $x$ is homotopically trivial in $X$.
\end{dfn}

Note that $U$ need not be simply connected itself.

\begin{thm}[Existence of covering spaces]
Let $X$ be connected, locally path connected, and semi-locally
simply connected. Then for every $G< \pi_1 (X,x_0)$ there is a
covering space $(\wt X,p)$ of $X$ and $\wt x_0\in\wt X$
such that $p_* (\pi_1(\wt X,\wt x_0)) =G$. Furthermore, $(\wt X,p)$
is unique  up to isomorphism.
\end{thm}

\begin{dfn}[universal cover]
\index{universal cover}
\index{cover!universal}
A connected, locally path connected cover is called {\em universal\/}
if  and only if  its fundamental group is trivial.
\end{dfn}

\begin{cor}
Every connected, locally path connected, semi-locally simply
connected space has a universal covering space.
\end{cor}

\begin{exercise}
Find universal covers for the Klein bottle, torus, and projective plane.
Then show explicitly that ${\mathcal{C}}(\wt X,p) \cong \pi_1 (X)$.
\end{exercise}

\section{Theorems about groups}

The algebra of a fundamental group tells us something about the
topology of its corresponding space. It is also possible to use topology
to study algebraic groups. One can construct a complex that represents
any finitely generated group, and use the fundamental group and covering spaces to deduce
properties of the group.


\begin{thm} 
Every tree is simply connected.
\end{thm}

\begin{thm} Let $G$ be a graph, and $T$ be a maximal tree in $G$.
Then if $\{e_1,\ldots e_n \}$ is the set of edges that are not in $T$,
$\pi_1(G)=F_n$, the free group on $n$ generators; and there is a system of
generators that are in one-to-one correspondence with the
edges $\{e_1,\ldots e_n \}$.
\end{thm}

\begin{cor}
A subgroup $H$ of a free group $F_n$ is always a free group.
\end{cor}

\begin{question}
Describe a regular $k$-fold cover $\wt X$ of a wedge of $n$-circles.
What is the number of generators of $\pi_1(\wt{X})$, given $k$ and $n$? What does this
tell us about the normal
subgroups of finite index of the free group with $n$ generators?
\end{question}


\begin{exercise}{\rule{0cm}{0.1cm}}\newline\vspace*{-0.5cm}
\be
\item Let $F$ be a free group on $n$ letters. Let $G\subset F$ be of finite index $k$
and contain $7$ free generators. What can the value of $n$ be?
\item Let $F$ be a free group on $n$ letters. Let $G\subset F$ be of finite index $k$
and contain $4$ free generators. What can the value of $n$ be?
\item Let $F$ be a free group on $n$ letters. Let $G\subset F$ be of finite index $k$
and contain $24$ free generators. What can the value of $n$ be?
\ee
\end{exercise}

\begin{exercise}
Let $S_k$ be the set of all $k$-fold covers of $\K$, the Klein Bottle.
\be
\item Describe $S_k$.
\item Describe all subgroups of $\pi_1(\K)$ of finite index.
\ee
\end{exercise}

\begin{exercise}
Let $F$ be the closed orientable surface of genus $2$, and $G=\pi_1(F)$.
Show that all subgroups of $G$ of finite index $k$ are isomorphic.
\end{exercise}