 \chapter{Topology}\label{top chap}\index{topology|(}
 \markboth{\quad\thechapter\ Topology\hfill}{}

The word topology in the context of this course has two quite different
meanings:

\paragraph{``Point-set topology''} Slogan: a topological space is a ``metric
space without a metric''. In analysis, this idea leads to a fairly minor
generalisation of the definition of metric space, but the definition of
topology has applications in other areas of math, where it turns out to be
logical or algebraic in content. I give the abstract definition and some
examples of topological spaces that are definitely not metric. This is an
important ingredient in all advanced math (algebra, analysis, arithmetic,
geometry, logic, etc.). Topology has lots of advantages even when the only
spaces of interest are metric spaces. It provides, in particular, a simple
rigorous language for ``sufficiently near'' without epsilons and deltas.

 \paragraph{``Rubber-sheet geometry''}
 \index{winding number|(}
 \index{M\"obius strip}
The abstract language gives us tools to study spaces that are geometric in
origin, such as the torus and the M\"obius strip. Geometric concepts in
topology include the winding number and the number of holes of a surface.

\bigskip

Here is a sample of the results proved in this chapter:

 \begin{enumerate}

 \item\index{homeomorphism} If $f\colon S^1\to\Si\subset \R^2$ is
bijective and continuous, then the inverse map $f\1\colon\Si\to S^1$
is also continuous; that is, $f$ is a {\em homeomorphism}. Joke:
topology is geometry in which
 \[
 \heartsuit=0.
 \]
Imagine trying to prove this from first principles! The point is that
$f$ can be very complicated, and $f\1$ might not be given by any
simple function.

 \item The cylinder is different from the M\"obius strip. \index{M\"obius
 strip}

 \item The winding number:\index{winding number|)} let
 $\fie\colon[0,1]\to\R^2\setminus(0,0)$ be a continuous map with
 $\fie(0)=\fie(1)$. Then the number of times $\fie$ winds around the origin
 is not changed by deforming the loop\index{loop} continuously; in other
 words, the winding number is a {\em homotopy invariant} of the map $\fie$.
 \end{enumerate}

 \section{Definition of a topological space}\label{defn top}
 \markright{\hfill\thesection\ Definition of a topological space\quad}
 Let $X$ be a set. A {\em topology} on $X$ is a collection $\sT$ of subsets
 of~$X$ satisfying the following three axioms:

 \begin{itemize}

 \item {\bf Finite intersection:}
 $U_1,\dots,U_n\in\sT\implies U_1\cap\cdots\cap U_n\in\sT$;

 \item {\bf Arbitrary union:}
 $U_\la\in\sT$ for $\la\in\La$ $\implies \Cup_{\la\in\La}U_\la\in\sT$, where
$\La$ is an arbitrary indexing set;

 \item {\bf Conventions on empty set:}\index{empty set}
$\emptyset,X\in\sT$.
 \end{itemize}

A {\em topological space} is a pair $X,\sT$ consisting of a set $X$ and a
topology $\sT=\sT_X$ on it. $U\in\sT$ is called an {\em open set}\index{open
set|(} of the topology $\sT$. We often speak of the topological space $X$
and its open sets $U$, omitting $\sT$ from the notation when it is clear
what topology is intended. $V\subset X$ is {\em closed}\index{closed} if its
complement is open; the topology could be specified equally well by the
collection of closed sets, which enjoys finite union and arbitrary
intersection.\index{intersection} If $Z\subset X$, the {\em closure} of $Z$,
denoted $\overline{Z}$, is the intersection of all closed sets containing
$Z$. By the arbitrary intersection property of closed sets, $\overline{Z}$
is closed; it clearly contains $Z$. A {\em neighbourhood\/} of a point $x\in
X$ is any subset $V\subset X$ containing an open set containing~$x$.

We will see presently that if $X$ is a metric space then there is a natural
choice of open sets of $X$ which form a topology. Here are some simpler examples.

 \paragraph{Example~1} Let $X=\{P_1,P_2,P_3\}$ be a set consisting of
3 points, and
 \[
\sT_X=\{\emptyset,\{P_1\},\{P_1,P_2\},X\}.
 \]
Then $\{P_1\}$ is open, but every neighbourhood of $P_2$ contains $P_1$, and
every neighbourhood of $P_3$ contains both $P_1,P_2$.

\paragraph{Example~2} There are two extreme topologies defined on any set
$X$. The {\em discrete topology}\index{discrete topology} has every subset
open. The {\em indiscrete topology}\index{indiscrete topology} has no open
sets except $\emptyset$ and $X$ itself.

\paragraph{Example~3} The {\em cofinite
topology}\label{cofinite}\index{cofinite topology} on an infinite set $X$ is
the topology for which the open sets are $\emptyset$ or the complements of
finite sets; that is, $U\subset X$ is open if and only if either
$U=\emptyset$ or $X\setminus U$ is finite; it is obvious that this satisfies
finite intersection and arbitrary union. In this topology, if $x\in U$ and
$y\in V$ are neighbourhoods of any two points then $U\cap V$ is also the
complement of a finite set, and hence nonempty.

\section{Motivation from metric spaces}\label{metric motiv}
\markright{\hfill\thesection\ Motivation from metric spaces\quad} Let
$(X,d)$ and $(Y,d')$ be metric spaces (see Appendix~\ref{app!metric} if you
need reminding what this means) and $f\colon X\to Y$ a map. By definition,
$f$ is {\em continuous} if \begin{quote} for every $x\in X$ and for any
given $\ep>0$, there exists $\de>0$ such that $d(x,y)<\de\implies
d'(f(x),f(y))<\ep$. \end{quote} The intuitive meaning is clear without
epsilons and deltas: if $x\in X$ is any given point, I can guarantee that
$f(y)$ is arbitrarily close to $f(x)$ by forcing $y$ to be sufficiently
close to $x$.

The idea of topology on a space is to break up the definition of continuity
into two steps. First use the metric to derive the open sets and
neighbourhoods of points; then describe continuity in terms of open sets.

\paragraph{Definition} If $(X,d)$ is a metric space, a set $U\subset X$ is a
{\em neighbourhood} of $x$ if $B(x,\ep)\subset U$ for some $\ep$. Here
$B(x,\ep)$ is the open ball\index{ball} of radius $\ep$ centred at $x$; if
you cannot guess the formal definition, look in Appendix~\ref{app!metric}. A
set $U\subset X$ is {\em open} if it is a neighbourhood of every one of its
points, that is, for all $x\in U$, $B(x,\ep)\subset U$ for some $\ep$. The
open sets $U$ of $X$ form a topology on $X$, the {\em metric
topology}\index{metric!topology} of $(X,d)$. (See~\exref{metric_top_ex}.)

 \paragraph{Equivalent conditions} Standard easy result on metric spaces:
 \begin{align*}
 \hbox{$f$ is continuous}\; &\iff\;
 \begin{array}{l}
 \hbox{$\forall\ x\in X$ and $\forall$ neighbourhood $V\subset Y$,}\\
 \hbox{$f\1V\subset X$ is a neighbourhood of $x$}
 \end{array} \\[6pt]
&\iff\; \hbox{$\forall$ open $V\subset Y$, $f\1V\subset X$ is open.}
 \end{align*}

In other words, the ``epsilon-delta'' definition of continuity for metric
spaces can be replaced by an equivalent condition which involves only open
sets of the metric topology. I will adopt this equivalent condition
in~\ref{homeo cont} to define continuity for a map between arbitrary
abstract topological spaces.

The idea of a topological space is a natural abstraction and generalization
of the idea of a metric space. When going from a metric space $(X,d)$ to the
corresponding topological space, we forget the metric, and keep only the
notion of neighbourhoods, or equivalently open sets.\index{open set|)}
There are several advantages. In the context of metric spaces, closeness means
that the distance $d(x,y)$ is small. But just as some things in life have
a value that cannot be expressed as a sum of money, in some contexts
closeness cannot always be expressed as a distance measured as a real number.
In particular, the following three properties are forced on metric spaces by
definition, but are optional for topological spaces.
 \begin{enumerate}

 \item Symmetry: in a metric space, $x$ is close to $y$ if and only
if~$y$ is close to~$x$.

 \item Hausdorff property:\index{Hausdorff property} given two points
$x\ne y\in X$, there exist disjoint open sets\index{open set} $x\in U$
and $y\in V$ (see Figure~\ref{hausdorff fig}).
 \begin{figure}[ht]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}a}
 \centering
 \leavevmode\hbox{\epsfysize=4.9 true cm\epsffile{top_smilies.eps}}
 \caption{Hausdorff property}\index{Hausdorff property}
 \label{hausdorff fig}
 \end{figure}

 \item Countable neighbourhoods: given a point $x\in X$ of a metric
space, consider the family $B_n=B(x,\frac{1}{n})$. Then $B_n$ are
neighbourhoods of $x$; they are countable in number; every
neighbourhood of $x$ contains a $B_n$; and $\Cap B_n=\{x\}$. This can
be used in convergence arguments in analysis (see \exref{seqcont ex}).

 \end{enumerate}

The idea of having the open sets\index{open set} specified as the
basic construction is of course more abstract and less intuitive than
definitions in first analysis or metric spaces courses, but
abstractness has its own advantages. In many cases, the spaces I am
interested in may actually be metric spaces, but I may not really
care about the distances, just in what it means for $d(x,y)\ll1$. For
example, if you think of the circle $S^1\subset\R^2$ as the
identification space obtained by glueing together the ends of the
interval $[0,1]$, then $S^1$ is a metric space, with metric
 \[
 d_{S^1}(P,Q)=\min
 \begin{pmatrix}
 d_{[0,1]}(P,Q),\quad d_{[0,1]}(0,P)+d_{[0,1]}(Q,1),\\
 d_{[0,1]}(0,Q)+d_{[0,1]}(P,1)
 \end{pmatrix},
 \]
which is a fairly tedious expression to work with; but I really do not
care about the metric, only the system of arbitrarily small
neighbourhoods of points. A small neighbourhood of any point other
than the ``seam'' $P_0$, the image of the endpoints $0,1$, is given
by $(x-\ep,x+\ep)$ from the interior of the interval. For $P_0$, you
glue together small neighbourhoods of the glued endpoints:
$[0,\ep)\cup (1-\ep,1]$; see Figure~\ref{S1}.

 \begin{figure}
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}b}
 \centering
 \leavevmode\hbox{\epsfysize=3.5 true cm\epsffile{top_create_S1.eps}}
 \caption{$S^1=[0,1]$ with the ends identified}
 \label{S1}
 \end{figure}

As a final example, note that the discrete topology\index{discrete
topology} on any set $X$, defined in \ref{defn top}, Example 2, is
metric: just set $d(x,y)=1$ for every $x\ne y$. On the other hand,
the indiscrete topology is not metric.

 \section{Continuous maps and homeomorphisms}\label{homeo cont}
 \index{homeomorphism}
 \markright{\hfill\thesection\ Continuous maps and
homeomorphisms\quad}
 \subsection{Definition of a continuous map}\label{def_cont_sect}
 If $X$ and $Y$ are topological spaces, a map $f\colon X\to Y$ is
{\em continuous} if
 \[
 \hbox{$f\1(U)\subset X$ is open\index{open set} for every open
 $U\subset Y$.}
 \]
 Notice that I am already omitting mention of the topologies $\sT_X$
and $\sT_Y$. To use the language literally, I should have said the
following: let $X,\sT_X$ and $Y,\sT_Y$ be topological spaces, then $f$
is continuous
if
 \[
 U\in\sT_Y\implies f\1(U)\in\sT_X.
 \]

 \paragraph{Example~1} If $X$ is any set with the discrete topology of
\ref{defn top}, Example 2, then every map $X\to Y$ from $X$ to any
topological space is continuous. If $X$ has the indiscrete
topology\index{indiscrete topology}, then every map $Y\to X$ from any
topological space to $X$ is continuous.

 \paragraph{Example~2} Consider an infinite field $k$ with the
cofinite topology\index{cofinite topology} on it (see
\ref{defn top}, Exam\-ple 3). Let $f\colon k\to k$ be a polynomial
map given by $a\mapsto f(a)$, where $f$ is a polynomial in one
variable. Then $f$ is continuous. For $U\subset k$ is open\index{open
set} if and only if $U=\emptyset$ or $U$ is the complement of a finite
set, say $U=k\setminus\{b_1,\dots,b_n\}$; then $f(x)=b_i$ has at most
$\deg f$ solutions, so that $f\1(U)$ is also the complement of a
finite set.

 \subsection{Definition of a homeomorphism}\label{def_of_homeom}
 \index{homeomorphism}
 A map $f\colon X\to Y$ is a {\em homeomorphism} if $f$ is bijective,
and both $f$ and $f\1$ are continuous. This means that
 \[
f\colon X\bij Y\qquad\hbox{and}\qquad\sT_X\bij\sT_Y,
 \]
 or in other words, $f$ is an isomorphism of all the structure there
is. $X$ and $Y$ are homeomorphic, written $X\simeq Y$, if there
exists a homeomorphism\index{homeomorphism} $f\colon X\to Y$.

 \paragraph{Example~3} An open interval $(a,b)$ is homeomorphic to the
real line, $(a,b)\simeq\R$. For example, the map
 \[
 f\colon (0,1)\to\R \quad \hbox{defined by}\quad
 f(x)=\frac{-1}{x}+\frac{1}{1-x}=\frac{2x-1}{x(1-x)}
 \]
 is a homeomorphism, illustrated on Figure~\ref{k}.\index{homeomorphism}

 \begin{figure}[ht]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}a}
 \centering
 \leavevmode\hbox{\epsfysize=8 true cm\epsffile{top_interval_R.eps}}
 \caption{$(0,1)\simeq\R$}
 \label{k}
 \end{figure}

\paragraph{Example~4}\index{homeomorphism} The square is homeomorphic to the
circle in $\R^2$. To see this, put the square inside the circle and project
out from an interior point (see Figure~\ref{l}). A similar radial projection
argument shows also that the full square is homeo\-morphic to the
closed\index{closed} disc\index{disc} $\{x^2+y^2\le 1\}\subset\R^2$.
In\index{homeomorphism!criterion} Theorem~\ref{homeo crit} below I show that
if $f\colon S^1\to\R^2$ is any one-to-one and continuous map (that is, a
simple closed curve) then $f$ is a homeo\-morphism.
 \begin{figure}
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}b}
 \centering
 \leavevmode\hbox{\epsfysize=4.6 true cm\epsffile{top_square_circle.eps}}
 \caption{Squaring the circle}
 \label{l}
 \end{figure}

\paragraph{Example~5}\index{isometry} If $(X, d)$ and $(Y, d')$ are metric
spaces and $f\colon X\to Y$ is an isometry, then $f$ is a homeomorphism.
Note however a map $f$ can set up a homeomorphism between (the metric
topologies of) metric spaces {\em without} being an isometry, as in
Examples~3 and~4 above. Being homeomorphic is a much coarser relation on
metric spaces than being isometric. Example~\ref{comp} discusses this issue
from a slightly different point of view.
 \subsection{Homeomorphisms and the Erlangen program}\label{sec!separation}
 \index{Erlangen program}
The group $\Homeo(X)$ of self-homeomorphisms is a transformation
group\index{transformation group} of the topological space $X$
(compare~\ref{transf as group}). In the framework of the Erlangen program of
Section~\ref{sec!erlan}, topology can be viewed as the study of properties
invariant under $\Homeo(X)$.

The homeomorphism group of $X=\R$ is already an uncomfortably large infinite
group, and its action mixes up the points of $\R$ like anything, so at first
sight it seems hard to imagine how any invariant properties can survive.
However, such properties do exist; one example is {\em between-ness}, or
{\em separation}, derived from the order relation of $\R$: a homeomorphism
$f$ takes three real numbers $x,y,z$ with $y$ between $x$ and $z$ into
$f(x),f(y),f(z)$ with $f(y)$ between $f(x)$ and $f(z)$; this follows at once
from the intermediate value theorem.

If a geometry has lines which are homeomorphic copies of the real line $\R$,
then the separation property can be formulated in the geometry: a point cuts
a line into two disconnected subsets, and hence it makes sense to ask
whether a point $Q$ on a line {\em lies between} two other points $P,R$.
Euclidean and hyperbolic geometry are examples where this property holds. In
contrast, the lines (great circles) of spherical geometry have the topology
of the circle $S^1$, so they have the ``no separation'' property: cutting a
point leaves behind a set which is still connected. See~\ref{sec!history}
for the historic significance of this issue.

 \subsection{The homeomorphism problem}\label{homeo prob}
 \index{homeomorphism!problem|(}
The following $5$ spaces are not homeomorphic\index{homeomorphism} (for
proofs, please be patient until~\ref{sec!using}):
 \begin{enumerate}
 \renewcommand{\labelenumi}{\theenumi}
 \renewcommand{\theenumi}{(\arabic{enumi})}

 \item the closed\index{closed} interval $[a,b]$;

 \item the open interval $(a,b)\simeq\R$;

 \item the circle $S^1$;

 \item the plane $\R^2$;

 \item the sphere\index{sphere $S^2$} $S^2\subset\R^3$.
 \end{enumerate}

The examples here and in~\ref{def_of_homeom} illustrate an important general
point. If you want to prove that two given topological spaces $X$ and $Y$
are homeo\-morphic,\index{homeomorphism} then it is your job to supply a
homeo\-morphism $f\colon X\to Y$, for example by a geometric construction;
or at least, to prove that one exists. On the other hand, to prove that $X$
and $Y$ are {\em not} homeomorphic, you need to find some property of spaces
that are the same for homeomorphic spaces, but different for $X$ and $Y$.
This is called the ``homeomorphism problem''.

The next few sections introduce some basic notions of topology and use them
to prove assertions of this type. Algebraic topology\index{algebraic
topology} has as one of its main aims to develop systematic invariants of
topological spaces that can be used to prove that spaces are not
homeomorphic, notably the fundamental group\index{fundamental!group}
$\pi_1(X,x_0)$ and homology groups $H_i(X,\Z)$; but in this book I work only
with very simple ideas. \index{homeomorphism!problem|)}

\section{Topological properties}\label{top prop}
\markright{\hfill\thesection\ Topological properties\quad} Some properties
of a topological space depend only on the topology. A {\em topological
property} of topological spaces is a property that can be expressed in terms
of points and open sets\index{open set} only. {\em Homeomorphisms preserve
topological properties.}\index{homeomorphism}

For example, if $X$ is a metric space, then {\em bounded} is not a
topological property: it depends on distance ($d(x,y)\le K$ for some $K$),
and not just on the topology. Thus $(a,b)\simeq\R$ (see Figure~\ref{k}), but
the left-hand side is bounded, while the right-hand side is not.

 \subsection{Connectedness}\label{subsec!connected}

A topological space $X$ is {\em connected},\index{connected} if it
cannot be written as a disjoint union of two nonempty open\index{open
set} subsets; that is, there does not exist any decomposition
 \[
 X=U_1\sqcup U_2\quad\hbox{with $U_1,U_2$ open},
 \]
where $\sqcup$ denotes disjoint union.

A {\em path}\index{path} in a space $X$ is a continuous map
$\fie\colon[0,1]\to X$; $X$ is {\em path
connected}\index{path!connected} if for any two points $x_1,x_2\in
X$, there exists a path $\fie$ with $\fie(0)=x_1$ and $\fie(1)=x_2$
(that is, any two points can be joined by a path).

 \begin{figure}
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}b}
\centering
 \leavevmode\hbox{\epsfysize=4.6 true cm\epsffile{top_path.eps}}
\caption{Path connected set}\index{path!connected}
\label{m}
 \end{figure}

Connected and path connected are both topological properties, since only
open sets\index{open set} and continuous maps appear in their definitions.
Thus given two spaces $X,Y$, if $X\simeq Y$ then $X$ and $Y$ are either both
(path) connected or both (path) disconnected.

 \begin{lem} \begin{enumerate} \renewcommand{\labelenumi}{\theenumi}
\renewcommand{\theenumi}{(\arabic{enumi})}
 \item The interval $[0,1]$ is connected.\index{connected}

 \item A path connected set is connected.

 \end{enumerate} \end{lem}

 \begin{pf} For (1), suppose $[0,1]=U_1\sqcup U_2$ with opens
$U_1,U_2\subsetneq [0,1]$. Say $0\in U_1$, and consider
 \[
 z=\sup\bigl\{x \bigm| [0,x]\subset U_1 \bigr\},
 \]
where $\sup$ is least upper bound from your first analysis course. The sup
exists by the completeness axiom of the reals. If $z\in U_1$, then because
$U_1$ is open, there is a neighbourhood of $z$ in $U_1$, that is,
$[z,z+\ep)\subset U_1$ for some $\ep>0$, so $z$ is not an upper bound. If
$z\in U_2$, there is a neighbourhood of $z$ in $U_2$, so an interval
$(z-\ep, z]$ disjoint from $U_1$ and so $z-\ep$ is a strictly smaller lower
bound, which also contradicts the definition of $z$ as sup. (The proof is
the same as that of the intermediate value theorem in a first analysis
course.)

To show (2), suppose $X$ is path connected\index{path!connected} and
$X=U_1\sqcup U_2$ with opens $U_1,U_2\subsetneq X$. Then choose $x\in U_1$
and $y\in U_2$ and apply the definition of path
connected,\index{path!connected} so that there is a continuous map
$\fie\colon [0,1]\to X$ with $\fie(0)=x$ and $\fie(1)=y$. Then
$[0,1]=\fie\1(U_1)\sqcup\fie\1(U_2)$ is a disjoint union, with both
$\fie\1(U_1)$ and $\fie\1(U_2)$ open and nonempty, which contradicts (1).
\QED \end{pf}

If $X$ is any topological space, define a relation on $X$ by setting $x\sim
y$ if and only if there is a connected\index{connected!component} subset $U$
of $X$ containing $x,y$. It is clear that $\sim$ is symmetric and reflexive,
and a bit of thought tells you that it is also transitive, hence it is an
equivalence relation. Equivalence classes of $\sim$ are called {\em
components} of the topological space $X$.

\paragraph{Remark} The property used to define a path connected space
corresponds to our usual perception of ``connectedness'': you can get from
point $A$ to point $B$ using an unbroken ``path''. In the context of surface
travel, mainland Eurasia forms a path connected space but the United States
does not: you cannot get from New York to Alaska without crossing Canada or
going by air or sea. However, in the context of general topological spaces,
connectedness as defined above, without reference to paths, is preferable.
By the Lemma, path connectedness implies this more general form of
connectedness. Similar remarks apply to components: the definition is a
natural extension of the obvious notion under which the connected components
of the United Kingdom include mainland Britain and mainland Northern
Ireland, along with any number of smaller islands around the coast.

 \subsection{Compactness}\label{comp} \index{compact|(}

The space $X$ is {\em compact} if
\begin{quote} for every cover
$X=\Cup_{\la\in\La}U_\la$ of $X$ by an arbitrary collection of opens
$U_\la$,\index{open set} there exists a finite number of indexes
$\la_1,\dots,\la_n\in\La$ such that $X=\Cup_{i=1}^n U_{\la_i}$.
\end{quote}
(Slogan: {\em every open cover has a finite subcover.}) This property
manifestly depends only on open sets.

A sequence of points $a_1,a_2,\dots$ in a topological space $X$ {\em
converges} to a limit $l\in X$, written $a_{i_n}\to l$, if for any
neighbourhood $U$ of $l$, the $a_i$ are eventually all in $U$. In other
words,
 \begin{quote}
 for every open set $U$ of $X$ with $l\in U$, there exists
 $n_0$ such that $a_i\in U$ for all $i\ge n_0$.
 \end{quote}
In other words, $a_1,a_2,\dots$ tend to $l\in X$ if, {\em for any measure of
closeness}, the $a_i$ are {\em eventually all close} to $l$.

The space $X$ is {\em sequentially compact}\index{sequentially compact|(} if
every sequence has a convergent subsequence, that is, for every infinite
sequence $a_1,a_2,\dots$ of points of $X$, there exists a point $x\in X$ and
a sequence $i_1,i_2,\dots$ of indexes such that $a_{i_n}\to x$. (Slogan:
{\em every sequence has a convergent subsequence.})

The following statement relates these notions to each other and to more
familiar ones in {\em metric spaces}.

 \begin{prop}
 \index{closed!and bounded|(}
 \begin{enumerate}\renewcommand{\labelenumi}{\theenumi}
\renewcommand{\theenumi}{(\arabic{enumi})}

 \item For $V$ a subset of\/ $\R^n$ with its usual (Euclidean) metric,
 \[
 \hbox{$V$ is closed and bounded}\ \iff\ \hbox{$V$ is sequentially
 compact.}
 \]
 \item For $X$ any metric space and $V\subset X$ a subset,
 \[
 \hbox{$V$ is sequentially compact}\ \iff\ \hbox{$V$ is compact}.
 \]
 \end{enumerate}
 \end{prop}

Here is a brief discussion of where you can find this in the literature:
Compactness is the subject of Sutherland \cite{Su}, Chapter~5. The statement
that a closed bounded subset of $\R^n$ is compact is the Heine--Borel
theorem,\index{Heine--Borel theorem} proved in \cite{Su}, Theorem~5.3.1 for
$n=1$, and in general (by reducing to the case $n=1$) in Theorem~5.7.1.
Compact implies sequentially compact (in a metric space) is proved in
\cite{Su}, Theorem~7.2.6. The other way round, sequentially compact implies
compact (in a metric space) is proved in \cite{Su}, Chapter~7. The proof is
a bit tricky, but Sutherland breaks it up into 3 self-contained steps, each
of which takes a half-page. (See also, for example, Rudin~\cite{Ru},
2.31--2.40.)

This is not primarily a course on foundational stuff in metric spaces, and I
take a commonsense approach: when I am working in a metric space, I use
compact or sequentially compact more-or-less interchangeably. With general
topological spaces, the language of compactness is more natural and more
convenient.\index{sequentially compact|)}

\paragraph{Example} Consider the {\em $n$-sphere}\index{sphere $S^n$}
 \[
 S^n = \{(x_1,\dots,x_n)\in\R^n \mid x_1^2+\cdots+x_n^2 = 1\}.
 \]
You have already seen two different metrics on $S^n$: one is the Euclidean
distance\index{Euclidean!distance} of
points on $S^n\subset\R^n$, and the other one is the spherical
distance\index{spherical!distance} $d(\bx,\by)=\arccos (\bx\cdot\by)$
(see~\ref{defn of spherical} and compare Ex.~\ref{ch!hyp}.\ref{ex!Snsetup}).
However, points are close to each
other in one of the metrics if and only if they are close in the
other; said differently, the metric topologies given by the two
metrics are {\em the same}. Under the Euclidean metric inherited from $\R^n$,
the set $S^n$ is bounded (distance $1$ from the origin) and closed (clearly)
so $S^n$ is compact by (1) of the Proposition.

 \subsection{Continuous image of a compact space is compact}\label{im compact}
 \begin{prop} Let $X$, $Y$ be topological spaces and $f\colon X\to Y$
a surjective continuous map. Then if $X$ is compact, so is $Y$.
 \end{prop}

 \begin{pf} You just have to write out the definitions: if $Y=\Cup
V_\la$, an arbitrary union of open sets, let $U_\la=f\1(V_\la)$. Then
$U_\la$ is open, and $X=\Cup U_\la$. Therefore there exists a finite
set of indexes $\la_1,\dots,\la_n$ such that $X=\Cup_{i=1}^n
U_{\la_i}$. Finally,
 \[
Y=f(X)=\Cup_{i=1}^n f(U_{\la_i})=\Cup_{i=1}^n V_{\la_i}.\QED
 \]
 \end{pf}

Pretty easy wasn't it? This shows what a convenient property compactness
is. Compare the result in analysis: a continuous function
$f\colon[a,b]\to\R$ is bounded and attains its bound. This is hard to
prove from first principles, but is really easy once you have
established the definition of compactness, and proved
Proposition~\ref{comp}.

The notion of compactness is a powerful tool, and you should learn to
use it, even if you put off studying the proofs until later. A
typical use is the kind of ``continuity implies uniform continuity''
argument used all over the place in analysis. If $f\colon[a,b]\to\R$
is continuous, then given $\ep>0$, for all $x\in[a,b]$, you can force
$f(x')$ that close to $f(x)$ by squeezing $x'$ within $\de$ of $x$;
here $\de$ depends on $x$, but compactness allows you to choose one
$\de$ that works uniformly for all $x\in[0,1]$.\index{compact|)}

There is a famous Bertrand Russell quotation about the advantages of
the axiomatic method: {\em they are the advantages of theft over
honest labour}. You must either understand a proof of the
Heine--Borel theorem (e.g., Sutherland~\cite{Su}, Theorem~5.7.1), or
take it on trust as an axiom and accept the advantages.

 \subsection{An application of topological properties}\label{sec!using}

The notions set up so far are already enough to give a proof of
the statement in~\ref{homeo prob}. For example, the topological nature of
connectedness implies that $(a,b)\not\simeq[a,b]$, because any point
disconnects the left-hand side. In more detail, if $x\in(a,b)$ is any
point then it disconnects $(a,b)$ into two disjoint open intervals
$(a,x)$ and $(x,b)$; if $\fie\colon [a,b]\to(a,b)$ were a
homeomorphism,\index{homeomorphism} then $\fie(a)=x\in(a,b)$ would be
an interior point, so $(a,b)\setminus x$ would be disconnected,
whereas $[a,b]\setminus\{a\}=(a,b]$ is connected.\index{connected}
For exactly similar reasons,
 \begin{align*}
 S^1\not\simeq[a,b], \R^2, S^2&\quad\hbox{(any 2 points disconnect
 l.h.s.)}\\
 \R\not\simeq\R^2&\quad\hbox{(any point disconnects l.h.s.)}\\
 [a,b]\not\simeq\R^2 \ \hbox{or}\ S^2&
 \quad\hbox{(any 3 points disconnect l.h.s.)}
 \end{align*}

To complete the argument, note that
 \[
 [a,b],S^1,S^2\not\simeq(a,b),\R,\R^2;
 \]
 because all the spaces on the left-hand side are
compact,\index{compact!} and all those on the right are not.

 \section{Subspace and quotient topology}\label{sub and quot}
 \index{subspace topology}\index{quotient topology|(}
 \markright{\hfill\thesection\ Subspace and quotient topology\quad}
 If $X$ is a topological space and $Z\subset X$ a subset, write
$i\colon Z\into X$ for the inclusion map, that is $i(z)=z\in X$ for
every $z\in Z$. Then the {\em subspace topology}\index{subspace
topology} of $Z$ is the topology whose open sets\index{open set} are
of the form $U\cap Z$, where $U$ is an open of $X$. If $X$ is a
metric space with the topology defined by the metric $d$, then the
subspace topology of $Z$ is also metric, defined by the same metric
restricted to $Z$. This definition of the topology of $Z$ has $U\cap
Z=i\1(U)$ as open sets, so that the inclusion map $i$ is continuous.
It has no other opens, so it is the topology with the {\em fewest}
open sets needed to make $i$ continuous.

\medskip

Now let $X$ be a set and $\sim$ an equivalence relation on $X$.
Consider the set $Y=X/{\sim}$ of equivalence classes of $\sim$. That
is, in $Y$, if I write $\xbar$ for the class of $x$, I have
$\xbar=\ybar$ if and only if $x\sim y$, so that $Y$ is obtained by
identifying or ``glueing together'' points $x$ and $y$ when $x\sim
y$. Every surjective map $f\colon X\to Y$ of $X$ to a set $Y$ is
obtained in this way, by just declaring $\sim$ to be the relation
$x\sim y\iff f(x)=f(y)$.

Now suppose that $X$ is a topological space, and let $\sim$ and
$f\colon X\onto Y=X/{\sim}$ be as before. The {\em quotient topology}
of $Y$ has open sets defined by
 \[
 \hbox{$U\subset Y$ is open}\iff \hbox{$f\1(U)$ is open in $X$}.
 \]
It is easy to see that this satisfies the axioms for a topology.
Clearly $f$ is continuous, and this is the topology with the {\em most}
open sets\index{open set} for which $f$ is continuous. It often
happens that the quotient topology of $Y$ is not a metric topology,
as we see presently.

As above, let $X$ be a topological space, and $\sim$ an equivalence
relation.

 \begin{prop} The quotient space $Y=X/{\sim}$ has the following
properties:
 \begin{enumerate}\renewcommand{\labelenumi}{\theenumi}
\renewcommand{\theenumi}{(\arabic{enumi})}

 \item There is a continuous map $f\colon X\to Y$ such that
 \[
 x\sim y\implies f(x)=f(y)
 \]
 (that is, $f$ is constant on equivalence classes of\/ $\sim$).

 \item Given a space $Z$ and a continuous map $g\colon X\to Z$ that
is constant on equivalence classes of\/ $\sim$, there exists a unique
continuous map $h\colon Y\to Z$ such that $g=h\circ f$.
 \end{enumerate}
 \end{prop}

 \begin{pf} (1) comes from the definition as I discussed above.

(2) Given $g$, the map $h$ must take $f(x)\in Y$ to $g(x)$. In other words,
an element of $Y$ is an equivalence class $[x]$ of elements of $X$ under
$\sim$, so choose $x$ in that class, and set $h([x])=g(x)$. This is well
defined because of the assumption that $g$ is constant on equivalence
classes. Why is $h$ continuous? For $U\subset Z$ open, $g\1(U)$ is open in
$X$, so that $f\1(h\1(U))$ is open in $X$, and $h\1(U)$ is open in $Y$ by
definition of the quotient topology of $Y$. \QED \end{pf}

This property of the topological space $Y$ and the quotient map $f\colon
X\to Y$ is called a {\em universal mapping property}\index{universal mapping
property} or {\em UMP}. Constructions throughout abstract math can be
specified in terms of UMPs: you say what you want to do (in this case, find
a continuous map that is constant on equivalence classes), and then ask for
the solution of a UMP. In the present case, the universal mapping property
says that $f$ does not do anything that is not forced by the conditions that
$f$ is constant on equivalence classes of $\sim$, and is continuous. In
other words, $f$ identifies exactly the equivalence classes of $\sim$, and
makes no more identifications, and $Y$ has the {\em most} open
sets\index{open set} subject to $f$ being continuous. It is interesting to
analyse the above proof to see that this is exactly what is required to make
$h$ well defined and continuous.

 \section{Standard examples of glueing}\label{std exs}
 \markright{\hfill\thesection\ Standard examples of glueing\quad}
 The quotient topology on $X/{\sim}$ provides the definition of
``glueing'', the space obtained from $X$ by glueing together points
$x\sim y$. Here I discuss some basic examples;
see~\exref{ex!squarequot}-\exref{ex!genus2} below for more.

 \paragraph{Example~0} $S^1=[0,1]/{\sim}$ where $\sim$ glues the
endpoints (see Figure~\ref{S1}).

 \paragraph{Example~1} Let $X$ be the unit square $[0,1]\times[0,1]$.
The M\"obius strip\index{M\"obius strip} $M$ is defined by glueing
some of the sides of $X$ as in Figure~\ref{Moebius fig}. More
formally, consider the following equivalence relations on $X$:
 \[
(x,y)\sim(x',y')\iff
\left\{
 \begin{array}{l}
\hbox{either $(x,y)=(x',y')$,}\\
\hbox{or $x=0,x'=1$ and $y=1-y'$,}\\
\hbox{or vice versa,}
 \end{array}
\right.
 \]
 and define the M\"obius strip\index{M\"obius strip} $M$ by
$M=X/{\sim}$, with the quotient topology. By definition of the quotient
topology, a point on the glued line has a neighbourhood obtained from
neighbourhoods of its two inverse images in $X$.

 \begin{figure}
 \renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}a}
 \centering
 \leavevmode\hbox{\epsfysize=4.7 true cm\epsffile{top_mobius.eps}}
 \caption{The M\"obius strip $M$}\index{M\"obius strip}
 \label{Moebius fig}
 \end{figure}

 \paragraph{Example~2} The cylinder $S^1\times[0,1]$ is obtained by
glueing the unit square $[0,1]\times[0,1]$ as in Figure~\ref{cylinder
fig}.

 \begin{figure}
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}b}
 \centering
 \leavevmode\hbox{\epsfysize=2.6 true cm\epsffile{top_cylinder.eps}}
 \caption{The cylinder $S^1\times[0,1]$}
 \label{cylinder fig}
 \end{figure}

 \paragraph{Example~3} The torus\index{torus} $T\simeq S^1\times S^1$
is obtained from the unit square $[0,1]\times[0,1]$ by the glueing of
Figure~\ref{torus fig}. By definition of the quotient topology, the 4
corners of the square correspond to a point of the torus, and a
neighbourhood of it is obtained from neighbourhoods of the four
corners in $X$. You can regard this as a surface of rotation in
$\R^3$, or the surface in $\R^4$ given by $x_1^2+y_1^2=x_2^2+y_2^2=1$.

 \begin{figure}
 \renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}c}
 \centering
 \leavevmode\hbox{\epsfysize=8.5 true cm\epsffile{top_torus.eps}}
 \caption{The torus}\index{torus}
 \label{torus fig}
 \end{figure}
 \index{torus}
 \paragraph{Example~4} The surface with $g$ handles. The picture is as in
Figure~\ref{surf fig}: you get it by starting from $S^2$, marking
$2g$ distinct points on $S^2$, cutting out small discs around these,
and glueing back in $g$ small cylinders. See~\exref{ex!genus2} as well
as~\ref{sec!trichotomy} for further discussion.

 \begin{figure}
 \renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}d}
 \centering
 \leavevmode\hbox{\epsfysize=4.8 true cm\epsffile{top_glue_handles.eps}}
 \caption{Surface with $g$ handles}\index{genus}
 \label{surf fig}
 \end{figure}
 \index{quotient topology|)}

Notice that all these spaces can easily be made into metric spaces,
but you do not really gain anything by doing so.

 \begin{prop} The M\"obius strip\index{M\"obius strip} $M$, the
cylinder $N=S^1\times[0,1]$ and the torus\index{torus} $T$ are not
homeomorphic.\index{homeomorphism} \end{prop}

I can almost prove this now, though I relegate one crucial statement
to the end of the chapter. The proof consists of the following steps:

 \paragraph{Step~1: Main claim} Points of the boundary $\partial
M\subset M$ and $\partial N\subset N$ are distinguished from points
of the interior by their topological properties.

 \begin{figure}[ht]
 \renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}e}
 \centering
\input{top_in_out.pstex_t}
 \caption{Boundary and interior points}
 \label{r}
 \end{figure}

 \paragraph{Step~2} Therefore, if there exists a homeomorphism
$\fie\colon M\to N$, it must map $\partial M$ to $\partial N$, and
the restriction must define a homeomorphism $\partial M\simeq\partial
N$.\index{homeomorphism}

 \paragraph{Step~3}\index{path!connected} $\partial M$ is path
connected, whereas $\partial N$ is disconnected; hence a
homeo\-morphism $M\simeq N$ as in Step~2 cannot exist. In the same
way, $\partial T=\emptyset$, so that $T\not\simeq M$ and $T\not\simeq
N$.

\medskip

Given the main claim, Steps~2--3 are obvious, and the point is
therefore to understand Step~1. How do I distinguish points of the
interior of a surface from points on the boundary? The point is that
every small neighbourhood $U\setminus P$ of an interior point $P$
contains a small punctured disc $D^*$ about $P$; the {\em punctured
disc}\index{punctured disc $D^*$} is the topological space
$\{0<x^2+y^2<1\}\subset\R^2$. On the other hand, if $P$ is a boundary
point, it has an arbitrarily small neighbourhood homeomorphic to a closed
half-disc,\index{homeomorphism} that can be written in polar
coordinates
 \[
 U\simeq \bigl\{(r,\theta)\bigm|0\leq r<1,\ \theta\in[-\pi/2,\pi/2]\bigr\}
 \]
with $P$ at the centre of the half-disc. Hence $U\setminus P$ is
homeomorphic to
 \[
 U\setminus P\simeq\bigl\{(r,\theta)\bigm|0< r<1,\ \theta\in[-\pi/2,\pi/2]\bigr\}
 \]
which in turn is homeomorphic to a closed disc with parts of the boundary
removed, as on Figure~\ref{r}. Hence the essential content of telling
interior and boundary points apart consists in showing that the punctured disc
$D^*$ is not homeomorphic to the disc $D$.\index{homeomorphism}
Think it through yourself to see whether you find this statement intuitive;
see~\ref{fun thm}, Corollary~1 for the proof.

 \section{Topology of $\protect\PP^n_\R$}\label{Top Pn}
 \index{topology!of $\PP^n$}
 \markright{\hfill\thesection\ Topology of $\protect\PP^n_\R$\quad}
 Recall \ref{Definition of Pn}: projective $n$-space, as a set, is
defined to be the set of lines of $\R^{n+1}$ through the origin, or
in other words, the quotient of $\R^{n+1}\setminus \{0\}$ by the
equivalence relation which identifies $\bx$ with $\la\bx$ for $\la\ne
0$. The topology of $\PP^n$ is the quotient topology of
$\R^{n+1}\setminus\{0\}$. This section considers various ways of
looking at this topology.

Write $S^n=\{\bx\in\R^{n+1}\,|\,\sum x_i^2=1\}\subset\R^{n+1}$ for the
$n$-sphere.\index{sphere $S^n$} Obviously $S^n$ meets every line of
$\R^{n+1}$ through 0 in a pair of antipodal points. Therefore, as a set,
$\PP^n_\R=S^n/\pm$, where $\pm$ is the equivalence relation identifying
antipodal points of the sphere (that is, pairs $\pm \bx$ of opposite
points). The topology of $\PP^n$ coincides with the quotient
topology\index{quotient topology} of $S^n/\pm$; indeed, a subset of the
lines through 0 is open in $\R^{n+1}\setminus0$ if and only if its
intersection with $S^n$ is open\index{open set} in the subspace
topology\index{subspace topology} of $S^n$. Note that $S^n\subset\R^{n+1}$
is closed and bounded hence compact\index{compact!} (Example~\ref{comp});
thus $\PP^n$, being the continuous image of a compact space, is also compact
by the tautological Proposition~\ref{im compact}. This was one of the
motivations for constructing projective space discussed in~\ref{Pncomp}.

There are many ways of understanding the quotient, by choosing a
closed subset of $S^n$ that picks out just one of each pair of
antipodal points for a big open subset and then glueing around the
boundary: for example, the closed northern hemisphere of $S^n$
contains one of each pair of antipodal points, except that I still
have to identify antipodal points of the equatorial sphere
$S^{n-1}$.\index{sphere $S^n$}

In the case $n=2$, we can do the following:
view $S^2$ as the union of 3 pieces, a cap around the north pole, a
band around the equator, and a cap around the south pole (see
Figure~\ref{aa}). Every point in the southern cap is equivalent to a point in
the northern cap, so the southern cap is not needed. Now cut the
equatorial band into its front and back halves; as before, every
point in the back half is equivalent a point in the front half, so this
piece is also not needed. Now $\pm$ glues together the left and
right intervals of the front half to give a M\"obius
strip;\index{M\"obius strip} this glueing is the same as in
Figure~\ref{Moebius fig}. The northern cap is a disc,\index{disc}
with boundary a circle; the M\"obius strip also has boundary a
circle, and $\PP^2$ is obtained by glueing these two pieces together
along their boundaries. Note that this is an abstract construction: you cannot
do it in $\R^3$ without allowing self-crossing.

It is an interesting exercise to see the components of this
construction as the result of cutting $\PP^2$ along a line and
along a conic. See \exref{ex!p2r}.

 \begin{figure}
 \renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}}
 \centering
 \leavevmode\hbox{\epsfysize=6 true cm\epsffile{top_projective_space.eps}}
 \caption{Topology of $\PP^2_\R$: M\"obius strip with a disc
 glued in}\index{M\"obius strip}
 \label{aa}
 \end{figure}

 \section{Nonmetric quotient topologies}\label{nonmetric}
 \markright{\hfill\thesection\ Nonmetric quotient topologies\quad}
 \paragraph{Example~1 (the mousetrap topology)}\index{mousetrap
topology|(} $X=\{P,Q\}$ is a space with only $2$ points and open sets
 \[
 \sT_X=\Big\{\emptyset,\{P\},X\Big\}.
 \]
Here $P$ is an open point, but not $Q$. Every neighbourhood of $Q$
(there is only one) contains $P$. In terms of convergence, the
constant sequence $P,P,\dots$ converges both to $P$ and to $Q$
(please check this as an instant exercise; refer back to~\ref{comp} for
the definition of convergence if needed). This implies, of course,
that the topology of $X$ is not metric. $X$ is a quotient
topology\index{quotient topology}: introduce the equivalence relation
$\sim$ on $\C$ defined by
 \[
 x\sim y\iff x=\la y\quad\hbox{with $\la\in\C$, $\la\ne0$.}
 \]
 Then there are only two equivalence classes, $Q=[0]$ and
$P=[\la\in\C\setminus\{0\}]$; $\{P\}$ is obviously open while $\{Q\}$ is not.
The point is that if you are at $0$ then any arbitrarily small perturbation
takes you into a nonzero number; that is, viewed from $Q$, the point
$P$ is infinitely close. But if you are at a nonzero number $\la$, all the
points in a small neighbourhood are also nonzero, so viewed from $P$,
the point $Q$ is far away. Being zero is an unstable, or closed
condition; being nonzero is a stable or open condition.

 \begin{figure}
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}a}
 \centering
\input{top_mousetrap.pstex_t}
 \caption{The mousetrap topology}
 \label{s}
 \end{figure}

I call this the {\em mousetrap topology} (Figure~\ref{s}) because if you are at $Q$
(outside the trap), it is no distance at all to get into the
trap.\index{mousetrap topology|)} But if you are at $P$ (inside the
trap), then it is a long way out. Thus the content of the topology is
more logical than geometric.

There are many equivalence relations of interest with this kind
of behaviour. One example is the equivalence relation on $\R$ with
 \[
 \{x\in\R\mid x>0\},\quad\{0\},\quad\{x\in\R\mid x<0\}
 \]
as its $3$ equivalence classes.

 \paragraph{Example~2: Quadratic forms}\index{quadratic form}
 A similar but more substantial example: consider quadratic forms
$q(x,y)=ax^2+2bxy+cy^2$ on $\R^2$. There is a coordinate change that
puts $q(x,y)$ in one of the $6$ normal forms:
 \[
 q_1=x^2+y^2, \enspace
 q_2=x^2-y^2, \enspace
 q_3=-x^2-y^2, \enspace
 q_4=x^2, \enspace
 q_5=-x^2, \enspace \hbox{or} \enspace q_0=0.
 \]
 All the quadratic forms on $\R^2$ are parametrised by
$(a,b,c)\in\R^3$, corresponding to the symmetric matrix $A=
\begin{pmatrix} a&b\\b&c\\ \end{pmatrix}$. Now introduce the
equivalence relation on $\R^3$ corresponding to a coordinate change:
 \[
 A\sim B\; \iff\;\hbox{$\exists M\in\GL(2,\R)$ such that $A=\tr MBM$.}
 \]
 (Here $\GL(2,\R)$ is the group of $2\times2$ invertible matrixes.)
This means exactly that I consider quadratic forms up to change of
basis. So there are exactly the 6 classes, the strata of
Figure~\ref{light cone fig}. The quotient topology\index{quotient topology}
on the set
 \[
 X=\R^3/{\sim}\hbox{}=\bigl\{q_1,q_2,q_3,q_4,q_5,q_0\bigr\}
 \]
has open sets
 \[
 \{q_1\},\quad\{q_2\},\quad\{q_3\},
 \quad\{q_4,q_1,q_2\},\quad\{q_5,q_2,q_3\},\quad X
 \]
 and their unions. For example, every neighbourhood of $q_4$ contains
$q_1,q_2$.
 \begin{figure}[t]
 \renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}b}
 \centering
\input{top_qforms.pstex_t}
 \caption{Equivalence classes of quadratic forms $ax^2+2bxy+cy^2$}
 \label{light cone fig}
 \end{figure}

 \section{Basis for a topology}\label{basis}
 \index{basis for a topology|(}
 \markright{\hfill\thesection\ Basis for a topology\quad}
 This is a formal idea for constructing topologies. Let $\sB$ be a
collection of subsets of $X$. Then $\sB$ is a {\em basis for a
topology} if it satisfies the three axioms

 \begin{enumerate}

 \item Finite intersections:
 \[
 U_1,\dots,U_n\in\sB\implies U_1\cap\cdots\cap U_n\in\sB.
 \]

 \item Involves every point. For all $x\in X$ there exists $U\in\sB$
such that $x\in U$.

 \item Empty convention:\index{empty set} $\emptyset\in\sB$.
 \end{enumerate}

 \begin{constr} If $\sB$ is a basis for a topology, the family of subsets
 \[
 \sT=\Bigl\{\Cup_{\la\in\La} U_\la : U_\la\in \sB,
 \La\mbox{\rm\ arbitrary index set}\Bigr\}
 \]
of $X$ is a topology on $X$, the topology {\rm generated} by $\sB$.
 \end{constr}

 \begin{pf} This is entirely formal. $X\in\sT$ using 2 and the
construction. $\sT$ is closed under arbitrary unions by construction.
To show that $\sT$ is closed under finite intersections, note that
 \[
 \Bigl(\Cup_{\la\in\La} U_\la\Bigr)\cap
 \Bigl(\Cup_{\mu\in M} U_\mu\Bigr)
 \ =\ \Cup_{\la,\mu}U_\la\cap U_\mu. \QED
 \]
 \end{pf}

I can save time by listing only a basis for the topology, rather than
by saying what all the open sets are. The idea here is that a
topology is specified by the neighbourhoods of each point (because an
open set is determined by the condition that it is a neighbourhood of
each of its points). In turn it is enough to specify any system of
sufficiently small neighbourhoods of each point.\index{basis for a
topology}

 \paragraph{Example 1} In \ref{nonmetric}, Example~2, I described
the quotient topology\index{quotient topology} on $X=\R^3/{\sim}$ by
telling you that its open sets\index{open set} are unions of
 \[
 \{q_1\},\quad\{q_2\},\quad\{q_3\},\quad\{q_4,q_1,q_2\},\quad\{q_5,q_2,q_3\},\quad\{q_0,q_1,q_2,q_3,q_4,q_5\}.
 \]

 \paragraph{Example 2} Let $X,d$ be a metric space, and
 \[
 \sB=\bigl\{B(x_1,\ep_1)\cap\cdots\cap B(x_n,\ep_n)\bigr\}
 \]
 be the set of finite intersection of open balls $B(x,\ep)=\{y\mid
d(x,y)<\ep\}$. Then $\sB$ is a basis for a topology $\sT$, the
usual metric topology. \index{metric!topology}

 \paragraph{Example 3: Profinite topology of an infinite
group}\index{profinite topology} Another more substantial example.
Take any group $G$; recall that a subgroup $H\subset G$ is normal (written
$H\vartriangleleft G$ ) if $gH=Hg$ for every $g\in G$, that is, its
right and left cosets coincide. A normal subgroup $H\vartriangleleft
G$ of finite index $n$ is the kernel of a surjective homomorphism
$G\to\Ga$ to a finite group of order $n$. For example, if $G=\Z$ then every normal
subgroup of finite index is just $n\Z$ for some integer $n$.

Let $G$ be a group, with $e\in G$ the identity element. Then there is
a topology on $G$ such that:

 \begin{enumerate}
 \renewcommand{\labelenumi}{\theenumi}
 \renewcommand{\theenumi}{(\alph{enumi})}

 \item normal subgroups $H\vartriangleleft G$ of finite index form a set of
sufficiently small neighbourhoods of $e$;

 \item the right translation\index{translation!map} maps $r_g\colon
G\to G$ defined by $f\mapsto fg$ are homeo\-morphisms.
 \end{enumerate}

It follows from (a) and (b) that a set of sufficiently small neighbourhoods of
any $g\in G$ are given by cosets $gH$, where the $H$ are as in (a).
So take
 \[
 \sB=\{\emptyset\}\cup\{\hbox{cosets of normal subgroups of finite
index}\}
 \]
 as a basis for a topology. I check that this is a basis by going
through the three axioms. Indeed, $\emptyset,G\in\sB$. Also if
$H_1,\dots,H_n$ are normal subgroups of finite index then so is
$H_1\cap\cdots\cap H_n$, clearly, and if $g_1H_1,\dots,g_nH_n$ are
their cosets then either $g_1H_1\cap\cdots\cap g_nH_n=\emptyset$, or
$\exists g\in g_1H_1\cap\cdots\cap g_nH_n$, in which case
 \[
 g_1H_1\cap\cdots\cap g_nH_n=gH_1\cap\cdots\cap gH_n
 =g(H_1\cap\cdots\cap H_n).
 \]
 The topology generated by this basis is called the {\em profinite
topology}\index{profinite topology} of $G$. Note that if
$H\vartriangleleft G$ is a normal subgroup of finite index then its
cosets form a partition of $G$ by finitely many disjoint open sets.
Therefore any of these cosets is also closed.

\begin{rem} Profinite topologies on groups have lots of applications
in algebra and number theory. For example, in number theory, you may
want to solve an equation $f(x,y)=0$ in $\Z$, knowing that you can
solve it modulo all $N$. Another example occurs in Galois theory. The
idea is that if $k\subset L$ is an infinite Galois field extension,
the finite extension fields $k\subset K\subset L$ correspond to
subgroups of finite index in the infinite Galois group
$\Gal(L/k)$. The Galois group $\Gal(L/k)$ is automatically profinite,
in the sense that it is defined by its finite quotient groups.
 \end{rem}

 \section{Product topology}\label{prod top}
 \index{product topology|(}
 \markright{\hfill\thesection\ Product topology\quad}
 Let $X$ and $Y$ be topological spaces; I show how to put a topology
on $X\times Y$. Take the set of subsets
 \[
 \sB=\{U\times V\subset X\times Y\}
 \quad\hbox{with $U\subset X$ and $V\subset Y$ open}.
 \]
 Then
 \[
 (U_1\times V_1)\cap(U_2\times V_2)=(U_1\cap U_2)\times(V_1\cap V_2)
 \]
gives the finite intersection property; the other two axioms are
obvious, so $\sB$ is a basis for a topology\index{basis for a
topology|)} on $X\times Y$. The {\em product topology} on $X\times Y$
is defined to be the topology generated by $\sB$.

If $X$ and $Y$ are metric spaces, it is easy to see that the product
topology is the topology defined by any of the metrics
$\max(d_X,d_Y)$, $d_X+d_Y$, $\sqrt{d_X^2+d_Y^2}$, etc.\ (see
Figure~\ref{u}). It follows that for $n, m$ positive integers, the product
topology on $\R^n\times\R^m$ is the same as the metric topology on $\R^{n+m}$.
For example, on $\R^2=\R\times \R$, the sets
$(a_1,b_1)\times(a_2,b_2)$ provide arbitrarily small open sets, but
obviously not all open sets are of this form.

 \begin{figure}[ht]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}}
 \centering
 \leavevmode\hbox{\epsfysize=4.3 true cm\epsffile{top_different_shapes.eps}}
 \caption{Balls for product metrics}
 \label{u}
 \end{figure} \index{product topology|)}

 \section{The Hausdorff property}\label{Hausdorff}
 \index{Hausdorff property|(}
 \markright{\hfill\thesection\ The Hausdorff property\quad}
 A topological space is {\em Hausdorff\/}\footnote{Felix Hausdorff
(1868--1942) was the originator of many of the basic ideas of metric
and topological spaces, and the author of a famous and influential
book {\em Grundz\"uge der Mengenlehre}. He was Professor at the
University of Bonn until he was forced out as a Jew in 1935. He
committed suicide in January 1942, together with several members of his
family, to avoid being sent to a Nazi internment camp.} if for all
$x\ne y\in X$, there exist disjoint open sets $U,V\subset X$ with
$x\in U$, $y\in V$. (See Figure~\ref{hausdorff fig}.) This is clearly
another topological property.

If $X$ is Hausdorff then every point $x\in X$ is closed: for
if $x\ne y$ there exists an open set containing $y$ and not $x$, and
therefore $X\setminus x$ is open. This is a weaker separation axiom,
 \[
 \forall x\ne y\in X, \ \hbox{$\exists$ an open set $U$ containing
 $y$ and not $x$}
 \]
called Hausdorff's $T_1$ condition. (The Hausdorff condition on $X$
introduced here is sometimes also called $T_2$.)

\paragraph{Example 1}
A metric space $X$ is automatically Hausdorff: just choose
$\ep>0$ with $\ep<\half d(x,y)$ and set $U=B(x,\ep)$, $V=B(y,\ep)$.

\paragraph{Example 2}
 Examples~1 and~2 of \ref{nonmetric} are clearly not Hausdorff.
The cofinite topology\index{cofinite topology} of an infinite
set $X$ (\ref{cofinite}, Example~2) is not Hausdorff either: a nonempty open
set is the complement of a finite set, so the intersection of any two
open sets is again the complement of a finite set, so nonempty. Thus
these are certainly not metric topologies.

\paragraph{Example 3}
A topology on a finite set $X$ is Hausdorff if and only if it
is the discrete topology.\index{discrete topology} Indeed, if $X$ is
Hausdorff then any point $x\in X$ is closed, so every subset of $X$
is closed.

\medskip

 \begin{prop} A topological space $X$ is Hausdorff if and only the
diagonal
 \[
\De_X=\{(x,x)\mid x\in X\}\subset X\times X
 \]
is closed\index{closed!diagonal|(} in the product topology of
$X$.\index{product topology}
 \end{prop}

% \begin{figure} \renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}} \centering\leavevmode\hbox{\epsfysize=4.9 true cm\epsffile{top_diag.eps}} \caption{Closed diagonal}\label{fig!closeddiag} \end{figure}

 \begin{pf} Note first that for any subsets $U,V\subset X$,
 \[
 U\times V\cap\De_X=\{(x,x)\mid x\in U\cap V\},
 \]
in other words, $U\times V\cap\De_X$ is just the diagonal embedding of $U\cap V$
into $X\times X$.

A point of $X\times X\setminus\De_X$ is just a pair $(x,y)$ with
$x\ne y$. Consider the problem of finding an open neighbourhood $W$
of $(x,y)$ in the product topology\index{product topology} such that
$W\cap\De_X=\emptyset$. By definition of the product topology, an
arbitrary small neighbourhood of $(x,y)$ is $U\times V$ with
$U,V\subset X$ open and $x\in U$, $y\in V$. Now by the
first remark, $U\times V\cap\De_X=\emptyset$ if and only if
$U\cap V=\emptyset$.
%(See Figure~\ref{fig!closeddiag}).

Since $\De_X$ is closed\index{closed!diagonal|)} if and only if
$X\times X\setminus\De_X$ is open, this happens if and only if for
every $(x,y)$ with $x\ne y$ there exists open sets $U,V\subset X$
open, with $x\in U$, $y\in V$ and $U\cap V=\emptyset$. \QED
\end{pf}

 \section{Compact versus closed}\label{comp v closed}
 \index{compact!versus closed|(}
 \markright{\hfill\thesection\ Compact versus closed\quad}
 \begin{prop} Let $X$ be a topological space, and $Y\subset X$
a subset with the subspace topology.\index{subspace topology}
 \begin{enumerate}
\renewcommand{\labelenumi}{\theenumi}
\renewcommand{\theenumi}{(\roman{enumi})}

 \item If\/ $X$ is a compact topological space and\/ $Y\subset X$ is
closed, then $Y$ is also compact.

 \item If\/ $X$ is Hausdorff and\/ $Y\subset X$ is compact, then $Y$ is
closed.

 \item In particular, if\/ $X$ is compact and Hausdorff, then $Y\subset
X$ is compact if and only if it is closed.
 \end{enumerate}
 \end{prop}

 \begin{pf} (i) Suppose that $V_\la$ for $\la\in\La$ are open subsets
of $Y$, in the subspace topology,\index{subspace topology} such that
$Y=\Cup V_\la$. Then by definition of the subspace topology \ref{sub
and quot}, for each $\la$ there exists an open set $U_\la$ of $X$
such that $V_\la=Y\cap U_\la$. Now also $X\setminus Y$ is open, by
the assumption that $Y$ is closed. Therefore $X=\Cup
U_\la\cup(X\setminus Y)$ is an open cover of $X$. By definition of
compactness, a finite cover will do, say $X=\Cup_{i=1}^n
U_{\la_i}\cup(X\setminus Y)$, and then obviously $Y=\Cup_{i=1}^n
V_{\la_i}$.

(ii) Fix $x\in X\setminus Y$. For every $y\in Y$, using the Hausdorff
assumption on $X$, choose disjoint open sets $U_y$ and $V_y$ with
$x\in U_y$ and $y\in V_y$. By construction, $y\in V_y$, so that
$Y\subset\Cup V_y$, or equivalently $Y=\Cup(Y\cap V_y)$. But since
$Y$ is compact, a finite number of the open sets $Y\cap V_y$ cover it,
and hence there is a finite set of $V_{y_i}$ with
$Y\subset\Cup_{i=1}^n V_{y_i}$. Set $U=\Cap_{i=1}^n U_{y_i}$, which
is a finite intersection of opens, therefore open. Since $U_y\cap
V_y=\emptyset$ for each $y$, it follows that $U\cap\Cup_{i=1}^n
V_{y_i}=\emptyset$, and in particular $U\cap Y=\emptyset$.
(See Figure~\ref{sep_closed_point}.) This
proves that for any $x\notin Y$, there exists an open set $U$
containing $x$ disjoint from $Y$, and therefore $Y$ is closed.\QED
\end{pf}\index{Hausdorff property|)}

 \setcounter{figure}{\value{section}}\addtocounter{figure}{-1}
 \begin{figure}\centering
 \leavevmode\hbox{\epsfysize=4.9 true cm\epsffile{top_circles.eps}}
 \caption{Separating a point from a compact subset} \label{sep_closed_point}
 \end{figure}

 \section{Closed maps}\label{closed} \index{closed!map|(}
 \markright{\hfill\thesection\ Closed maps\quad}
 A map $f\colon X\to Y$ between topological spaces is {\em closed}, if
$f(V)\subset Y$ is closed for every closed set $V\subset X$.

 \paragraph{Example 1} Consider the closed interval $[a,b]\subset\R$.
Then the second projection $\pi\colon [a,b]\times\R\to\R$ is a closed map.

 \begin{pf} Start with a closed set $V\subset[a,b]\times\R$ and a point
$x\in\overline{\pi(V)}$ of the closure of $\pi(V)$. Take a closed interval $I$
containing $x$, and restrict attention to the second projection
 \[
B=[a,b]\times I\to I
 \]
Then $B$ is closed and bounded\index{closed!and bounded|)} in $\R^2$,
so compact (see Proposition~\ref{comp}); hence $V\cap B$ is compact by
Proposition~\ref{comp v closed}~(i). Therefore by Proposition~\ref{im
compact}, $f(V\cap B)$ is a compact subset of $I$, therefore closed
in $I$. Therefore $x\in\pi(V)$, and $\pi(V)$ is closed. \QED \end{pf}
\index{compact!versus closed|)}

 \begin{figure}
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}a}
 \centering
 \leavevmode\hbox{\epsfysize=4 true cm\epsffile{top_closedmap.eps}}
 \caption{Closed map}
 \label{z}
 \end{figure}

 \paragraph{Example 2} The projection to the $x$-axis $\R^2\to\R$ is not
closed. For consider the hyperbola $C:(xy=1)$; it is closed in $\R^2$, but its
image in $\R$ is $\R\setminus0$.

 \begin{figure}
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}b}
 \centering
 \leavevmode\hbox{\epsfysize=4.9 true cm\epsffile{top_nonclosed_map.eps}}
 \caption{Nonclosed map}
 \label{y}
 \end{figure}

 \begin{prop} If $X$ is compact and $Y$ Hausdorff\index{Hausdorff property|(}
then any continuous map $f\colon X\to Y$ is closed. \end{prop}

 \begin{pf} $V\subset X$ closed implies $V$ compact by
Proposition~\ref{comp v closed}~(i). Therefore $f(V)$ is compact by
Proposition~\ref{im compact}, and $f(V)\subset Y$ is
closed\index{closed!map|)} by Proposition~\ref{comp v closed}~(ii).
\QED \end{pf}

 \section{A criterion for homeomorphism}\label{homeo crit}
 \index{homeomorphism!criterion|(}
 \markright{\hfill\thesection\ A criterion for homeomorphism\quad}
 Let $X$ and $Y$ be topological spaces and $f\colon X\to Y$ a map. I claim
that
 \[
 \hbox{$f$ is a homeomorphism}\iff \hbox{$f$ is bijective,
 continuous, and closed.}
 \]
$\implies$ is of course clear. If $f$ is bijective, then $f$
closed\index{closed!map} means exactly that $f\1$ is continuous: for
$U\subset X$ open gives $X\setminus U$ closed, which implies that
$f(X\setminus U)$ is closed; but $f(X\setminus U)=Y\setminus f(U)$
because $f$ is bijective, so $f(U)$ is open, that is, $f\1$ is
continuous.
 \begin{thm}[$\protect\heartsuit=0$] If\/ $X$ is compact and\/ $Y$
Hausdorff\index{Hausdorff property}, then a
continuous bijective map $f\colon X\to Y$ is a homeomorphism. \end{thm}

 \begin{pf} $f$ is closed\index{closed!map} by Proposition~\ref{closed}.\QED \end{pf}
 \index{homeomorphism!criterion|)}

 \paragraph{Example} A {\em simple closed curve} in $\R^2$ is a
continuous map $f\colon [0,1]\to\R^2$ that is one-to-one except for
$f(0)=f(1)$. Write $\sim$ for the equivalence relation that glues the
endpoints of the interval as in Figure~\ref{S1}. Clearly $f$ defines
a continuous one-to-one map $f'\colon [0,1]/{\sim}=S^1\to\R^2$. I
claim that $f'\colon S^1\to f'(S^1)$ is a
homeomorphism.\index{homeomorphism} Indeed, it is a continuous
one-to-one map from a compact space $S^1$ to a
Hausdorff\index{Hausdorff property|)} space $f'(S^1)\subset\R^2$.
This proves that $\heartsuit=0$.

 \section{Loops and the winding number}\label{wind}
 \index{loop}\index{winding number|(}
 \markright{\hfill\thesection\ Loops and the winding number\quad}
 Let $D=\bigl\{(x,y)\in\R^2 \bigm|x^2+y^2<1\bigr\}$ be the unit
disc\index{disc} in $\R^2$ and $D^*=D\setminus(0,0)$ the punctured
disc.\index{punctured disc $D^*$} This final section will answer the
following question, left open in the proof of Proposition~\ref{std exs}:

\paragraph{Question} How can we tell that $D^*$ is not
homeomorphic\index{homeomorphism} to $D$?

\paragraph{Answer}
$D$ is {\em simply connected}\/:\index{simply connected} any
loop\index{loop} in $D$ (starting and ending at $P_0$, say) can be contracted
in $D$ to the constant loop;\index{loop}\index{contractible loop|(} on the
other hand, a loop in $D^*$ has a {\em winding number $n$} around the
puncture $(0,0)$, and the loop\index{loop} can be contracted if and
only if $n=0$.

\bigskip

The intuitive picture is clear: think of taking a dog on a long lead
for a walk in a park having a tall pole in the middle. In classical
math, the winding number\index{winding number} $n$ is the ambiguity
of $2\pi n$ in the functions $\arcsin x$ and $\arccos x$ and the
ambiguity of $2\pi ni$ in the complex function $\log z$. The content
of the following sections is the first step in the theory of the
fundamental group\index{fundamental!group} $\pi_1(X,P_0)$ in algebraic
topology;\index{algebraic topology} Theorem~\ref{htpy} on the winding
number is equivalent to the statement that $\pi_1(D^*,P_0)=\Z$.

 \subsection{Paths, loops and families}\label{IxI}
 \index{path}
 \index{loop}
 Recall that a path\index{path} in a topological space $X$ is a
continuous map $\fie\colon[0,1]\to X$, written $t\mapsto\fie(t)$. Fix
a {\em base point} $P_0\in X$. A {\em loop}\index{loop} in $X$ {\em
based at} $P_0$ is a path starting and ending at $P_0$; in other
words, a continuous map $f\colon[0,1]\to X$ such that
$f(0)=f(1)=P_0$. These are called {\em based loops}\index{based loop}
(as opposed to {\em free loops} where we insist that $f(0)=f(1)$, but
allow this to be any point in $X$). A loop\index{loop} is allowed to
cross over itself any number of times, or even to stop for a while or
go back along itself.

\index{family of paths|(}\index{loop}\index{path}
A {\em family of paths (or loops)} $(\fie^{(s)})$ depending on a
parameter $s\in[0,1]$ is just an indexed family of paths (or loops),
one for each $s\in[0,1]$. Write $I_t$ for the interval $[0,1]$ of the
path parameter $t$, and $I_s$ for the interval $[0,1]$ of the family
parameter $s$.

\paragraph{Tentative definition} Let $X$ be a metric space. A family of paths
$(\fie^{(s)})$ is {\em continuous} at $s$ if for every $\ep>0$, there exists a
$\de$ such that
 \[
 |s-s'|<\de\implies d(\fie^{(s)}(t),\fie^{(s')}(t))<\ep\quad\hbox{for all
 $t\in[0,1]$.}
 \]
We say that $(\fie^{(s)})$ is a {\em continuous family of
paths}\index{continuous family of paths|(} if it is continuous at all
$s\in[0,1]$. The definition applies in exactly the same way to a family of
based loops,\index{based loop} except that I insist that
$\fie^{(s)}(0)=\fie^{(s)}(1)=P_0$ for every $s$.

\medskip

Note that the continuity assumption is uniform in $t$ (the same $\de$ is
supposed to guarantee closeness for all $t$). The hard thing is to understand
why the definition just given is the right one. The point is that to say that
the path $\fie^{(s)}$ moves just a little, we have to guarantee that every
step $\fie^{(s)}(t)$ for fixed $t$ should move just a little, bounded in~$t$
(compare \exref{uniform}).

\begin{lem} Corresponding to a family of paths\index{family of paths|)}
$(\fie^{(s)})$, consider the map
 \[
 \Fie\colon I_s\times I_t=[0,1]\times[0,1]\to X\quad\hbox{given by}\quad\Fie(s,t)=\fie^{(s)}(t).
 \]
Then $(\fie^{(s)})$ is a continuous family of paths\index{continuous family of
paths} if and only if\/ $\Fie$ is continuous. See Figure~\ref{IxI fig}.

 \begin{figure}
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}a}
 \centering
 \leavevmode\hbox{\epsfysize=4 true cm\epsffile{top_define_maponsquare.eps}}
 \caption{Continuous family of paths}
 \label{IxI fig}
 \end{figure}

 \end{lem}

\begin{rem} Notice that $\Fie$ continuous is a topological property. The point
of the lemma is that it makes the notion of continuous family of paths purely
topological. If $X$ is a topological space, the ``uniform'' definition of a
continuous family of paths is not applicable (it depends on the metric in
$X$); in the Definition below I define a family of paths $\fie^{(s)}$ to be
continuous by the property that $\Fie$ is continuous. \end{rem}

 \begin{pf} $\implies$ A standard ``divide the $\ep$ in two'' argument. Suppose
given $(s_0,t_0)\in I_s\times I_t$ and $\ep>0$. First, because $\fie^{(s_0)}$
is continuous, there exists $\de$ such that
 \[
d(t,t_0)<\de\implies d(\fie^{(s_0)}(t),\fie^{(s_0)}(t_0))<\ep/2.
 \]
 Next, because $\fie^{(s)}$ is a continuous family of
paths\index{continuous family of paths|)} at $s_0$, there exists a
$\de$ such that
 \[
d(s,s_0)<\de\implies d(\fie^{(s)}(t),\fie^{(s_0)}(t))<\ep/2\quad\hbox{for all
$t$.}
 \]
 Therefore $\max\{d(s,s_0),d(t,t_0)\}<\de$ implies both of these inequalities,
so that
 \begin{eqnarray*}
 \lefteqn{d((s,t),(s_0,t_0))=\max\{d(s,s_0),d(t,t_0)\}<\de\implies}\\
 && d(\Fie(s,t),\Fie(s_0,t_0))\le
 d(\fie^{(s_0)}(t),\fie^{(s_0)}(t_0))+
 d(\fie^{(s)}(t),\fie^{(s_0)}(t))<\ep.
 \end{eqnarray*}
This proves $\Fie$ is continuous as a function of $(s,t)$.

 $\Longleftarrow$ In this direction, I have to use compactness of
$I_t$ to get uniformity in $t$. If $\Fie$ is continuous, each
$\fie^{(s)}\colon I_t\to X$ is obviously continuous. I fix some
$s_0\in I_s$, and try to prove that
$(\fie^{(s)})$ is a continuous family of paths at $s_0$. Suppose
given $\ep>0$. Start by working in a neighbourhood of a fixed $t\in
I_t$.

Then because $\Fie$ is continuous at $(s_0,t)$, there exists some
$\de$ (possibly depending on $t$) such that
 \[
d((s,t'),(s_0,t))<\de\implies d(\fie^{(s)}(t'),\fie^{(s_0)}(t))<\ep/2.
 \]
 Therefore $d(s,s_0)<\de$ and $d(t',t)<\de$ implies that
$\fie^{(s)}(t')$ is close to $\fie^{(s_0)}(t)$ is close to
$\fie^{(s_0)}(t')$. In other words, for all $t'$, there is a $\de$
neighbourhood of $t$,
 \[
d(s,s_0)<\de\implies d(\fie^{(s)}(t'),\fie^{(s_0)}(t'))<\ep.
 \]

Now I have proved that every point of the $t$-interval has a $\de$
neighbourhood with this property; by compactness the $t$-interval is covered
by finitely many of these, and by taking $\de$ to be the minimum of finitely
many $\de_i$ I get $\fie^{(s)}(t)$ close to $\fie^{(s_0)}(t)$ for all $t$ and
all $s$ close to $s_0$. \QED
\end{pf}

 \begin{defn}
 \index{based loop|(}
 Let $X$ be a topological space and $P_0\in X$ a base point. A family
of loops $\fie^{(s)}$ in $X$ based at $P_0$ is {\em continuous}, if
the map
 \[
 \Fie\colon[0,1]\times[0,1]\quad
 \hbox{defined by $\Fie(s,t)=\fie^{(s)}(t)$}
 \]
is continuous. A loop $\fie\colon[0,1]\to X$ based at $P_0$ is {\em
contractible}\index{contractible loop} in $X$, if there is a continuous family
of loops joining $\fie$ to the constant loop $\fie_0$ (obviously,
$\fie_0(t)=P_0$ for all $t$). A path connected space\index{path!connected} $X$
is {\em simply connected}\index{simply connected} if every loop in $X$ (with
every possible base point, though see~\exref{ex!basept}) is contractible.
\end{defn}

A homeomorphism\index{homeomorphism} $f\colon X\to Y$ takes paths and
continuous families of paths in $X$ into paths and continuous families of
paths in $Y$. In particular, being simply connected is a topological property.

\paragraph{Example} Every loop\index{loop} in the unit disc\index{disc}
$D\in\R^2$ is contractible.\index{contractible loop|)} This is obvious on a
sheet of paper; formally, it is best to use vector notation: if $\bx_0$ is the
base point, and $\fie(t)=\bx_t$ is the loop\index{loop} then
$\Phi(s,t)=\bx_0+s(\bx_t-\bx_0)$ gives a continuous family of paths connecting
$\fie$ to the constant path at $\bx_0$. The point is just that $D$ is convex;
the same argument gives the same conclusion for any convex subset of $\R^n$.
\index{based loop|)}

 \subsection{The winding number}\label{sec!winddef}
 \index{winding number}

To discuss the winding number formally, I use ordinary Cartesian
coordinates $(x,y)$ on the disc\index{disc} $D$, and polar coordinates
$(r,\theta)$ on the punctured disc $D^*$.\index{punctured disc $D^*$}
Note that $r>0$, and that polar coordinates do not really work at the
origin. The two coordinate systems are related by the usual rules
$x=r\sin\theta$, $y=r\cos\theta$.

What values do we allow for $\theta$? Since $\sin$ and $\cos$ are
periodic with period $2\pi$, the right answer is an equivalence class
of $\R$ modulo $2\pi\Z$. Note that every equivalence class of
$\R/2\pi\Z$ has a unique representative $\theta\in[0,2\pi)$; in applications
$\theta\in(-\pi,\pi]$ may be more convenient. If
you want $\theta$ to be unique, you should insist that
$(x,y)\ne(0,0)$, and choose the representative $\theta\in[0,2\pi)$.
But if you want $\theta$ to vary continuously with $(x,y)$, you
should arrange that $(x,y)$ stays well away from $(0,0)$ and choose
$\theta\in\R$.

\begin{prop} Suppose that the base point $P_0$ is in the $x$-axis
(so that $\theta=0$ is a possible choice). Let $\fie\colon[0,1]\to
D^*$ be a path with $\fie(0)=P_0$. Then there exist unique continuous
functions $r\colon[0,1]\to\R_+$ and $\Theta\colon[0,1]\to\R$ such that
 \[
\fie(t)=(r(t),\Theta(t))\quad\hbox{for all $t\in[0,1]$}.
 \]
If $\fie$ is a loop\index{loop}, then the end point is $\fie(1)=P_0$;
hence the value $\Theta(1)$ is of the form $2\pi n$ for some integer $n$.
 \end{prop}

\begin{defn} The integer number $n$ in the expression $\Theta(1)=2\pi n$
is the {\em winding number}\index{winding number} of the loop $\fie$,
written $n=\nu(\fie)$.
\end{defn}

 \begin{pf} Write $\fie(t)=(x(t), y(t))$ and set $r(t)=\sqrt{x(t)^2+y(t)^2}$
 for $t\in[0,1]$. Clearly $r(t)$ is continuous and strictly positive. Since
 $[0,1]$ is compact,\index{compact|(} $r(t)$ is bounded above and below by
 some $R,\rho>0$. Define
 \[
 \fie_1\colon[0,1]\to S^1\quad\hbox{by}\quad\fie_1(t)=
 \left(\frac{x(t)}{r(t)},\frac{y(t)}{r(t)}\right).
 \]
 Then $\fie_1$ is continuous, because $x,y$ and $r$ are, and $r(t)$
is bounded away from $0$.

Now $\fie_1(t)\in S^1$ is certainly of the form
$(\sin\theta,\cos\theta)$ for some $\theta=\theta(t)\in\R$. The
problem is that $\theta(t)$ is determined up to addition of multiples
of $2\pi$, and we have to choose the value for each
$t$ to make the function continuous.

Clearly the map $e\colon\R\to S^1$ defined by
 \[
e\colon\theta\mapsto(\sin\theta,\cos\theta)
 \]
defines a homeomorphism\index{homeomorphism} of any open interval
$(a,b)\subset\R$ of length $b-a<2\pi$ onto an open sector of the circle $S^1$
(similarly for closed). To prove the proposition, it is enough to chop up
$[0,1]$ into finitely many short intervals $U_i$ so that $\fie_1$ maps each
$U_i$ into such a sector, then take a suitable branch of $e\1$ on each of
these.

To do this very explicitly, cover $D^*$ by a number of overlapping open radial
sectors. To be definite, say, the ``top'' and ``bottom'' $200^\circ$ sectors
 \begin{figure}
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}b}
 \centering
 \leavevmode\hbox{\epsfysize=6.5 true cm\epsffile{top_cover_disc.eps}}
 \caption{$D^*$ covered by overlapping open radial sectors}
 \label{overlap sectors}
 \end{figure}
 \[
 \De_+:-10^\circ<\theta<190^\circ,\quad \De_-:170^\circ<\theta<370^\circ.
 \]
as in Figure~\ref{overlap sectors} (or make your own choice). Let me write
$\ep=10^\circ=\pi/18$, so that the sector intervals are $(0-\ep,\pi+\ep)$ and
$(\pi-\ep,2\pi+\ep)$. Then $\R$ is divided up into countably many intervals
 \[
 I^l_+=(2l\pi-\ep,(2l+1)\pi+\ep)\quad\hbox{and}\quad
 I^l_-=((2l-1)\pi-\ep,2l\pi+\ep)
 \]
for $l\in\Z$, in such a way that the restriction of $e$ to each interval
$I^l_\pm$ is a homeomorphism\index{homeomorphism} $e^l_\pm\colon
I^l_\pm\to\De_\pm$.

For every $t\in[0,1]$, the image $\fie_1(t)\in D^*$ is in one of the
$\De_\pm$. Since $\fie_1$ is continuous, $\fie_1\1(\De_\pm)$ is open, so there
exists a neighbourhood $U(t)\subset[0,1]$ of $t$ with
$\fie_1(U(t))\subset\De_\pm$. I can assume that each of the $U(t)$ is an open
interval of $[0,1]$ (except the first and last, which are half-open
intervals). The $U(t)$ form an open cover of $[0,1]$, so by compactness it has
a finite subcover. It follows that I can choose a cover of $[0,1]$ by a finite
number of overlapping open intervals
 \[
 [0,1]=\Cup_{i=0}^m U_i,\qquad
 \begin{array}{l}
\hbox{with $U_0=[0,b_1)$, $U_i=(a_i,b_{i+1})$,
$U_n=(a_m,1]$, and}\\
\hbox{$0<a_1<b_1<a_2<\cdots<b_{m-1}<a_n<b_m<1$,}
 \end{array}
 \]
 such that $\fie_1(U_i)\subset\De_\pm$. (For each $U_i$, if there is
any doubt, make the choice of $\pm$ at the outset.)
 \begin{figure}
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}c}
\centering
 \leavevmode\hbox{\epsfysize=2.5 true cm\epsffile{top_cover_interval.eps}}
\caption{Overlapping intervals}
\label{overlap intervals}
 \end{figure}

Now since $e^l_\pm\colon I^l_\pm\to\De_\pm$ is a
homeomorphism,\index{homeomorphism} we clearly define $\Theta$ over
$U_i\subset\De_\pm$ to be $(e^l_\pm)\1\circ\fie_1$, and the only
remaining question is the choice of $l$. First, $\fie(0)=P_0$ has
$\theta=0$ by assumption, so that either $U_0\subset\De_+$ or
$U_0\subset\De_-$. In the first case, choose $I^0_+$, in the second choose
$I^0_-$. These are forced by the requirement that $\Theta(0)=0$. Next,
suppose by induction that $\Theta$ is defined and continuous on $U_0\cup
U_1\cup\cdots\cup U_{i-1}$. The initial point $a_i$ of $U_i$ is in the
overlap with $U_{i-1}$, so that $\Theta$ is already defined there. This
determines the choice of $I^l_\pm$.
\QED \end{pf}

 \subsection{Winding number is constant in a family}\label{htpy}
 \index{winding number}

 \begin{thm} Let $(\fie^{(s)})$ be a continuous family of loops
$\fie^{(s)}\colon[0,1]\to D^*$. Then the winding number of the loop $\fie^{(s)}$
is constant (independent of $s$). In particular
$\nu(\fie^{(0)})=\nu(\fie^{(1)})$. \end{thm}

 \begin{pf} Write $\nu(\fie)$ for the winding number of a loop $\fie$. The point
is to show that $\nu(\fie)$ depends continuously on the path
$\fie\colon[0,1]\to D^*$.

For some value $s$, suppose that $\nu(\fie^{(s)})=n$. I claim that there is a
neighbourhood $V_s=(s-\de,s+\de)$ such that $\nu(\fie^{(s')})=n$ for all $s'\in
V_s$. In other words, the subset
 \[
\Si_n = \bigl\{s \bigm|\nu(\fie^{(s)})=n \bigr\}\subset [0,1]
 \]
 is open. This claim proves the theorem, because the interval $[0,1]$ is
connected, and is a disjoint union of the open sets $\Si_n$, therefore only one
value of $n$ occurs.

First, as in the proof of Proposition~\ref{wind}, I normalise all the paths by
dividing by the factor $r^{(s)}(t)$, so that each $\fie^{(s)}$ maps to $S^1$.
The normalisation factor is bounded away from $0$ because $I_s\times
I_t=[0,1]\times[0,1]$ is compact and $\Fie\colon I_s\times I_t\to D^*$ is
continuous. Thus I assume from now on that $\fie^{(s)}\colon[0,1]\to S^1$.

Recall the construction of Proposition~\ref{wind} for $\fie^{(s)}$. There is
a cover of $[0,1]=I_t$ by a finite chain of overlapping open intervals
$U_i=(a_i,b_{i+1})$ such that $\fie^{(s)}_1(U_i)\subset\De_\pm$. After this, the
map $\Theta$ just lifts $\De_\pm$ to $I^{n}_\pm$, where the value of $n$ is
determined inductively by the already known value of the starting point
$\Theta(a_i)$.

Now I choose slightly bigger ``top'' and ``bottom'' sectors $\De'_\pm$ of
$S^1$; to be explicit, choose
 \[
\De'_+:-20^\circ<\theta<200^\circ,\quad \De'_-:160^\circ<\theta<380^\circ,
 \]
 or in the previous notation $\De'_+=(0-2\ep,\pi+2\ep)$, etc. As far as
$\fie^{(s)}$ is concerned, nothing has changed: I still have
$\fie^{(s)}_1(U_i)\subset\De_\pm\subset\De'_\pm$, and the construction
of $\Theta$ can be made equally well with the bigger intervals.

However, by the definition of continuous family of loops, there exists a small
neighbourhood $s\in V_s\subset[0,1]$ such that also
$\fie^{(s')}(U_i)\subset\De'_\pm$ for all $s'\in V_s$. Thus I can use the same
collection of intervals $U_i$ to construct the argument function $\Theta^{(s')}$
of $\fie^{(s')}$ for all $s'\in V_s$.

Then $\Theta^{(s')}(t)$ on $V_s\times U_i$ is equal to the composite
$e_\pm^n\circ\Fie(s',t)$, and hence it is a continuous function of
$(s',t)\in V_s\times U_i$. It follows that $\Theta^{(s')}(t)$ is a
continuous function of $s'\in V_s$ for any $t$. In particular,
$\Theta^{(s')}(1)$ is a continuous function of $s'\in V_s$. However,
it is an integer multiple of $2\pi$. Therefore it is constant for
$s'\in V_s$. This proves the claim. \QED \end{pf}

 \subsection{Applications of the winding number}
 \label{fun thm}
 \index{winding number}
 \begin{corn} The punctured disc\index{punctured disc $D^*$} $D^*$ is not
homeomorphic\index{homeomorphism} to the disc $D$.
 \end{corn}

 \begin{pf} By Theorem~\ref{htpy}, a loop $\fie$ in $D^*$ of winding
number $\ne0$ is not contractible.\index{contractible loop} On the other
hand, every loop in $D$ is contractible (Example~\ref{IxI}). The
property that a loop\index{based loop|(} is contractible is a
topological property, so is preserved by
homeomorphism.\index{homeomorphism} Therefore there does not exist a
homeomorphism between $D$ and $D^*$. \QED \end{pf}

\paragraph{Remark} The same proof shows that the punctured disc $D^*$ is not
homeomorphic to the disc $D$ with some of its boundary added, since loops in
the latter are still contractible. This concludes the proof of the Main Claim
in Proposition~\ref{std exs}: a boundary point of a surface is
topologically different from an interior point.

 \begin{corn}[``Fundamental theorem of algebra'']
 \index{algebraically closed field|(}
 \index{fundamental!theorem of algebra} Let
 \[
f(z)=z^n+a_{n-1}z^{n-1}+\cdots+a_1z+a_0
 \]
 be a polynomial of degree $n\ge1$ in $z$, with complex coefficients
$a_i\in\C$. Then there exists a complex number $\ze$ such that
$f(\ze)=0$. In other words, $\C$ is algebraically closed.
 \end{corn}

 \begin{pf} Write $\C^*=\C\setminus\{0\}$. Obviously $\C^*$ is
homeomorphic\index{homeomorphism} to $D^*$, so that the definition and properties
of the winding number\index{winding number|)} apply also to $\C^*$.

I first give the proof forgetting the small detail of the base point $P_0$, then
explain how to patch this up. For $K\in\R$, $K\ge0$, define
 \[
\fie_K\colon[0,1]\to\C\quad\hbox{by $t\mapsto f(K\exp(2\pi it))$.}
 \]
 If $\fie_K(t)=0$ for some $K$ and some $t$ then $f(\ze)=0$ for
$\ze=K\exp(2\pi it)$. Assume by contradiction that this never
happens. Then $\fie_K\colon[0,1]\to\C^*$ is a continuous family of
loops in $\C^*$. When $K=0$ it is the constant loop: $\fie_0(t)=a_0$
for all $t$. When $K\gg0$ it has winding number $n$. Indeed, if
$K>1+\sum_{i=0}^{n-1}|a_i|$ the term $z^n$ in $f(z)$ is bigger than
all the other terms put together, so that the loop looks like
$K^n(\sin nt+i\cos nt)$ plus a smaller error term that does not allow
the path to reach to the origin.

However, by Theorem~\ref{htpy}, if we assume that $\fie_K$ maps
$[0,1]$ to $\C^*$, the winding number must be constant, independent
of $K$. This a contradiction. Therefore, sometimes $f(z)=0$.

The proof just given does not work as it stands, because
Theorem~\ref{htpy} dealt only in based loops.\index{based loop} There
are several ways of dealing with this; one method would be to reprove
Theorem~\ref{htpy} without base points, or to prove that the winding
number does not depend on the choice of a base point.

An easy ad hoc method is to define a new family of paths $\fie_K$ starting from
the base point $P_0=a_0$ in the following way: we spend the first 1/3 of the
time in the interval $[0,1]$ plodding out from $f(0)=a_0$ to $f(K)=\fie_K(0)$
along the path $f(\R)$; then we pursue the loop $\fie_K$ at 3 times the original
speed, returning to $f(K)=\fie_K(1)$ at time $t=2/3$; then we spend the final
1/3 of the time returning from $f(K)$ to $f(0)$ by retracing our steps along
the same path $f(\R)$. The new path has the same winding number as the old,
because any change in the argument $\theta$ made in plodding out to $f(K)$
is exactly cancelled when we retraced our steps. The details are easy to work
out. \QED \end{pf}
 \index{based loop|)}
 \index{algebraically closed field|)}

 \section*{Exercises to Chapter~\thechapter}
 \addcontentsline{toc}{section}{Exercises to Chapter~\thechapter}
 \markright{\hfill Exercises to Chapter~\thechapter\quad}
 \begin{enumerate}
\renewcommand{\labelenumi}{\thechapter.\theenumi.}

\item\label{metric_top_ex} Let $(X,d)$ be a metric space. Check that
Definition~\ref{metric motiv} does indeed define a topology on $X$; in other
words, check that the set $\sT$ of open sets in the metric sense is a
topology. \begin{hint} Use the triangle inequality. \end{hint}

 \item[] Questions on point-set topology:

 \item\label{cont ex} $X,Y,Z$ are topological spaces and $f\colon
X\to Y$, $g\colon Y\to Z$ continuous maps. Prove that $g\circ f$ is
continuous. Count the lines of your proof, and compare with the same
proof in a standard analysis or metric spaces course.

 \item\label{conv ex} $X$ is a metric space with metric topology
$\sT_X$. Prove that a sequence of points $a_i\in X$ converges to $l$
in the sense of the metric if and only if it converges in the sense
of topology as in~\ref{comp}.

 \item\label{seqcont ex} By definition, a sequence of points
$\{x_i\}_{i=1,2,\dots}$ converges to $x\in X$ in a topological space
if every neighbourhood $U$ of $x$ contains all but finitely many of
the $x_i$. Let $X,Y$ be topological spaces and $f\colon X\to Y$
continuous.
 \begin{enumerate}

 \item Prove that $\{x_i\}$ converge to $x$ implies $\{f(x_i)\}$
converge to $f(x)$. That is, ``continuity implies sequential
continuity'' for topological spaces.

 \item Conversely, prove that for a metric space $X$, this
convergence for all sequences implies that $f$ is continuous. In
other words, ``sequential continuity implies continuity'' for metric
spaces.

 \item Now let $X$ be a topological space, not necessarily metric, in
which every point $x\in X$ has a countable basis of neighbourhoods
(referred to in \ref{metric motiv}). Prove sequential continuity
implies continuity.

 \item Prove that if $X$ is an uncountable set with the cofinite topology
(\ref{cofinite}, Example~2), then there does not exist a countable
basis for the neighbourhoods of $x\in X$.

 \item (harder) Find a topological space and a map $f\colon Y\to X$
which is sequentially continuous but not continuous.
 \end{enumerate}

 \item $X$ is a metric space, $x,y\in X$ and $a_1,a_2,\dots$ a
sequence of points of $X$. Which of the following are topological
properties?
 \begin{enumerate}

 \item $X\setminus x$ is disconnected.

 \item $a_i\to x$ as $i\to\infty$.

 \item $x$ is in the closure of $\{y\}$.

 \item $a_i$ is a Cauchy sequence.

 \item The ball\index{ball} $B(x,1)$ is compact.

 \item Every neighbourhood of $x$ is a countable set.

 \item The closure of the ball $B(x,1)$ is connected.

 \item For every compact subset $V\subset X$, the complement
$X\setminus V$ is disconnected.\index{connected}
 \end{enumerate}
For each statement, give a proof or a counterexample, or both.

 \item How many capital letters of the alphabet are there up to
homeo\-mor\-phism\index{homeomorphism} in a typeface without knobs on,
such as
 \[
\hbox{\textsf {ABCDEFGHIJKLMNOPQRSTUVWXYZ}}?
 \]
 Scrabble players do it with $\textsf K$ and $\textsf Q$.

 \item $X$ and $Y$ are topological spaces and $f\colon X\to Y$ a
continuous surjective map. Prove that if $X$ is sequentially
compact,\index{sequentially compact} so is $Y$. \begin{hint} consider
a sequence in $Y$ and use the stated properties of $f$ and $X$.
Compare the proof of Proposition~\ref{comp}.\end{hint}

 \item Prove that a continuous function $f\colon X\to\R$ on a compact
space $X$ is bounded, and achieves its bounds. \begin{hint} to get
bounded, just say balls, lots of balls,\dots as before. Let $K=\sup
f(X)\in\R$, which exists by the completeness axiom. By contradiction
assume that $f(x)\ne K$ for all $x\in X$; consider the open sets
$U_\ep\subset X$ defined by $U_\ep=\bigl\{x \bigm|f(x)\le
K-\ep\bigr\}$.\end{hint} \index{compact|)}

 \item Prove that a continuous function $f\colon[a,b]\to\R$ is
uniformly continuous. \begin{hint} for a given $\ep$, the definition
of continuity gives balls $B(x,\de_x)$, \dots\end{hint}

 \item\label{sspace ex} $X$ is a topological space and $Y\subset X$ a
subset with the subspace topology; prove that every {\em
closed}\index{closed} subset of $Y$ is of the form $Y\cap V$ with $V$
closed in $X$.

 \item\label{submetric ex} $X$ is a metric space and $Y\subset X$ a
subset. Prove that the following two topologies on $Y$ are identical:
 \begin{enumerate}

 \item Take the metric topology $\sT_X$ and the subspace topology
$\sT_{Y,1}$ on $Y$.

 \item Restrict the metric $d_X$ to $Y$ to get a metric $d_Y$, then
take the metric topology $\sT_{Y,2}$ on $Y$ corresponding to $d_Y$.
 \end{enumerate}

 \item Find all the possible topologies on a set $\{x,y\}$ with two
points.

 \item Study the possible topologies on a finite set.
 \begin{enumerate}

 \item If a topological space is not $T_1$ (see \ref{Hausdorff}) then
there exist $x\ne y$ such that the constant sequence $y,y,\dots$
converges to $x$. That is, $x$ is in the closure of the set $\{y\}$.

 \item Write $x\mathbin{C}y$ if $x$ is in the closure of $y$, and
think of this as a relation between $x$ and $y$. Prove that
$\mathbin{C}$ is a transitive relation.

 \item Define the relation $x\mathbin{R}y$ by
 \[
 x\mathbin{R}y \;\iff\; \hbox{$x\mathbin{C}y$ and $y\mathbin{C}x$}.
 \]
 Prove that $\mathbin{R}$ is an equivalence relation.

 \item Let $Y\subset X$ be an equivalence class of $R$; prove that the
subspace topology on $Y$ is the indiscrete topology\index{indiscrete
topology} (no opens other than $\emptyset$ and $X$).

 \item (harder) Use steps (a--d) to describe all possible topologies
on a finite set $Y$.
 \end{enumerate}

 \item Let $X$ be a topological space, $\sim$ an equivalence relation
on $X$ and $Y=X/{\sim}$ the quotient topological space. Think of the
relation $\sim$ as the subset
 \[
Z(\sim)=\bigl\{(x,y)\bigm|x\sim y \bigr\}\subset X\times X
 \]
 where $X\times X$ is given the product topology.\index{product
topology}
 \begin{enumerate}

 \item By imitating the proof of Proposition~\ref{Hausdorff}, prove
that $Y$ is Hausdorff\index{Hausdorff property} if and only if
$Z({\sim})\subset X\times X$ is closed.

 \item Let $Z\subset X\times X$ be the {\em closure} of the diagonal,
considered as a relation ($x\sim y$ if and only if $(x,y)\in Z$);
describe what $x\sim y$ means in terms of neighbourhoods of $x$ and $y$,
and prove that $\sim$ is an equivalence relation.

 \item Prove that $X$ has a continuous map $f\colon X\to X'$ to a
Hausdorff\index{Hausdorff property} space which has the
UMP\index{universal mapping property} for such maps.
 \end{enumerate}

 \item[] Exercises on surfaces

 \item Write down equations for a torus,\index{torus} a solid torus
and a M\"obius strip\index{M\"obius strip} in terms of Cartesian
coordinates $(x,y,z)$ or cylindrical polar coordinates $(r,\theta,z)$
for $\R^3$. \begin{hint} you get a torus by rotating a circle about
an axis outside it, and a M\"obius strip by letting a diameter of the
circle rotate simultaneously to get $1,3,5,\dots$
halftwists.\end{hint}

 \item Prove that $S^2\setminus\{\hbox{2 points}\}$ is
homeomorphic\index{homeomorphism} to the cylinder $S^1\times\R$.
\begin{hint} Let the two points be the poles $N$ and $S$, and think of
Mercator's projection.\end{hint}

 \item Using Figure~\ref{aa}, prove the following statements:
\begin{enumerate}
\item\label{ex!p2r} If $L=\PP^1$ is the line obtained from the equatorial
circle, then $\PP^2\setminus L$ is topologically a disc\index{disc} (the
upper half-sphere), and a neighbourhood of $L$ in $\PP^2$ is a M\"obius
strip.\index{M\"obius strip}

\item If $Q=\{x^2+y^2=z^2\}\subset\PP^2$ is a conic curve, then
$\PP^2\setminus Q$ consists of two pieces, one a M\"obius
strip\index{M\"obius strip} and the other a disc\index{disc}; a
neighbourhood of $Q$ in $\PP^2$ is a cylinder.
\end{enumerate}
Draw pictures illustrating the following statement: cutting $\PP^2$ along a
line is like cutting a M\"obius strip along its central curve, whereas
cutting $\PP^2$ along a conic is like cutting a M\"obius strip along the
curve trisecting the width of the strip.\index{topology!of $\PP^n$}

 \index{quotient topology|(}
 \item\label{ex!squarequot} In~\ref{std exs}, I obtained the
{M\"obius strip}\index{M\"obius strip}, the cylinder and the torus from a
square by glueing its edges in a particular fashion. In
Figure~\ref{fig!moresquare}, I give two other glueing rules.
\begin{enumerate}

 \item Show that the first pattern builds a surface homeomorphic to
the projective plane $\PP^2$.

 \item Show that the second pattern corresponds to a surface that you can
build in two steps, first glueing a cylinder as in Figure~\ref{torus fig}
and then identifying the circles at the ends, carefully remembering their
orientation. This surface is called the {\em Klein
bottle}\index{Klein bottle}. It shares with $\PP^2$ the property that it
cannot be embedded in $\R^3$ without self-crossing.
 \end{enumerate}

 \begin{figure}
 \renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}a}
 \centering
% \leavevmode\hbox{\epsfysize=8.5 true cm\epsffile{top_torus.eps}}
 \input{top_moresquares.pstex_t}
 \caption{Glueing patterns on the square}
 \label{fig!moresquare}
 \end{figure}

\item\label{ex!genus2} The left panel of Figure~\ref{fig!octagon} shows a
surface with two\index{genus} handles, with a set of circles marked on
its surface, in analogy with the last panel of Figure~\ref{torus fig}.
Verify that cutting the surface along the marked circles leads to the octagon
on the right of Figure~\ref{fig!octagon}, with the edges identified as shown.
Hence conversely, glueing the octagon with the given pattern leads to a
surface with two handles!

 \begin{figure}
 \renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}b}
 \centering
% \leavevmode\hbox{\epsfysize=8.5 true cm\epsffile{top_torus.eps}}
 \input{top_genus2.pstex_t}
 \caption{The surface with two handles and the octagon}
 \label{fig!octagon}
 \end{figure}
 \index{quotient topology|)}

 \item[] Exercises on loops\index{loop}

 \item\label{uniform} Draw the graph of the function
 \[
 f^{(s)}(t)= \begin{cases} 4t/s & \hbox{for $0\le t\le s/4$}, \\
 2-4t/s & \hbox{for $s/4\le t\le s/2$}, \\
 0 & \hbox{for $s/2\le t\le1$}.
 \end{cases}
 \]
 Here $s\in(0,1]$. Have you seen anything like this before? Set
$f^{(0)}(t)=0$, and prove the following:
 \begin{enumerate}

 \item for any fixed $s\in[0,1]$ the formula
$\fie^{(s)}(t)=(f^{(s)}(t),t)$ defines a path
$\fie^{(s)}\colon[0,1]\to\R^2$ (i.e., it is continuous);

 \item for fixed $t\in[0,1]$ the map $s\mapsto\fie^{(s)}(t)$ is
continuous;

 \item $\fie^{(s)}$ is not a continuous family of paths in $\R^2$ in
the sense of Definition~\ref{IxI};

 \item $\fie^{(s)}$ is something you would not do to a dog lead;

 \item $\Fie(s,t)=f^{(s)}(t)$ is not a continuous function of $s,t$
near $(0,0)$.
 \end{enumerate}
The point of the question is to justify the tentative definition in \ref{IxI},
in particular to convince you of the requirement for uniformity in~$t$.

 \item\label{ex!basept}\index{path!connected}\index{contractible
loop} Suppose that $X$ is a path connected topological space and pick
two points $P_0,Q_0\in X$. Prove that all loops in $X$ based at $P_0$
are contractible if and only if all loops in $X$ based at $Q_0$ are
contractible. \begin{hint} compare the end of~\ref{fun thm}.\end{hint}
 \end{enumerate}
 \index{topology|)}

% \renewcommand{\labelenumi}{\theenumi}
% \renewcommand{\theenumi}{(\arabic{enumi})}
