\magnification=\magstep1
\newfam\msbfam
\font\tenmsb=msbm10 \textfont\msbfam=\tenmsb
\font\sevenmsb=msbm7 \scriptfont\msbfam=\sevenmsb
\def\Bbb#1{{\fam\msbfam\relax#1}}
\def\N{{\Bbb N}}
\def\Q{{\Bbb Q}}
\def\R{{\Bbb R}}
\def\Z{{\Bbb Z}}
\def\barf{{\overline{f}}}
\def\barK{{\overline{K}}}
\def\barS{{\overline{S}}}
\def\barU{{\overline{U}}}
\def\barV{{\overline{V}}}
\def\scra{{\cal A}}
\def\scrb{{\cal B}}
\def\scrc{{\cal C}}
\def\scrd{{\cal D}}
\def\scre{{\cal E}}
\def\scrf{{\cal F}}
\def\scrg{{\cal G}}
\def\scri{{\cal I}}
\def\scrk{{\cal K}}
\def\scrm{{\cal M}}
\def\scrn{{\cal N}}
\def\scro{{\cal O}}
\def\scrp{{\cal P}}
\def\scrq{{\cal Q}}
\def\scrr{{\cal R}}
\def\scrs{{\cal S}}
\def\scru{{\cal U}}
\def\scrv{{\cal V}}
\def\scrw{{\cal W}}
\def\scrts{{\cal T\cal S}}
\def\scrpts{{\cal P\cal T\cal S}}
\def\scrfb{{\cal F\cal B}}
\def\scrns{{\cal N\cal S}}
\def\scrtns{{\cal T\cal N\cal S}}
\def\scrdd{{\cal D\cal D}}
\def\hatc{{\widehat{c}}}
\def\hatf{{\widehat{f}}}
\def\hatg{{\widehat{g}}}
\def\hath{{\widehat{h}}}
\def\hatE{{\widehat{E}}}
\def\hatX{{\widehat{X}}}
\def\hatalpha{{\widehat{\alpha}}}
\def\hatbeta{{\widehat{\beta}}}
\def\hatgamma{{\widehat{\gamma}}}
\def\hatalphabeta{{\widehat{\alpha\|\beta}}}
\def\tildef{{\widetilde{f}}}
\def\tildeg{{\widetilde{g}}}
\def\tildeh{{\widetilde{h}}}
\def\tildez{{\widetilde{z}}}
\def\tildeW{{\widetilde{W}}}
\def\tildeX{{\widetilde{X}}}
\def\tildeY{{\widetilde{Y}}}
\def\tildeZ{{\widetilde{Z}}}
\def\Cl{{\rm Cl}}
\def\Int{{\rm Int}}
\def\Bd{{\rm Bd}}
\def\Hom{{\rm Hom}}
\def\Arr{{\rm Arr}}
\def\Dom{{\rm Dom}}
\def\Tar{{\rm Tar}}
\def\dom{{\rm dom}}
\def\tar{{\rm tar}}
\def\Obj{{\rm Obj}}
\def\id{{\rm id}}
\def\Id{{\rm Id}}
\def\catofsets{\{\hbox{Sets}\}}
\def\catofrings{\{\hbox{Rings}\}}
\def\catofgps{\{\hbox{Groups}\}}
\def\set{{{\rm set}}}
\def\SL{{{\rm SL}}}
\def\Stab{{{\rm Stab}}}
\def\det{{{\rm det}}}
\def\supp{{{\rm supp}}}
\centerline{\bf Notes for Math 8301, Manifolds and Topology, Fall 2004}
\centerline{\bf The Fundamental Group and Covering Spaces}
\vskip.1in\noindent
{\it Definition.}
$B_\delta(x)=(x-\delta,x+\delta)\subseteq\R$.
\vskip.1in\noindent
{\it Definition.}
A function $f:\R\to\R$ is {\bf continuous at $x$} if,
for all $\epsilon>0$, there exists $\delta>0$
such that $f(B_\delta(x))\subseteq B_\epsilon(f(x))$,
{\it i.e.}, $B_\delta(x)\subseteq f^{-1}(B_\epsilon(f(x)))$,
\vskip.1in\noindent
{\it Note:} For $\phi:A\to B$ and $B_0\subseteq B$,
we define $\phi^{-1}(B_0)=\{a\in A\,|\,\phi(a)\in B_0\}$.
\vskip.1in\noindent
{\it Definition.}
A function $f:\R\to\R$ is {\bf continuous}
if, for all $x\in\R$, we have: $f$ is continuous at $x$.
\vskip.1in\noindent
{\it Question:} How do we define continuity for $f:\R^m\to\R^n$?
How about $f:S^n\to\R^m$, where
$S^n:=\{x\in\R^{n+1}\,|\,d(x,0)=1\}$?
How about $f:S^n\to S^m$?
Let's try to formulate a general definition
of continuity that will include all these definitions
as special cases.
\vskip.1in\noindent
{\it Definition.}
An {\bf$N$-space} consists of
\itemitem{(1)} a set $X$ and
\itemitem{(2)} a function $x\mapsto\scrn_x:X\to
\{\hbox{subsets of }\{\hbox{subsets of }X\}\}$
\noindent
such that
\itemitem{(A)} for all $x\in X$, we have: $\scrn_x\ne\emptyset$; and
\itemitem{(B)} for all $x\in X$, for all $N\in\scrn_x$,
we have: $x\in N$.
\vskip.1in\noindent
{\it Example.}
$X=\R$, $\scrn_x=\{B_\delta(x)\,|\,\delta>0\}$.
\vskip.1in\noindent
{\it Definition.}
For all $x\in X$, an {\bf ``$N$-neighborhood''} of $x$
is an element of $\scrn_x$.
\vskip.1in\noindent
{\it Definition.}
Let $X$ and $Y$ be $N$-spaces and let $f:X\to Y$
be a function.
For a point $x\in X$,
we say that $f:X\to Y$ is {\bf $N$-continuous at $x$} if,
for any $N$-neighborhood $V$ of $f(x)$,
there exists an $N$-neighborhood $U$ of $x$
such that $f(U)\subseteq V$, {\it i.e.}, $U\subseteq f^{-1}(V)$.
We say that $f$ is {\bf $N$-continuous} if $f$ is $N$-continuous
at every point $x\in X$.
\vskip.1in\noindent
{\it Definition.}
A category consists of
\itemitem{(1)} a class $\scrc$ of {\bf objects};
\itemitem{(2)} for all $C,C'\in\scrc$,
a set $\Hom(C,C')$ of {\bf morphisms} or {\bf arrows}
from $C$ to $C'$;
\itemitem{(3)} for all $C,C',C''\in\scrc$,
a {\bf composition} function
$$\Hom(C,C')\times\Hom(C',C'')\to\Hom(C,C'');\qquad\hbox{and}$$
\itemitem{(4)} for all $C\in\scrc$, an {\bf identity}
arrow $\id_C\in\Hom(C,C)$
\noindent
such that
\itemitem{(A)} for all $C,C'\in\scrc$, for all $f\in\Hom(C,C')$,
we have $f\circ\id_C=\id_{C'}\circ f=f$;
\itemitem{(B)} for all $C,C',C'',C'''\in\scrc$,
for all $f\in\Hom(C,C')$,
for all $g\in\Hom(C',C'')$,
for all $h\in\Hom(C'',C''')$, we have
$(f\circ g)\circ h=f\circ(g\circ h)$;
\noindent
and probably other properties.
\vskip.1in\noindent
{\bf EXERCISE 1A:} Look up, and write out the defintion of a category.
\vskip.1in\noindent
{\it Example.}
$\scrns:=\{N\hbox{-spaces}\}$.
That is, $\scrns$ is the category of $N$-spaces,
together with $N$-continuous maps, together with the usual
composition of maps, together with the usual identity maps.
\vskip.1in\noindent
{\bf EXERCISE 1B:}
\itemitem{(1)} For all $X\in\scrns$,
show that $x\mapsto x:X\to X$ is $N$-continuous.
\itemitem{(2)} For all $X,Y,Z\in\scrns$,
for all $N$-continuous $f:X\to Y$, for all $N$-continuous $g:Y\to Z$,
show that $g\circ f:X\to Z$ is $N$-continuous.
\vskip.1in\noindent
Let $\catofsets$ denote the category of sets
(and functions). For any $X\in\scrns$,
let $X_\set$ be the underlying set of $X$.
For $X,Y\in\scrns$, for $f\in\Hom(X,Y)$,
let $f_\set:X_\set\to Y_\set$ be the underlying
set function of $f$.
Then $X\mapsto X_\set:\scrns\to\catofsets$
is an example of a {\bf functor}.
(Really, we should say
$X\mapsto X_\set$ together with $f\mapsto f_\set$
forms a functor.)
Note that, for $X,Y,Z\in\scrns$,
for $f\in\Hom(X,Y)$, for $g\in\Hom(Y,Z)$,
we have $(g\circ f)_\set=g_\set\,\circ\,f_\set$.
This is one of the basic properties of a functor.
\vskip.1in\noindent
{\bf EXERCISE 1C:} Look up, and write down the
definition of a functor from one category to another.
Be sure to note that some functors are covariant,
while others are contravariant.
\vskip.1in\noindent
{\it Example.}
Let $\catofrings$ denote the category
of (not necessarily commutative) rings with unit,
together with unit-preserving ring homomorphisms.
Let $\catofgps$ denote the category of groups,
with group homomorphisms.
Then $(R,\cdot,+)\mapsto(R,+)$ is a functor
from $\catofrings$ to $\catofgps$ (and even takes
values in the subcategory of Abelian groups).
Then $(R,\cdot,+)\mapsto(\{\hbox{units in }R\},\cdot)$
is another functor from $\catofrings$ to $\catofgps$.
\vskip.1in\noindent
{\it Definition.}
Let $\scrc$ be any category.
Then there is an identity functor $\Id_\scrc:\scrc\to\scrc$
defined by $\Id_\scrc(C)=C$. (Note: This defines the
functor on objects, and it is left to you you to guess
what the functor does on arrows.) (Note: Well, okay, I'll
tell you this time -- it leaves each arrow fixed,
{\it i.e.}, for all $f:C\to C'$ in $\scrc$,
we define $\Id_\scrc(f)=f$. In general, though,
it is an expectation that you should be able to figure
this kind of thing out, but always feel free to ask
if it's unclear.)
\vskip.1in\noindent
Fix, until END OF DISCUSSION \#1, some $X\in\scrns$.
Let's say we've identified some points of $X$ that are
``happy''.
\vskip.1in\noindent
Let $0\in X$.
If we say {\bf ``points close to $x$ are happy''}
or {\bf ``points sufficiently close to $x$ are happy''}
or {\bf ``all points sufficiently close to $x$ are happy''},
we mean that there is an $N$-neighborhood $U$ of $x$
such that, for all $u\in U$, we have: $u$ is happy.
Now suppose that $y$ is sufficiently close to $x$
that it is guaranteed to be happy.
That is, suppose that $y\in U$.
{\it Question:}
Does it automatically follow that all points sufficiently
close to $y$ are guaranteed to be happy by being in $U$?
In other words, can we conclude that there
is a neighborhoos $V$ of $y$ such that $V\subseteq U$.
\vskip.1in\noindent
{\it Answer:} No. For example, let $X_\set=\R$, and for all
$x\in X$, let $\scrn_x:=\{[x-\delta,x+\delta]\,|\,\delta>0\}$.
Then $X=(X_\set,x\mapsto\scrn_x)$ is an $N$-space.
Suppose that $[-1,1]$ is exactly the set of happy points.
Let $x:=0, y:=-1$. Then $U=[-1,1]$ is an $N$-neighborhood of $x$
and $y\in U$, but there is no $N$-neighborhood $V$ of $y$
such that $V\subseteq U$.
So even though $y$ is happy, we have no ``room to maneuver''
because the kinds of $N$-neigborhoods in $X$ are, somehow,
very {\it unforgiving},
in the sense that, if you have a point that is in an $N$-neighborhood,
and you then perturb it ever so slightly, you may find yourself
outside the $N$-neighborhood.
\vskip.1in\noindent
{\it Definition.}
We say that an $N$-space $X$ is {\bf forgiving}
if, for all $x\in X$, for any $N$-neighborhood $U$ of $x$,
for any $y\in U$, there is an $N$-neighborhood $V$ of $y$
such that $V\subseteq U$.
\vskip.1in\noindent
END OF DISCUSSION \#1
\vskip.1in\noindent
Fix, until END OF DISCUSSION \#2, some $X\in\scrns$.
Let's say we've identified some points of $X$ that are
``happy'' and some point that are ``dull''.
So $X$ breaks up into happy dull points,
unhappy dull points, happy sharp points and unhappy sharp points.
Let $x\in X$ and assume that all points sufficiently close
to $x$ are happy.
That is, there is an $N$-neighborhood $U$ of $x$
such that every element of $U$ is happy.
Assume, furthermore that all points sufficiently close
to $x$ are dull.
That is, there is an $N$-neighborhood $V$ of $x$
such that every element of $V$ is happy.
Does it automatically follow that there is
a neighborhood $W$ of $x$ such that
$W\subseteq U\cup V$, which guarantees that all
points sufficiently close to $x$ are both happy and dull?
The answer is no. It's not an assigned problem, but you should
think about constructing a counterexample.
\vskip.1in\noindent
{\it Definition.}
An $N$-space $X$ is {\bf intersection-stable}
if, for all $x\in X$, for any $N$-neighborhoods $U$ and $V$ of $x$,
there is an $N$-neighborhood $W$ of $x$ such that $W\subseteq U\cap V$.
\vskip.1in\noindent
END OF DISCUSSION \#2
\vskip.1in\noindent
{\it Definition.}
An $N$-space is said to be {\bf topological}
if it is both forgiving and intersection-stable.
\vskip.1in\noindent
{\it Definition.}
Let $\scrtns$ denote the category of
topological $N$-spaces (with $N$-continuous maps).
\vskip.1in\noindent
{\it Definition.}
Let $X$ be a set. A {\bf topology on $X$}
is a subset $\scro$ of $\{\hbox{subsets of }X\}$
such that
\itemitem{(A)} $\emptyset,X\in\scro$;
\itemitem{(B)} for all $\scrs\subseteq\scro$,
we have $\cup\scrs\in\scro$; and
\itemitem{(C)} for all finite $\scrf\subseteq\scro$,
we have $\cap\scrf\in\scro$.
\vskip.1in\noindent
{\it Definition.}
A {\bf topological space} consists of
\itemitem{(1)} a set $X$; and
\itemitem{(2)} a topology on $X$.
\vskip.1in\noindent
{\it Definition.}
An {\bf open subset} of $X$ is an element of $\scro$.
A subset of $X$ is said to be {\bf open} or {\bf open in $X$}
if it is an open subset of $X$.
\vskip.1in\noindent
{\it Definition.}
Let $X$ and $Y$ be topological spaces
and let $f:X\to Y$ be a function defined on their underlying sets.
We say that $f$ is {\bf continuous} if,
for any open $V\subseteq Y$, we have that $f^{-1}(V)$ is open in $X$.
\vskip.1in\noindent
What motivates the definition of topological space and of
this definition of continuity?
\vskip.1in\noindent
{\it Definition.}
Let $\scrts$ denote the category of topological spaces,
with continuous maps.
\vskip.1in\noindent
{\it Definition.}
Define a functor $\scrf:\scrtns\to\scrts$ by
$$\scrf((X,x\mapsto\scrn_x))=
(X,\{\cup\scrs\,|\,\scrs\subseteq\cup_{x\in X}\scrn_x\}).$$
Here, I leave it to you to guess what $\scrf$ does on arrows.
Note that $\{\hbox{open sets in }\scrf X\}$ is
exactly the closure of $\{N\hbox{-neighborhoods of points of }X\}$
under union.
Define a functor $\scrg:\scrts\to\scrtns$ by
$$\scrg((Y,\scro))=
(Y,y\mapsto\{U\in\scro\,|\,y\in U\}).$$
\vskip.1in\noindent
{\bf EXERCISE 1D:} Show, for all $X\in\scrtns$,
that $\scrf X\in\scrts$. (You must show that
the open subsets of $\scrf X$ are closed under
finite intersection and arbitrary union, and that
both $\emptyset$ and $X$ are open in $\scrf X$.)
\vskip.1in\noindent
{\bf EXERCISE 1E:} Show, for all $Y\in\scrts$,
that $\scrg Y\in\scrtns$.
\vskip.1in\noindent
{\bf EXERCISE 1F:} Show, for all $Y\in\scrts$,
that $\scrf\scrg Y=Y$. (You must show that a
subset of $Y$ is open in $\scrf\scrg Y$
iff it is open in $Y$.)
\vskip.1in\noindent
The preceding exercise asks you to prove
that $\scrf\scrg=\Id_\scrts$.
We now show that $\scrg\scrf X$ is not necessarily
equal to $X$.
That is, we show that $\scrf\scrg\ne\Id_\scrtns$:
\vskip.1in\noindent
{\it Example.}
Let
$X:=(\R,x\mapsto\{(x-\delta,x+\delta)\,|\,\delta>0\})$.
Then the set of open subsets of $\scrf X$ is
exactly the closure of
$\{(x-\delta,x+\delta)\,|\,x\in\R,\delta>0\}$
under union.
In particular, $(-1,1)\cup(2,4)$ is open in $\scrf X$.
Then $(-1,1)\cup(2,4)$ is an $N$-neighborhood of $0$
in $\scrg\scrf X$.
However it is {\it not} an $N$-neighborhood of $0$
in $X$.
Therefore $\scrg\scrf X\ne X$.
\vskip.1in\noindent
{\it Definition.}
An arrow (or morphism) $f:C\to C'$
in a category $\scrc$ is an {\bf isomorphism}
if there exists an arrow $g:C'\to C$
such that $f\circ g=\id_{C'}$ and such that $g\circ f=\id_C$.
\vskip.1in
That is, an isomorphism is simply an invertible arrow.
\vskip.1in\noindent
{\it Definition.}
In the category $\scrts$,
an isomorphism will be called a {\bf homeomorphism}.
In the category $\scrtns$,
an isomorphism will be called an {\bf$N$-homeomorphism}.
\vskip.1in\noindent
{\it Definition.}
Let $\scra$ and $\scrb$ be functors
from a category $\scrc$ to a category $\scrd$.
A {\bf natural transformation} $\tau:\scra\to\scrb$
associates to each object $C\in \scrc$
an arrow $\tau_C:\scra C\to\scrb C$,
provided that this association has the property that:
for any arrow $f:C\to C'$ in $\scrc$,
we have
$(\scrb f)\circ\tau_C=\tau_{C'}\circ(\scra f)$.
\vskip.1in
Note that, for their to be a natural transformation from
$\scra$ to $\scrb$, the domains of $\scra$ and $\scrb$ must
agree. Moreover, their targets must agree as well.
Let $\scra$ and $\scrb$ be functors from a category
$\scrc$ to a category $\scrd$.
It may be helpful to picture $\scra\scrc$ as a subcategory
inside of $\scrd$.
In some sense, $\scra$ is a ``parametric'' subcategory of $\scrd$,
where the objects are parametrized by objects in $\scrc$.
Similarly, $\scrb\scrc$ is a parametric subcategory of $\scrd$.
With this intution,
a transformation $\tau:\scra\to\scrb$ may be
thought of as a parametric family of
arrows in $\scrd$, one for each object in $\scrc$.
For each $C\in\scrc$, the corresponding arrow runs from
$\scra C$ to $\scrb C$.
Moreover, for each arrow in $\scrc$, we now get a diagram
in $\scrd$ (with four objects and four arrows).
If these diagrams all commute, then
we say that the transformation is ``natural''.
\vskip.1in\noindent
{\it Example.}
Let $\scrv$ be the category of
finite dimensional vector spaces, and linear transformations.
Let $\scri:\scrv\to\scrv$ be the (covariant) identity functor.
Let $\scrd:\scrv\to\scrv$ be the (contravariant) functor
defined by $\scrd(V)=V^*:=\Hom(V,\R)$.
Let $\scrdd:\scrv\to\scrv$ be the (covariant) functor
$\scrdd:=\scrd\circ\scrd$ obtained by composing $\scrd$
with itself. Then, for all $V\in\scrv$, we have $\scrdd(V)=V^{**}$.
For all $V\in\scrv$, let $\tau_V:V\to V^{**}$
be defined by $(\tau_Vv)(l)=l(v)$.
(Note: The map $\tau_Vv:V^*\to\R$ is sometimes called
the ``evaluational at $v$''.)
Then $\tau:\scri\to\scrdd$ is a natural tranformation,
as Exercise 2B will verify.
By contrast, the next exercise shows that there
is no natural transformation from $\scri$ to $\scrd$,
even though, for all $V\in\scrv$, the vector spaces
$V$ and $V^*$ are isomorphic, since they have the same dimension.
\vskip.1in\noindent
{\bf EXERCISE 2A:} Let $V:=\R^2$ and let $f:V\to V^*$
be an isomorphism. For any linear map $g:V\to V$,
let $g^*:V^*\to V^*$ be defined by $g^*(l)=l\circ g$.
Show that there exists an isomorphism $g:V\to V$ such that
$f\ne g^*\circ f\circ g$.
\vskip.1in\noindent
{\bf EXERCISE 2B:} As in the example above,
for any real vector space $V$,
let $\tau_V:V\to V^{**}$ be defined by $(\tau_Vv)(l)=l(v)$.
Let $W$ and $X$ be real vector spaces
and let $g:W\to X$ be a linear transformation.
Define $g^*:X^*\to W^*$ by $g^*(l)=l\circ g$
and define $g^{**}:W^{**}\to X^{**}$
by $g^{**}(l)=l\circ g^*$.
Show that $g^{**}\circ\tau_W=\tau_X\circ g$.
\vskip.1in\noindent
{\it Definition.}
Let $\scrc$ be a category.
Then the {\bf arrow category of $\scrc$}, $\Arr(\scrc)$,
is a category whose objects are the arrows in $\scrc$,
and for which, for all pairs of arrows
$f:C_1\to C_2$, $f':C'_1\to C'_2$ in $\scrc$,
we define $\Hom(f,f')$ to be the collection of all those
$$(g_1,g_2)\in\Hom(C_1,C'_1)\times\Hom(C_2,\times C'_2)$$
such that $g_2\circ f=f'\circ g_1$.
\vskip.1in
We leave it as an exercise to guess what the definitions
of compositions and identity arrows in $\Arr(\scrc)$.
\vskip.1in\noindent
{\it Definition.}
Given a category $\scrc$,
we define two functors $\Dom_\scrc,\Tar_\scrc:\Arr(\scrc)\to\scrc$
by: for all $f:D\to T$ in $\scrc$, we set
$\Dom_\scrc(f)=D$ and $\Tar_\scrc(f)=T$.
\vskip.1in
With this terminology in place, if $\scra,\scrb:\scrc\to\scrd$
are two functors, then a natural transformation $\mu:\scra\to\scrb$
is equivalent to a functor $\scrm:\scrc\to\Arr(\scrd)$
satisfying both $\Dom_\scrd\circ\scrm=\scra$ and
$\Tar_\scrd\circ\scrm=\scrb$.
\vskip.1in
Now recall the functors $\scrf:\scrtns\to\scrts$
and $\scrg:\scrts\to\scrtns$ defined above.
Recall that $\scrf\scrg$ is the identity functor on $\scrts$,
but that $\scrg\scrf$ is not the identity functor on $\scrtns$.
We will make clear next a sense in which $\scrf\scrg$
is ``close'' to the identity on $\scrtns$.
\vskip.1in\noindent
{\it Definition.}
Let $\scrc$ and $\scrd$ be categories
and let $\scra:\scrc\to\scrd$ be a functor.
Define a natural transformation $\iota^\scra:\scra\to\scra$
by $\iota^\scra_C:=\id_{\scra C}:\scra C\to\scra C$.
This natural transformation is called
{\bf the identity on $\scra$}.
\vskip.1in\noindent
{\it Definition.}
Let $\scrc$ and $\scrd$ be categories,
let $\scrp,\scrq,\scrr:\scrc\to\scrd$ be functors
and let $\alpha:\scrp\to\scrq$ and $\beta:\scrq\to\scrr$
be natural transformations.
We define a natural transformation $\beta\alpha:\scrp\to\scrr$ by
$(\beta\alpha)_C=\beta_C\circ\alpha_C$.
\vskip.1in\noindent
{\it Definition.}
We say that two functors $\scra,\scrb:\scrc\to\scrd$
are {\bf equivalent} (and write $\scra\sim\scrb$)
if there exist natural transformations
$\tau:\scra\to\scrb$ and $\mu:\scrb\to\scra$
such that $\mu\tau=\iota^\scra$ and $\tau\mu=\iota^\scrb$.
\vskip.1in
Intuitively, if two functors are equivalent, then they have
the same ``information'' in them. If I'm an expert on one of
the functors and you're an expert on the other, then we'll
probably spend our days quibbling over notation, but will
be well aware that we both know the same stuff.
\vskip.1in\noindent
{\bf EXERCISE 2C:} Let $\scra,\scrb:\scrc\to\scrd$ be functors.
Let $\tau:\scra\to\scrb$ be a natural transformation.
Suppose, for all $C\in\scrc$,
that $\tau_C:\scra C\to\scrb C$ is an isomorphism.
Show that there is a unique natural transformation $\mu:\scrb\to\scra$
such that $\mu\tau=\iota^\scra$ and $\tau\mu=\iota^\scrb$.
\vskip.1in\noindent
{\bf EXERCISE 2D:} Let $\scra,\scrb:\scrc\to\scrd$ be functors.
Let $\tau:\scra\to\scrb$ and $\mu:\scrb\to\scra$
be natural transformations
such that $\mu\tau=\iota^\scra$ and $\tau\mu=\iota^\scrb$.
Show, for all $C\in\scrc$, that $\tau^C:\scra C\to\scrb C$
is an isomorphism.
\vskip.1in\noindent
{\bf EXERCISE 2E:} Recall the functors
$\scrf:\scrtns\to\scrts$ and $\scrg:\scrts\to\scrtns$
defined above.
Recall that $\scrf\scrg=\Id_\scrts$
but that $\scrg\scrf\ne\Id_\scrtns$.
Prove that $\scrg\scrf\sim\Id_\scrtns$.
\vskip.1in\noindent
{\it Definition.}
A category $\scrc$ is said to be {\bf equivalent}
to a category $\scrd$ (write $\scrc\sim\scrd$)
if there are functors $\scrp:\scrc\to\scrd$
and $\scrq:\scrd\to\scrc$
such that $\scrp\scrq\sim\Id_\scrd$
and $\scrq\scrp\sim\Id_\scrc$.
\vskip.1in
Intuitively, if two categories are equivalent, then they have
the same ``information'' in them. If I'm an expert on one of
the categories and you're an expert on the other, then we'll
probably spend our days quibbling over notation, but will
be well aware that we both know the same stuff.
The preceding exercise asks you to demonstrate that
$\scrtns$ is equivalent to $\scrts$.
Which is better, $\scrtns$ or $\scrts$?
In some sense, neither, since they're equivalent.
On the other hand,
with a little thought, you will readily see that
$\scrf:\scrtns\to\scrts$ is not ``one-to-one''.
To almost any topological space Y, there are several
topological $N$-spaces whose images under $\scrf$
are all equal to $Y$.
For example, $(\R,x\mapsto\{(x-\delta,x+\delta)\})$
and $(\R,x\mapsto\{(x-2\delta,x+\delta)\})$ have the
same image under $\scrf$.
So, in some sense, $\scrtns$ is redundant, with
more copies of each space than one really needs.
From this perspective, $\scrts$ is better than $\scrtns$.
Finally, a few broad words about mathematics.
Typically a working mathematician develops expertise in
a category, {\it e.g.}, the category $\scrts$ of topological spaces.
A fundamental question in topology is to determine, given two topological
spaces, whether they are homeomorphic, but, more generally,
whatever your ``category of choice'' may be, you'll want
to be able to form many examples of objects in it, and, in
many cases to show that two such examples are not isomorphic
in the category.
If one can form a functor $\scrf$ that goes from your category of
choice $\scrc$ to some other category $\scrd$,
and if tools are available for distinguishing objects in $\scrd$,
then, given $C,C'\in\scrc$,
if you are able to show that $\scrf C$ and $\scrf C'$
are not isomorphic in $\scrd$, then you can immediately
conclude that $C$ and $C'$ are not isomorphic in $\scrc$.
(See the next exercise.)
For example, there is a functor called ``first homology'',
and denoted $H_1$ which runs from topological spaces to
Abelian groups.
Given two topological spaces, if one is able to show that
their first homology groups are not isomorphic,
then the spaces are not homeomorphic.
This kind of argument will be central to much of this course!
\vskip.1in\noindent
{\bf EXERCISE 2F:} Show that functors carry isomorphisms
to isomorphisms. That is, show that, if $\scrf:\scrc\to\scrd$
is a functor, if $f:C\to C'$ is an isomorphism in $\scrc$,
then $\scrf f:\scrf C\to\scrf C'$ is an isomorphism in $\scrd$.
In the next definition, we make the convention
that $\cup\emptyset=\emptyset$.
\vskip.1in\noindent
{\it Definition.}
Let $X$ be a set and let
$\scrs\subseteq
\{\hbox{subsets of }X\}$.
The {\bf union closure of $\scrs$} is
$\{\cup\scra\,|\,\scra\subseteq\scrs\}$.
The {\bf finite intersection closure in $X$ of $\scrs$} is
$\{X\}\cup\{\cap\scra\,|\,\scra\subseteq\scrs, 0<|\scra|<\infty\}$.
\vskip.1in\noindent
{\it Definition.}
Let $X$ be a set and let $\scro$ be a topology on $X$.
Let $\scrb\subseteq
\{\hbox{subsets of }X\}$.
We say that $\scrb$ is a {\bf basis} for $\scro$
if the union closure of $\scrb$ is $\scro$.
We say that $\scrb$ is a {\bf subbasis} for $\scro$
if the finite intersection closure in $X$ of $\scrb$
is a basis for $\scro$.
\vskip.1in
Note that any basis for a topology is a subbasis.
\vskip.1in\noindent
{\bf EXERCISE 2G:}
Let $X$ be a set and
let $\scrs\subseteq
\{\hbox{subsets of }X\}$.
Show that $\scrs$ is a subbasis of a topology on $X$.
(That is, show that the union closure of the
finite intersection closure in $X$ of $\scrs$ is a topology
on $X$.)
\vskip.1in\noindent
{\bf EXERCISE 2H:}
Let $X$ be a set and let
$\scrb\subseteq
\{\hbox{subsets of }X\}$.
Show that $\scrb$ is a basis of a topology on $X$
if and only if, both of the following conditions hold:
\itemitem{(1)} $\cup\scrb=X$; and
\itemitem{(2)} for all $B,B'\in\scrb$,
for all $x\in B\cap B'$,
there exists $B''\in\scrb$
such that $x\in B''\subseteq B\cap B'$.
\vskip.1in\noindent
{\it Definition.}
Let $\scro$ and $\scro'$ be topologies on a set $X$.
We say that $\scro$ is {\bf finer} (or {\bf stronger})
than $\scro'$ if $\scro\supseteq\scro'$.
We say that $\scro$ is {\bf coarser} (or {\bf weaker})
than $\scro'$ if $\scro\subseteq\scro'$.
\vskip.1in\noindent
{\it Remark.}
Let $X$ and $I$ be sets
and, for all $i\in I$, let $\scro_i$ be a topology
on $X$.
Then $\displaystyle{{\mathop\bigcap_{i\in I}}\,\scro_i}$
is a topology on $X$.
\vskip.1in
The following is a restatement of the preceding remark.
\vskip.1in\noindent
{\it Remark.}
Let $\Xi\subseteq
\{\hbox{topologies on }X\}$.
Then $\cap\Xi$ is a topology on $X$.
The preceding two remarks are asserting that,
given a collection of topologies on a set,
there exists a unique topology which is
finest among:
\centerline{those topologies coarser than every topology
in the collection.}
\vskip.1in\noindent
{\it Question:} How about the other way around?
That is, given a collection of topologies on a set,
does there there exist a unique topology which is
coarsest among:
\centerline{those topologies finer than every topology
in the collection.}
\vskip.1in\noindent
{\it Answer:}
We decided in class that the answer is yes.
This can probably be argued using Zorn's lemma, but
more directly, one can simply take the union of the
topologies in the collection and then
take the intersection of all the topologies
containing that union.
By the preceding two remarks,
that intersection is again a topology.
Yet another way to say the same thing is that
you take the union of the topologies in the
collection, then take the finite intersection
closure of that union,
then take the union closure of {\it that}.
\vskip.1in
{\it Definition.}
Let $X$ be a set and let
$$\scrs\subseteq
\{\hbox{subsets of }X\}.$$
Then the {\bf topology generated by $\scrs$}
is the coarsest topology
among those topologies containing $\scrs$.
Note that,
if $\scrb$ is a basis or subbasis of a topology $\scro$ on a set $X$,
then $\scro$ is the topology generated by $\scrb$.
\vskip.1in\noindent
{\it Definition.}
Let $X$ and $Y$ be sets, let $f:X\to Y$ be a function and let
$$\scrs\subseteq
\{\hbox{subsets of }Y\}.$$
Then we define $f^*\scrs:=\{f^{-1}(S)\,|\,S\in\scrs\}$.
\vskip.1in\noindent
{\bf EXERCISE 2I:}
Let $X$ and $Y$ be sets, let $f:X\to Y$ be a function
and let $\scro$ be a topology on $Y$.
Show that $f^*\scro$ is a topology on $X$.
\vskip.1in\noindent
{\it Remark.}
Let $X$ and $Y$ be sets, let $f:X\to Y$ be a function
and let $\scro$ be a topology on $Y$.
Then $f^*\scro$ is the coarsest topology making $f$ continuous.
\vskip.1in\noindent
{\it Definition.}
Let $Y$ be a topological space and let $X\subseteq Y$.
Then the {\bf inherited topology on $X$}
(or {\bf relative topology on $X$}
or {\bf subspace topology on $X$}) is:
$$\{X\cap U\,|\,U\hbox{ is open in }Y\}.$$
\vskip.1in\noindent
{\it Remark.}
Let $Y$ be a topological space and let $X\subseteq Y$.
Let $i:X\to Y$ be the inclusion, defined by $i(x)=x$.
Let $\scro:=\{\hbox{open subsets of }Y\}$.
Then $i^*\scro$ is the inherited topology on $X$.
It is the coarsest topology making $i$ continuous.
\vskip.1in\noindent
{\it Definition.}
Let $I$ be a set and, for all $i\in I$,
let $X_i$ be a topological space.
Let $X:=\displaystyle{{\mathop\prod_{i\in I}}\,X_i}$.
For all $i\in I$, let
$p_i:X\to X_i$ be the $i$th coordinate projection map.
For all $i\in I$,
let $\scro_i$ be the set of all open subsets of $X_i$.
Then the {\bf product topology} on $X$
is the topology generated by
$\displaystyle{{\mathop\bigcup_{i\in I}}\,p_i^*\scro_i}$.
\vskip.1in
Note that the product topology is the coarsest among those
that make all the projection maps $p_i$ continuous.
In general, whenever a product of topological spaces is formed,
we will give it the product topology, unless otherwise specified.
In particular, when $X$ and $Y$ are topological spaces,
$X\times Y$ is a topological space as well.
\vskip.1in\noindent
{\it Definition.} Let $X$ be a topological space, let $Y$ be a set
and let $f:X\to Y$ be a function. The {\bf quotient topology}
on $Y$ is the finest making $f$ continuous.
(See Exercise 3A below.)
\vskip.1in
Now let's start making some topological spaces!
\vskip.1in\noindent
{\it Definition.}
Let $n\ge0$ be an integer.
The {\bf standard topology} on $\R^n$
is the topology generated by $\{B_r(x)\,|\,x\in\R^n,r>0\}$.
\vskip.1in\noindent
{\it Definition.}
Let $n\ge0$ be an integer.
We define $S^n:=\{x\in\R^{n+1}\,|\,d(x,0)=1\}$.
The {\bf standard topology} on $S^n$
is the topology inherited from $\R^{n+1}$.
\vskip.1in\noindent
{\it Definition.}
If $X$ is a topological space, and if $x\in X$,
then an {\bf open neighborhood of $x$ in $X$}
is an open subset $U$ of $X$ such that $x\in U$.
If $X$ is a topological space, and if $S\subseteq X$,
then an {\bf open neighborhood of $S$ in $X$}
is an open subset $U$ of $X$ such that $S\subseteq U$.
\vskip.1in\noindent
{\it Definition.}
To say that a topological space $X$ is {\bf Hausdorff}
means: for all $x,x'\in X$, if $x\ne x'$,
then there are open neighborhoods $U$ of $x$ and $U'$ of $x'$ in $X$
such that $U\cap U'=\emptyset$.
\vskip.1in\noindent
{\it Definition.}
Let $S$ be a subset of a topological space $X$.
We say that $S$ is {\bf closed in $X$}
if $X\backslash S$ is open in $X$.
\vskip.1in
Note that $\{\hbox{closed subsets of }X\}$
is closed under finite union and arbitrary intersection.
\vskip.1in\noindent
{\it Definition.}
Let $X$ be a topological space
and assume, for all $x\in X$, that $\{x\}$ is closed in $X$.
We say that $X$ is {\bf regular} if the following condition holds:
For all $x_0\in X$, for any closed subset $C_0$ of $X$,
if $x_0\ne C_0$, then there exist open nieghborhoods
$U$ and $V$ of $x_0$ and $C_0$ in $X$
such that $U\cap V=\emptyset$.
We say that $X$ is {\bf normal} if the following condition holds:
For any closed subsets $C,C'$ of $X$,
if $C\cap C'=\emptyset$, then there exist open nieghborhoods
$U$ and $U'$ of $C$ and $C'$ in $X$
such that $U\cap U'=\emptyset$.
\vskip.1in\noindent
{\it Definition.}
Let $X$ be a topological space and let $S\subseteq X$.
The {\bf closure} of $S$ in $X$
is the intersection of all closed subsets of $X$ that contain $S$,
and is denoted $\barS$ or $\Cl_X(S)$.
The {\bf interior} of $S$ in $X$
is the union of all open subsets of $X$ contained in $S$,
and is denoted $S^\circ$ or $\Int_X(S)$.
The {\bf boundary} of $S$ in $X$
is $\barS\backslash S^\circ$,
and is denoted $\partial S$ or $\Bd_X(S)$.
\vskip.1in
Since an intersection of closed sets is closed and
a union of open sets is open, it follows that the
closure of $S$ is the smallest closed set containing $S$,
while the interior of $S$ is the largest open set
contained in $S$.
Let $S$ be a subset of a topological space $X$.
We say that $S$ is {\bf dense} in $X$ if $\Cl_X(S)=X$.
\vskip.1in\noindent
{\it Definition.}
We say that a topological space is {\bf discrete}
if all of its subsets are open.
We say that a topological space $X$ is {\bf indiscrete}
if its only open sets are $\emptyset$ and $X$.
\vskip.1in\noindent
{\it Definition.}
Let $X$ be a Hausdorff topological space.
We say $X$ is {\bf compact}
if, for any $\scrs\subseteq\{\hbox{open subsets of }X\}$,
we have:
\itemitem{($*$)} if $\cup\scrs=X$, then there is a finite
subset $\scrf$ of $\scrs$ such that $\cup\scrf=X$.
\vskip.1in
That is, a Hausdorff topological space is said to be
compact if every open cover has a finite subcover,
in the following terminology:
\vskip.1in\noindent
{\it Definition.}
Let $X$ be a topological space.
An {\bf open cover} of $X$
is a subset $\scru$ of $\{\hbox{open subsets of }X\}$
such that $\cup\scru=X$.
Given two covers $\scru,\scrv$ of $X$,
we say that $\scru$ is a {\bf subcover} of $\scrv$
if $\scru\subseteq\scrv$.
\vskip.1in\noindent
{\it Question:}
Is $S^2$ homeomorphic to $\R^2$?
\vskip.1in\noindent
{\it Fact.}
$S^2$ is compact, whereas $\R^2$ is not.
\vskip.1in\noindent
{\it Answer to preceding question:} No.
\vskip.1in\noindent
{\it Definition.}
Let $X$ be a Hausdorff topological space.
Then $X$ is said to be {\bf locally compact}
if, for all $x\in X$,
there exists an open neighborhood $U$ of $x$ in $X$
such that $\barU$ is compact.
\vskip.1in\noindent
{\it Example.}
$\R$ is locally compact, and
any finite product of locally compact topological spaces
is again locally compact.
Consequently, for all integers $n\ge1$,
$\R^n$ is locally compact.
However infinite products of locally compact
topological spaces are generally {\it not}
locally compact, {\it e.g.},
$\displaystyle{{\mathop\prod_{i=1}^\infty}\,\R}$
is not locally compact.
\vskip.1in\noindent
{\it Next Question:}
Is $\R$ homeomorphic to $\R^2$?
\vskip.1in\noindent
{\bf EXERCISE 2J:}
For all $p,q\in\R^2$,
show that $\R^2\backslash\{p\}$
is homeomorphic to $\R^2\backslash\{q\}$.
\vskip.1in\noindent
{\bf EXERCISE 2K:}
Let $X$ and $Y$ be topological spaces and let $x\in X$.
Assume that $X$ is homeomorphic to $Y$.
Show that there exists $y\in Y$
such that $X\backslash\{x\}$ is homeomorphic to $Y\backslash\{y\}$.
\vskip.1in\noindent
{\it Definition.}
Let $S$ be a subset of a topological space $X$.
We say that $S$ is {\bf clopen in $X$}
if $S$ is both closed an open in $X$.
\vskip.1in\noindent
{\it Definition.}
We say that a topological space $X$ is {\bf connected}
if it has no clopen sets other than $\emptyset$ and $X$.
A topological space is {\bf disconnected} if it is not connected.
\vskip.1in\noindent
{\it Fact.} $\R\backslash\{0\}$ is disconnected,
while $\R^2\backslash\{(0,0)\}$ is connected.
\vskip.1in\noindent
{\it Remark.} $\R$ is not homeomorphic to $\R^2$.
\vskip.1in\noindent
{\it Proof:}
Assume, for a contradiction that $\R$ and $\R^2$
are homeomorphic.
By Exercise 2K, choose $y\in\R^2\backslash\{(0,0)\}$
such that $\R\backslash\{0\}$ is homeomorphic
to $\R^2\backslash\{y\}$.
By Exercise 2J,
$\R^2\backslash\{y\}$ is homeomorphic to
$\R^2\backslash\{(0,0)\}$.
Then $\R\backslash\{0\}$ is homeomorphic
to $\R^2\backslash\{(0,0)\}$.
However, by the fact above, $\R\backslash\{(0\}$ is disconnected,
while $\R^2\backslash\{(0,0)\}$ is connected. {\bf QED}
\vskip.1in
In the above proof, we used, without comment, the fact
that if two topological spaces are homeomorphic and
if one is connected, than the other is as well.
That is, ``connectedness is a homeomorphism invariant''.
In fact, for a property of topological spaces to be
useful, it is typically a homeomorphism invariant, and
we will leave it as an implicit exercise to verify that
the properties we may introduce below are, in fact,
homeomorphism invariants.
Typically, such arguments are straightforward.
\vskip.1in\noindent
{\it Next Question:}
Is $\R^2$ homeomorphic to $\R^3$?
\vskip.1in
Note that both $\R^2\backslash\{(0,0)\}$
and $\R^3\backslash\{(0,0,0)\}$ are connected.
Nevertheless, some variant of the argument given above
will work to show that the answer to the above question is ``no''.
\vskip.1in\noindent
{\it Definition.}
A {\bf pointed topological space} consists of
\itemitem{(1)} a topological space $X$; and
\itemitem{(2)} a point $x\in X$.
\vskip.1in\noindent
We call $x$ the {\bf basepoint} of the pointed topological
space $(X,x)$.
\vskip.1in\noindent
{\it Definition.}
The category of pointed topological spaces
(with basepoint preserving continuous maps)
is denoted $\scrpts$.
The forgetful functor $(X,x)\mapsto X\scrpts\to\scrts$
is denoted by $\scrfb$.
An isomorphism in $\scrpts$ is called
a {\bf basepoint-preserving homeomorphism}.
\vskip.1in
The next Fact is nontrivial, but we will
assume it for now, and prove it in the next few lectures.
\vskip.1in\noindent
{\it Fact.}
There is a functor $\pi_1:\scrpts\to\{\hbox{groups}\}$
such that:
\itemitem{(1)} for all $x\in X:=\R^2\backslash\{(0,0)\}$,
we have that $\pi_1(X,x)$ is isomorphic to the additive
group of integers; and
\itemitem{(2)} for all $y\in Y:=\R^3\backslash\{(0,0,0)\}$,
we have that $\pi_1(Y,y)$ is isomorphic to the trivial group.
\vskip.1in\noindent
{\it Remark.}
For all topological spaces $X,Y$,
for all $x\in X$,
if $X$ is homeomorphic to $Y$
then there exists $y\in Y$ such that
$(X,x)$ is basepoint-preserving homeomorphic to $(Y,y)$.
\vskip.1in\noindent
{\it Remark.}
Let $X:=\R^2\backslash\{(0,0)\}$
and $Y:=\R^3\backslash\{(0,0,0)\}$.
Then $X$ and $Y$ are not homeomorphic.
\vskip.1in\noindent
{\it Proof:}
Say for a contradiction that $X$ is homeomorphic to $Y$.
Choose $x\in X$.
By the second to last Remark,
choose $y\in Y$
such that $(X,x)$ is basepoint-preserving homeomorphic to $(Y,y)$.
Let $\pi_1:\scrpts\to\{\hbox{groups}\}$ be the functor described
in the preceding Fact.
Then, by Exercise 2F, $\pi_1(X,x)$ is isomorpic to $\pi_1(Y,y)$, in
the category of groups.
We say that $\pi_1(X,x)$ is the {\bf fundamental group of $X$}
(with respect to $x$).
Then, by the preceding fact, the additive group of integers
is isomorphic to the trivial group, contradiction. {\bf QED}
\vskip.1in\noindent
{\it Corollary.} $\R^2$ is not homeomorphic to $\R^3$.
\vskip.1in
Given the preceding Remark, the preceding Corollary is
argued exactly as in the proof that $\R$ and $\R^2$
are not homeomorphic.
Our goal now becomes to define and study the functor
$\pi_1:\scrpts\to\{\hbox{groups}\}$ of the preceding Fact.
\vskip.1in\noindent
{\it Definition.}
A space $X$ is {\bf locally compact} if every point of
$X$ has an open neighborhood whose closure in $X$ is compact.
\vskip.1in\noindent
{\it Definition.}
Let $X$ and $Y$ be topological spaces and assume that
$X$ locally compact.
Recall that
$\Hom(X,Y)$ is the set of continuous maps from $X$ to $Y$.
Let $\scrk$ denote the set of all compact subsets of $X$.
Let $\scro$ denote the set of all open subsets of $Y$.
For all $K\in\scrk$, for all $U\in\scro$, let
let $\scrw(K,U):=\{f\in\Hom(X,Y)\,|\,f(K)\subseteq U\}$.
The topology on $\Hom(X,Y)$ generated by the subbasis
$\{\scrw(K,U)\,|\,K\in\scrk,U\in\scro\}$
is called the {\bf compact-open} topology on $\Hom(X,Y)$.
We let $C(X,Y)$ denote the topological space
whose underlying set is $\Hom(X,Y)$ and whose
topology is the compact-open topology.
\vskip.1in
One has the general feeling that a space is
locally compact if it is ``almost'' finite dimensional.
In particular, if $I$ is a set and, for all $i\in I$,
we have a Hausdorff topological space $X_i$,
then $\displaystyle{{\mathop\prod_{i\in I}}\,X_i}$
is locally compact iff
$\{i\in I\,|\,X_i\hbox{ is noncompact}\}$ is finite.
The topological space $C(\R,\R)$ does not feel
anywhere near finite dimensional, so one would
guess that it is not locally compact, and that is,
in fact true:
\vskip.1in\noindent
{\it Remark.}
$C(\R,\R)$ is not locally compact.
\vskip.1in\noindent
{\it Sketch of proof:}
Define $f:\R\to\R$ by $f(x)=0$. Then $f\in C(\R,\R)$.
Let $U$ be an open neighborhood of $f$ in $C(\R,\R)$.
Assume, for a contradiction, that $\barU$ is compact.
Choose $\epsilon>0$ and a compact subset $K\subseteq\R$
such that
$$U_0:=\{g\in C(\R,\R)\,|\,|g|<\epsilon\hbox{ on }K\}\subseteq U.$$
(It is an unassigned exercise to prove existence of such
an $\epsilon$ and $K$.)
For all integers $j\ge1$,
let $g_j:\R\to\R$ be defined by $g_j(x)=[\epsilon/2][\sin(jx)]$;
then $g_j\in U_0$, so $g_j\in U$.
We leave it as an unassigned exercise to show
that $g_j$ has no convergent subsequence in $C(\R,\R)$.
After all, to what could these functions possibly converge?
(Recall that, in any compact Hausdorff topological
space, any sequence has a convergent subsequence.)
{\bf QED}
\vskip.1in\noindent
{\it Fact.} Let $X$ and $Y$ be locally compact topological spaces.
For all $f\in C(X,C(Y,Z))$, define $\alpha_f\in C(X\times Y,Z)$
by $\alpha_f(x,y)=(f(x))(y)$.
Then $f\mapsto\alpha_f:C(X,C(Y,Z))\to C(X\times Y,Z)$
is a homeomorphism.
\vskip.1in
If $X$ is a locally compact topological space
and if $(Y,d)$ is a metric space,
then the {\bf $d$-uniform on compacta topology}
on $\Hom(X,Y)$ is the topology generated by the basis
$\{f\in\Hom(X,Y)\,|\,\forall k\in K, d(f(k),f_0(k))<\epsilon\}$,
where $f_0$ ranges over $\Hom(X,Y)$,
where $K$ ranges over compact subsets of $X$ and
where $\epsilon$ ranges over the positive real numbers.
It is a fact that, for the topology on $Y$ generated
by the basis
$$\{\hbox{open balls with respect to the metric }d\},$$
the compact-open topology on $\Hom(X,Y)$ agrees
with the $d$-uniform on compacta topology on $\Hom(X,Y)$.
\vskip.1in\noindent
{\it Definition.}
Let $X$ be a topological space and let $I:=[0,1]$.
Let $x,x'\in X$.
A {\bf path in $X$ from $x$ to $x'$}
is an element $\gamma\in C(I,X)$
such that both $\gamma(0)=x$ and $\gamma(1)=x'$.
A {\bf loop in $X$ at $x$}
is a path from $x$ to $x$.
\vskip.1in\noindent
{\it Definition.}
Let $X$ be a topological space and let $I:=[0,1]$.
Let $x_0,x_1\in X$
and let
$$P_{x_0}^{x_1}:=\{\gamma\in C(I,X)\,|\,\gamma(0)=x_0,\gamma(1)=x_1\}$$
be the topological space of paths in $X$ from $x_0$ to $x_1$.
(Give $P_{x_0}^{x_1}$ the inherited topology from $C(I,X)$.)
Let $\gamma_0,\gamma_1\in P_{x_0}^{x_1}$.
An {\bf endpoint fixed homotopy from $\gamma_0$ to $\gamma_1$}
is a path in~$P_{x_0}^{x_1}$ from $\gamma_0$ to $\gamma_1$.
We say that $\gamma_0$ and $\gamma_1$
are {\bf endpoint fixed homotopic} if there is an
endpoint fixed homotopy from $\gamma_0$ to $\gamma_1$.
\vskip.1in
Following the preceding Fact,
we may equivalently define an endpoint fixed homotopy from
$\gamma_0$ to $\gamma_1$ to be a continuous $H:I\times I\to X$
such that
\itemitem{(1)} for all $s\in I$, we have $H(s,0)=x_0$;
\itemitem{(2)} for all $s\in I$, we have $H(s,1)=x_1$;
\itemitem{(3)} for all $t\in I$, we have $H(0,t)=\gamma_0(t)$; and
\itemitem{(4)} for all $t\in I$, we have $H(1,t)=\gamma_1(t)$.
We have some point set topology exercises:
\vskip.1in\noindent
{\bf EXERCISE 3A:}
Let $X$ be a topological space, let $Y$ be a set
and let $f:X\to Y$ be a function.
Let $\scro:=\{U\subseteq Y\,|\,f^{-1}(U)\hbox{ is open in }X\}$.
Show that $\scro$ is the finest topology on $Y$ making $f$ continuous.
\vskip.1in\noindent
{\bf EXERCISE 3B:}
Let $X$ be a topological space,
let $S\subseteq X$ and let $x_0\in X$.
Show that $x_0\in\partial S$ iff the following
condition holds:
For all open neighborhoods $U$ of $x_0$ in $X$,
we have $U\cap S\ne\emptyset\ne U\backslash S$.
\vskip.1in\noindent
{\bf EXERCISE 3C:}
Let $X$ be a topological space and let $S\subseteq X$.
Show that $X\backslash(S^\circ)=\overline{X\backslash S}$.
\vskip.1in\noindent
{\bf EXERCISE 3D:}
Let $X$ be a topological space
and let $U$ be an open subset of $X$.
Show that $(\partial U)^\circ=\emptyset$.
That is, show that $\Int_X(\Bd_X(U))=\emptyset$.
\vskip.1in\noindent
{\it Definition.}
Let $(X,x)$ be a pointed topological space.
Define an equivalence relation $\sim$ on $P_x^x(X)$
by: For all $\gamma,\gamma'\in P_x^x(X)$,
$\gamma\sim\gamma'$ means that $\gamma$ and $\gamma'$
are endpoint fixed homotopic,
{\it i.e.}, that $P_\gamma^{\gamma'}(P_x^x(X))\ne\emptyset$.
We define $\pi_1(X,x):=(P_x^x(X))/\sim$.
\vskip.1in
Note that this defines a functor
$\pi_1:\{\hbox{pointed topological spaces}\}\to\{\hbox{sets}\}$.
Eventually we'll redefine $\pi_1$ as a functor
$\pi_1:\{\hbox{pointed topological spaces}\}\to\{\hbox{groups}\}$,
and the current $\pi_1$ will be the compostion with that
redefined $\pi_1$, followed by the forgetful functor
$\{\hbox{sets}\}\to\{\hbox{groups}\}$.
For the moment, however, this definition will suffice for our
purposes.
Recall that, in any topological space $Z$,
if we define an equivalence relation $\cong$ on $Z$
by $z\cong z'$ iff $P_z^{z'}(Z)\ne\emptyset$,
then the equivalence classes of $\cong$ are called
{\bf path components of $Z$}.
With that terminology, $\pi_1(X,x)$ is simply
the set of path components in $P_x^x(X)$.
\vskip.1in\noindent
{\bf EXERCISE 3E:}
Given an arrow $f:(X,x)\to(Y,y)$ in the category
$$\{\hbox{pointed topological spaces}\},$$
define an arrow
$\pi_1(f):\pi_1(X,x)\to\pi_1(Y,y)$
in $\{\hbox{sets}\}$.
\vskip.1in
As is typical with covariant functors,
we use $f_*$ as an abbreviation
for $\pi_1(f)$.
(For {\it contra}variant functors,
$f^*$ is typical as an abbreviation
for the result of applying the functor to
an arrow $f$.)
Recall that we aim to prove that $\R^2$ and $\R^3$ are
not homeomorphic.
We already observed that it suffices to show that
$X:=\R^2\backslash\{(0,0)\}$ and $Y:=\R^3\backslash\{(0,0,0)\}$
are not homeomorphic.
It now suffices to show, for all
$x\in X$ and all $y\in Y$,
that $|\pi_1(X,x)|\ge2$, while $|\pi_1(Y,y)|=1$.
\vskip.1in\noindent
{\it Definition.}
We say that a topological space $X$ is path
connected if $X$ has only one path component,
{\it i.e.}, if the following condition
holds: for all $x,x'\in X$, we have $P_x^{x'}(X)\ne\emptyset$.
We will argue, in Exercise 3G, that if $X$ is path
connected, then, for any $x,x'\in X$,
we have that $\pi_1(X,x)$ and $\pi_1(X,x')$ are
bijective, {\it i.e.}, isomorphic in the category $\{\hbox{sets}\}$.
Eventually this will be refined to saying that
$\pi_1(X,x)$ and $\pi_1(X,x')$ are
isomorphic in the category $\{\hbox{groups}\}$.
\vskip.1in\noindent
{\it Definition.}
Let $x,x',x''\in X\in\scrts$.
For all $\gamma\in P_x^{x'}(X)$,
for all $\gamma'\in P_{x'}^{x''}(X)$,
we define
$\gamma\|\gamma'\in P_x^{x''}(X)$ by
$$(\gamma\|\gamma')(t)=\cases{
\gamma(2t), &if $t\in[0,1/2]$;\cr
\gamma'(2t-1), &if $t\in[1/2,1]$.\cr}$$
\vskip.1in
The path $\gamma\|\gamma'$ is called
the {\bf concatenation} of $\gamma$ with $\gamma'$.
\vskip.1in\noindent
{\it Definition.}
Let $x,x'\in X\in\scrts$.
For all $\gamma\in P_x^{x'}(X)$,
we let $[\gamma]$ denote
the endpoint fixed homotopy class of $\gamma$,
{\it i.e.}, we define
$$[\gamma]:=\{\delta\in P_x^{x'}(X)\,|\,
P_\gamma^\delta(P_x^{x'}(X))\ne\emptyset\}.$$
\vskip.1in\noindent
{\bf EXERCISE 3F:}
Let $a,b,c,d\in X\in\scrts$.
Let $\gamma\in P_a^b(X)$,
$\delta\in P_b^c(X)$ and
$\epsilon\in P_c^d(X)$.
Show that
$[\gamma\|(\delta\|\epsilon)]=[(\gamma\|\delta)\|\epsilon]$.
\vskip.1in
Exercise 3F asserts that concatenation is
``associative up to homotopy''.
Note that it is not associative,
{\it i.e.}, note that it frequently happens that
$\gamma\|(\delta\|\epsilon)\ne(\gamma\|\delta)\|\epsilon$.
\vskip.1in\noindent
{\it Definition.}
Let $a,b\in X\in\scrts$.
For all $\gamma\in P_a^b(X)$,
we define $\overleftarrow\gamma\in P_b^a(X)$ by
$(\overleftarrow\gamma)(t)=\gamma(1-t)$.
\vskip.1in\noindent
{\bf EXERCISE 3G:}
Let $a,b\in X\in\scrts$ and let $\gamma\in P_a^b(X)$.
Show that
$$[\delta]\mapsto[\overleftarrow\gamma\|\delta\|\gamma]:
\pi_1(X,a)\to\pi_1(X,b)$$
is a {\it well-defined} bijection.
\vskip.1in
Part of the the preceding exercise is to show
that, if $[\delta]=[\delta']$, then
$[\overleftarrow\gamma\|\delta\|\gamma]=
[\overleftarrow\gamma\|\delta'\|\gamma]$.
This is what is mean by proving ``well-definedness''.
One summarizes Exercise 3G by saying
that ``$\pi_1$ is essentially independent of the basepoint''.
This will even be true after redefining $\pi_1$ in a group
theoretic way.
We can define a category of ``pathed topological spaces'',
each object of which consists of a topological space $X$,
together with a path $[0,1]\to X$. Let
$$\scra,\scrb:\{\hbox{pathed topological spaces}\}\to\{\hbox{sets}\}$$
be defined by
$$\scra(X,\gamma)=\pi_1(X,\gamma(0)),\qquad\qquad
\scrb(X,\gamma)=\pi_1(X,\gamma(1)).$$
Then the argument of Exercise 3G
shows that $\scra$ and $\scrb$ are equivalent functors,
and this is the key point of that exercise.
So, for $x_0,x_1\in X$,
the bijection $\pi_1(X,x_0)\to\pi_1(X,x_1)$
is ``natural up to choosing a path connecting
the basepoints $x_0$ and $x_1$''.
Note that, since $X:=\R^2\backslash\{(0,0)\}$
and $Y:=\R^3\backslash\{(0,0,0)\}$ are both path-connected,
it now suffices to show that there exist $x\in X$ and $y\in Y$
such that $|\pi_1(X,x)|\ge2$ and such that $|\pi_1(Y,y)|=1$.
(Before, we needed to show this for all $x\in X$, $y\in Y$,
so the problem is now formally easier.)
\vskip.1in\noindent
{\it Definition.}
Let $I:=[0,1]$.
Let $X,Y\in\scrts$ and let $f,g:X\to Y$ be continuous.
A {\bf homotopy} from $f$ to $g$ is a continuous
map $H:I\times X\to Y$
such that $H(0,\cdot)=f$ and $H(1,\cdot)=g$.
We say that $f$ is {\bf homotopic} to $g$,
if there exists a homotopy from $f$ to $g$.
\vskip.1in\noindent
{\bf EXERCISE 3H:}
Let $X,Y\in\scrts$, let $f,g:X\to Y$ be continuous and let $x\in X$.
Assume that $f$ is homotopic to $g$.
As usual,
let $f_*:=\pi_1(f):\pi_1(X,x)\to\pi_1(Y,f(x))$
and let $g_*:=\pi_1(g):\pi_1(X,x)\to\pi_1(Y,g(x))$.
Prove that there is a bijection
$$b:\pi_1(Y,f(x))\to\pi_1(Y,g(x))$$
such that $b\circ f_*=g_*$.
(Hint: If $H$ is a homotopy from $f$ to $g$,
then $H(\cdot,x)$ is a path from $f(x)$ to $g(x)$.
The needed bijection then comes from Exercise 3G.)
\vskip.1in
One summarizes Exercise 3H by the buzzphrase:
``Homotopic maps induce the same map on $\pi_1$.''
However, to be technically precise one needs to remember
that the two maps on $\pi_1$, namely $f_*$ and $g_*$,
yield different basepoints on $Y$, namely $f(x)$ and $g(x)$.
A precise interpretation of this buzzphrase must
take this into account.
\vskip.1in\noindent
{\it Definition.}
Let $X,Y\in\scrts$.
We say that $X$ is {\bf homotopy equivalent} to $Y$
(or that $X$ and $Y$ {\bf have the same homotopy type})
if there are continuous maps
$f:X\to Y$ and $g:Y\to X$
such that both of
the following conditions hold:
\itemitem{(1)} $g\circ f:X\to X$ is homotopic to $\id_X:X\to X$; and
\itemitem{(2)} $f\circ g:Y\to Y$ is homotopic to $\id_Y:Y\to Y$.
\vskip.1in\noindent
{\bf EXERCISE 3I:}
Let $X,Y\in\scrts$ and assume that both $X$ and $Y$ are path-connected.
Assume that $X$ is homotopy equivalent to $Y$.
Show, for any $x\in X$ and any $y\in Y$,
that $\pi_1(X,x)$ is bijective to $\pi_1(Y,y)$,
{\it i.e.}, that $\pi_1(X,x)$ is isomorphic to $\pi_1(Y,y)$
in the category $\{\hbox{sets}\}$.
\vskip.1in
Once again, the preceding exercise will strengthen to saying
that $\pi_1(X,x)$ and $\pi_1(Y,y)$ are isomorphic in the category
$\{\hbox{groups}\}$, once we've redefined the functor $\pi_1$.
\vskip.1in\noindent
{\bf EXERCISE 3J:}
Let $n\ge2$ be an integer.
Show that $S^{n-1}:=\{x\in\R^n\,|\,d(0,x)=1\}$
is homotopy equivalent to $Q:=\R^n\backslash\{0\}$.
(Hint: Use the maps
$p\mapsto p:S^{n-1}\to Q$ and $q\mapsto q/(d(0,q)):Q\to S^{n-1}$.
\vskip.1in
Given the last two exercises,
our goal is now to show that there exist
$a\in S^1$ and $b\in S^2$
such that $|\pi_1(S^1,a)|\ge2$
and such that $|\pi_1(S^2,b)|=1$.
\vskip.1in\noindent
{\it Definition.}
Let $X$ be a topological space and $x\in X$.
We say that $X$ is {\bf $x$-avoidable}
if, for all $a,b\in X\backslash\{x\}$,
for all $\gamma\in P_a^b(X)$,
there exists $\gamma'\in P_a^b(X\backslash\{x\})$
such that $\gamma$ is endpoint fixed homotopic to $\gamma'$.
\vskip.1in\noindent
{\bf EXERCISE 4A:}
Let $n\ge2$ be an integer.
Let $U$ be an open, convex, nonempty subset of $\R^n$.
Let $X$ be a topological space and assume
that $X$ is homeomorphic to $U$.
Show, for all $x\in X$, that $X$ is $x$-avoidable.
(Hint: Start by showing, for all $a,b\in U$,
for all $\gamma,\gamma'\in P_a^b(U)$,
that $\gamma$ is endpoint fixed homotopic to $\gamma'$.)
\vskip.1in\noindent
{\it Lemma.}
Let $X$ be a topological space and let $x\in X$.
Assume that $\{x\}$ is a closed subset of $X$.
Let $U$ be an open neighborhood of $x$ in $X$.
Assume that $U$ is $x$-avoidable.
Then $X$ is $x$-avoidable.
\vskip.1in\noindent
{\it Proof:}
Let $a,b\in X\backslash\{x\}$
and let $\gamma\in P_a^b(X)$.
We wish to show, for some $\gamma\in P_a^b(X\backslash\{x\})$,
that $\gamma$ is endpoint fixed homotopic to $\gamma'$.
Let $V:=X\backslash\{x\}$.
Then $\{U,V\}$ is an open cover of $X$.
Let $\delta>0$ be a Lebesgue number for
the open cover $\{\gamma^{-1}(U),\gamma^{-1}(V)\}$ of $[0,1]$.
Choose $s_0,\ldots,s_n\in[0,1]$
such that $0=s_0